Class 12 Commerce Maths - Three Dimensional Geometry

(-2, -1), (-1, -4) and (-4,-1)

No, they are not collinear as they do not lie on one single straight line.

Find the values of $p$ so the line $\cfrac{1-x}{3}=\cfrac{7y-14}{2p}=\cfrac{z-3}{2}$ and $\cfrac{7-7x}{3p}=\cfrac{y-5}{1}=\cfrac{6-z}{5}$ are at right angles.

The given equations can be written in the standard form as

$\cfrac{x-1}{-3}=\cfrac { y-2 }{ \cfrac { 2p }{ 7 } } =\cfrac{z-3}{2}$ and

$\cfrac { x-1 }{ \cfrac { -3p }{ 7 } } =\cfrac{y-5}{1}=\cfrac{z-6}{-5}$
The direction ratios of the lines are

$-3, \cfrac{-3p}{7}, 2$ and

$\cfrac {-3p}{7}, 1, -5$ respectively.
Two lines with direction ratios

${a}_{1}, {b}_{1}, {c}_{1}$ and

${a}_{2}, {b}_{2}, {c}_{2}$ , are perpendicular to each other, if

${a}_{1}{a}_{2}+{b}_{1}{b}_{2}+{c}_{1}{c}_{2}=0$ .

$\therefore$

$(-3)\left( \cfrac { -3p }{ 7 } \right)+\left( \cfrac { -3p }{ 7 } \right) .(1)+(2).(-5)=0$

$\Rightarrow$

$\cfrac{9p}{7}+\cfrac{2p}{7}=10$

$\Rightarrow$

$11p=70$

$\Rightarrow p=\cfrac{70}{11}$

Find the direction cosines of the vector joining the points $A (1, 2, -3)$ and $B (-1, -2, 1)$ directed from $A$ to $B$ .

The given points are

$A (1, 2, -3)$ and

$B (-1, -2, 1)$ .

$\therefore \vec {AB}=(-1-1)\hat {i}+(-2-2)\hat {j}+\left \{1-(-3)\right \}\hat {k}$

$\therefore \vec {AB}=-2\hat {i}-4\hat {j}+4\hat {k}$

$\therefore |\vec {AB}|=\sqrt {(-2)^2+(-4)^2+4^2}=\sqrt {4+16+16}=\sqrt {36}=6$

Hence, the direction cosines of

$\vec {AB}$ are

$\left (-\dfrac {2}{6}, -\dfrac {4}{6}, \dfrac {4}{6}\right )=\left (-\dfrac {1}{3}, -\dfrac {2}{3}, \dfrac {2}{3}\right )$ .

If $O$ be the origin and the coordinates of $P$ be $(1,2,-3)$ , then find the equation of the plane passing through $P$ and perpendicular to $OP$ .

The coordinates of the points

$O$ and

$P$ are

$(0,0,0)$ and

$(1,2,-3)$ respectively.
Therefore, the direction ratios of

$OP$ are

$(1-0)=1,(2-0)=2$ and

$(-3-0)=-3$
Thus the direction ratios of normal are

$1,2$ and

$-3$ and the point

$P$ is

$(1,2,-3)$ .
Thus, the equation of the required plane is

$1(x-1)+2(y-2)-3(z+3)=0$

$\Rightarrow$

$x+2y-3z-14=0$

If a line makes angles $90^{\circ}, 135^{\circ}, 45^{\circ}$ with the $x, y$ and $Z-\text{axes}$ respectively, find its direction cosines.

Let the direction cosines of the line be

$l, m, n$ .

$l=\cos {{90}^{o}}=0$

$m=\cos {{135}^{o}}=-\cfrac { 1 }{ \sqrt { 2 } }$

$n=\cos {{45}^{o}}=\cfrac { 1 }{ \sqrt { 2 } }$
Therefore, the direction cosines of the line are

$0, -\cfrac{1}{\sqrt {2}}$ and

$\cfrac{1}{\sqrt {2}}$ .

Find the equation plane passing $(a, b, c)$ and parallel to the plane $\vec {r} \cdot (\hat {i} + \hat {j} + \hat {k}) = 2$ .

Any plane to parallel to the plane

$\vec { r } .\left( \hat { i } +\hat { j } +\hat { k } \right) =2$ is for the form

$\vec { r } .\left( \hat { i } +\hat { j } +\hat { k } \right) =\lambda .....(1)$
The plane passes through the point

$(a,b,c)$ . Therefore, the position vector

$\vec { r }$ of this point is

$a\hat { i } +b\hat { j } +c\hat { k }$ .
Therefore, equation (1) becomes

$(a\hat { i } +b\hat { j } +c\hat { k } ).(\hat { i } +\hat { j } +\hat { k } )=\lambda$

$\Rightarrow$

$a+b+c=\lambda$
Substituting

$\lambda=a+b+c$ in equation (1) we obtain

$\vec { r } .\left( \hat { i } +\hat { j } +\hat { k } \right) =a+b+c......(2)$
This is the vector equation of the required plane.
Substituting

$\vec { r } =x\hat { i } +y\hat { j } +z\hat { k }$ in equation (2), we obtain

$(x\hat { i } +y\hat { j } +z\hat { k } ).(\hat { i } +\hat { j } +\hat { k } )=a+b+c$

$\Rightarrow$

$x+y+z=a+b+c$

Find the direction cosines of the vector $\hat {i}+2\hat {j}+3\hat {k}$ .

Let

$r=i+2j+3k$ be the given vector

The magnitude of vector

$r$ is

$|r|=\sqrt{1^2+2^2+3^2}=\sqrt{14}$

Let the directional cosines of vector

$r$ be

$cos \alpha , cos \beta , cos \gamma$

We have

$cos \alpha =\dfrac{r.i}{|r|}=\dfrac{1}{\sqrt{14}}$

We have

$cos \beta =\dfrac{r.j}{|r|}=\dfrac{2}{\sqrt{14}}$

We have

$cos \gamma =\dfrac{r.k}{|r|}=\dfrac{3}{\sqrt{14}}$

Therefore the directional cosines of given vector are

$\dfrac{1}{\sqrt{14}} , \dfrac{2}{\sqrt{14}}$ and

$\dfrac{3}{\sqrt{14}}$

Find the angle between the lines whose direction ratios are $a,b,c$ and $b-c, c-a, a-b$

The angle

$\theta$ between the lines with direction cosines

$a,b,c$ and

$b-c, c-a, a-b$ is given by,

$\cos { \theta } =\left| \cfrac { a(b-c)+b(c-a)+c(a-b) }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } } \sqrt { { \left( b-c \right) }^{ 2 }+{ \left( c-a \right) }^{ 2 }+{ \left( a-b \right) }^{ 2 } } } \right|$

$\Rightarrow \cos { \theta } =0$

$\Rightarrow \theta=\cos ^{ -1 }{ 0 }$

$\Rightarrow \theta ={ 90 }^{ o }$
Thus, the angle between the lines is

${90}^{o}$ .

Find the vector and the Cartesian equations of the lines that pass through the origin and $(5, -2, 3)$

The required line passes through the origin. Therefore, its position vector is given by,

$\vec { a } =\vec { 0 } ......(1)$
The direction ratios of the line through origin and

$(5, -2, 3)$ are

$(5-0)=5, (-2-0)=-2, (3-0)=3$
The line is parallel to the vector given by the equation,

$\vec { b } =5\hat { i } -2\hat { j } +3\hat { k }$
The equation of the line in vector form through a point with position vector

$\vec { a }$ and parallel to

$\vec { b }$ is ,

$\vec { r } =\vec { a } +\lambda \vec { b }$ ,

$\lambda \in R$

$\Rightarrow$

$\Rightarrow \vec { r } =\vec { 0 } +\lambda \left( 5\hat { i } -2\hat { j } +3\hat { k } \right)$

$\Rightarrow \vec { r } =\lambda \left( 5\hat { i } -2\hat { j } +3\hat { k } \right)$
The equation of the line through the point

$({x}_{1}, {y}_{1}, {z}_{1})$ and direction ratios

$a, b, c$ is given by,

$\cfrac { x-{ x }_{ 1 } }{ a } =\cfrac { y-{ y }_{ 1 } }{ b } =\cfrac { z-{ z }_{ 1 } }{ c }$
Therefore, the equation of the required line in the Cartesian form is

$\cfrac{x-0}{5}=\cfrac{y-0}{-2}=\cfrac{z-0}{3}$

$\Rightarrow$

$\cfrac{x}{5}=\cfrac{y}{-2}=\cfrac{z}{3}$

Show that the points $A (1, 2, 7), B (2, 6, 3)$ and $C (3, 10, -1)$ are collinear .

The given points are

$A (1, 2, 7), B (2, 6, 3)$ and

$C (3, 10, -1)$ .

$\therefore \vec {AB}=(2-1)\hat {i}+(6-2)\hat {j}+(3-7)\hat {k}=\hat {i}+4\hat {j}-4\hat {k}$

$\vec {BC}=(3-2)\hat {i}+(10-6)\hat {j}+(-1-3)\hat {k}=\hat {i}+4\hat {j}=4\hat {k}$

$\vec {AC}=(3-1)\hat {i}+(10-2)\hat {j}+(-1-7)\hat {k}=2\hat {i}+8\hat {j}-8\hat {k}$

$|\vec {AB}|=\sqrt {1^2+4^2+(-4)^2}=\sqrt {1+16+16}=\sqrt {33}$

$|\vec {BC}|=\sqrt {1^2+4^2+(-4)^2}=\sqrt {1+16+16}=\sqrt {33}$

$|\vec {AC}|=\sqrt {2^2+8^2+8^2}=\sqrt {4+64+64}=\sqrt {132}=2\sqrt {33}$

$\therefore |\vec {AC}|+|\vec {AB}|+|\vec {BC}|$

Hence, the given points

$A, B$ , and

$C$ are collinear.

The equation of a line is $5x-3=15y+7=3-10z$ . Write the direction cosines of the line.

$5x-3=15y+7=3-10z\\ \Rightarrow \dfrac { x-\dfrac { 3 }{ 5 } }{ \dfrac { 1 }{ 5 } } =\dfrac { y-(-\dfrac { 7 }{ 15 } ) }{ \dfrac { 1 }{ 15 } } =\dfrac { z-\dfrac { 3 }{ 10 } }{ -\dfrac { 1 }{ 10 } }$ .

Comparing this with the standard equation of straight line:

$\dfrac { x-{ a }_{ 1 } }{ { b }_{ 1 } } =\dfrac { y-{ a }_{ 2 } }{ { b }_{ 2 } } =\dfrac { z-{ a }_{ 3 } }{ { b }_{ 3 } }$ , we get,

${ b }_{ 1 }=\dfrac { 1 }{ 5 } { ,b }_{ 2 }=\dfrac { 1 }{ 15 } { ,b }_{ 3 }=-\dfrac { 1 }{ 10 }$ .

The direction cosines are the components of the unit vector

$\hat { b } =\dfrac { { b }_{ 1 }\hat { i } +{ b }_{ 2 }\hat { j } +{ b }_{ 3 }\hat { k } }{ \sqrt { { b }_{ 1 }^{ 2 }+{ b }_{ 2 }^{ 2 }+{ b }_{ 3 }^{ 2 } } } =\dfrac { \dfrac { 1 }{ 5 } \hat { i } +\dfrac { 1 }{ 15 } \hat { j } -\dfrac { 1 }{ 10 } \hat { k } }{ \dfrac { 7 }{ 30 } } =\dfrac { 6 }{ 7 } \hat { i } +\dfrac { 2 }{ 7 } \hat { j } -\dfrac { 3 }{ 7 } \hat { k }$ .

So, the direction cosines are

$\dfrac { 6 }{ 7 } ,\dfrac { 2 }{ 7 } ,\dfrac { -3 }{ 7 }.$

The acute angle between the planes $2x-y+z=6$ and $x+y+2z=3$ is _____.

Two planes given:

$P_1: 2x-y+z=6$

$P_2: x+y+2z=3$

Normals are:

$\vec{n_1} : 2\hat i-\hat j+\hat k$

$\vec{n_2}: \hat i+\hat j+2\hat k$

Acute angle will be

$\theta$ and

$\theta = \cos^{-1} \cfrac{\vec{n_1}\cdot\vec{n_2}}{|\vec{n_1}||\vec{n_2}|}$

$\Rightarrow \theta = \cos^{-1} \left(\cfrac12\right) = 60˚$

Write the direction cosines of $x$ -axis.

Say

${ (x }_{ 1 },0,0),{ (x }_{ 2 },0,0)$ be two points on x axis

DR's of a line is given by subtracting position vectors.

DR's of x axis is

${ (x }_{ 1 }-{ x }_{ 2 },0,0)$

DC's of any vector are obtained by dividing the vector with it's magnitude

${ |x }_{ 1 }-{ x }_{ 2 }|={ ({ x }_{ 1 }-{ x }_{ 2 }) }^{ 2 }+{ 0 }^{ 2 }+{ 0 }^{ 2 }$

$DC's=\dfrac { { (x }_{ 1 }-{ x }_{ 2 })\hat { i } }{ { |x }_{ 1 }-{ x }_{ 2 }| } =\pm (1,0,0)$

DC's of x axis are

$(1,0,0),(-1,0,0)$

Determine the direction cosines of the line through the points $(2,0,3)$ and $(4,1,1)$

If we have two points

$P({ x }_{ 1 },y_{ 1 },{ z }_{ 1 })$ and

$Q({ x }_{ 2 },y_{ 2 },z_{ 2 })$ ,then the dc’s of the line segment joining these two points are

$\frac { ({ x }_{ 2 }-{ x }_{ 1 }) }{ PQ } ,\frac { ({ y }_{ 2 }-{ y }_{ 1 }) }{ PQ } ,\frac { ({ z }_{ 2 }-{ z }_{ 1 }) }{ PQ } \quad$

The distance between given points

$=\sqrt { { (4-2) }^{ 2 }+{ (1-0) }^{ 2 }+{ (1-3) }^{ 2 } } =3$

$\Rightarrow$ the directional cosines

$=\dfrac { (4-2) }{ 3 } ,\dfrac { (1-0) }{ 3 } ,\dfrac { (1-3) }{ 3 }$

$\Rightarrow$ dc's

$= \dfrac { 2 }{ 3 } ,\dfrac { 1 }{ 3 } ,\dfrac { -2 }{ 3 }$

The Cartesian equation of a line is: $2x-3=3y+1=5-6z$ . Find the vector equation of a line passing through $(7, -5, 0)$ and parallel to the given line.

Given line is

$L: \dfrac{x-3/2}{1/2}=\dfrac{y+1/3}{1/3}=\dfrac{z-5/6}{-1/6}$

That means

$\vec{v}=(1/2,1/3,-1/6)$

$\therefore$ Vector equation of a line passing through

$\vec{r_0}=P_0(7,-5,0)$ and parallel to

$L($ Means parallel to vector

$\vec{v})$

$\rightarrow L_1:\space \vec{r}=\vec{r_0}+\lambda \vec{v}$

$\vec{r}=(7,-5,0)+\lambda(1/2,1/3,-1/6)$

Find the acute angle between the lines passing through (-3 , -1 , 0), (2, -3 , 1) and (1 , 2 , 3),(-1 ,4 , -2) respectively.

Let the given points are

$\begin{array}{l} A\left( { -3,-1,0 } \right) \\ B\left( { 2.-3.1 } \right) \\ C\left( { 1,2,3 } \right) \\ D\left( { -1,4,-2 } \right) \end{array}$

Direction ratios of line

$AB$ are

$2+3,-3+1,1-0$

$5,-2,1$

Let

${a_1} = 5,{b_1} = - 2,{c_1} = 1$

Direction ratio opf the line

$CD$ are

$\begin{array}{l} { a_{ 2 } }=-1-1=-2 \\ { b_{ 2 } }=4-2=2 \\ { c_{ 2 } }=-2-3=-5 \end{array}$

let

$O$ be the acute angle between line

$CD$ and

$AB$

then

$\begin{array}{l} \cos \theta =\left| { \dfrac { { { a_{ 1 } }{ a_{ 2 } }+{ b_{ 1 } }{ b_{ 2 } }+{ c_{ 1 } }{ c_{ 2 } } } }{ { \sqrt { { a^{ 2 } }_{ 1 }+{ b_{ 1 } }^{ 2 }+{ c_{ 1 } }^{ 2 } } \sqrt { { a_{ 2 } }^{ 2 }+{ b_{ 2 } }^{ 2 }+{ c_{ 2 } }^{ 2 } } } } } \right| \\ \cos \theta =\left| { \dfrac { { -10-4-5 } }{ { \sqrt { 25+4+1 } \sqrt { 4+4+25 } } } } \right| \\ \cos \theta =\dfrac { { -19 } }{ { \sqrt { 30 } \sqrt { 33 } } } \\ \cos \theta =\dfrac { { 19 } }{ { \sqrt { 3\times 10 } \sqrt { 3\times 11 } } } \\ \cos \theta =\dfrac { { 19 } }{ { 3\sqrt { 110 } } } \\ \theta ={ \cos ^{ -1 } }\left( { \dfrac { { 19 } }{ { 3\sqrt { 110 } } } } \right) \end{array}$

Find direction cosine line $x=3z+2,y=2-5z$ .

Given line is

$x=3z+2,y=2-5z$

$\therefore \cfrac { x-2 }{ 3 } =z;\cfrac { y-2 }{ -5 }$

$\therefore \cfrac { x-2 }{ 3 } =\cfrac { y-2 }{ -5 } =\cfrac { z }{ 1 }$
Now compate with standard equation
Direction of line

$\overline { l } =\left( 3,-5,1 \right) \quad$

$\left| \overline { l } \right| =\sqrt { 9+25+1 } =\sqrt { 35 }$

$\therefore$ Direction cosines of given line

$=\cfrac { \overline { l } }{ \left| \overline { l } \right| } =\cfrac { \left( 3,-5,1 \right) }{ \sqrt { 35 } } =\left( \cfrac { 3 }{ \sqrt { 35 } } ,\cfrac { -5 }{ \sqrt { 38 } } ,\cfrac { 1 }{ \sqrt { 35 } } \right)$

If $A\left( { x }_{ 1 },{ y }_{ 1 },{ z }_{ 1 } \right)$ and $B\left( { x }_{ 2 },{ y }_{ 2 },{ z }_{ 2 } \right)$ are two points such that the direction cosines of $\bar { AB }$ are $l$ , $m$ , $n$ then
$l=\dfrac { { x }_{ 2 }-{ x }_{ 1 } }{ \left| \bar { AB } \right| }$ , $m=\dfrac { { y }_{ 2 }-{ y }_{ 1 } }{ \left| \bar { AB } \right| }$ , $n=\dfrac { { z }_{ 2 }-z_{ 1 } }{ \left| \bar { AB } \right| }$ .

Direction cosines of

$\overline{AB}$ is the cosines of angles made by the line with the positive direction of

$x$ - axis,

$y$ -axis and

$z$ - axis respectively.

Let,

$\alpha$ ,

$\beta$ ,

$\gamma$ be the angles made by the line with the positive direction of

$x$ -axis,

$y$ -axis and

$z$ -axis.

Then

$l=\cos \alpha=\large{\frac{x_2-x_1}{\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}}}=\frac{(x_2-x_1)}{|\overline{AB}|}.$

and

$m=\cos \beta=\large{\frac{y_2-y_1}{\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}}}=\frac{(y_2-y_1)}{|\overline{AB}|}.$

and

$n=\cos \alpha=\large{\frac{z_2-z_1}{\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}}}=\frac{(z_2-z_1)}{|\overline{AB}|}.$

Find the direction collinear of a line which makes equal angles with the positive co-ordinate axes.

$\alpha =\beta =\gamma$
W.K.T

$\cos ^{ 2 }{ \alpha } +\cos ^{ 2 }{ \alpha } +\cos ^{ 2 }{ \alpha } =1\quad$

$3\cos ^{ 2 }{ \alpha } =1\quad$

$\cos ^{ 2 }{ \alpha } =\cfrac { 1 }{ 3 }$

$\cos { \alpha } =\cfrac { 1 }{ \sqrt { 3 } }$

$\therefore$ direction cosines are

$\cfrac { 1 }{ \sqrt { 3 } } .\cfrac { 1 }{ \sqrt { 3 } } .\cfrac { 1 }{ \sqrt { 3 } }$

Find the length of the perpendicular drawn from the origin to the plane $2x-3y+6z+21=0$ .

Given, the equation of the plane is

$2x-3y+6z+21=0$

Therefore, the length of the perpendicular drawn from the origin

$(0,0)$ to this plane

$=\left |\dfrac{2.0-3.0+6.0+21}{\sqrt{(2)^2+(-3)^2+(6)^2}}\right |=\left |\dfrac{21}{\sqrt{49}}\right |$

$=\dfrac{21}{7}=3$ units

Find the value of $\beta$ so that the line $\dfrac {x - 2}{6} = \dfrac {y - 1}{\beta} = \dfrac {z + 5}{-4}$ is perpendicular to the plane $3x - y - 2z = 7$ .

For a line and a plane,

angle

$sin\theta = \left| \dfrac{a_1a_2 + b_1b_2 + c_1c_2} {\sqrt{a_{1}^{2}+{b_{1}^{2}+{c_{1}^{2}}}}\sqrt{a_{2}^{2}+{b_{2}^{2}+{c_{2}^{2}}}}}\right|$

$(a_1 , b_1 , c_1) = (6 , \beta , -4)$

$(a_2 , b_2 , c_2) = (3 , -1 , -2)$

if they are perpendicular,

then,

$\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$

$\Rightarrow \dfrac{6}{3} = \dfrac{\beta}{-1} = \dfrac{-4}{-2}$

$\Rightarrow \beta = -2$

Find the direction cosines of the line $\dfrac{{x + 2}}{2} = \dfrac{{2y - 5}}{3};z = - 1$ . Also, find the vector equation of the line.

Equation of planes given,

${ P }_{ 1 }:\cfrac { x+2 }{ 2 } =\cfrac { 2y-5 }{ 3 } \Rightarrow 3x+6=4y-10\Rightarrow 3x-4y+16=0\\ { P }_{ 2 }:z=-1$

The required line is line of intersection of

${ P }_{ 1 }$ &

${ P }_{ 2 }$

let

$(l,m,n)$ be directions of required line,it is perpendicular to the normals of

${ P }_{ 1 }$ and

${ P }_{ 2 }$

$\therefore 3l-4m+0n=0\\ \therefore 0l+0m+n=0$

From cramers rule,

$\cfrac { l }{ -4 } =\cfrac { -m }{ 3 } =\cfrac { n }{ 0 }$

$\therefore$ Directions are

$(-4,-3,0)$

Direction cosines of line are

$(\cfrac { 4 }{ 5 } ,\cfrac { 3 }{ 5 } ,0)$

One point on the line

$\overrightarrow { a }$ is

$0\hat { i } +4\hat { j } -1\hat { k }$

Equation of line,

$\overrightarrow { r } =\overrightarrow { a } +\lambda \overrightarrow { b } \\ \overrightarrow { r } =4\hat { j } -\hat { k } +\lambda (\cfrac { 4 }{ 5 } \hat { i } +\cfrac { 3 }{ 5 } \hat { j } )$

Find the vector equation of the line through $A(3,4,-7)$ and $B(6,-1,1)$ . Also find the cartesian form.

$\overrightarrow a = (3,4, - 7) = 3\widehat i + 4\widehat j - 7\widehat k$

$\overrightarrow b = (1, - 1,6) = \widehat i - \widehat j + 6\widehat k$

$\overrightarrow r = \overrightarrow a + \lambda \overrightarrow b$

$\overrightarrow r = \overrightarrow a + \lambda (\overrightarrow b - \overrightarrow a )$

$\overrightarrow r = 3\widehat i + 4\widehat j - 7\widehat k + \lambda ( - 2\widehat i - 5\widehat j + 13\widehat k)$

$z\widehat i + y\widehat j + z\widehat k = (3 - 2\lambda )\widehat i + (4 - 5\lambda )\widehat j + ( - 7 + 13\lambda )\widehat k$

$x = 3 - 2\lambda$ ,

$y = 4 - 5\lambda$ ,

$z = - 7 + 13\lambda$

Solve for

$\lambda$ ,

$\dfrac{{x - 3}}{{ - 2}} = \dfrac{{y - 4}}{{ - 5}} = \dfrac{{3 + 7}}{{13}}$

Show that the lines whose d.c.s are given by $l + m + n = 0 , 2mn + 3ln - 5lm = 0$ are perpendicular to each other.

First relation

$l=(-m-n)..........................(1)$

putting the value of

$l$ in the relation

$2mn+3(-m-n)n-5(-m-n)m=0$

$\begin{array}{l} 5{ m^{ 2 } }+4mn-3{ n^{ 2 } }=0 \\ 5{ \left( { \dfrac { m }{ n } } \right) ^{ 2 } }+4\left( { \dfrac { m }{ n } } \right) -3=0..........\left( 2 \right) \end{array}$

suppose

$l_1,m_1,n_1$ and

$l_2,m_2,n_2$ are the direction cosines of the two line, then the roots of the equation (2) are

$\dfrac{m_1}{n_1}$ and

$\dfrac{m_2}{n_2}$

nOW,

Product of the roots

$=\dfrac{m_1}{n_1}.\dfrac{m_2}{n_2}=-\dfrac{3}{5}$

then,

$\dfrac{m_1m_2}{3}=\dfrac{n_1n_2}{-5}$ -----

$(3)$

Again from

$(1),\,\,n=-1-m$ and putting this value of

$n$ in

$(2)$ equation, we get

$2m(-l-m)+3l(-l-m)-5lm=0$

and,

$3(\dfrac{l}{m})^2+10(\dfrac{l}{m})+2=0$

$\dfrac{l_1}{m_1}.\dfrac{l_2}{m_2}=\dfrac{2}{3}$

then,

$\dfrac{l_1l_2}{2}=\dfrac{m_1m_2}{3}$

From

$(3)$ and

$(4)$

$\dfrac{l_1l_2}{2}=\dfrac{m_1m_2}{3}=\dfrac{n_1n_2}{-5}=k$ (consider.)

$l_1l_2+m_1m_2+n_1n_2=(2+3-5)k=0$

$k=0$

therefore lines are at right angle.

If a line is making same angle $\theta$ with the positive direction of $X-,Y-$ and $Z-$ axes, then show that $\theta=\cos^{-1}(\dfrac {1}{\sqrt {3}}$ .

Direction cosine of the line are

$l=\cos {\theta},\;m=\cos {\theta},\;n=\cos {\theta}$

$l^2+m^2+n^2=1$

$\cos^2 {\theta}+\cos^2{\theta}+\cos^2 {\theta}=1$

$cos^2 {\theta}=\dfrac{1}{3}$

${\theta}=\cos^{-1} (\dfrac{1}{\sqrt3})$

Find the equation of the line in vector and in cartesion form that passes through the point with position vector $2\hat {i}-\hat{j}+4\ \hat {k}$ and is in the direction $\hat {i}+2\hat {j}-\hat {k}$ .

$Vector\quad equation\quad of\quad a\quad line\quad is:$

$\overline { r } =\overline { a } +\lambda \overline { b }$

$where\quad \overline { r } \quad is\quad a\quad point\quad on\quad the\quad line,\overline { a } \quad is\quad a\quad known$

$point\quad lying\quad on\quad line,and\quad \overline { b } \quad is\quad the\quad line\quad to\quad which$

$the\quad given\quad line\quad is\quad parallel,$

$\implies\quad \overline { r } =(2\overset { \wedge }{ i } -\overset { \wedge }{ j } +4\overset { \wedge }{ k } )+\lambda (\overset { \wedge }{ i } +2\overset { \wedge }{ j } -\overset { \wedge }{ k } )$

$For\quad cartesian\quad form:$

$\because The\quad given\quad line\quad passes\quad through\quad (2,-1,4)\quad and\quad direction.$

$no.\quad is\quad a=1,b=2,c=-1$

$Equation\quad is:\quad \cfrac { x-x1 }{ a } =\cfrac { y=y1 }{ b } =\cfrac { z-z1 }{ c }$

$Cartesian\quad equation,$

$\implies\quad \cfrac { x-2 }{ 1 } =\cfrac { y-(-1) }{ 2 } =\cfrac { z-4 }{ -1 }$

$\implies\quad \cfrac { x-2 }{ 1 } =\cfrac { y+1 }{ 2 } =\cfrac { z-4 }{ -1 }$

Find the direction cosines of the lines , connected by the relations: l + m + n = 0 and 2lm + 2ln - mn = 0.

Let

$(l,m,n)$ be direction cosines.

$\therefore { l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 }=1\\ { ({ l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 }) }^{ 2 }-2(lm+mn+nl)=1\quad \quad \\ -2lm-2mn-2nl=1\quad \quad (\because l+m+n=0)\\ -3mn=1\quad \quad (\because 2lm+2ln=mn)\\ mn=\cfrac { -1 }{ 3 } \quad -(i)\\ l+m+n=0\quad \Rightarrow l=-m-n\quad -(ii)\\ 2lm+2ln-mn=0\\ 2m(-m-n)+2(-m-n)n-mn=0\\ -[2{ m }^{ 2 }+2mn+2mn+2{ n }^{ 2 }]=mn\\ -2{ m }^{ 2 }-5mn-2{ n }^{ 2 }=0\\ -2{ m }^{ 2 }-2{ n }^{ 2 }=5\times \cfrac { -1 }{ 3 } \quad \quad (\because mn=\cfrac { -1 }{ 3 } )\\ { m }^{ 2 }+{ n }^{ 2 }=\cfrac { 5 }{ 6 } \quad \quad -(iii)\\ { l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 }=1\\ { l }^{ 2 }+\cfrac { 5 }{ 6 } =1\quad \Rightarrow l=\pm \cfrac { 1 }{ \sqrt { 6 } }$

$\therefore m+n=\mp { l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 }=1$ (from equation

$2$ )

${ m }^{ 2 }+{ n }^{ 2 }=\cfrac { 5 }{ 6 }$ (from equation

$3$ )

$mn=\cfrac { -1 }{ 3 }$ (from equation

$1$ )

$m+n=\cfrac { -1 }{ \sqrt { 6 } } \quad (l=\cfrac { 1 }{ \sqrt { 6 } } )\\ mn=\cfrac { -1 }{ 3 } \\ \cfrac { -1 }{ 3n } +n+\cfrac { 1 }{ \sqrt { 6 } } =0\\ { n }^{ 2 }+\cfrac { n }{ \sqrt { 6 } } -\cfrac { 1 }{ 3 } =0\\ n=\cfrac { 1 }{ \sqrt { 6 } } ,\cfrac { -\sqrt { 6 } }{ 3 } \\ \therefore m=\cfrac { -1 }{ 3n } \\ \Rightarrow m=\cfrac { -\sqrt { 6 } }{ 3 } ,\cfrac { 1 }{ \sqrt { 6 } } \\ m+n=\cfrac { 1 }{ \sqrt { 6 } } \quad (l=\cfrac { -1 }{ \sqrt { 6 } } )\\ mn=\cfrac { -1 }{ 3 } \\ \cfrac { -1 }{ 3n } +n-\cfrac { 1 }{ \sqrt { 6 } } =0\\ { n }^{ 2 }-\cfrac { n }{ \sqrt { 6 } } -\cfrac { 1 }{ 3 } =0\\ n=\cfrac { -1 }{ \sqrt { 6 } } ,\cfrac { \sqrt { 6 } }{ 3 } \\ \therefore m=\cfrac { -1 }{ 3n } \\ \Rightarrow m=\cfrac { \sqrt { 6 } }{ 3 } ,\cfrac { -1 }{ \sqrt { 6 } }$

$\therefore$ Possible values of

$(l,m,n)$ are,

$(\cfrac { 1 }{ \sqrt { 6 } } ,\cfrac { 1 }{ \sqrt { 6 } } ,\cfrac { -\sqrt { 6 } }{ 3 } ),(\cfrac { 1 }{ \sqrt { 6 } } ,\cfrac { -\sqrt { 6 } }{ 3 } ,\cfrac { 1 }{ \sqrt { 6 } } ),(\cfrac { -1 }{ \sqrt { 6 } } ,\cfrac { -1 }{ \sqrt { 6 } } ,\cfrac { \sqrt { 6 } }{ 3 } )$

and

$(\cfrac { -1 }{ \sqrt { 6 } } ,\cfrac { \sqrt { 6 } }{ 3 } ,\cfrac { -1 }{ \sqrt { 6 } } )$

If the lines through the points $(4, 1, 2)$ and $(5, k, 0)$ is parallel to the line through the points $(2, 1, 1)$ and $(3, 3, 1)$ , find $k$ .

Here ,

the points are (4,1,2) and (5,k,0) is parallel to the line (2,1,1) and (3,3,1)

from the point, (4,1,2) and (5,k,0).----------equation (1) $\begin{array}{l} =\sqrt { { { (5-4) }^{ 2 } }+{ { (k-1) }^{ 2 } }+{ { (0-2) }^{ 2 } } } \\ =\sqrt { 1+{ { (k-1) }^{ 2 } }+4 } \\ =\sqrt { 5+{ { (k-1) }^{ 2 } } } \end{array}$ $\begin{array}{l} now, \\ \frac { { 5-4 } }{ { \sqrt { 5+{ { (k-1) }^{ 2 } } } } } =\frac { { k-1 } }{ { \sqrt { 5+{ { (k-1) }^{ 2 } } } } } =\frac { { -2 } }{ { \sqrt { 5+{ { (k-1) }^{ 2 } } } } } \end{array}$

and

from the point of, (2,1,1) and (3,3,1)----------equation (2)

$\begin{array}{l} =\sqrt { { { (3-2) }^{ 2 } }+{ { (3-1) }^{ 2 } }+{ { (1-1) }^{ 2 } } } \\ =\sqrt { 1+4 } \\ =\sqrt { 5 } \end{array}$

now,

$\begin{array}{l} \Rightarrow \frac { { 5-4 } }{ { \sqrt { 5+{ { (k-1) }^{ 2 } } } } } =\frac { 1 }{ { \sqrt { 5 } } } \\ \begin{array} { \Rightarrow 5=5+(k-1) } \\ { k=1 } \end{array} \end{array}$

so, the value of k is 1.

If the d.c's of a line are $\left( {\dfrac{1}{c}\,,\,\dfrac{1}{c}\,,\,\dfrac{1}{c}} \right)$ , find c.

Consider the problem

As we know

Sum of square of direction ratios is equal to

$1$ .

Therefore,

$\dfrac{1}{{{c^2}}} + \dfrac{1}{{{c^2}}} + \dfrac{1}{{{c^2}}} =1$

$3\dfrac{1}{c^2}=1$

$c^2=3$

$c=\pm \sqrt3$

So, the value of

$c=\pm \sqrt 3$

find the angle between the lines whose direction cosines satisfy the equation $\ell + m + n = 0,\,\,{\ell ^2} + {m^2} - {n^2} = 0$

Consider the problem

Given

$l+m+n=0$ ---- (i)

$n=-(l+m)$

And

${\ell ^2} + {m^2} - {n^2} = 0$ ---- (ii)

Put value of

$n$ in equation (ii)

${l^2} + {m^2} - {l^2} - {m^2} - 2ml = 0$

$2ml = 0$

Either

$m=0$ or

$l=0$

Now, first

put

$m=0$ in (i)

then

$l=-n$

therefore

direction ratios

$(l,m,n)=(0,1,-1)$

Now, find

$b_1,b_2$

${b_1}.{b_2} = (1,0, - 1).(0,1, - 1)=0+0+1=1$

$\begin{array}{l} |{ b_{ 1 } }|=\sqrt { { 0^{ 2 } }+{ 1^{ 2 } }+{ { (-1) }^{ 2 } } } =\sqrt { 2 } \\ |{ b_{ 2 } }|=\sqrt { { 0^{ 2 } }+{ 1^{ 2 } }+{ { (-1) }^{ 2 } } } =\sqrt { 2 } \end{array}$

Then

$\begin{array}{l} cos\theta =\dfrac { { \overrightarrow { { b_{ 1 } } } .\overrightarrow { { b_{ 2 } } } } }{ { |\overrightarrow { { b_{ 1 } } } ||\overrightarrow { { b_{ 2 } } } | } } \\ \cos \theta =\dfrac { 1 }{ { \sqrt { 2 } \sqrt { 2 } } } =\dfrac { 1 }{ 2 } \\ \theta =\dfrac { \pi }{ 3 } \end{array}$

Find the co-ordinate of the pt. $p$ , if $\ell \left( PQ \right) =21$ , $Q\equiv \left( -2,1,-8 \right)$ and directional cosines of $PQ$ are $(6/7), (2/7), (3/7)$ ?

$l(PQ)=21\quad ,\quad Q=(-2,1,-8)\\ (l,m,n)=\cfrac { 6 }{ 7 } ,\cfrac { 2 }{ 7 } ,\cfrac { 3 }{ 7 }$

$\therefore$ Equation of line

$PQ\Rightarrow \cfrac { x+2 }{ \cfrac { 6 }{ 7 } } =\cfrac { y-1 }{ \cfrac { 2 }{ 7 } } =\cfrac { z+8 }{ \cfrac { 3 }{ 7 } }$

$\therefore$ Let

$P$ be

$(\cfrac { 6 }{ 7 } r-2,\cfrac { 2 }{ 7 } r+1,\cfrac { 3 }{ 7 } r-8)$

$\therefore { (\cfrac { 6 }{ 7 } r-2+2) }^{ 2 }+{ (\cfrac { 2 }{ 7 } r+1-1) }^{ 2 }+{ (\cfrac { 3 }{ 7 } r-8+8) }^{ 2 }={ 21 }^{ 2 }={ PQ }^{ 2 }\\ \cfrac { (36+4+9) }{ 49 } { r }^{ 2 }={ 21 }^{ 2 }\\ \Rightarrow r=21,-21$

$\therefore$ Co-ordinates of

$P$ can be

$(16,7,1)$ &

$(-20,-5,-17)$

A line makes angle $\theta_{1}, \theta_{2}, \theta_{3}, \theta_{4}$ with the diagonals of the cube. Show that $\cos^{2}\theta_{1} + \cos^{2}\theta_{2} + \cos^{2}\theta_{3} + \cos^{2}\theta_{4} = \dfrac{4}{3}$ ?

Let side of cube be 'a'

Dr's of a diagonal OG = a ,a ,a

Dr's of a diagonal CD = a ,a , - a

Dr's of a diagonal BE = a , - a ,a

Dr's of a diagonal AF = - a ,a ,a

$now\,\,Dc's\,\,of\,diagonal\,\,OG = \left( {\frac{1}{{\sqrt 3 }},\,\,\frac{1}{{\sqrt 3 }}\,,\,\frac{1}{{\sqrt 3 }}} \right)$

$now\,\,Dc's\,\,of\,diagonal\,\,CD = \left( {\frac{1}{{\sqrt 3 }},\,\,\frac{1}{{\sqrt 3 }}\,,\,\frac{ -1}{{\sqrt 3 }}} \right)$

$now\,\,Dc's\,\,of\,diagonal\,\,BE = \left( {\frac{1}{{\sqrt 3 }},\,\,\frac{ - 1}{{\sqrt 3 }}\,,\,\frac{1}{{\sqrt 3 }}} \right)$

$now\,\,Dc's\,\,of\,diagonal\,\,AF = \left( {\frac{ - 1}{{\sqrt 3 }},\,\,\frac{1}{{\sqrt 3 }}\,,\,\frac{1}{{\sqrt 3 }}} \right)$

lets dcs of given line which is making

${ \theta _{ 1 } },{ \theta _{ 2 } },{ \theta _{ 3 } },{ \theta _{ 4 } },$ with the diagonal be l, m, n now applying between two angle

$\begin{array}{l} cos{ \theta _{ 1 } }=l\left( { \frac { 1 }{ { \sqrt { 3 } } } } \right) +m\, \left( { \frac { 1 }{ { \sqrt { 3 } } } } \right) \, +n\, \left( { \frac { 1 }{ { \sqrt { 3 } } } } \right) \to \left( 1 \right) \\ cos{ \theta _{ 2 } }=l\left( { \frac { 1 }{ { \sqrt { 3 } } } } \right) -m\, \left( { \frac { 1 }{ { \sqrt { 3 } } } } \right) \, +n\, \left( { \frac { 1 }{ { \sqrt { 3 } } } } \right) \to \left( 2 \right) \\ cos{ \theta _{ 3 } }=l\left( { \frac { 1 }{ { \sqrt { 3 } } } } \right) +m\, \left( { \frac { 1 }{ { \sqrt { 3 } } } } \right) \, -n\, \left( { \frac { 1 }{ { \sqrt { 3 } } } } \right) \to \left( 3 \right) \\ cos{ \theta _{ 4 } }=-l\left( { \frac { 1 }{ { \sqrt { 3 } } } } \right) +m\, \left( { \frac { 1 }{ { \sqrt { 3 } } } } \right) \, +n\, \left( { \frac { 1 }{ { \sqrt { 3 } } } } \right) \to \left( 4 \right) \end{array}$

squaring and adding equation (1), (2) , (3) (4)

$co{s^2}{\theta _1} + co{s^2}{\theta _2} + co{s^2}{\theta _3} + co{s^2}{\theta _4} =$

$\frac { 1 }{ 3 } \left[ { { l^{ 2 } }+{ m^{ 2 } }+{ n^{ 2 } }+2lm+2mn+2\ln { + } { l^{ 2 } }+{ m^{ 2 } }+{ n^{ 2 } }+2lm-2mn-2\ln { + } { l^{ 2 } }+{ m^{ 2 } }+{ n^{ 2 } }-2lm-2mn+2\ln { + } { l^{ 2 } }+{ m^{ 2 } }+{ n^{ 2 } }-2lm+2mn-2\ln { } } \right]$

$\frac { 1 }{ 3 } \left[ { 4\left( { { l^{ 2 } }+{ m^{ 2 } }+{ n^{ 2 } } } \right) } \right]$

$\frac { 1 }{ 3 } \left( 4 \right) \, \, \, \, \, \, \to \left( { { l^{ 2 } }+{ m^{ 2 } }+{ n^{ 2 } } } \right) =1$

$\frac { 4 }{ 3 }$

$co{ s^{ 2 } }{ \theta _{ 1 } }+co{ s^{ 2 } }{ \theta _{ 2 } }+co{ s^{ 2 } }{ \theta _{ 3 } }+co{ s^{ 2 } }{ \theta _{ 4 } }=\frac { 4 }{ 3 }$

1130131_1109841_ans_4be60d08ceee4b978680411093812945.JPG

Find the vector equation of the line joining $(1,2,3)$ and $(-3,4,3)$ and show pependicular to the z-axis

Vector equation passing through

$(1, 2, 3)(-3, 4, 3)$ is

$\bar{r}=\bar{a}+\lambda (b-a)$

$a=i+2j+3k, b=-3i+4j+3k$

$\bar{r}=(\bar{i}+2j+3k)+\lambda(-3i+4j+3k-i-2j-3k)$

$\therefore$ Equation of line

$=(ii+2j+3k)+\lambda (-4i+2j)$

To prove perpendicular to z-axis.

$a_1a_2+b_1b_2+c_1c_2=0$

$-4i+zj$

$a_1=4, b_1=2, c_1=0$

Direct rations

$\Rightarrow 0i+0j+1k$

$a_2=, b_2=0, c_2=1$

$\therefore a_1a_2+b_1b_2+c_1c_2=0+0+0=0$

Hence proved.

The cartesian equation of a line is $\dfrac{{x + 3}}{2} = \dfrac{{y - 5}}{4} = \dfrac{{z + 6}}{2}$ find the vector equation of the line?

Consider the Cartesian equation,

$\dfrac{x+3}{2}=\dfrac{y-5}{4}=\dfrac{z+6}{2}$

Now,

$\overrightarrow{a}=-3\widehat{i}+5\widehat{j}-6\widehat{k}$

$\overrightarrow{b}=2\widehat{i}+4\widehat{j}+2\widehat{k}$

Vector equation,

$\overrightarrow{r}=\overrightarrow{a}+\lambda .\overrightarrow{b}$

$=-3\widehat{i}+5\widehat{j}-6\widehat{k}+\lambda \left( 2\widehat{i}+4\widehat{j}+2\widehat{k} \right)$

This is the required equation.

If the d.c's of a line are $\left( {\frac{1}{c},\,\frac{1}{c},\,\frac{1}{c}} \right)$ , find c.

d.c's line

$\to \,\frac{1}{c},\frac{1}{c},\frac{1}{c}$

${\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = 1$

$\frac{1}{{{c^2}}} + \frac{1}{{{c^2}}} + \frac{1}{{{c^2}}} = 1$

$\frac{3}{{{c^2}}} = 1$

${c^2} = 3$

$c = \pm \sqrt 3$

the value of c is

$\pm \sqrt 3$

If a line makes angles $\alpha,\beta,\gamma$ with positive directions of X,Y,Z-axes, what is the value of ${sin}^{2}{\alpha}\ +\ {sin}^{2}{\beta} +\ {sin}^{2}{\gamma}$ ?

Given , A line make a angle

$\alpha ,\beta ,\chi ,$ with x , y , z axis , then

let length of line be

$\left| r \right|$ x , y , z be the intercept mode by line on the three coordinates axis

$\cos \alpha = \frac{x}{r}\,\,\,\,\,\,\cos \beta = \frac{y}{r}\,\,\,\,\cos \chi = \frac{z}{r}$

${\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\chi = \frac{{{r^2}}}{{{r^2}}}$

${\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\chi = 1$

$1 - {\sin ^2}\alpha + 1 - {\sin ^2}\beta + 1 - {\sin ^2}\chi = 1$

$\therefore \,\,\,{\sin ^2}\alpha + {\sin ^2}\beta + {\sin ^2}\chi = 2$

so answer is 2

1127987_1109184_ans_130f47075f7a4058a2fa7684fc82fb39.jpg

Find the direction ratio of the line $\dfrac{x-1}{2}=3y=\dfrac{2z+3}{4}$

$\cfrac { x-1 }{ 2 } =\cfrac { y }{ \cfrac { 1 }{ 3 } } =\cfrac { z+\cfrac { 3 }{ 2 } }{ 2 }$

$\therefore$ Directions of line

$=(2,\cfrac { 1 }{ 3 } ,2)$

If a line makes angle $\left( {\alpha ,\,\beta ,\,\gamma } \right)$ with the co-ordinates axes. Prove that ${\sin ^2}\alpha \, + \,{\sin ^2}\beta \, - \,{\cos ^2}\gamma \, = \,1$

$\begin{matrix} l=\cos \alpha \\ m=\cos \beta \\ n=\cos \gamma \\ \end{matrix}$

Now

$\begin{matrix} { l^{ 2 } }+{ m^{ 2 } }+{ n^{ 2 } }=1 \\ { \cos ^{ 2 } }\alpha +{ \cos ^{ 2 } }\beta +{ \cos ^{ 2 } }\gamma =1 \\ 1-{ \sin ^{ 2 } }\alpha +1-{ \sin ^{ 2 } }\beta +{ \cos ^{ 2 } }\gamma =1 \\ 2-{ \sin ^{ 2 } }\alpha -{ \sin ^{ 2 } }\beta +{ \cos ^{ 2 } }\gamma =1 \\ { \sin ^{ 2 } }\alpha +{ \sin ^{ 2 } }\beta +1-{ \cos ^{ 2 } }\gamma -2=0 \\ { \sin ^{ 2 } }\alpha +{ \sin ^{ 2 } }\beta -{ \cos ^{ 2 } }\gamma -1=0 \\ { \sin ^{ 2 } }\alpha +{ \sin ^{ 2 } }\beta -{ \cos ^{ 2 } }\gamma =1 \\ \end{matrix}$

Find the equation of line passing through $(5, 0, 5)$ & $(2, 1, 3)$ . Also show that $(5, 0, 5), (2, 1, 3)$ & $(-4, 3, -1)$ are collinear.

Equation of line passing through

$(5,0,5)$ and

$(2,1,3)$ is:

$\cfrac { x-5 }{ 3 } =\cfrac { y-0 }{ -1 } =\cfrac { z-5 }{ 2 } \\ (x,y,z)\equiv (-4,3,-1)\\ \cfrac { -4-5 }{ 3 } =\cfrac { 3 }{ -1 } =\cfrac { -1-5 }{ 2 } =-3$

$\therefore$ point

$(-4,3,-1)$ satisfies equation of line hence the three points

$(5,0,5),(2,1,3)$ &

$(-4,3,-1)$ are collinear as they fall on one same line.

Find the vector equation of the line joining points $2\hat i + \hat j + 3\hat k$ and $- 4\hat i + 3\hat j - \hat k$ .

Consider the given points be

$A$ and

$B$

$\begin{array}{l} \overrightarrow { OA } =2\hat { i } +\hat { j } +3\hat { k } \\ \\ \overrightarrow { OB } =-4\hat { i } +3\hat { j } -\hat { k } \end{array}$

Therefore, coordinates of point

$A$ and

$B$ are

$(2,1,3)$ and

$(-4,3,-1)$ respectively.

Direction ratios of the line

$AB$ are proportional to

$(-4-2),(3-1),(-1-3)$

i.e.

$3,-1,2$

Thus vector equation of the line is

$\vec r=\vec a+\lambda \vec b$

$\vec r=(2\hat i+\hat j +3\hat k)+\lambda (3\hat i-\hat j+2\hat k)$

Find the measure of the angle between two lines if their direction cosines $\ell, m, n$ satisfy $\ell+m-n=0, \ell^{2}+m^{2}-n^{2}=0$ .

given ocs of two line satisfy

l + m - n = 0 equation (1)

${l^2} + {m^2} - {n^2} = 0\,\,equation\,(2)$

to add angle between from (1) l = n - m in equation (2)

$\begin{array}{l} \Rightarrow { \left( { n-m } \right) ^{ 2 } }+{ m^{ 2 } }-{ n^{ 2 } }=0 \\ { n^{ 2 } }+{ m^{ 2 } }-2nm+{ m^{ 2 } }-{ n^{ 2 } }=0 \\ 2{ m^{ 2 } }-2nm=0 \\ 2m\left( { m-n } \right) =0 \end{array}$

m = 0 equation (3) and m - n = 0 equation (4)

l + m - n = 0

0 + m - n = 0

$\dfrac{l}{{0 + 1}} = \dfrac{m}{0} = \dfrac{n}{1}$

l:m:n = 1 : 0 : 1

equation 1 and 4

l + m - n = 0

0 + m - n = 0

$\dfrac{l}{0} = \dfrac{m}{{ - \left( { - 1} \right)}} = \dfrac{n}{1}$

l : m : n = 0 : 1 : 1

$\cos \theta = \dfrac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {{a_1}^2 + {b_1}^2 + {c_1}^2} \sqrt {{a_2}^2 + {b_2}^2 + {c_2}^2} }}$

$= \dfrac{{\left( 1 \right)\left( 0 \right) + \left( 0 \right)\left( 0 \right) + \left( 1 \right)\left( 1 \right)}}{{\sqrt {1 + 0 + 1} \sqrt {0 + 1 + 1} }} = \dfrac{1}{2}$

$\cos \theta = \dfrac{1}{2} \Rightarrow \theta = \dfrac{\pi }{2}$

The angle between the lines whose direction cosines satisfy the equations $l+m+n=0$ and $l^2=m^2+n^2$ is?

Angle between two lines is:

$\begin{array}{l} \cos \theta =\dfrac { { { a_{ 1 } }{ a_{ 2 } }+{ b_{ 1 } }{ b_{ 2 } }+{ c_{ 1 } }{ c_{ 2 } } } }{ { \sqrt { { a_{ 1 } }^{ 2 }+{ b_{ 1 } }^{ 2 }+{ c_{ 1 } }^{ 2 } } \sqrt { { a_{ 2 } }^{ 2 }+{ b_{ 2 } }^{ 2 }+{ c_{ 2 } }^{ 2 } } } } \\ l+m+n=0.......(1) \\ l=-\left( { m+n } \right) \\ { \left( { m+n } \right) ^{ 2 } }={ l^{ 2 } }........(2) \\ { m^{ 2 } }+{ n^{ 2 } }+2mn={ m^{ 2 } }+{ n^{ 2 } } \\ \left[ { { l^{ 2 } }={ m^{ 2 } }+{ n^{ 2 } },\, given } \right] \\ 2mn=0 \\ When\, m=0\Rightarrow l=-n \\ Hence,\left( { l,m,n } \right) \, is\, \left( { 1,0,-1 } \right) \\ When\, n=0,\, then\, l=-m \\ Hence,\, \left( { l,m,n } \right) \, is\, \left( { 1,0,-1 } \right) \\ \cos \theta =\dfrac { { 1+0+0 } }{ { \sqrt { 2 } \sqrt { 2 } } } =\dfrac { 1 }{ 2 } \Rightarrow \theta =\dfrac { \pi }{ 3 } \end{array}$

Find the direction cosines of the vector $\vec{r}=(6\hat{i}+2\hat{j}-3\hat{k})$ .

Given

$\vec{r} = (6\hat{i} + 2\hat{j} - 3\hat{k})$

The direction ratios of this vector are

$6, 2, -3$ .

The direction cosines are

$\dfrac{6}{\sqrt{6^2 + 2^2 + 3^2}}, \dfrac{2}{\sqrt{49}}, \dfrac{-3}{\sqrt{49}}$

$\Rightarrow \pm \dfrac{6}{7}, \pm \dfrac{2}{7}$

Given three points whose position are vectors $x\hat i + y\hat j + z\hat k$ and $- \hat i - \hat j$ . Find the condition for the points to be collinear.

1130450_1112816_ans_bd0f2f0191004d44adf90262896dfb2a.jpg

Using vectors, find the value of $\lambda$ such that the points $(\lambda, -10, 3), (1, -1, 3)$ and $(3, 5, 3)$ are collinear.

Let the given points be

$\begin{array}{l} A\left( { \lambda ,-10,3 } \right) ,\, B\left( { 1,-1,3 } \right) \, and\, C\left( { 3,5,3 } \right) \\ \overrightarrow { AB } =\left( { \widehat { i } -\widehat { j } +3\widehat { k } } \right) -\left( { \lambda \widehat { i } -10\widehat { j } +3\widehat { k } } \right) =\left( { 1-\lambda } \right) \widehat { i } +9\widehat { j } \\ \overrightarrow { AC } =\left( { 3\widehat { i } +5\widehat { j } +3\widehat { k } } \right) -\left( { \lambda \widehat { i } -10\widehat { j } +3\widehat { k } } \right) =\left( { 3-\lambda } \right) \widehat { i } +15\widehat { j } \\ If\, the\, points\, A,B,C\, are\, collinear,\, then\, \overrightarrow { AB } =K\overrightarrow { AC } \, for\, some\, scalar\, K \\ \left( { 1-\lambda } \right) \widehat { i } +9\widehat { j } =K\left[ { \left( { 3-\lambda } \right) \widehat { i } +15\widehat { j } } \right] \\ 1-\lambda =K\left( { 3-\lambda } \right) \, \, \, and\, 9=15K \\ 1-\lambda =\frac { 3 }{ 5 } \left( { 3-\lambda } \right) \\ 5-5\lambda =9-3\lambda \\ 2\lambda =-4 \\ \lambda =-2 \end{array}$

Find the vector equation of the line passing through the points $2\hat i + 3\hat j + \hat k$ and parallel to the vector $4\hat i - 2\hat j + 3\hat k$ .

Consider the problem

Let,

$\begin{array}{l} \vec { a } =2\hat { i } +3\hat { j } +\hat { k } \\ \\ \vec { b } =4\hat { i } -2\hat { j } +3\hat { k } \end{array}$

Therefore, vector equation of the line is

$\vec r=\vec a+\lambda \vec b$

because the line is passing through point A and is parallel to line B

$=(2\hat i+3\hat j+\hat k)+\lambda (4\hat i-2\hat j+3\hat k)$

Show that points $A\left( a,b+c \right), B\left( b,c+a \right), C\left( c,a+b \right)$ are collinear.

The points are A(a,b+c), B(b,c+a), C(c,a+b)

If the area of triangle is zero then the points are called collinear points.

if three points (

$x_1,y_1$ ), (

$x_2,y_2$ ), (

$x_3,y_3$ ) are collinear then :

$[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$ =0

$[a(c+a-a-b)+b(a+b-b-c)+c(b+c-c-a)]=0$

$[ac-ab+ab-bc+bc-ac]=0$

The points A(a,b+c), B(b,c+a), C(c,a+b) are collinear.

The number of straight line that are equally inclined to the three dimensional co- ordinate axes , is

$4$ lines.
1241161_1156794_ans_70d5afbe5b6b4e47a76c50827a38042a.jpg

If a unit vector $\overrightarrow { a }$ makes angles $\cfrac{\pi}{3}$ with $\hat { i }$ , $\cfrac{\pi}{4}$ with $\hat { j }$ and an acute angle $\theta$ with $\hat { k }$ , then find $\theta$ and hence, the components of $\overrightarrow { a }$

Let

$\alpha =\cfrac { \pi }{ 3 } ,\beta =\cfrac { \pi }{ 4 } ,\gamma =\theta$ be the angles made with

$\hat { i } \hat { j } \hat { k }$ axes.

$\therefore \cos { ^{ 2 }\alpha } +\cos { ^{ 2 }\beta } +\cos { ^{ 2 }\gamma } =1\\ \Rightarrow \cos { ^{ 2 }\gamma } =1-\cos { ^{ 2 }\cfrac { \pi }{ 3 } } -\cos { ^{ 2 }\cfrac { \pi }{ 4 } } \\ \Rightarrow \cos { ^{ 2 }\gamma } =\cfrac { 1 }{ 4 } \\ \Rightarrow \gamma =\cfrac { \pi }{ 3 }$

$\therefore$ the acute angle

$\theta$ that it makes with

$\hat { k }$ is

$\cfrac { \pi }{ 3 }$ .

Hence,components of

$\overrightarrow { a }$ along

$\hat { i } ,\hat { j } ,\hat { k }$ are

$\hat { i } ,\hat { j } ,\hat { k }$

$\Rightarrow \cfrac { 1 }{ 2 } ,\cfrac { 1 }{ \sqrt { 2 } } ,\cfrac { 1 }{ 2 }$

Find the equation of plane passing through the point $(3,2,4)$ and perpendicular to the $X-$ axis.

$\begin{array}{l} \overrightarrow { a } =\left( { 3\widehat { i } +2\widehat { j } +4\widehat { k } } \right) \\ \overrightarrow { n } =\widehat { i } \\ Equation\, \, of\, plane. \\ \left( { \overrightarrow { r } -\overrightarrow { a } } \right) .\overrightarrow { n } =0 \\ \left( { \overrightarrow { r } -\left( { 3\widehat { i } +2\widehat { j } +4\widehat { k } } \right) } \right) .\overrightarrow { i } =0 \end{array}$

The direction cosines of a line are $\dfrac{1}{2}, \dfrac{1}{2}, k$ and $k > 0$ , then $k=?$

Given directional cosines

$l,m,n$ are

$\cfrac{1}{2},\cfrac{1}{2},k$

$l^2+m^2+n^2=1\\ \cfrac{1}{4}+\cfrac{1}{4}+k^2=1\\ \cfrac{1}{2}+k^2=1\\ k^2=\dfrac{1}{2}\\ k=\dfrac{1}{ \sqrt2} \quad or\quad \dfrac{-1}{\sqrt2}$

Given that

$k>0$

$\therefore k=\dfrac{1}{\sqrt2}$

The cartasian equation of a line is $\dfrac {x-5}{3}=\dfrac {y+4}{7}=\dfrac {z-6}{2}$ . Find its vector equation.

The given Cartesian equation of the line is

$\dfrac{{x - 5}}{3} = \dfrac{{y + 4}}{7} = \dfrac{{z - 6}}{2}$

That is,

$\dfrac{{x - 5}}{3} = \dfrac{{y - \left( { - 4} \right)}}{7} = \dfrac{{z - 6}}{2}$

Therefore,

the line passes through the point whose position vector

$\vec a = 5\hat i - 4\hat j + 6\hat k$

And, parallel to the vector

$\vec b = 3\hat i + 7\hat j + 2\hat k$

Thus, the equation is

$\vec r=\vec a+\lambda \vec b$

Therefore,

$\vec r = 5\hat i - 4\hat j + 6\hat k + \lambda \left( {3\hat i + 7\hat j + 2\hat k} \right)$

where

$\lambda$ is a parametre.

Show that the points $A(-2\hat{i}+3\hat{j}+5\hat{k}), B(\hat{i}+2\hat{j}+3\hat{k})$ and $C(7\hat{i}-\hat{k})$ are collinear.

$\begin{array}{l} A\left( { -2,3,5 } \right) B\left( { 1,2,3 } \right) C\left( { 7,0,-1 } \right) \\ \overrightarrow { AB } =\left( { 3,-1,-2 } \right) \\ \overrightarrow { AC } =\left( { 9,-3,-6 } \right) \\ \overrightarrow { AC } =3\overrightarrow { AB } \\ { { Im } }plies\, A,B\, and\, C\, are\, collinear \end{array}$

Show that the points with position vectors $\overrightarrow{a}+\overrightarrow{b}, \overrightarrow{a}-\overrightarrow{b}$ and $\overrightarrow{a}+k\overrightarrow{b}$ are collinear for all values of $k$ .

If the three points are

$P,Q,R$ respectively, then

$\overrightarrow{PQ}\times \overrightarrow{PR}=\left(\overrightarrow{OQ}-\overrightarrow{OP}\right)\times \left(\overrightarrow{OR}-\overrightarrow{OP}\right)$

$=\left(\overrightarrow{a}-\overrightarrow{b}-\overrightarrow{a}-\overrightarrow{b}\right)\times \left(\overrightarrow{a}+\hat{k}\overrightarrow{b}-\overrightarrow{a}-\overrightarrow{b}\right)$

$=\left(-2\overrightarrow{b}\right)\times \left(\left(k-1\right)\overrightarrow{b}\right)$

$=0$ since

$\overrightarrow{b}\times \overrightarrow{b}=0$

Find the equation of plane passing through the point $(2-1,3)$ and perpendicular to the $X-$ axis.

$\begin{array}{l} \overrightarrow { a } =2\widehat { i } -\widehat { j } +3\widehat { k } \\ \overrightarrow { n } =\widehat { i } \\ Equation\, \, of\, plane. \\ \left( { \overrightarrow { r } -\overrightarrow { a } } \right) .\overrightarrow { n } =0 \\ \left( { \overrightarrow { r } -\left( { 2\widehat { i } -\widehat { j } +3\widehat { k } } \right) } \right) .\overrightarrow { i } =0 \\ x-2=0 \end{array}$

Show that
$A(1,-2,-8),B(5,0,-2)$ and $C(11,3,7)$ are collinear.

The given points are A(

$1,-2,-8$ ), B(

$5,0,-2$ ), C(

$11,3,7$ )

$\begin{array}{l} \therefore \overrightarrow { AB } =\left( { 5-1 } \right) \widehat { i } +\left( { 0+2 } \right) \widehat { j } +\left( { -2+8 } \right) \widehat { k } =4\widehat { i } +2\widehat { j } +6\widehat { k } \\ \overrightarrow { BC } =\left( { 11-5 } \right) \widehat { i } +\left( { 3-0 } \right) \widehat { j } +\left( { 7+2 } \right) \widehat { k } =6\widehat { i } +3\widehat { j } +9\widehat { k } \\ \overrightarrow { AC } =\left( { 11-1 } \right) \widehat { i } +\left( { 3+2 } \right) \widehat { j } +\left( { 7+8 } \right) \widehat { k } =10\widehat { i } +5\widehat { j } +15\widehat { k } \\ \left| { \overrightarrow { AB } } \right| =\sqrt { { 4^{ 2 } }+{ 2^{ 2 } }+{ 6^{ 2 } } } =\sqrt { 16+4+36 } =\sqrt { 56 } =2\sqrt { 14 } \\ \left| { \overrightarrow { BC } } \right| =\sqrt { { 6^{ 2 } }+{ 3^{ 2 } }+{ 9^{ 2 } } } =\sqrt { 36+9+81 } =\sqrt { 126 } =3\sqrt { 14 } \\ \left| { \overrightarrow { AC } } \right| =\sqrt { { { 10 }^{ 2 } }+{ 5^{ 2 } }+{ { 15 }^{ 2 } } } =\sqrt { 100+25+225 } =\sqrt { 350 } =5\sqrt { 14 } \\ \therefore \left| { \overrightarrow { AC } } \right| =\left| { \overrightarrow { AB } } \right| +\left| { \overrightarrow { BC } } \right| \end{array}$

Thus, the given points A,B, and C are collinear.

If the direction cosines of a line are $\left(\dfrac{1}{2},0,n\right)$ and $n<0$ . Then find $n$ .

$\cos^2\alpha+\cos^2\beta +\cos^2\gamma =1$

$\dfrac{1}{4}+0+n^2=1$

$n^2=\dfrac{3}{4}$

$n=\pm \dfrac{\sqrt{3}}{2}$

But

$n < 0$

So,

$n=\dfrac{-\sqrt{3}}{2}$ .

Determine whether the points are collinear.
$P(-2,3),B(1,2),C(4,1)$

Slope of PB

$= \dfrac{2-3}{1-(-2)}$

$= \dfrac{-1}{3}$

Slope of BC

$= \dfrac{1-2}{4-1}$

$= \dfrac{-1}{3}$

Since slope are Equal

Points are colliner.

1118052_1163953_ans_473a7f3173004324a0cea96285fda818.jpeg

The position vectors $\vec {a},\vec {b},\vec {c}$ of three points satisfy the relation $2\vec {a}+7\vec {b}+5\vec {c}=\vec {0}$ . Are these points collinear?

$2\vec{a}+7\vec{b}+5\vec{c} = 0$

$\therefore 7\vec{b} = -5\vec{c}-2\vec{a}$

$\therefore \vec{b} = \dfrac{5\vec{c}-2\vec{a}}{7}$

$\therefore \vec{b} = -\dfrac{(5\vec{c}+2\vec{a})}{7}$

Above indicates

$\vec{b}$ divides

$\vec{c}$ and

$\vec{a}$ internally

in ratio

$5:2$

Hence points are collinear.

1118091_1163999_ans_0d3227a307c64587ae93b9f16f048b9f.jpeg

If a line makes angles $\alpha ,\beta$ & $\gamma$ with $OX,OY$ & $OZ$ respectively Prove that: (i) $\sin ^{ 2 }{ \alpha } +\sin ^{ 2 }{ \beta } +\sin ^{ 2 }{ \gamma } =2$

$\begin{array}{l} { \cos ^{ 2 } }\alpha +{ \cos ^{ 2 } }\beta +{ \cos ^{ 2 } }\gamma =1 \\ 1-{ \sin ^{ 2 } }\alpha +1-{ \sin ^{ 2 } }\beta +1-{ \sin ^{ 2 } }\gamma =1 \\ 3-1={ \sin ^{ 2 } }\alpha +{ \sin ^{ 2 } }\beta +{ \sin ^{ 2 } }\gamma \\ 2={ \sin ^{ 2 } }\alpha +{ \sin ^{ 2 } }\beta +{ \sin ^{ 2 } }\gamma \end{array}$

The direction ratios of a vector are $2, -3, 4$ . Find its direction cosines

Direction Ratios are

$2,-3,4$ .

Direction cosines.

$l=\dfrac{2}{\sqrt{2^{2}+(-3)^{2}+4^{2}}}=\dfrac{2}{\sqrt{29}}$

$m=\dfrac{-3}{\sqrt{2^{2}+(-3)^{2}+4^{2}}}=\dfrac{-3}{\sqrt{29}}$

$n=\dfrac{4}{\sqrt{2^{2}+(-3)^{2}+4^{2}}}=\dfrac{4}{\sqrt{29}}$

Obtain the equation of the plane which passes through (3, 4, -5) and (1, 2, 3) and parallel to Z axis.

Vector in plane

$= (1, 2, 3) - (3, 4, 5)$

$= (-2, -2, -2)$
vector correction for 2 axis

$(0, 0, 1)$
Normal vector =

$\begin{vmatrix} \hat{J} & \hat{i} & \hat{k} \\ -2 & -2 & -2 \\ 0 & 0 & 1 \end{vmatrix}$

$= \dot{J} (-2) - \hat{i} (-2) + 0$

$= -2\hat{J} + 2 \hat{i}$
Equation of plane

$-2 (x - 3) + 2(y - 4) + 0 = 0$

$-2x + 6 + 2y - 8 = 0$

$-2x + 2y - 2 = 0$

$- x + y - 1 = 0$

$x - y + 1 = 0$

Prove that the points $A(4,-3,-1), B(5,-7,6)$ and $C(3,1,-8)$ are collinear.

$\overrightarrow{AB} = (5-4)\hat{i}+(-7+3)\hat{j}+(6+1)\hat{k}$

$=\hat{i}-4\hat{j}+7\hat{k}$

$\overrightarrow{AC} =(3-4)\hat{i}+(1+3)\hat{j}+(-8+1)\hat{k}$

$=-\hat{i}+4\hat{j}-7\hat{b}$

$=-(\hat{i}-4\hat{j}+7\hat{k})$

$\overrightarrow{AB} =(-1) \overrightarrow{AC}$

$\therefore A,B,C$ are collinear

Find the angle between the pairs of lines with direction ratios proportional to
(i) $5 , - 12,13 \text { and } - 3,4,5$
(ii) $2,2,1 \text { and } 4,1,8$
(iii) $1,2 , - 2 \text { and } - 2,2,1$
(iv) $a , b , c \text { and } b - c , c - a , a - b$

1627153_1209633_ans_e47c7d645c9343c18ccacc8327cbbd19.jpg

Find the value of p so the line $\dfrac{1-x}{3}=\dfrac{7y-14}{2p}=\dfrac{z-3}{2}$ and $\dfrac{7-7x}{3p}=\dfrac{y-5}{1}=\dfrac{6-z}{5}$ are at right angles.

The given equations can be written as

$\dfrac{x-1}{-3}=\dfrac{y-2}{\dfrac{2p}{7}}=\dfrac{z-3}{2}$ and

$\dfrac{x-1}{-\dfrac{3p}{7}}=\dfrac{y-5}{1}=\dfrac{z-6}{-5}$

If the given lines are at right angles, then,

$-3(-\dfrac{3p}{7})+\dfrac{2p}{7}(1)+2(-5)=0$

$\dfrac{9p}{7}+\dfrac{2p}{7}-10=0$

$\dfrac{11p}{7}=10$

$11p=70$

$p=\dfrac{70}{11}$

If a line makes $\theta _ { 1 } , \theta _ { 2 } , \theta _ { 3 }$ angles with the co-ordinates axes, then prove that $\cos 2 \theta _ { 1 } + \cos 2 \theta _ { 2 } + \cos 2 \theta _ { 3 } + 1 = 0$

We know,

${ \cos }^{ 2 }{ \theta }_{ 1 }+{ \cos }^{ 2 }{ \theta }_{ 2 }+{ \cos }^{ 2 }{ \theta }_{ 3 }=1$

$\Rightarrow { 2\cos }^{ 2 }{ \theta }_{ 1 }+{ 2\cos }^{ 2 }{ \theta }_{ 2 }+{ 2\cos }^{ 2 }{ \theta }_{ 3 }=2$

$\Rightarrow { 2\cos }^{ 2 }{ \theta }_{ 1 }-1+2{ \cos }^{ 2 }{ \theta }_{ 2 }-1+2{ \cos }^{ 2 }{ \theta }_{ 3 }-1=2-1-1-1$

$\Rightarrow { \cos2\theta }_{ 1 }+{ \cos2\theta }_{ 2 }+{ \cos2\theta }_{ 3 }=-1$

$\Rightarrow { \cos2\theta }_{ 1 }+{ \cos2\theta }_{ 2 }+{ \cos2\theta }_{ 3 }+1=0$

Proved.

Show that each of the given three vectors is a unit vector:
$\frac{1}{7}(2\bar{i}+3\bar{j}+6\bar{k},)$ $\frac{1}{7}(3\bar{i}+6\bar{j}+\bar2{k}),$ $\frac{1}{7}(6\bar{i}+2\bar{j}+3\bar{k})$

$\dfrac{1}{7}|(2\widehat{i}+3\widehat{j}+6\widehat{k})|=\dfrac{1}{7}\sqrt{4+9+36}=\dfrac{7}{7}=1$

$\dfrac{1}{7}|3\widehat{i}+6\widehat{j}+\widehat{k}|=\dfrac{1}{7}\sqrt{9+36+4}=\dfrac{7}{7}=1$

$\dfrac{1}{7}|6\widehat{i}+2\widehat{j}+3\widehat{k}|=\dfrac{1}{7}\sqrt{36+4+9}=\dfrac{7}{7}=1$

Hence, each of these vectors is a unit vector.

Find the vector equation of the line through A(3,4,-7) and B(6,-1,1).

The line passes through A(3, 4, -7) and B(6,-1, 1).

Hence, the equation of line is

$\overrightarrow{r}= \overrightarrow{a}+\lambda(\overrightarrow{b}-\overrightarrow{a})$

$\overrightarrow{r}$ =

$3\widehat{i} +4\widehat{j}-7\widehat{k}$

$+\lambda$

$(3\widehat{i}-5\widehat{j}+8\widehat{k})$

If a line makes angles $90^{o}, 135^{o}, 45^{o}$ with x, y and z-axis respectively, find its direction cosines.

Let direction cosines of the line be l, m and n.

$l=\cos 90^{o}=0$

$m=\cos 135^{o}=\dfrac{-1}{\sqrt{2}}$

$n=\cos 45^{o}=\dfrac{1}{\sqrt{2}}$

Therefore, the direction cosines of the line are

$0, \dfrac{-1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}}$ .

Find the angles at which the normal vector to the plane $4x+8y+z=5$ is inclined to the coordinate axes.

Let

$\overrightarrow n$ be a vector normal to the plane. Since direction ratios of normal to the plane are proportional to 4, 8, 1.

Therefore,

$\overrightarrow n = 4\widehat{i}+8\widehat{j}+\widehat{k}$

Direction cosines of

$\overrightarrow n$ are

$\dfrac{4}{9}, \dfrac{8}{9}, \dfrac{1}{9}$

Let

$\alpha, \beta, \gamma$ be the angles made by

$\overrightarrow r$ with x-axis, y-axis and z-axis respectively.

Then,

$\alpha = cos^{-1}(\dfrac{4}{9}), \beta = cos^{-1}(\dfrac{8}{9}), \gamma = cos^{-1}(\dfrac{1}{9})$

Write the equation of the plane passing through the point $\left({x}_{1},{y}_{1},{z}_{1}\right)$ and having $a, b, c$ , as direction ratios of its normal.

Equation of plane passing through the point

$(x_1 , y_1 , z_1)$ and (a,b,c)

and having direction ratio of its normal.

$\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$

1194113_1368823_ans_2bae3b655c2943499383f64829734fc6.JPG

If $\vec{PO}+\vec{OQ}=\vec{QO}+\vec{OR}$ , prove that the points $P, Q, R$ are collinear.

$\overrightarrow{PO} + \overrightarrow{OQ}=\overrightarrow{QO}+\overrightarrow{OR}$

$\Rightarrow \overrightarrow{O} - \overrightarrow{P}+ \overrightarrow{q} - \overrightarrow{O} = \overrightarrow{O} - \overrightarrow{q} + \overrightarrow{r} - \overrightarrow{O}$

$\Rightarrow\overrightarrow{q} - \overrightarrow{p} = \overrightarrow{r}-\overrightarrow{q}$

$\Rightarrow 2\overrightarrow{q}= \overrightarrow{p} + \overrightarrow{r}$

$\Rightarrow \overrightarrow{q} = \dfrac{\overrightarrow{p} + \overrightarrow{r}}{2}$

$\Rightarrow Q$ is mid-point of

$PR$

$\Rightarrow P.Q,R$ are collinear.

A ray makes angles $\dfrac{\pi}{3},\dfrac{\pi}{3}$ with $\overrightarrow { OX }$ and $\overrightarrow { OY }$ respectively. Find the angle made by it with $\overrightarrow { OZ }$ .

$\cos^2\alpha +\cos^2\beta +\cos^2\gamma =1$

$\cos^2\alpha =$ angle with

$\vec{ox}$

$\cos^2\beta =$ angle with

$\vec{oy}$

$\cos^2\gamma =1$ angle with

$\vec{oz}$

$\left(\dfrac{1}{\alpha}\right)^2+\left(\dfrac{1}{\alpha}\right)^2+\cos^2y=1$

$\cos^2y=\dfrac{2}{4}$

$\cos y=\pm\dfrac{1}{\sqrt{2}}$

So, angle made with

$\vec{oz}=45^o$ .

Find the equation of the line passing through the point $(-1, 2, 1)$ and parallel to the line $\dfrac {2x - 1}{4} = \dfrac {3y + 5}{2} = \dfrac {2 - z}{3}$ .

Parallel vector to the given line is

$(4\hat{i}+2\hat{j}+3\hat{k})$

And, Position vector of one point on the required line is

$(-\hat{i}+2\hat{j}+\hat{k})$

Thus, Vector equation of line

$\rightarrow \vec{r} = (-\hat{i}+2\hat{j}+\hat{k}) + \lambda(4\hat{i}+2\hat{j}+3\hat{k})$ where

$\lambda \in Z$

Find the vector equation of the coordinates planes.

$1.$ XY plane is perpendicular to z-axis and passes through origin, So its equation is ,

$(\vec{r}-\vec{0}).(\vec{z})=0$

$2$ . YZ plane is perpendicular to x-axis and passes through origin, So its equation is ,

$(\vec{r}-\vec{0}).(\vec{x})=0$

$3.$ ZX plane is perpendicular to y-axis and passes through origin, So its equation is ,

$(\vec{r}-\vec{0}).(\vec{y})=0$

Find the equation of the line joining the points $(-1,3)$ and $(4,-2)$ .

Here, two points are

$(x_{1},y_{1})=(-1,3)$ and

$(x_{2},y_{2})=(4.-2)$

So, the equation of the line in two-point form is :

$y-y_1=\left(\dfrac{y_2-y_1}{x_2-x_1}\right)(x-x_1)$

$\Rightarrow y-3=\dfrac{3-(-2)}{-1-4}(x+1)$

$\Rightarrow y-3=-x-1$

$\therefore x+y-2=0$

The foot of the perpendicular drawn from the origin to the plane is $(4,2,6)$ . Find the equation of the plane.

Since, the foot of the perpendicular to the plane is $A(4,2,6)$ . Therefore $(4,2,6)$ is the point on the plane.

So, equation of the plane passing through the point

$(4 ,2, 6)$ is:

$a(x – 4) + b(y – 2) + c(z – 6) = 0.$

Now, the direction ratios of the perpendicular line

$OA = 4 – 0, 2 – 0, 6 – 0$ , i.e.,

$4, 2, 6.$

Therefore, the required plane is:

$4(x – 4) + 2(y – 2) + 6(z – 6) = 0$

i.e,

$4x + 2y + 6z = 56$ .

Find a normal to the plane $x+2y+3z-6=0$

Equation of the plane is

$x+2y+3z−6=0$

Dividing with

$\sqrt{{1}^{2}+{2}^{2}+{3}^{2}}=\sqrt{1+4+9}=\sqrt{14}$

$\dfrac{1}{\sqrt{14}}x+\dfrac{2}{\sqrt{14}}y+\dfrac{3}{\sqrt{14}}z-\dfrac{6}{\sqrt{14}}=0$

which is the normal form of the plane.

Find the direction cosines if the passing through two points $(-2,4,-5)$ and $(1,2,3).$

The direction ratios of the line passing through

$A(-2,4,-5)$ and

$B(1,2,3)$ are proportional to

$1+2,2-4,3+5$ or,

$3,-2,8.$

So, its direction cosines are

$\dfrac{3}{\sqrt{9+4+64}},\dfrac{-2}{\sqrt{9+4+64}},\dfrac{8}{\sqrt{9+4+64}}$ or

$\dfrac{3}{\sqrt{77}},\dfrac{-2}{\sqrt{77}},\dfrac{8}{\sqrt{77}}$

Find the equation of the plane passing through the point $(2,3,1)$ given that the direction ratios of normal to the plane are proportional to $5, 3,2$

Let the equation of a plane passing through the point

$(2,3,1)$ is

$a(x-2)+b(y-3)+c(z-1)=0$ ....(1) where

$a,b,c$ are the d.rs of the normal to the plane of (1).

Given

$a,b, c$ are proportional to

$5,3,2$ .

Then

$a=5k,b=3k,c=2k$

Putting these values in (1) we get,

$5(x-2)+3(y-3)+2(z-1)=0$

or,

$5x+3y+2z=21$ .

This is the required equation of the plane.

If a line has direction ratio $2,-1,-2,$ determine its direction cosines.

The direction ratio of the line are proportional to

$2,-1,-2.$ So, its direction cosines are

$\dfrac{2}{\sqrt{2^{2}+(-1)^{2}+(-2)^{2}}},\dfrac{-1}{\sqrt{2^{2}+(-1)^{2}+(-2)^{2}}},\dfrac{-2}{\sqrt{2^{2}+(-1)^{2}+(-2)^{2}}}$ or,

$\dfrac{2}{3},\dfrac{-1}{3},\dfrac{-2}{3}$

Let $l_i,m_i,n_i$ $i = 1,2,3$ be the direction cosines of three mutually perpendicular vectors in space. Show that $A A ^ { ' } = I _ { 3 } ,$ where $A = \left[ \begin{array} { l l l } { l _ { 1 } } & { m _ { 1 } } & { n _ { 1 } } \\ { l _ { 2 } } & { m _ { 2 } } & { n _ { 2 } } \\ { l _ { 3 } } & { m _ { 3 } } & { n _ { 3 } } \end{array} \right].$

1394933_1475598_ans_cb8a65379c544772921f0bd3ab0d6fb6.jpg

Find the vector equation of the line passing through the point $(1, 2, -4)$ and parallel to the line $\dfrac {x - 3}{4} = \dfrac {y - 5}{2} = \dfrac {z + 1}{3}$ .

Parallel vector to the given line is

$(4\hat{i}+2\hat{j}+3\hat{k})$

And, Position vector of one point on the required line is

$(\hat{i}+2\hat{j}-4\hat{k})$

Thus, Vector equation of line

$\rightarrow \vec{r} = (\hat{i}+2\hat{j}-4\hat{k}) + \lambda(4\hat{i}+2\hat{j}+3\hat{k})$ where

$\lambda \in Z$

Find the vector equation of a plane which is at 42 unit distance from the origin and which is normal to the vector $2\hat{i}+\hat{j}-2\hat{k}$

$\hat{n}$ is a unit vector along the normal and p i the length of the perpendicular from origin to the plane, then the vector equation of the plane

$\hat{r}.\hat{n}=p$
Hence,

$\bar{n}=2\hat{i}+\hat{j}-2\hat{k}$ and

$p=42$

$\therefore \left | \bar{n} \right |=\sqrt{2^{2}+1^{2}+(-2)^{2}}$

$=\sqrt{9}$

$=3\hat{n}=\dfrac{\bar{n}}{\left | N \right |}$

$=\dfrac{1}{3}\left ( 2\hat{i}+\hat{j}-2\hat{k} \right )$

$\therefore$ the vector equation of the required plane is

$\bar{r}\left [ \dfrac{1}{3}\left ( 2\hat{i}+\hat{j}-2\hat{k} \right ) \right ]=42$

$i.e.,\bar{r}\left ( 2\hat{i}+\hat{j}=2\hat{k} \right )=126$

Write the vector equation of a line whose Cartesian equations are $\cfrac{x-5}{3}=\cfrac{y+4}{7}=\cfrac{6-z}{2}$

1567265_1707834_ans_90f396d508bf4a55b8b7bc97e20f75a5.jpg

Find the vector equation of the line passing through the point vector $\hat{i}+\hat{j}+\hat{k}$ and $2\hat{i}-\hat{j}+\hat{k}$

Let

$\bar{b}=\hat{i}+\hat{j}+k\hat{k}$ and

$\bar{c}=2\hat{i}-\hat{j}+\hat{k}$
The vector perpendicular to the vectors

$\bar{b}$ and

$\bar{c}$ is given by

$\vec{b}\times \vec {c}$

Since, the line is perpendicular to the

$\bar{b}$ and

$\bar{c}$ it is parallel to

$\bar{b}\times \bar{c}$ .

The vector equation of the line passing through

$A(\bar{a})$

$\bar{b}\times \bar{c}$ is

$\bar{r}=\bar{a}+\left ( \bar{b}\times \bar{c} \right )$

Where

$\lambda$ is scalar

Here,

$\bar{a}=\hat{i}=\hat{j}+\hat{k}$

Here , the vector equation of the required line is

$\bar{r}=\left ( \hat{i}+\hat{j}-3\hat{k} \right )+\lambda (2\hat{i}+\hat{j}-3\hat{k})$

Show that the points $A(2,3,4),B(-1,-2,1)$ and $C(5,8,7)$ are collinear.

D.r's of

$AB$ are

$(3+2),(−6−4),(−8−7)$ ie.,

$5,−10,−15$ ie.,

$1,−2,−3$

Dr's of

$AC$ are

$(1+2),(−2−4),(−2−7)$ ie.,

$3,−6,−9$ ie.,

$1,−2,−3$

$AB∥AC$ and both have a common end point

$A$

Hence

$A,B,C$ are collinear

Find the equation of the line passing through the point $(2, -1, 3)$ and parallel to the line $\vec {r} = (\hat {i} - 2\hat {j} + \hat {k}) + \lambda (2\hat {i} + 3\hat {j} - 5\hat {k})$ .

Parallel vector to the given line is

$(2\hat{i}+3\hat{j}-5\hat{k})$

And, Position vector of one point on the required line is

$(2\hat{i}-\hat{j}+3\hat{k})$

Thus, Vector equation of line

$\rightarrow \vec{r} = (2\hat{i}-\hat{j}+3\hat{k}) + \lambda(2\hat{i}+3\hat{j}-5\hat{k})$ where

$\lambda \in Z$

Find the vector equation of the line passing through the point having position vector $-2\hat{i} + \hat{j}+\hat{k}$ and parallel to vector $4\hat{i}-\hat{j}+2\hat{k}$

The vector equation of the line passing through

$\bar{a}$ and parallel to the vector

$\bar{b}$ is

$\bar{r}=\bar{a}+\lambda \bar{b}$ ,
where

$\lambda$ is scalar .

$\therefore$ the vector equation of the line passing through the point having position vector

$-2\hat{i} + \hat{j}+\hat{k}$ and parallel to vector

$4\hat{i}-\hat{j}+2\hat{k}$ is

$\bar{r} = (-2\hat{i}+\hat{j}+\hat{k})+\lambda (4\hat{i}-\hat{j}+2\hat{k})$

The direction ratios of two lines are 1,-3, 4 and -2, 0,The direction cosines of a line perpendicular to both the lines are

Direction ratio of

$1^{st}$ line is 1, -3, 4
direction ratio of

$2^{nd}$ line is -2, 0, 6
so direction ratio of line perpendicular to both is cross product of direction ratio of

$1^{st}$ &

$2^{nd}$ line

$=$

$\begin{vmatrix} i& J & K\\ 1 & -3 &4 \\ -2 & 0 & 6\end{vmatrix}=\hat (-18)-\hat j(14)+K(-6)$
& the direction cosine is

$(\frac {-18}{\sqrt {(-18)^2+(-14)^2+(-6)^2}}, \frac {-14}{\sqrt {(-18)^2+(-14)^2+(-6)^2}}, \frac {-6}{\sqrt {(-18)^2+(-14)^2+(-6)^2}})$

State True or False.
The following position vectors are collinear.
$a-2b+3c$ ,
$2a+3b-4c$ ,
$-7b+10c$ .

Test for collinearity is that thier cross product should be zero.

$a-2b+3c$

$2a+3b-4c$

$-7b+10c$

$\left| \begin{matrix} 1 & -2 & 3 \\ 2 & 3 & -4 \\ 0 & -7 & 3 \end{matrix} \right| =1\times (3\times3-(-7)\times(-4))-(-2)(2\times3-0)+3(2\times-7-0)=-19+12-42=-1$

They are not collinear as the cross product is non-zero.

Test the co-linearity of the set of points with the following position vectors: If they are co-linear then write 1 otherwise write 0.
$5a+4b+2c, \ 6a+2b-c, \ 7a+b-c.$

Considering

$a,b,c$ as unit vectors.

Therefore, for the vectors to be collinear, each pair of vectors from the above set of two vectors should have their vector product as 0.

Therefore,

$(5a+4b+2c)\times (6a+2b-c)$

$=-8a+17b-14c$

$\neq 0$

Since

$5a+4b+2c$ and

$6a+2b-c$ are non collinear.

Hence the above set of point all together are non-collinear.

Hence the answer is 0.

The directions cosines of the joins of the following pairs of vectors :
(i) $(6,3,2), (5,1,4)$ be $\dfrac{k}{3}, \dfrac{m}{3}, -\dfrac{m}{3}$
(ii) $(3,-4,7), (0,2,5)$ be $\dfrac{n}{7},\dfrac{l}{7},\dfrac{m}{7}$
Find $k+m+n+l$ ?

For $P\left( { x }_{ 1 }{ y }_{ 1 }{ z }_{ 1 } \right)$ and $Q\left( { x }_{ 2 }{ y }_{ 2 }{ z }_{ 2 } \right)$

Direction cosines are $\displaystyle \dfrac { { x }_{ 2 }-{ x }_{ 1 } }{ PQ } ,\dfrac { { y }_{ 2 }-{ y }_{ 1 } }{ PQ } ,\dfrac { { z }_{ 2 }-{ z }_{ 1 } }{ PQ }$

Now for $P\left( 6,3,2 \right) ,Q\left( 5,1,4 \right)$

$PQ=\sqrt { { \left( 6-5 \right) }^{ 2 }+{ \left( 3-1 \right) }^{ 2 }+{ \left( 2-4 \right) }^{ 2 } } =\sqrt { 1+4+4 } =3$

And direction cosine $\displaystyle \dfrac { 6-5 }{ 3 } ,\dfrac { 3-1 }{ 3 } ,\dfrac { 2-4 }{ 3 } =\dfrac { 1 }{ 3 } ,\dfrac { 2 }{ 3 } ,\dfrac { -2 }{ 3 }$

And for $A\left( 3,-4,7 \right) ,B\left( 0,2,5 \right)$

$AB=\sqrt { { \left( 3-0 \right) }^{ 2 }+{ \left( -4-2 \right) }^{ 2 }+{ \left( 7-2 \right) }^{ 2 } } =\sqrt { 9+36+4 } =\sqrt { 49 } =7$

Direction cosine is $\displaystyle \dfrac { 3-0 }{ 7 } ,\dfrac { -4-2 }{ 7 } ,\dfrac { 7-5 }{ 7 } =\dfrac { 3 }{ 7 } ,\dfrac { -6 }{ 7 } ,\dfrac { 2 }{ 7 }$

$\therefore k+m+n+l=1+2+3-6=0$

If $A(1, 2, 1), B(2, 6, 2)$ and $C(\lambda, 14, 4)$ are collinear, then value of $\lambda$ is

Equation of line

$AB$ is given by,

$\cfrac{x-1}{1}=\cfrac{y-2}{4}=\cfrac{z-1}{1}$
For

$A,B, C$ to be collinear,

$C$ must also lie on the line

$AB$ ,

$\Rightarrow \cfrac{\lambda-1}{1}=\cfrac{14-2}{4}=\cfrac{4-1}{1}$

$\Rightarrow \lambda -1=3$

$\Rightarrow \lambda = 4$

Find the direction cosines of the line $\displaystyle{\frac{x + 2}{2}}$ = $\displaystyle{\frac{2y - 7}{6}}$ = $\displaystyle{\frac{5 - z}{6}}$ . then the vector equation of the line through the point A(1, 2, 3) and parallel to the given line can be expressed as $\vec{r} = \hat{i} + 2\hat{j}+ 3\hat{k} + \lambda (2\hat{i} +3\hat{j} - 6\hat{k} )$ , where $\lambda$ is real.
If true enter 1 else enter 0

Given line is

$\displaystyle{\frac{x + 2}{2}}$ =

$\displaystyle{\frac{2y - 7}{6}}$ =

$\displaystyle{\frac{5 - z}{6}}$

$\rightarrow \displaystyle{\frac{x + 2}{2}}$ =

$\displaystyle{\frac{y - 7/2}{3}}$ =

$\displaystyle{\frac{z - 5}{-6}}$
Direction ratios of the line are

$2,3,-6$
If

$a,b,c$ are the direction ratios of a line then direction cosines are

$\displaystyle\frac { a }{ \sqrt { a^{ 2 }+b^{ 2 }+c^{ 2 } } } ,\displaystyle\frac { b }{ \sqrt { a^{ 2 }+b^{ 2 }+c^{ 2 } } } ,\displaystyle\frac { c }{ \sqrt { a^{ 2 }+b^{ 2 }+c^{ 2 } } }$

$\therefore$ direction cosines are

$\displaystyle\frac { 2 }{ 7 } ,\displaystyle\frac { 3 }{ 7 } ,\displaystyle\frac { -6 }{ 7 }$
And the line passing through

$(1,2,3)$ and parallel to the given line is

$\displaystyle{\frac{x - 1}{2}}$ =

$\displaystyle{\frac{y - 2}{3}}$ =

$\displaystyle{\frac{z - 3}{-6}}$

The vector equation of the line can be written as :

$\vec{r} = \hat{i} + 2\hat{j}+ 3\hat{k} + \lambda (2\hat{i} +3\hat{j} - 6\hat{k} )$ , where

$\lambda$ is real.

The cartesian equation of a line is $\dfrac {x - 5}{3} = \dfrac {y + 4}{7} = \dfrac {z - 6}{2}$ write its vector form.

The Cartesian equation of a line is

$\cfrac{x-5}{3}=\cfrac{y+4}{7}=\cfrac{z-6}{2}......(1)$
The given line passes through the point

$(5, -4, 6)$ . The position vector of this point is

$\vec { a } =5\hat { i } -4\hat { j } +6\hat { k }$ .
Also, the direction ratios of the given line are

$3, 7$ and

$2$ .
This means that the line is in the direction of vector

$\vec { b } =3\hat { i } +7\hat { j } +2\hat { k }$ .
It is known that the line through position vector

$\vec { a }$ and in the direction of the vector

$\vec { b }$ is given by the equation,

$\vec { r } =\vec { a } +\lambda \vec { b }$ ,

$\lambda \in R$

$\Rightarrow$

$\vec { r } =\left( 5\hat { i } -4\hat { j } +6\hat { k } \right) +\lambda \left( 3\hat { i } +7\hat { j } +2\hat { k } \right)$
This is the required equation of the given line in vector form.

If a line has the direction ratios $-18, 12,- 4$ , then what are its direction cosines?

If a line has direction ratios of

$-18, 12$ and

$-4$ , then its direction cosines are

$\cfrac { -18 }{ \sqrt { { \left( -18 \right) }^{ 2 }+{ \left( 12 \right) }^{ 2 }+{ \left( -4 \right) }^{ 2 } } } , \cfrac { 12 }{ \sqrt { { \left( -18 \right) }^{ 2 }+{ \left( 12 \right) }^{ 2 }+{ \left( -4 \right) }^{ 2 } } } , \cfrac { -4 }{ \sqrt { { \left( -4 \right) }^{ 2 }+{ \left( 12 \right) }^{ 2 }+{ \left( -4 \right) }^{ 2 } } }$

i.e.,

$\cfrac{-18}{22}, \cfrac{12}{22}, \cfrac{-4}{22}\equiv \cfrac{ 9}{11}, \cfrac{6}{11}, \cfrac{-2}{11}$

Thus, the direction cosines are

$-\cfrac{9}{11}, \cfrac{6}{11}$ and

$\cfrac{-2}{11}$ .

The angle between the following pair of lines:
$\cfrac{x}{2}=\cfrac{y}{2}=\cfrac{z}{1}$ and $\cfrac{x-5}{4}=\cfrac{y-2}{1}=\cfrac{z-3}{8}$ is $\theta = \cos^{-1}\dfrac{a}{3}$ then $a=$

The drs of given two lines are

$2,2,1$ and

$4,1,8$

Let the angle between those two lines be

$\theta$

We have

$cos\theta=\frac{2 \times 4+2 \times 1+1 \times 8}{3 \times 9}=\frac{18}{27}=\frac{2}{3}$

Therefore

$\theta = cos^{-1}{\frac{2}{3}}$

Find the equation of the line in vector and in cartesian form that passes through the point with position vector $2\hat {i} - \hat {j} + 4\hat {k}$ and is in the direction $\hat {i} + 2\hat {j} - \hat {k}$ .

It is given that the line passes through the point with position vector,

$\vec { a } =2\hat { i } -\hat { j } +4\hat { k } ...(1)$

$\vec { b } =\hat { i } +2\hat { j } -\hat { k } ...(2)$
It is known that a line through a point with position vector

$\vec { a }$ and parallel to

$\vec { b }$ is given by the equation,

$\vec { r } =\vec { a } +\lambda \vec { b }$

$\Rightarrow$

$\vec { r } =2\hat { i } -\hat { j } +4\hat { k } +\lambda \left( \hat { i } +2\hat { j } -\hat { k } \right)$
This is the required equation of the line in vector form.

$\Rightarrow \quad x\hat { i } -y\hat { j } +z\hat { k } =(\lambda +2)\hat { i } +(2\lambda -1)\hat { j } +(-\lambda +4)\hat { k }$
Eliminating

$\lambda$ , we obtain the Cartesian form equation as

$\cfrac{x-2}{1}=\cfrac{y+1}{2}=\cfrac{z-4}{-1}$

Find the Cartesian equation of the line which passes through the point $(-2, 4, -5)$ and parallel to the line given by $\cfrac{x+3}{3}=\cfrac{y-4}{5}=\cfrac{z+8}{6}$

It is given that the line passes through the point

$(-2, 4, -5)$ and is parallel to

$\cfrac{x+3}{3}=\cfrac{y-4}{5}=\cfrac{z+8}{6}$ ....(1)
The direction ratios of the line (1) are

$3, 5$ and

$6$ .

The required line is parallel to equation (1).

Therefore, its direction ratios are

$3k, 5k$ and

$6k$ , where

$k\ne 0$ .

It is known that the equation of the line through the point

$({x}_{1}, {y}_{1}, {z}_{1})$ and with direction ratios,

$a, b, c$ is given by

$\cfrac { x-{ x }_{ 1 } }{ a } =\cfrac { y-{ y }_{ 1 } }{ b } =\cfrac { z-{ z }_{ 1 } }{ c }$

Therefore, the equation of the required line is,

$\cfrac{x+2}{3k}=\cfrac{y-4}{5k}=\cfrac{z+5}{6k}$

$\Rightarrow$

$\cfrac{x+2}{3}=\cfrac{y-4}{5}=\cfrac{z+5}{6}$

Show that the points $(-2, 3, 5), (1, 2, 3)$ and $(7, 0, -1)$ are collinear.

Let points

$(-2, 3, 5), (1, 2, 3)$ and

$(7, 0, -1)$ be denoted by

$P, Q$ and

$R$ respectively.

Points

$P, Q$ and

$R$ are collinear if they lie on a line.

$PQ=\displaystyle \sqrt{\left ( 1+2 \right )^{2}+\left ( 2-3 \right )^{2}+\left ( 3-5 \right )^{2}}$

$=\displaystyle \sqrt{\left ( 3 \right )^{2}+\left ( -1 \right )^{2}+\left ( -2 \right )^{2}}$

$=\displaystyle \sqrt{9+1+4}$

$=\displaystyle \sqrt{14}$

$QR=\displaystyle \sqrt{\left ( 7-1 \right )^{2}+\left ( 0-2 \right )^{2}+\left ( -1-3 \right )^{2}}$

$=\displaystyle \sqrt{\left ( 6 \right )^{2}+\left ( -2 \right )^{2}+\left ( -4 \right )^{2}}$

$=\displaystyle \sqrt{36+4+16}$

$=\displaystyle \sqrt{56}$

$=\displaystyle 2\sqrt{14}$

$PR=\displaystyle \sqrt{\left ( 7+2 \right )^{2}+\left ( 0-3 \right )^{2}+\left ( -1-5 \right )^{2}}$

$= \displaystyle \sqrt{\left ( 9 \right )^{2}+\left ( -3 \right )^{2}+\left ( -6 \right )^{2}}$

$=\displaystyle \sqrt{81+9+36}$

$=\displaystyle \sqrt{126}$

$=\displaystyle 3\sqrt{14}$

Here

$PQ + QR = \displaystyle \sqrt{14}+2\sqrt{14}$

$=3 \sqrt{14}$

$=PR$

$\Rightarrow PQ+QR=PR$

Hence points

$P(-2, 3, 5), Q(1, 2, 3)$ and

$R(7, 0, -1)$ are collinear

Find the direction cosines of a line which makes equal angles with the coordinate axes.

Let the direction cosines of the line make an angle

$a$ with each of the coordinate axes.

$\therefore$

$l=\cos {a}, m=\cos {a}, n=\cos {a}$

We know

${l}^{2}+{m}^{2}+{n}^{2}=1$

$\Rightarrow$

$\cos ^{ 2 }{ \alpha } +\cos ^{ 2 }{ \alpha } +\cos ^{ 2 }{ \alpha } =1$

$\Rightarrow$

$3\cos ^{ 2 }{ \alpha } =1$

$\Rightarrow$

$\cos ^{ 2 }{ \alpha } =\cfrac{1}{3}$

$\Rightarrow$

$\cos {\alpha}=\pm \cfrac{1}{\sqrt 3}$

Thus, the direction cosines of the line which is equally inclined to the coordinate axes are

$\pm \cfrac{1}{\sqrt 3}, \pm \cfrac{1}{\sqrt 3}$ and

$\pm \cfrac{1}{\sqrt 3}$ .

$x+y+z=1$

Given,

$x+y+z=1......(1)$
The direction ratios of normal are

$1,1$ and

$1$ .
Now

$\sqrt { { \left( 1 \right) }^{ 2 }+{ \left( 1 \right) }^{ 2 }+{ \left( 1 \right) }^{ 2 } } =\sqrt { 3 }$
Therefore, the direction cosines of the normal are

$\cfrac { 1 }{ \sqrt { 3 } } ,\cfrac { 1 }{ \sqrt { 3 } }$ and

$\cfrac { 1 }{ \sqrt { 3 } }$

and the distance of plane from the origin is

$=\left|\cfrac { 0+0+0-1}{ \sqrt { 1^2+1^2+1^2 } } \right|=\dfrac{1}{\sqrt 3}$

Find the equation of the line which passes through the point $(1, 2, 3)$ and is parallel to the vector $(3\hat {i} + 2\hat {j} - 2\hat {k}$ .

It is given that the line passes through the point

$A(1,2,3)$ .
Therefore, the position vector through

$A$ is

$\vec { a } =\hat { i } +2\hat { j } +3\hat { k }$

$\vec { b } =3\hat { i } +2\hat { j } -2\hat { k }$
It is known that the line which passes through point

$A$ and parallel to

$\vec { b }$ is given by

$\vec { r } =\vec { a } +\lambda \vec { b }$ , where

$\lambda$ is a constant.

$\Rightarrow$

$\vec { r } =\hat { i } +2\hat { j } +3\hat { k } +\lambda \left( 3\hat { i } +2\hat { j } -2\hat { k } \right)$
This is the required equation of the line.

Find the vector and the Cartesian equations of the line that passes through the points $(3, -2, -5), (3, -2,6)$

Let the line passing through the points

$P(3,-2,-5)$ and

$Q(3, -2,6)$ be

$PQ$ . Since

$PQ$ passes through

$P(3, -2, -5)$ , its position vector is given by,

$\vec { a } =3\hat { i } -2\hat { j } -5\hat { k }$
The direction ratios of

$PQ$ are given by,

$(3-3)=0, (-2+2)=0, (6+5)=11$
The equation of the vector in the direction of

$PQ$ is

$\vec { b } =0\hat { i } -0\hat { j } +11\hat { k } =11\hat { k }$
The equation of

$PQ$ in vector form is given by

$\vec { r } =\vec { a } +\lambda \vec { b } ,\lambda \in R$

$\Rightarrow \vec { r } =\left( 3\hat { i } -2\hat { j } -5\hat { k } \right) +11\lambda \hat { k } \quad$
The equation of

$PQ$ in Cartesian form is

$\cfrac { x-{ x }_{ 1 } }{ a } =\cfrac { y-{ y }_{ 1 } }{ b } =\cfrac { z-{ z }_{ 1 } }{ c }$

$\Rightarrow \cfrac{x-3}{0}=\cfrac{y+2}{0}=\cfrac{z+5}{11}$

Find the distance of the point $(-1,-5,-10)$ from the point of intersection of the line $\vec { r } =\left( 2\hat { i } -\hat { j } +2\hat { k } \right) +\lambda \left( 3\hat { i } +4\hat { j } +2\hat { k } \right)$ and the plane $\vec { r } .\left( \hat { i } -\hat { j } +\hat { k } \right) =5$

Hint : To find the intersection of line and plane put the value of r from equation of line into the equation of plane

Step 1:

The equation of the given line is

$\vec { r } =\left( 2\hat { i } -\hat { j } +2\hat { k } \right) +\lambda \left( 3\hat { i } +4\hat { j } +2\hat { k } \right) .....(1)$

The equation of the given plane is

$\vec { r } .\left( \hat { i } -\hat { j } +\hat { k } \right) =5........(2)$

Step 2 :Substituting the value of $\vec { r }$ in equation $(2)$ , we get,

$\left[( 2\hat { i } -\hat { j } +2\hat { k } \right) +\lambda \left( 3\hat { i } +4\hat { j } +2\hat { k }) \right][\hat{i}-\hat{j}+\hat{k}]=5$

$\implies \left[( 2+3\lambda)\hat { i } +(4\lambda-1)\hat { j } +(2+2\lambda)\hat { k } \right][\hat{i}-\hat{j}+\hat{k}]=5$

$\implies (2+3\lambda)-(4\lambda-1)+(2\lambda+2)=5$

$\implies \lambda =0$
Thus the point of intersection of the given line and the plane is,

$\vec { r } =2\hat { i } -\hat { j } +2\hat { k } =(2,-1,2)$
This means that the position vector of the point of intersection of the line and plane is given by the coordinates

$(2,-1,2)$ . The point is

$(-1,-5,-10)$ .

Hence, distance

$d$ between the points,

$(2,-1,2)$ and

$(-1,-5,-10)$ is

$d=\sqrt { { \left( -1-2 \right) }^{ 2 }+{ \left( -5+1 \right) }^{ 2 }+{ \left( -10-2 \right) }^{ 2 } } =\sqrt { 9+16+144 } =\sqrt { 169 } =13$

Find the coordinates of the foot of the perpendicular drawn from the origin to the plane $5y + 8 = 0$ .

Let the coordinates of the foot of perpendicular

$P$ from the origin to the given plane be

$({x}_{1}, {y}_{1}, {z}_{1})$ .
We have,

$5y+8=0$

$\Rightarrow$

$0x-5y+0z=8......(1)$
The direction ratios of the normal are

$0,-5$ and

$0$

$\therefore$

$\sqrt { 0+{ \left( 5 \right) }^{ 2 }+0 } =5$
Dividing both sides of equation (1) by

$5$ , we obtain

$-y=\cfrac{8}{5}$
This equation is of the form

$lx+my+nz=d$ , where

$l.m.n$ are the direction cosines of normal to the plane and

$d$ is the distance of normal from the origin.
The coordinates of the foot of the perpendicular are given by

$(ld.md,nd)$
Therefore, the given coordinates of the foot of the perpendicular are

$\left( 0,-1\left( \cfrac { 8 }{ 5 } \right) ,0 \right)$ i.e.,

$\left( 0,-\cfrac { 8 }{ 5 } ,0 \right)$

that passes through the point $(1,4,6)$ and the normal vector to the plane is $\hat { i } -2\hat { j } +\hat { k }$

The position vector of point

$(1,4,6)$ is

$\vec { a } =\hat { i } +4\hat { j } +6\hat { k }$

The normal vector

$\vec { n }$ perpendicular to the plane is

$\vec { n } =\hat { i } -2\hat { j } +\hat { k }$

The vector equation of the plane is given by

$\left( \vec { r } -\vec { a } \right) .\vec { n} =0$

$\Rightarrow$

$\left[ \vec { r } -(\hat { i } +4\hat { j } +6\hat { k } ) \right] .\left( \hat { i } -2\hat { j } +\hat { k } \right) =0.......(1)$

Where

$\vec { r }$ is the position vector of any point

$P(x,y,z)$ in the plane.

And given by,

$\vec { r } =x\hat { i } +y\hat { j } +z\hat { k }$

Therefore, the equation(1) becomes

$\Rightarrow$

$\left[ (x-1)\hat { i } +(y-4)\hat { j } +(z-6)\hat { k } \right] .\left( \hat { i } -2\hat { j } +\hat { k } \right) =0$

$\Rightarrow$

$(x-1)-2(y-4)+(z-6)=0$

$\Rightarrow$

$x-2y+z+1=0$

that passes through the point $(1,0,-2)$ and the normal to the plane is $\hat { i } +\hat { j } -\hat { k }$

The position vector of point

$(1,0,-2)$ is

$\vec { n}$ perpendicular to the plane is

$\vec { n } =\hat { i } +\hat { j } -\hat { k }$ .
The vector equation of the plane is given by,

$\left( \vec { r } -\vec { a } \right) \cdot \vec { n } =0$

$\Rightarrow$

$\left[ \vec { r } -(\hat { i } -2\hat { k } ) \right] .\left( \hat { i } +\hat { j } -\hat { k } \right) =0.......(1)$

The position vector of any point

$P(x,y,z)$ in the plane is given by,

$\vec { r } =x\hat { i } +y\hat { j } +z\hat { k }$

Therefore, equation (1) becomes

$\left[ \left( x\hat { i } +y\hat { j } -z\hat { k } \right) -\left( \hat { i } -2\hat { k } \right) \right] .\left( \hat { i } +\hat { j } -\hat { k } \right) =0$

$\Rightarrow$

$\left[ (x-1)\hat { i } +y\hat { j } +(z+2)\hat { k } \right] .\left( \hat { i } +\hat { j } -\hat { k } \right) =0$

$\Rightarrow$

$(x-1)+y-(z+2)=0$

$\Rightarrow$

$x+y-z-3=0$

$\Rightarrow$

$x+y-z=3$

Show that points
A (a, b + c), B (b, c + a), C (c, a + b) are collinear

Area of triangle ABC is,

$\Delta =\begin{vmatrix}a&b+c&1\\b&c+a&1\\c&a+b&1\end{vmatrix}$

$\quad =\begin{vmatrix}a+b+c&b+c&1\\a+b+c&c+a&1\\a+b+c&a+b&1\end{vmatrix}$ Using

$C_1\to C_1+C_2$

$\quad =(a+b+c)\begin{vmatrix}1&b+c&1\\1&c+a&1\\1&a+b&1\end{vmatrix}=0$
Since

$C_1=C_2$
Hence points A, B, and C are collinear.

In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.
$x + y + z = 1$ .

Let the coordinates of the foot of perpendicular

$P$ from the origin to the plane be

$({x}_{1}, {y}_{1}, {z}_{1})$

$x+y+z=1........(1)$
The direction ratios of the normal are

$1,1$ and

$1$

$\therefore$

$\sqrt { { \left( 1 \right) }^{ 2 }+{ \left( 1 \right) }^{ 2 }+{ \left( 1 \right) }^{ 2 } } =\sqrt { 3 }$
Dividing both sides of equation (1) by

$\sqrt 3$ , we obtain

$\cfrac { 1 }{ \sqrt { 3 } } x+\cfrac { 1 }{ \sqrt { 3 } } y+\cfrac { 1 }{ \sqrt { 3 } } z=\cfrac { 1 }{ \sqrt { 3 } }$
This equation is of the form

$lx+my+mz=d$ , where

$l,m,n$ are the direction cosines of normal to the plane and

$d$ is the distance of normal from the origin.
The coordinates of the foot of the perpendicular are given by

$(ld,md,nd)$
Therefore, the coordinates of the foot of the perpendicular are

$\left( \cfrac { 1 }{ \sqrt { 3 } } .\cfrac { 1 }{ \sqrt { 3 } } ,\cfrac { 1 }{ \sqrt { 3 } } .\cfrac { 1 }{ \sqrt { 3 } } ,\cfrac { 1 }{ \sqrt { 3 } } .\cfrac { 1 }{ \sqrt { 3 } } \right)$ i.e.,

$\left( \cfrac { 1 }{ 3 } ,\cfrac { 1 }{ 3 } ,\cfrac { 1 }{ 3 } \right)$

Find the equation of a line parallel to x-axis and passing through the origin

The line parallel to

$x$ -axis and passing through the origin is

$x$ -axis itself.

Let

$A$ be a point on

$x$ -axis.

Therefore, the coordinates of

$A$ are given by

$(a,0,0)$ , where

$a\in R$ .

Direction ratios of

$OA$ are

$(a-0)=a,0,0$ .

The equation of

$OA$ is given by,

$\cfrac{x-0}{a}=\cfrac{y-0}{0}=\cfrac{z-0}{0}$

$\cfrac{x}{1}=\cfrac{y}{0}=\cfrac{z}{0}=a$

Thus, the equation of line parallel to

$x$ -axis and passing through origin is

$\cfrac{x}{1}=\cfrac{y}{0}=\cfrac{z}{0}$ .

Show that the vector $\hat {i}+\hat {j}+\hat {k}$ is equally inclined to the axes $OX, OY$ and $OZ$ .

Let

$\vec {a}=\hat {i}+\hat {j}+\hat {k}$

Then,

$|\vec {a}|=\sqrt {1^2+1^2+1^2}=\sqrt 3$

Therefore, the direction cosines of

$\vec {a}$ are

$\left (\dfrac {1}{\sqrt {3}}, \dfrac {1}{\sqrt {3}}, \dfrac {1}{\sqrt {3}}\right )$ .

Now, let

$\alpha, \beta$ , and

$\gamma$ be the angles formed by

$\vec {a}$ with the positive direction directions of

$x, y$ , and

$z$ axes.

Then, we have

$\cos\alpha =\dfrac {1}{\sqrt {3}}, \cos\beta =\dfrac {1}{\sqrt {3}}, \cos\gamma =\dfrac {1}{\sqrt {3}}$ .

Hence, the given vector is equally inclined to axes

$OX, OY$ , and

$OZ$ .

Find the vector equation of a plane which is at a distance of $5$ units from the origin and its normal vector is $2\hat i - 3\hat j + 6\hat k$

When equation of a plane is

$ax + by + cz + d = 0$ , its normal vector is

$a\hat{i} + b\hat{j} + c\hat{k}$ and its distance from origin is

$\dfrac{|d|}{\sqrt{a^2 + b^2 + c^2}}$

Similarly,

$a = 2, b = -3, c = 6$ here.

Since

$5 = \dfrac{|d|}{\sqrt{4 + 9 + 36}}$

$\Rightarrow |d| = 35$

The equation of the plane can thus be

$2x - 3y + 6z \pm 35 = 0$

If a line makes angles $90^o, 60^0$ and $\theta$ with $x, y$ and $z-$ axis respectively, where $\theta$ is acute, then find $\theta$ .

$\alpha$ ,

$\beta$ and

$\gamma$ are the angles made by the line with the

$x, y$ and

$z$ axes respectively, then

$\cos ^{ 2 }{ \alpha } +\cos ^{ 2 }{ \beta } +\cos ^{ 2 }{ \gamma } =1$ .

Therefore

$\cos ^{ 2 }{ 90° } +\cos ^{ 2 }{ 60° } +\cos ^{ 2 }{ \theta } =1\\ \Rightarrow \cos { \theta } =\dfrac { \sqrt { 3 } }{ 2 } \Rightarrow \theta =30°$

Find the Cartesian equation of the line which passes through the point $(-2, 4, -5)$ and is parallel to the line $\displaystyle \frac{x+3}{3}=\frac{4-y}{5}=\frac{z+8}{6}$

Given line

$\dfrac{x+3}{3}=\dfrac{4-y}{5}=\dfrac{z+8}{6}$

$\dfrac{x+3}{3}=\dfrac{y-4}{-5}=\dfrac{z+8}{6}$

So comparing with general cartesian form of eq

$a=3,b=-5,c=6$

another line passing through

$P(-2,4,-5)$ we can assume it as position vector and parallel to given line so normal vector will be same

eq of line

$\dfrac{x+2}{3}=\dfrac{y-4}{-5}=\dfrac{z+5}{6}$

Find the vector equation of the plane passing through a point having position vector $3\hat {i} - 2\hat {j} + \hat {k}$ and perpendicular to the vector $4\hat {i} + 3\hat {j} + 2\hat {k}$

We know that the vector equation of the plane passing through a point

$A(\vec {a})$ and normal to

$\vec {n}$ is

$\vec {r}\cdot \vec {n} = \vec {a}\cdot \vec {n}$ .

Here,

$\vec {a} = 3\hat {i} - 2\hat {j} + \hat {k}$ and

$\vec {n} = 4\hat {i} + 3\hat {j} + 2\hat {k}$

$\therefore$ the equation of the required plane is

$\vec {r} \cdot (4\hat {i} + 3\hat {j} + 2\hat {k}) = (3\hat {i} - 2\hat {j} + \hat {k})\cdot(4\hat {i} + 3\hat {j} + 2\hat {k})$

$\Rightarrow \vec {r} \cdot (4\hat {i} + 3\hat {j} + 2\hat {k}) = 12 - 6 + 2$

$\Rightarrow \vec {r} \cdot (4\hat {i} + 3\hat {j} + 2\hat {k}) = 8$

This is the required vector equation.

If a line makes angles $90^{\circ}, 60^{\circ}$ and $30^{\circ}$ with the positive direction of $x, y$ and z-axes respectively, find its direction cosines

Let the angles made by the line with positive x axis, y axis and z axis be

$\alpha,\beta,\gamma$ then

$\alpha=90^{0}$ ,

$\beta=60^{0}$ and

$\gamma=30^{0}$ .

Therefore the direction cosines of the lines are

$(cos\alpha,cos\beta,cos\gamma)$

$=(0,\dfrac{1}{2},\dfrac{\sqrt{3}}{2})$

Find the direction cosines of the unit vector perpendicular to the plane $\overrightarrow { r } .\left( 6\hat { i } -3\hat { j } -2\hat { k } \right) +1=0$

Direction ratioes:

$(6, -3, -2)$

Hence Directions cosines will be:

$(\dfrac{6}{\sqrt{6^2+(-3)^2+(-2)^2}}, \dfrac{-3}{\sqrt{6^2+(-3)^2+(-2)^2}}, \dfrac{-2}{\sqrt{6^2+(-3)^2+(-2)^2}})$

$=(\dfrac{6}{7}, \dfrac{-3}{7}, \dfrac{-2}{7})$ Ans.

Find vector equation of line passing through the point whose position vector is
$3\hat { i } -4\hat { j } +\hat { k }$ and parallel to the vector $2\hat { i } +\hat { j } -3\hat { k }$ . Also write the equation in Cartesian form.

Here,

$\overrightarrow { a } =3\hat { i } -4\hat { j } +\hat { k }$
and

$\overrightarrow { b } =2\hat { i } +\hat { j } -3\hat { k }$

$\therefore$ Vector equation of the line passing through the point

$\overrightarrow { a }$ and parallel to the vector

$\overrightarrow { b }$ is given by,

$\overrightarrow { r } =\overrightarrow { a } +t\overrightarrow { b }$ where

$t$ is a scalar

$\overrightarrow { r } =\left( 3\hat { i } -4\hat { j } +\hat { k } \right) +t\left( 2\hat { i } +\hat { j } -3\hat { k } \right)$
Let

$\overrightarrow { r } =x\hat { i } +y\hat { j } +z\hat { k }$ , then

$x\hat { i } +y\hat { j } +z\hat { k } =\left( 3\hat { i } -4\hat { j } +\hat { k } \right) +t\left( 2\hat { i } +\hat { j } -3\hat { k } \right)$

$x\hat { i } +y\hat { j } +z\hat { k } =(3+2t)\hat { i } +(-4+t)\hat { j } +(1-3t)\hat { k } \quad$
Equating the coeffeicients of

$\hat { i } ,\hat { j}$ and

$\hat { k }$

$x=3+2t$

$y=-4+t$

$z=1-3t$

$\Rightarrow \cfrac { x-3 }{ 2 } =\cfrac { y+4 }{ 1 } =\cfrac { z-1 }{ -3 } =t$
These are the equation of the line in cartesian form.

If the points $(a,1),(1,2)$ and $(0,b+1)$ are collinear, then show that $\cfrac{1}{a}+\cfrac{1}{b}=1$

Area of the triangle from these points is

$0$

$\cfrac{1}{2}(2a+b+1+0-(1-0+(b+1))=0$

$\cfrac{1}{2}(2a+b+1-1-a(b+1))=0$

$2a+b+1-1-ab-a=0$

$a+b-ab=0$

$\cfrac{a}{ab}+\cfrac{b}{ab}=1$

$\cfrac{1}{b}+\cfrac{1}{a}=1$

Find the angle between the two lines whose direction cosines are given by equations $l+m+n=0$ and $2l+2m-nm=0$ .

Given:

$l+m+n=0$ ...........(i)

$2[l+m]-mn=0$ ...........(ii)

Putting the value of

$(l+m)$ in (ii)

$2(-n)-mn=0$

$-2n-mn=0$

$-n(2+m)=0$

either

$n=0$ or

$m=-2$

Putting the value of

$n$ is (i)

$l+m=0$

$l=-m$

We know that

$l^2+m^2+n^2=1$

$l^2+l^2+0=1$

$2l^2=1$

$l^2=\displaystyle\frac{1}{2}$

$l=\pm\displaystyle\frac{1}{\sqrt{2}}$

Similarly

$m=\pm\displaystyle\frac{1}{\sqrt{2}}$

Hence direction cosines of the lines

$\left(\displaystyle\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}, 0\right)$ and

$\left(-\displaystyle\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}, 0\right)$

Angle between the lines

$\cos\theta=l_1l_2+m_1m_2+n_1n_2$

$\cos\theta =\left[\displaystyle\frac{1}{\sqrt{2}}\right]\left[-\displaystyle\frac{1}{\sqrt{2}}\right]+\left(-\displaystyle\frac{1}{\sqrt{2}}\right)\left(\displaystyle\frac{1}{\sqrt{2}}\right)+0.0$

$\cos\theta =\left[-\displaystyle\frac{1}{2}-\frac{1}{2}\right]$

$\cos\theta =-1$

$\cos\theta =\cos\pi$

$\theta =\pi =180^o$ .

Write the vector equation of the line equation
$\cfrac { x-1 }{ 2 } =\cfrac { y-2 }{ 3 } =\cfrac { z-3 }{ 4 }$

If a line passes through a point

$\overrightarrow { a }$ and parallel to the vector

$\overrightarrow { r }$ then the vector form of the line can be written as

$\overrightarrow { a }+\lambda(\overrightarrow { r } )$ , where

$\lambda$ is a constant

The directional cosines of the given line are

$2,3,4$

Hence,

$2\hat{i}+3\hat{j}+4\hat{k}$ is parallel to the given line

The given line passes through

$\hat{i}+2\hat{j}+3\hat{k}$

Hence, the vector form of the line is given by

$(\hat { i } +2\hat { j } +3\hat { k } )+\lambda (2\hat { i } +3\hat { j } +4\hat { k } )$

If $l, m, n$ are the direction cosines of a line, then prove that $l^{2} + m^{2} + n^{2} = 1$ . Hence find the direction angle of the line with the $X$ axis which makes direction angles of $135^{\circ}$ and $45^{\circ}$ with $Y$ and $Z$ axes respectively

Let the line makes angle

$\alpha$ with x-axis,

$\beta$ with y-axis and

$\gamma$ with z-axis, then

$l^2+m^2+n^2=1$ or

$\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$

Here given

$\alpha=135^o$ and

$\beta = 45^o$ and

$\gamma=?$

$\cos^2135^0+\cos^245^o+\cos^2\gamma=1$

$\Rightarrow \dfrac{1}{2}+\dfrac{1}{2}+\cos^2\gamma=1$

$\Rightarrow \cos\gamma=0\Rightarrow \gamma=90^o$

Hence given line makes angle

$90^o$ with the z-axis

If from a point $Q(a, b, c)$ perpendiculars $QA$ and $QB$ are drawn to the $YZ$ and $ZX$ planes respectively, then find the vector equation of the plane $OAB$ .

Point

$Q(a,b,c)$ is given.

Perpendiculars from

$Q$ are dropped n

$YZ$ and

$ZX$ planes, having foot as

$A\ and\ B$ .

Coordinates of

$A$ are

$(0,b,c)$ and

$b$ are

$(a,0,c)$

Therefore,

$O = (0,0,0)$

$A = (0,b,c)$

$B = (a,0,c)$

$P = (x,y,z)$

$\vec{OA} = b\hat j + c\hat k$

$\vec{OB} = a\hat i + c\hat k$

Normal vector for the plane

$OAB$ is

$\vec{OA} \times \vec{OB} = (b\hat j + c\hat k)\times(a\hat i + c\hat k)$

$\vec n=-ab\hat k+bc\hat i+ac\hat j$

Equation of plane:

$(\vec{OP})\cdot\vec n = 0$

$\Rightarrow (x\hat i+y\hat j+z\hat k)\cdot(bc\hat i + ac\hat j-ab\hat k)$ =0

$\Rightarrow bcx + acy - abz = 0$

Drive the equation of a plane perpendicular to a given vector and passing through a given point both in vector and Cartesian form.

(Vector form)

Equation of a plane passing through point A whose position vector is

$\vec{a}$ and perpendicular to

$\vec{n}$ is given by

$(\vec{r}-\vec {a})\cdot \vec{n}=0$

(Cartesian form)

Equation of a plane passing through point

$(x_{1}, y_{1}, z_{1})$ and perpendicular to a line with direction ratios

$A,B,C$ is given by

$A(x-x_1)+B(y-y_1)+C(z-z_1)=0$

If $\cos \alpha, \cos \beta, \cos \gamma$ are direction-cosines of any straight line, then prove that:
$\cos 2\alpha + \cos 2\beta + \cos 2\gamma = -1$

$\cos \alpha, \cos \beta, \cos \gamma$ are direction-cosines of a straight line

$\therefore \cos^{2}\alpha + \cos^{2}\beta + \cos^{2}\gamma = 1$
now from

$L.H.S. = \cos{2}\alpha + \cos 2\beta + \cos 2\gamma$

$= 2\cos^{2}\alpha - 1 + 2\cos^{2}\beta - 1 + 2\cos^{2} \gamma - 1$

$= 2(\cos^{2}\alpha + \cos^{2}\beta + \cos^{2}\gamma) - 3$

$= 2\times 1 - 3$

$= 2 - 3$

$= -1= R.H.S.$

$L.H.S = R.H.S$

Find the equation to the plane through the point $(-1,3,2)$ and perpendicular to the planes $x+2y+2z=11$ and $3x+3y+2z=15$ .

Equation of plane passing through point

$(-1,3,2)$

$a(x+1)+b(y-3)+c(z-2)=0....(1)$

$x+2y+2z=11........(2)$
and

$3x++2z=15...(3)$

$\because$ plane (1) is perpendicular to plane (2) and (3)

$\therefore$ by condition or perpendicularity

${a}_{1}{a}_{2}+{b}_{1}{b}_{2}+{c}_{1}{c}_{2}=0$

$a+2b+2c=0......(4)$

$3a+3b+2c=0.....(5)$
by solving the equation (4) and (5)

$\cfrac { a }{ 4-6 } =\cfrac { b }{ 6-2 } =\cfrac { c }{ 3-6 } \quad$

$\cfrac { a }{ 2 } =\cfrac { b }{ -4 } =\cfrac { c }{ 3 }$
Let

$\cfrac { a }{ 2 } =\cfrac { b }{ -4 } =\cfrac { c }{ 3 } =K$

$\therefore$

$a=2K,b=-4K,c=3K$
Putting the value of a

$a,b,c$ in equation (1)

$2K(x+1)-4K(y-3)+3K*z-2)=0$

$K[2(x+1)-4(y-3)+3(z-2)]=0$

$2x+2-4y+12+3z-6=0$

$2x-4y+3z+8=0$
This is the required equation.

Find the equation of the plane passing through the point $(1, -2, 1)$ and perpendicular to the line joining the points $A(3, 2, 1)$ and $B(1, 4, 2)$ .

Vector equation of plane is

$(\vec { r } -\vec { a } ).\vec { n } =0$

Plane passes through

$(1,-2,1)$

$\Rightarrow \vec{a}=i-2j+k$

Plane is perpendiculat to line joing

$(3,2,1)$ and

$(1,4,2)$

$\Rightarrow \vec { n } =(1-3)i+(4-2)j+(2-1)k\\ \Rightarrow \vec { n } =-i+2j+k$

So the vector equation of the plane is

$\{ \vec { r } -(i-2j+k)\} .(-i+2j+k)=0$

Cartesian equation of plane is

$\{ xi+yj+zk-(i-2j+k)\} .(-i+2j+k)=0\\ \{ (x-1)i+(y+2)j+(z-1)k\} .(-i+2j+k)=0\\ -(x-1)+2(y+2)+(z-1)=0\\ -x+2y+z+4=0\\ x-2y-z-4=0$

The direction cosines of two lines are determined by the relation $l-5m+3n=0$ and $7l^2+5m^2-3n^2=0$ , find them.

Consider the given equation,

$l-5m+3n=0$

$l=5m-3n$ --------(1)

Substitute for

$l$ in

$7l^2+5m^2-3n^2=0$

$7(5m-3n)^2+5m^2-3n^2=0$

$7(25m^2+9n^2-30mn)+5m^2-3n^2=0$

$175m^2+63n^2-210mn+5m^2-3n^2$

$180m^2+60n^2-210mn$

$30(6m^2+2n^2-7mn)=0$

$6m^2+2n^2-7mn=0$

$(3m-2n)(2m-n)=0$

Hence,

$3m-2n=0$ or

$2m-n=0$

Case (1) :

$3m-2n=0$

$\implies m=\dfrac{2n}{3}$

Substitute for

$m$ in (1)

$l=5m-3n=5(\dfrac{2n}{3})-3n=\dfrac{n}{3}$

Hence, the direction cosines

$(l,m,n)$ is

$\left (\dfrac{n}{3},\dfrac{2n}{3},n \right )$

The direction ratio is proportional to

$(1,2,3)$ ----- {multiplying by

$3$ }

We have the formula for Direction cosines,

$(\dfrac{a}{\sqrt{a^2+b^2+c^2}}, \dfrac{b}{\sqrt {a^2+b^2+c^2}},\dfrac{c}{\sqrt {a^2+b^2+c^2}})$

$\therefore (\dfrac{1}{\sqrt{1^2+2^2+3^2}},\dfrac{2}{\sqrt{1^2+2^2+3^2}},\dfrac{3}{\sqrt{1^2+2^2+3^2}})=(\pm\dfrac{1}{\sqrt{14}},\pm\dfrac{2}{\sqrt{14}},\pm\dfrac{3}{\sqrt{14}})$ are the direction cosines

Case (2) :

$2m-n=0$

$\implies m=\dfrac{n}{2}$

Substitute for

$m$ in (1)

$l=5m-3n=5(\dfrac{n}{2})-3n=-\dfrac{n}{2}$

Hence, the direction cosines

$(l,m,n)$ is

$\left (-\dfrac{n}{2},\dfrac{n}{2},n \right )$

The direction ratio is proportional to

$(-1,1,2)$ ----- {multiplying by

$2$ }

We have the formula for Direction cosines,

$(\dfrac{a}{\sqrt{a^2+b^2+c^2}}, \dfrac{b}{\sqrt {a^2+b^2+c^2}},\dfrac{c}{\sqrt {a^2+b^2+c^2}})$

$\therefore (\dfrac{-1}{\sqrt{(-1)^2+1^2+2^2}},\dfrac{1}{\sqrt{(-1)^2+1^2+2^2}},\dfrac{2}{\sqrt{(-1)^2+1^2+2^2}})=(\pm\dfrac{1}{\sqrt{6}},\pm\dfrac{1}{\sqrt{6}},\pm\dfrac{2}{\sqrt{6}})$ are the direction cosines

Find the vector and Cartesian equation of the line that passes through the points (3 , -2 , -5) and (3 , -2 , 6).

$\begin{matrix} Let \\ A\equiv \left( { 3,\, -2,\, -5 } \right) ,\, \, \, \, B\equiv \left( { 3,\, -2,\, 6 } \right) \, \, and\, \, O=\left( { 0,\, 0,\, 0 } \right) \\ \overrightarrow { OA } =\overrightarrow { a } =3\widehat { i } -2\widehat { j } -5\widehat { k } \\ \overrightarrow { OB } =\overrightarrow { b } =3\widehat { i } -2\widehat { j } +6\widehat { k } \\ Eq{ u^{ n } }\, of\, the\, line\, through\, A\left( { \overrightarrow { a } } \right) \, and\, B\left( { \overrightarrow { b } } \right) \, is\, \\ \overrightarrow { r } =\overrightarrow { a } +\lambda \left( { \overrightarrow { b } -\overrightarrow { a } } \right) \\ =3\widehat { i } -2\widehat { j } -5\widehat { k } +\lambda \left( { 0\widehat { i } -0\widehat { j } +11\widehat { k } } \right) \\ \therefore \overrightarrow { r } =3\widehat { i } -2\widehat { j } -5\widehat { k } +\lambda \left( { 11\widehat { k } } \right) \\ Eq{ u^{ n } }\, in\, \, cartensian\, form \\ Eq{ u^{ n } }\, of\, AB\, is\, \dfrac { { x-3 } }{ { 3-3 } } =\dfrac { { y+2 } }{ { -2+2 } } =\dfrac { { z+5 } }{ { -5-6 } } \\ \Rightarrow \dfrac { { x-3 } }{ 0 } =\dfrac { { y+2 } }{ 0 } =\dfrac { { z+5 } }{ { -11 } } \\ \end{matrix}$

Find the equation of plane through the points $(2, 2, 1)$ and $(9, 3, 6)$ and perpendicular to the plane $2x+6y+6z=9$

As the required plane is perpendicular to

$2x+6y+6z=9$ , the normal vector

$\vec{n_1}$ of plane

$2x+6y+6z=9$ lies in the required plane.

The normal vector

$\vec{n}$ of the required plane can be found by taking cross product of vector of given points with

$\vec{n_1}$ .

$\therefore\vec{n}=\vec{n_1}\times(2\hat{i}+2\hat{j}+\hat{k}-(9\hat{i}+3\hat{j}+6\hat{k}))$

$\therefore\vec{n}=(2\hat{i}+6\hat{j}+6\hat{k})\times(-7\hat{i}-\hat{j}-5\hat{k})$

$\therefore\vec{n}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&6&6\\-7&-1&-5\end{vmatrix}$

$\therefore\vec{n}=(-30+6)\hat{i}-(-10+42)\hat{j}+(-2+42)\hat{k}$

$\therefore\vec{n}=-24\hat{i}-32\hat{j}-40\hat{k}$

Now, equation of required plane is

$\vec{n}\cdot\vec{r}=\vec{n}\cdot(2\hat{i}+2\hat{j}+\hat{k})$

$\therefore (-24\hat{i}-32\hat{j}-40\hat{k})\cdot(x\hat{i}+y\hat{j}+z\hat{k})=(-24\hat{i}-32\hat{j}-40\hat{k})\cdot(2\hat{i}+2\hat{j}+\hat{k})$

$\therefore -24x-32y-40z=-48-64-40$

$\therefore 3x+4y+5z=19$

This is the required equation.

Show that the points $(3, 3), (h, 0)$ and $(0, k)$ are collinear, if $\dfrac {1}{n} + \dfrac {1}{k} = \dfrac {1}{3}$ .

For collinearity

${ \left| { \begin{array} { *{ 20 }{ l } }1 & 3 & 3 \\ 1 & h & 0 \\ 1 & 0 & k \end{array} } \right| =0 }$

This implies that,

$[{1\left( {hk} \right) - 3\left( k \right) + 3\left( { - h} \right) = 0}$

${hk - 3k - 3h = 0}$

${hk = 3h + 3k}$

${1 = \frac{3}{k} + \frac{3}{h}}$

${\frac{1}{h} + \frac{1}{k} = \frac{1}{3}}$

Hence, these points are collinear.

The foot of the perpendicular drawn from the origin to the plane is $(4, -2, -5)$ , then find the vector equation of plane.

Let equation of the given point lying in the plane be

$\vec{r_0}$ .

Let equation of the normal be

$\vec{n}$ .

$\therefore \vec{n}=\vec{0}-(4\hat{i}-2\hat{j}-5\hat{k})=-4\hat{i}+2\hat{j}+5\hat{k}$

Now, equation of plane with normal

$\vec{n}$ and point

$\vec{n_0}$ is

$\vec{n}\cdot(\vec{r}-\vec{r_0})=0$ ,

where

$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$ is any general point in the plane.

$\implies (-4\hat{i}+2\hat{j}+5\hat{k})\cdot(\vec{r}-(4\hat{i}-2\hat{j}-5\hat{k}))=0$

$\implies -4x+2y+5z+16+4+25=0$

$\implies 4x-2y-5z=45$

This is the required equation of the plane.

Prove that the points $(4, 5, -5), (0, -11, 3)$ and $(2, -3, -1)$ are collinear.

Method-1:

$\begin{vmatrix}4&5&-5\\0&-11&3\\2&-3&-1\end{vmatrix}=4(11+9)+2(15-55)=80-80=0$

$\therefore$ the three points are collinear.

Method-2:

Distance formula in 3D:

Distance between two points

$(x_{1},y_{1},x_{1}) and (x_2,y_2,z_2)=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

Let us call the points as

$A(4,5,-5)$ ,

$B(0,-11,3)$ and

$C(2,-3,-1)$ .

Now, we can calculate the distances

$\vec{AB}$ ,

$\vec{BC}$ and

$\vec{AC}$ .

$\vec{AB}=\sqrt{4^2+16^2+8^2}=\sqrt{336}=4\sqrt{21}$

$\vec{BC}=\sqrt{2^2+8^2+4^2}=\sqrt{84}=2\sqrt{21}$

$\vec{AC}=\sqrt{2^2+8^2+4^2}=\sqrt{84}=2\sqrt{21}$

$\vec{BC}+\vec{AC}=\vec{AB}$ .

$\therefore$

$A,B$ and

$C$ lie on a straight line.

Thus, the points are proved to be collinear.

Find the position vector of a point $R$ which divides the line joining two points $P$ and $Q$ whose position vectors are $(2\vec {a} + \vec {b})$ and $(\vec {a} - 3\vec {b})$ externally in the ratio $1 : 2$ . Also, show that $P$ is the mid point of the line segment $RQ$ .

Let the position vector of

$P$ be

$\overrightarrow { OP } = 2\overrightarrow { a } +\overrightarrow { b }$

Position vector of

$Q$ be

$\overrightarrow { OQ } = \overrightarrow { a } -3\overrightarrow { b }$

Now it is given the point R divides the line PQ externally in the ratio

$1:2$

Therefore position vector of

$R = \dfrac { m\overrightarrow { OQ } -n\overrightarrow { OP } \quad }{ m-n }$

$=\dfrac { \left(1 \left( \overrightarrow { a } -3\overrightarrow { b } \right) \right) -2\left( 2\overrightarrow { a } +\overrightarrow { b } \right) \quad }{ 1-2 }$

$=3\overrightarrow { a } +5\overrightarrow { b }$

Midpoint of

$RQ = \frac { \overrightarrow { OQ } +\overrightarrow { OR } \quad }{ 2 }$

$=\dfrac { \left( \overrightarrow { a } -3\overrightarrow { b } \right) +( 3\overrightarrow { a } + 5\overrightarrow { b } ) \quad }{ 2 }$

$=\dfrac { ( 4\overrightarrow { a } +2 \overrightarrow { b } )}{ 2 }$

$= 2\overrightarrow { a } +\overrightarrow { b }$

$=\overrightarrow {OP}$

Hence,

$P$ is the mid point of segment

$RQ$

The direction cosines of $\bar { AB }$ are $-2,2,1$ . If $A\equiv (4,1,5)$ and $l(AB)=6$ units, find the coordinates of $B$ .

Given direction ratio of $AB= -2, 2,1$
Note= if any line have direction ratio = $(a, b, c)$ then $(ka, kb, kc)$ is also the direction ratio of that line..
$A=(x_{1}, y_{1}, z_{1} ) =(4, 1,5)$
Let $B=(x_{2}, y_{2}, z_{2})$
hence $(x_{2}-x_{1})=-2K,( y_{2}-y_{1})=2K,( z_{2}-z_{1})=K$ .........equ( $1$ )
hence $x_{2} =-2\times K+4$
$y_{2} =2\times K+1$
$z_{2} = K+5$
$l(AB) =6$ units
Apply distance formula $d =\sqrt{(x_{2}-x_{1})^2 +(y_{2}-y_{1})^2 +(z_{2}-z_{1})^2}$ ....... equ( $2$ )
using equation( $1$ ) and equ( $2$ )
$6=\sqrt{(-2\times K)^2 +(2 \times K)^2 +K^2}$
on solving this equation, we will get
$9\times K^2=36$
$K =2$ or $-2$
Now using equation( $1$ )
For $K=2$ $B=(0, 5, 7)$
And for $K=-2$ $B=(8, -3, 3)$

Find a vector $\vec { r }$ of magnitude $3\sqrt { 2 }$ units which makes and angle $\cfrac { \pi }{ 4 }$ and $\cfrac { \pi }{ 2 }$ with $y$ and z-axis respectively

$\overrightarrow { r } =x\hat { i } +y\hat { j } +2\hat { k } \\ \left| \vec { r } \right| =3\sqrt { 2 } \\ \sqrt { { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } } =3\sqrt { 2 } \\ { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }=18$

It makes $\cfrac { \pi }{ 4 }$ with yours

So, $\overrightarrow { r } .\hat { j } =\left| \vec { r } \right| \cos { \cfrac { \pi }{ 4 } } \\ y=\cfrac { 3\sqrt { 2 } }{ \sqrt { 2 } } =3$

It makes $\cfrac { \pi }{ 2 }$ with $z-axis$

So, $\overrightarrow { r } .\hat { k } =\left| \vec { r } \right| \cos { \cfrac { \pi }{ 2 } } \\z=0$

So, ${ x }^{ 2 }+{ 3 }^{ 2 }+{ 0 }^{ 2 }=18\\ { x }^{ 2 }+9=18\\ { x }^{ 2 }=9\\ x=\pm 3$

So, $\overrightarrow { r } =3\hat { i } +3\hat { j } /-3\hat { i } +3\hat { j }$

Find the vector equation of a line passing through the point $A$ whose position vector is $3\overline{i}+\overline{j}-\overline{k}$ and which is parallel to the vector $2\overline{i}-\overline{j}+2\overline{k}$ .If $P$ is a point of this line such that $AP=15$ ,find the position vector of $P$ .

Given:

$A\left( {3\vec i + \vec j - \vec k} \right)$ and

$P\left( {2\vec i - \vec j + 2\vec k} \right)$

The equation of line passing through

$\left( {3\vec i + \vec j - \vec k} \right)$ and parallel to

$\left( {2\vec i - \vec j + 2\vec k} \right)$

$\vec r = \left( {3\vec i + \vec j - \vec k} \right) + \lambda \left( {2\vec i - \vec j + 2\vec k} \right)$

Direction ratio of

$AP=(a-3,b-1,c+1)$ ,

$AP$ is parallel to

${2\vec i - \vec j + 2\vec k}$ then

$\frac{{a - 3}}{2} = \frac{{b - 1}}{{ - 1}} = \frac{{c + 1}}{2} = t$ (constant)

$AP=15$

$\sqrt { { { \left( { a-3 } \right) }^{ 2 } }+{ { \left( { b-1 } \right) }^{ 2 } }+{ { \left( { c+1 } \right) }^{ 2 } } } =15$

$\Rightarrow \sqrt { { { \left( { 2t+3-3 } \right) }^{ 2 } }+{ { \left( { -t+1-1 } \right) }^{ 2 } }+{ { \left( { 2t-1+1 } \right) }^{ 2 } } } =15$

$\Rightarrow \sqrt { 4{ t^{ 2 } }+{ t^{ 2 } }+4{ t^{ 2 } } } =15$

$\Rightarrow \sqrt { 9{ t^{ 2 } } } =15$

$\Rightarrow t=5$

$\begin{array}{l} a=2\times 5+3=13 \\ b=-5+1=4 \\ c=2\times 5-1=9 \end{array}$

Position vector of

$P = 13\vec i + 4\vec j + 9\vec k$

If $\vec a, \vec b$ and $\vec c$ are mutually perpendicular vectors of equal magnitudes, If the angles which the vector $2\vec a+\vec b+2\vec c$ makes with the vectors $\vec a$ is $\cos^{-1}\sqrt{\dfrac{2}{m}}$ . Find $m$

$\vec a.\vec b =\vec b.\vec c=\vec c.\vec a=0$

$|\vec a|=|\vec b|=|\vec c|=k$

$\vec c.(2 \vec a+\vec b+2\vec c)=|\vec c||2 \vec a+\vec b+ 2 \vec c|\cos \theta$

Where

$\theta$ is the angle between

$\vec c$ and

$2\vec a+\vec b+2\vec c$

$2 \vec a.\vec c+\vec b.\vec c+2\vec c.\vec c=k|2\vec a+\vec b+2\vec c|\cos \theta$

$[\vec a.\vec b=\vec b.\vec a]$

$2|\vec c|^2=k[2\vec a+\vec b+2\vec c]\cos \theta$

$2k^2=k[2\vec a+\vec b+2\vec c]\cos \theta$

$\cos \theta=\frac{2k}{|2\vec a+\vec b+2\vec c|}$

$\theta=\cos ^{-1}\frac{2k}{|2\vec a+\vec b+2\vec c|}$

$\cos ^{-1}\sqrt{\frac{2}{m}}=\cos ^{-1}\frac{2k}{|2\vec a+\vec b+2\vec c|}$

$\sqrt{\frac{2}{m}}=\frac{2k}{|2\vec a+\vec b+2\vec c|}$ ---- (i)

Now,

$\begin{array}{l} { \left| { 2\vec { a } +\vec { b } +2\vec { c } } \right| ^{ 2 } }=\left( { 2\vec { a } +\vec { b } +2\vec { c } } \right) .\left( { 2\vec { a } +\vec { b } +2\vec { c } } \right) \\ =4.\vec { a } .\vec { a } +2\vec { a } .\vec { b } +4.\vec { a } .\vec { c } +2.\vec { a } .\vec { b } +\vec { b } .\vec { b } +2.\vec { b } .\vec { c } +4.\vec { a } .\vec { c } +2.\vec { b } .\vec { c } +4.\vec { c } .\vec { c } \\ =4{ \left| { \vec { a } } \right| ^{ 2 } }+{ \left| { \vec { b } } \right| ^{ 2 } }+4{ \left| { \vec { c } } \right| ^{ 2 } } \\ =4{ k^{ 2 } }+{ k^{ 2 } }+4{ k^{ 2 } } \\ =9{ k^{ 2 } } \\ \left| { 2\vec { a } +\vec { b } +2\vec { c } } \right| =3k \end{array}$

Putting these values in (i)

$\sqrt{\frac{2}{m}}=\frac{2k}{3k}$

$\sqrt{\frac{2}{m}}=\frac{2}{3}$

$\frac{2}{m}=\frac{4}{9}$

$m=\frac{9}{2}$

Find the direction cosines (d.cs) of directed line $OP$ if coordinates of $P$ is $(2, 3, 7)$ , $O$ being the origin.

To find the direction cosines of the the line

$OP$ where co-ordinates of

$O$ and

$P$ are

$(0,0,0)$ and

$(2,3,7)$ respectively.

Now the direction co-sine of the line

$OP$ is

$\left(\dfrac{2-0}{\sqrt{2^2+3^2+7^2}},\dfrac{3-0}{\sqrt{2^2+3^2+7^2}},\dfrac{7-0}{\sqrt{2^2+3^2+7^2}}\right)=\left(\dfrac{2}{\sqrt{62}},\dfrac{3}{\sqrt{62}},\dfrac{7}{\sqrt{62}}\right)$ .

If a directed line $OP$ makes angles $45^o, 60^o$ & $60^o$ with $+ve$ direction of $x, y, z$ axis respectively then find d.cs of $OP$

Given the line

$OP$ makes angles

$45^o,60^o,60^o$ with the positive direction of

$x,y,z$ axis respectively.

Then the direction cosines (d.cs) of

$OP$ be respectively

$\cos 45^o, \cos 60^o, \cos 60^o$ i.e.

$\dfrac{1}{\sqrt{2}},\dfrac{1}{2},\dfrac{1}{2}$ .

Find the direction cosines of the sides of the triangle whose vertices are $(3,5,-4)$ and $(-1,1,2)$ and $(-5,-5,-2)$ what type of triangle is it.

Let,

$A(3, 5, - 4), B(- 1, 1, 2)$ and

$C(- 5, - 5, - 2)$

(i) Direction ratios of the line joining A and B are

$(-1-3), (1-5), (2-(-4)) = (-4, -4, 6)$

$\therefore AB = \sqrt{(-4)^2 + (-4)^2 + 6^2 } = \sqrt{16+16+36} = \sqrt{68} = 2\sqrt{17}$

The direction cosines are

$\dfrac{-4}{2\sqrt{17}}, \dfrac{-4}{2\sqrt{17}}, \dfrac{6}{2\sqrt{17}}$

$= \dfrac{-2}{\sqrt{17}}, \dfrac{-2}{\sqrt{17}}, \dfrac{3}{\sqrt{17}}$

(ii) Direction ratios of the line joining B and C are

$(-5-(-1)), (-5-1), (-2-2) = (-4, -6, -4)$

$\therefore BC = \sqrt{(-4)^2 + (-6)^2 + (-4)^2 } = \sqrt{16+36+16} = \sqrt{68} = 2\sqrt{17}$

The direction cosines are

$\dfrac{-4}{2\sqrt{17}}, \dfrac{-6}{2\sqrt{17}}, \dfrac{-4}{2\sqrt{17}}$

$= \dfrac{-2}{\sqrt{17}}, \dfrac{-3}{\sqrt{17}}, \dfrac{-2}{\sqrt{17}}$

(iii) Direction ratios of the line joining C and A are

$(3-(-5)), (5-(-5)), (-4-(-2)) = (8, 10, -2)$

$\therefore CA = \sqrt{(8)^2 + (10)^2 + (-2)^2 } = \sqrt{64+100+4} = \sqrt{168} = 2\sqrt{42}$

The direction cosines are

$\dfrac{8}{2\sqrt{42}}, \dfrac{10}{2\sqrt{42}}, \dfrac{-2}{2\sqrt{42}}$

$= \dfrac{4}{\sqrt{42}}, \dfrac{5}{\sqrt{42}}, \dfrac{-1}{\sqrt{42}}$

Two of the sides are equal, so the triangle is an isosceles triangle.

Can a line have direction angles as $45^o, 45^o$ & $60^o$ .

No, a line can't have the given direction angles.

Proof by contradiction:-

If possible let there exists a line which has the direction angles

$45^o,45^o,60^o$ .

Then its direction cosines will be

$(l,m,n)=(\cos 45^o,\cos45^o,\cos 60^o)=\left(\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}},\dfrac{1}{2}\right)$ .

The from the property of d.cs we must have

$l^2+m^2+n^2=1$ .

But in this case

$l^2+m^2+n^2$

$=\dfrac{5}{4}\ne 1$ .

So our assumption is wrong, there can't be any line having the given direction angles.

Find angle in degrees between the line joining the points $(2,1,3)$ and $(1,-1,2)$ and the line having direction ratios $2,1 ,-1$ .

Direction ratio of line passing through the points (2,1,3) and (1,-1,2)

are

$2-1,1-(-1),3-2=1,2,1$

Let angle between them be

$\theta$

Now,

$\cos\theta=\dfrac{2\times1+1\times2+1\times-1}{\sqrt{1^2+2^2+1^2}\sqrt{2^2+1^2+(-1)^2}}=\dfrac{3}{6}=\dfrac12\\ \Rightarrow \theta=cos^{-1}\dfrac12\Rightarrow\theta = 60^o$

Write the equation of the line passing through $(2 , -6, 3)$ & having D.R.'s $\propto$ $1, 5, 9$ .

The line passes through the points

$(2,-6,3)$

and direction ratios

$(1,5,9)$

If line passing through the point

$(a,b,c)$ and having direction ratios

$(p,q,r)$ then the equation of line will be,

$\dfrac{x-a}p=\dfrac{y-b}q=\dfrac{z-c}r$

From above data, equation will be;

$\dfrac{x-2}1=\dfrac{y+6}5=\dfrac{z-3}9$

$\Rightarrow 45(x-2)=45\left(\dfrac{y+6}5\right)=45\left(\dfrac{z-3}9\right)$

$\Rightarrow 45x-90=9y+54=5z-15$

Write the vector equation of the line $\dfrac {x - 5}{3} = \dfrac {y + 4}{7} = \dfrac {z - 6}{2}$ .

Cartesian equation

$=\dfrac{x-5}{3} = \dfrac{y+4}{7} = \dfrac{z-6}{2}$

Let

$\dfrac{x-5}{3} = \dfrac{y+4}{7} = \dfrac{z-6}{2} = \alpha$

$x = 3\alpha + 5, y = 7\alpha - 4, z = 2\alpha + 6$

Let

$(x, y, z)$ be position vectors.

$\therefore x\hat{i} + y\hat{j} + z\hat{k} = 5\hat{i} + (-4)\hat{j} + 6\hat{k} + \alpha (3\hat{i} + 7\hat{j} + 2\hat{k})$

vector form :

$5\hat{i} + (-4)\hat{j} + 6\hat{k} + \alpha (3\hat{i} + 7\hat{j} + 2\hat{k})$

If $P(x, y, z)$ is point in the space at a distance $r$ from the origin $O$ , then direction cosines of the line $OP$ are

Given

$|OP|=r$ , where

$O(0,0,0)$ is the origin and

$P(x,y,z)$ be any point in the space.

Now direction cosines of

$OP$ is

$\left(\dfrac{x-0}{|OP|},\dfrac{y-0}{|OP|},\dfrac{z-0}{|OP|}\right)$

$=\left(\dfrac{x}{r},\dfrac{y}{r},\dfrac{z}{r}\right)$ .

Show that the lines whose d.c's are given by $2l + 2m - n = 0,\,mn + nl + lm = 0$ are perpendicular to each other.

Consider the given equation.

$2l+2m-n=0$

$n=2m+2n$

Substitute this value in the second equation.

$m\left( 2l+2m \right)+l\left( 2l+2m \right)+lm=0$

$2lm+2{{m}^{2}}+2{{l}^{2}}+2lm+lm=0$

$2{{m}^{2}}+5lm+2{{l}^{2}}=0$

$\left( m+2l \right)\left( 2m+l \right)=0$

$m=-2l\ or\ -\dfrac{l}{2}$

Now, put $l=1$ .

$m=-2\ or\ -\dfrac{1}{2}$

$n=-2\ or\ 1$

Therefore,

${{d}_{1}}=\left( 1,-2,-2 \right)$

${{d}_{2}}=\left( 1,-\dfrac{1}{2},1 \right)$

Now,

${{d}_{1}}\cdot {{d}_{2}}=1\times 1+\left( -2 \right)\times \left( -\dfrac{1}{2} \right)+\left( -2 \right)\times \left( 1 \right)$

${{d}_{1}}\cdot {{d}_{2}}=1+1-2$

${{d}_{1}}\cdot {{d}_{2}}=0$

Hence, both the lines are perpendicular to each other.

If $l_1, m_1, n_1$ and $l_2, m_2, n_2$ are the direction cosines of two lines, then show that the direction cosines of the line perpendicular to them are proportional to $m_1n_2-m_2n_1$ , $n_1l_2-n_2l_1$ , $l_1m_2-l_2m_1$ .

Given:

Let ${l_1},{m_1},{n_1}$ and ${l_2},{m_2},{n_2}$ are the direction cosines of two given lines ${L_1}\, and \, {L_2}$ .

Let $\widehat{{n_1}}\;and\;\widehat {{n_2}}$ be the unit vectors along these lines ${L_1}\;and \, {L_2}$ .

$\vec{{n_1}} = {l_1}i + {m_1}j + {n_1}k \;and \; \vec{{n_2}} = {l_2}i + {m_2}j + {n_2}k$

The cross-product of two vectors $\widehat {{n_1}} \times \widehat {{n_2}} = \left| {\widehat {{n_1}}} \right| \cdot \left| {\widehat {{n_2}}} \right|\sin 90^\circ \hat{n}$

${L_1} \bot {L_2}$ , the angle between them is $90^\circ$

$\widehat {{n_1}} \times \widehat {{n_2}} = \hat{n}$

$\widehat {{n_1}} \times \widehat {{n_2}} = \hat{n} = \left| {\begin{array}{*{20}{c}} i&j&k\\ {{l_1}}&{{m_1}}&{{n_1}}\\ {{l_2}}&{{m_2}}&{{n_2}} \end{array}} \right|$

${\hat n} = \left( {{m_1}{n_1} - {m_2}{n_1}} \right)i - \left( {{l_1}{n_2} - {l_2}{n_1}} \right)j + \left( {{l_1}{m_2} - {l_2}{m_1}} \right)k$

Here, $$ \hat{n} $$ is a unit vector.

Thus, the direction cosines are $\left ({m_1}{n_1} - {m_2}{n_1} \right ),\left ({l_1}{n_2} - {l_2}{n_1} \right ),\left ({l_1}{m_2} - {l_2}{m_1} \right )$ .

If a line makes angles $90^{\circ}$ , $135^{\circ}$ , $45^{\circ}$ with the X, Y, and Z axes respectively, then find its product direction cosines.

Respective Direction cosines are

$\cos90^o,\cos135^o,\cos45^o$

Since,

$\cos90^o=0$

hence, The Product of three direction cosines

$=0$

The cartesian equation of a line are $3x - 1=6y + 2=1 - z$ . Find the direction ratios and writes its equation in vector form.

$L:3x-1=6y+2=1-z\\ \Rightarrow \cfrac { x-\cfrac { 1 }{ 3 } }{ \cfrac { 1 }{ 3 } } =\cfrac { y+\cfrac { 1 }{ 3 } }{ \cfrac { 1 }{ 6 } } =\cfrac { z-1 }{ -1 }$

Comparing with standard equation of line,

$\cfrac { x-{ x }_{ 1 } }{ l } =\cfrac { y-{ y }_{ 1 } }{ m } =\cfrac { z-{ z }_{ 1 } }{ n } \\ (l,m,n)=(\cfrac { 1 }{ 3 } ,\cfrac { 1 }{ 6 } ,-1)\quad ,(x,y,z)=(\cfrac { 1 }{ 3 } ,\cfrac { -1 }{ 3 } ,1)$

$\therefore$ Direction ratios are

$(\cfrac { 1 }{ 3 } ,\cfrac { 1 }{ 6 } ,-1)$

Equation of plane in vector form,

$\overrightarrow { r } =\cfrac { 1 }{ 3 } \hat { i } -\cfrac { 1 }{ 3 } \hat { j } +\hat { k } +\lambda (\cfrac { 1 }{ 3 } \hat { i } +\cfrac { 1 }{ 6 } \hat { j } -\hat { k } )$

If a line makes angles $\alpha, \beta, \gamma$ with coordinate axes, Find the value of $\sin^{2} \alpha + \sin^{2}\beta + \sin^{2} \gamma$

$\alpha, \beta, \gamma$ is the angle, a line makes with the co-ordinate axis,

$\therefore \cos ^2\alpha + \cos ^2\beta + \cos ^2\gamma = 1$

$\Rightarrow (1 - \sin ^2\alpha) + (1 - \sin ^2\beta) + (1 - \sin ^2\gamma) = 1$

$\Rightarrow 3 - (\sin ^2\alpha + \sin ^2\beta + \sin ^2\gamma) = 1$

$\therefore (\sin ^2\alpha + \sin ^2\beta + \sin ^2\gamma)= 2$

Find the angles between the lines whose direction cosines are given by the equations.
$l+m+n=0$
$l^2+m^2+n^2=0$

According to question,

$l + m + n = 0,\,\,\,\,{l^2} + {m^2} + {n^2} = 0$

so,

Angle between two lines is

$\cos \theta = \dfrac{{\overline {{b_1}} .{{\overline b }_2}}}{{\left| {{b_1}} \right|\,\left| {{b_2}} \right|}}$

$\begin{array}{l} \cos \theta =\dfrac { { \overline { { b_{ 1 } } } .{ { \overline { b } }_{ 2 } } } }{ { \left| { { b_{ 1 } } } \right| \, \left| { { b_{ 2 } } } \right| } } \\ step\, 1:\, l+m+n=0----(1) \\ \, \, \, \, \, \, \, \, \, \, \, \, l+m=-n \\ \, \, \, \, \, \, \, \, \, \, n=-(\, l+m) \\ and \\ \, \, \, \, \, \, \, \, \, { l^{ 2 } }+{ m^{ 2 } }+{ n^{ 2 } }=0-------(2) \\ \, \, \, \, \, \, \, \, \, { l^{ 2 } }+{ m^{ 2 } }+{ (-(\, l+m))^{ 2 } }=0 \\ \, \, \, \, \, \, { l^{ 2 } }+{ m^{ 2 } }+(-({ l^{ 2 } }+{ m^{ 2 } }+2lm))=0 \\ \, \, \, \, \, { l^{ 2 } }+{ m^{ 2 } }-{ l^{ 2 } }-{ m^{ 2 } }-2lm=0 \\ \, \, \, \, \therefore \, \, \, lm=0,\, \, \, \, \\ \, \, \, i.e\, either\, l=0,\, m=0 \\ \, \, \, \, \, let\, us\, put\, m=0\, \, in\, \, \, \, equ----(1) \\ \end{array}$

$\begin{array}{l} if\, \, m=0\, then\, \, \, \, l=-n\, \, \, \, \, \, \, \, \, (dr's\, \, is\, \, proportional) \\ direction\, \, ratio\, \, \, (l,m,n)=(1,0,-1)----({ b_{ 1 } } \\ \, \, \, \, \, \, put\, \, l=0,\, then\, \, we\, get\, \, m=-n \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, dr's\, (l,m,n).(0,1,-1)---({ b_{ { 2 } } } \\ steps\, 2: \\ \, \, \, \, \, \, \, \, \, { b_{ 1 } }{ b_{ 2 } }=(1,0,-1).(0,1,-1)=1 \\ \, \, \, \, \, \, \, \, \, \, \left| { { b_{ 1 } } } \right| =\sqrt { 1+01 } =\sqrt { 2\, } \, \, \, ,\, \, \, \left| { { b_{ 2 } } } \right| =\sqrt { 2\, } \\ steps3:\, \, \, \, \, \\ \, \, \, \, \, \, \, \cos \theta =\, \, \dfrac { { \overline { { b_{ 1 } } } \, \overline { { b_{ 2 } } } \, } }{ { \left| { { b_{ 1 } } } \right| \, \left| { { b_{ 2 } } } \right| } } =\dfrac { 1 }{ { \sqrt { 2\, } \, \sqrt { 2\, } } } =\dfrac { 1 }{ 2 } \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \therefore \, \, \theta =\, \, \dfrac { \pi }{ 3 } \end{array}$

Show that angles between any two diagonals $\theta= \cos^{-1} \left(\dfrac{1}{3}\right)$

Directions of

$OG,AE,CD,FB$

$OG=1,1,1$

$AE=-1,1,1$

$CD=1,1,-1$

$FB=-1,1,-1$

let angle between

$OG$ and

$AE$ be

$\therefore \cos { \theta } =\cfrac { \left| x \right| +\left| x \right| +\left| x \right| }{ \sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 }+{ 1 }^{ 2 } } \sqrt { { -1 }^{ 2 }+{ 1 }^{ 2 }+{ 1 }^{ 2 } } } =\cfrac { 1 }{ 3 }$

Clearly,

$\cos { \theta } =\cfrac { 1 }{ 3 }$ for any two diagonals.

$\therefore \theta =\cos ^{ -1 }{ (\cfrac { 1 }{ 3 } ) }$ is angle between any two diagonals.

1069465_1052919_ans_bd54c520d4ed493896cfe6416b62ecc6.png

If a line makes angles $\alpha, \beta, \gamma$ with coordinate axes, find
$\cos^{2}\alpha + \cos^{2}\beta + \cos^{2}\gamma +1$

If a line makes

$\alpha, \beta, \gamma$ with the co-ordinates axes,

$\therefore \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$

$\therefore \cos^2\alpha + \cos^2\beta + \cos^2\gamma - 1 = 0$

Hence, the answer is

$0$ .

Find the perpendicular distance from the origin to the plane passing through the points $(1, -2, 5), (0, -5, -1)$ and $(-3, 5,0)$ .

We can get two vectors in the plane by subtracting pairs of points in the

plane:

$\left[{\begin {matrix}1\\-2\\5\end {matrix}}\right]-\left[{\begin {matrix}0\\-5\\-1\end {matrix}}\right]=\left[{\begin {matrix}1\\3\\6\end {matrix}}\right]$

$\left[{\begin {matrix}-3\\5\\0\end {matrix}}\right]-\left[{\begin {matrix}0\\-5\\-1\end {matrix}}\right]=\left[{\begin {matrix}-3\\10\\1\end {matrix}}\right]$

The cross product of these two vectors will be in the unique direction orthogonal to both, and hence in the direction of the normal vector to the plane.

$\left[{\begin {matrix}1\\3\\6\end {matrix}}\right]$ x

$\left[{\begin {matrix}-3\\10\\1\end {matrix}}\right]=\left[{\begin {matrix}-57\\-19\\19\end {matrix}}\right]$

The Equation of Plane

$ax+by+cz+d=0$ where

$\left[{\begin {matrix}a\\b\\c\end {matrix}}\right]$ is the normal vector to the plane

Hence,

$\left[{\begin {matrix}-57\\-19\\19\end {matrix}}\right]$ is normal vector to the plane

$-57x-19y+19z+d=0$

Putting one of three points in this Equation we can get

$d$

$-57*0-5*(-19)+19*-1+d=0\Rightarrow d=76$

So, Equation of plane

$=-57x-19y+19z+76=0\Rightarrow -3x-y+z+4=0$

Distance of Origin from this plane

$=\dfrac{-3(0)-(0)+(0)+4}{\sqrt{3^2+1+1}}=\dfrac{4}{\sqrt{11}}$

A mirror and a source of light are situated at a origin $O$ and a point on $OX$ , respectively. A ray of light from the source strikes the mirror and is reflected . If the direction ratios of the normal to the plane are $1 ,-1, 1$ , then find the $DCs$ of the reflected ray.

Let the source of light be situated at

$A(a,0,0)$

Let

$OA$ be the incident ray and

$OB$ be the reflected ray

Let

$ON$ be the normal to the mirror at

$O$

So,

$\angle AON =\angle NOB=\dfrac{\theta}{2}$ (say) ...

$(1)$

$DR's$ of

$OA$ are

$a,0,0$ and so its

$DC's$ are

$1,0,0$

$DR's$ of

$ON$ are

$\dfrac{1}{\sqrt{3}},\dfrac{-1}{\sqrt{3}},\dfrac{1}{\sqrt{3}}$

So from

$(1)$ we can say that

$cos(\dfrac{\theta}{2})=\dfrac{1}{\sqrt{3}}$

Let

$l,m,n$ be the

$DC's$ of the reflected ray

$OB$

$\Rightarrow \dfrac{l+1}{2cos(\dfrac{\theta}{2})}=\dfrac{1}{\sqrt{3}}$ ........

$(2)$

$\Rightarrow \dfrac{m+0}{2cos(\dfrac{\theta}{2})}=\dfrac{-1}{\sqrt{3}}$ ........

$(3)$

$\Rightarrow \dfrac{n+0}{2cos(\dfrac{\theta}{2})}=\dfrac{1}{\sqrt{3}}$ ........

$(4)$

On substituting the value of

$cos(\dfrac{\theta}{2})=\dfrac{1}{\sqrt{3}}$ in

$(2),(3),(4)$ , we get

$\Rightarrow l=\dfrac{2}{3}-1$

$\Rightarrow l=\dfrac{-1}{3}$

$\Rightarrow m=\dfrac{-2}{3}$

$\Rightarrow n=\dfrac{2}{3}$

Therefore,

$DC's$ of the reflected ray are

$\dfrac{-1}{3},\dfrac{-2}{3},\dfrac{2}{3}$

Find the direction cosines of the sides of the triangle whose vertices are $(3, 5, -2),(-1, 1 , 2)$ and $(-5 , -5, -2)$ . What type of triangle is it?

Direction ratios of :

$AB=(4,4,-4)\\ BC=(8,10,0)\\ CA=(4,6,4)$

Directions of

$AB=(\cfrac { 1 }{ \sqrt { 3 } } ,\cfrac { 1 }{ \sqrt { 3 } } ,\cfrac { -1 }{ \sqrt { 3 } } )\\ BC=(\cfrac { 4 }{ \sqrt { 41 } } ,\cfrac { 5 }{ \sqrt { 41 } } ,0)\\ CA=(\cfrac { 2 }{ \sqrt { 17 } } ,\cfrac { 3 }{ \sqrt { 17 } } ,\cfrac { 2 }{ \sqrt { 17 } } )$

angle between,

$AB\& BC=\cos ^{ -1 }{ (\cfrac { 1 }{ \sqrt { 3 } } .\cfrac { 4 }{ \sqrt { 41 } } +\cfrac { 1 }{ \sqrt { 3 } } .\cfrac { 5 }{ \sqrt { 41 } } ) } =\cos ^{ -1 }{ (\cfrac { 9 }{ \sqrt { 123 } } ) } \\ BC\& CA=\cos ^{ -1 }{ (\cfrac { 4 }{ \sqrt { 41 } } .\cfrac { 2 }{ \sqrt { 17 } } +\cfrac { 5 }{ \sqrt { 41 } } .\cfrac { 3 }{ \sqrt { 17 } } ) } =\cos ^{ -1 }{ (\cfrac { 23 }{ \sqrt { 697 } } ) } \\ AB\& CA=\cos ^{ -1 }{ (\cfrac { 2 }{ \sqrt { 17 } } .\cfrac { 1 }{ \sqrt { 3 } } +\cfrac { 1 }{ \sqrt { 3 } } .\cfrac { 3 }{ \sqrt { 17 } } -\cfrac { 2 }{ \sqrt { 17 } } .\cfrac { 1 }{ \sqrt { 3 } } ) } =\cos ^{ -1 }{ (\cfrac { 3 }{ \sqrt { 51 } } ) }$

All angles different therefore triangle is scalene.

Show that the straight lines whose direction cosines are given by $2l+2m−n=0$ and $mn+nl+lm=0$ are at right angles.

$2l+2m=n$ ...............

$(1)$

put

$n$ in

$mn+nl+lm=0$ ......

$(2)$

$\Rightarrow 2lm+2{m}^{2}+2{l}^{2}+2lm+lm=0$

$\Rightarrow 2{m}^{2}+5lm+2{l}^{2}=0$

$\Rightarrow 2{m}^{2}+4lm+lm+2{l}^{2}=0$

$\Rightarrow 2m\left(m+2l\right)+l\left(m+2l\right)=0$

$\Rightarrow \left(m+2l\right)\left(2m+l\right)=0$

$\Rightarrow m=-2l\,\,,m=\dfrac{-l}{2}$

Put

$l=1\Rightarrow {d}_{1}=\left(1,-2,-2\right)$ and

${d}_{2}=\left(1,\dfrac{-1}{2},1\right)$

$\Rightarrow {d}_{1}{d}_{2}=0$

Hence the given lines are at right angles.

Find the equation of the plane if the perpendicular from the origin to the plane is (2, 3, -5).

Let the point of plane be

$p(2,3,-5)$

$\therefore$ the equation of plane through

$p(2,3,-5)$ is

$a\left( {n - 2} \right) + b\left( {y - 3} \right) + c\left( {z - \left( { - 5} \right)} \right) = 0\left( {0,0,0} \right)$

The disection ratios of the line through the point

$0(0,0,0)$ and

$(2,3,-5)$ are

$2-0,3-0,-5-0$

$\therefore$ the line op with disection -ratios

$4,3,2$ is normal to the plane

$(i)$

$\therefore$ equation (i) of plane becomes

$\begin{array}{l} 2\left( { x-2 } \right) +3\left( { y-3 } \right) +\left( { -5 } \right) \left( { z-\left( { -5 } \right) } \right) =0 \\ \Rightarrow 2x-2+3y-9-5z-25=0\, \, \left[ { \frac { a }{ 2 } =\frac { b }{ 3 } =\frac { c }{ { -5 } } } \right] \\ \Rightarrow 2x+3y-5z-38=0 \end{array}$

1202145_1062126_ans_b768c801b86e47968e7408c95313d66c.JPG

Find the vector equation of the line joining the points $2i+j+3k$ and $-4i+3j-k$ $(2i+j+3k) + \lambda (pi-2j+4k)$ then p =

1979592_1057236_ans_3d64d13a21d54a10b6f3a152d5b961c6.jpg

Find the equation of the line through the origin which intersect the line $\frac{{x - 3}}{2} = \frac{{y - 3}}{1} = \frac{z}{1}$ at an angle of $\dfrac{\pi}{2}$ and $\frac{{x - 3}}{1} = \frac{{y - 3}}{2} = \frac{z}{1}$ at an angle of $\dfrac{\pi}{2}$

Let

$a,b,c$ be of required line

From

$1st$ line

$\Rightarrow 2a+b+c= 0$ -------

$(1)$

From

$2nd$ line

$\Rightarrow a+2b+c=0$ -------

$(2)$

From

$(1), (2) \Rightarrow a : b : c = 1: 1: 3$

So the required line

$\dfrac{x}{1} = \dfrac{y}{1} = \dfrac{z}{3}$

Find the direction cosines of a line which makes equal angles with the positive coordinate axes.

Let angles made by line cos

coordinate axes be

$\alpha ,\beta ,\gamma$

$\therefore \alpha =\beta =\gamma\Rightarrow$ (Given)

We know that,

$\cos^{2}\alpha +\cos^{2}\beta +\cos^{2}\gamma =1\rightarrow$ Property

$\therefore \cos^{2}\alpha +\cos^{2}\alpha +\cos^{2} \alpha =1$

$\therefore 3\cos^{2}\alpha =1$

$\therefore \cos^{2}\alpha =\dfrac{1}{3}$

$\therefore \cos\alpha =\pm \dfrac{1}{\sqrt{3}}$

$\therefore$ direction cosines are

$\displaystyle \left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$

$\displaystyle \left(-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{3}}\right)$

Find the direction cosines of the line PQ joining the points $P(2, 3, 4)$ and $Q(2, 1, 1)$ .

$\begin{matrix} Direction\, \, \cos ines\, \, of\, \, the\, \, line \\ PQ\, \, is\, \, \left( { 0,-2,-3 } \right) \\ \end{matrix}$

Find the angle between the lines whose direction ratios are $3, 2, -6$ and $1, 2, 2$ .

$({ l }_{ 1 }m_{ 1 }n_{ 1 })=(3,2,-6)$

$({ l }_{ 2 }m_{ 2 }n_{ 2 })=(1,2,2)$

$\therefore \cos { \theta } =\cfrac { l_{ 1 }l_{ 2 }+m_{ 1 }m_{ 2 }+n_{ 1 }n_{ 2 } }{ \sqrt { l_{ 1 }^{ 2 }+m^{ 2 }_{ 1 }+n_{ 1 }^{ 2 } } \sqrt { l_{ 2 }^{ 2 }+m_{ 2 }^{ 2 }+n_{ 2 }^{ 2 } } }$

$=\cfrac { 3\times 1+2\times 2+(-6)\times 2 }{ \sqrt { 3^{ 2 }+2^{ 2 }+6^{ 2 } } \sqrt { 1^{ 2 }+2^{ 2 }+2^{ 2 } } } =\cfrac { -5 }{ 21 }$

$\therefore \theta =\cos ^{ -1 }{ \cfrac { -5 }{ 21 } } \Rightarrow \theta ={ 103.77 }^{ 0 }$

Show that the following poionts are collinear
(i) $A(3,2,-4), \ B(9,8,-10), \ C(-2,-3,1)$
(ii) $P(4,5,2),\ Q(3,2,4), \ R(5,8,0)$

$(i)A(3,2,-4),B(9,8,-10),C(-2,-3,1)$

$\therefore$ For

$A,B,C$ to be collinear,

$\lambda \overrightarrow { a } +\mu \overrightarrow { b } +\delta \overrightarrow { c } =0$ where

$\lambda ,\mu ,\delta \neq 0$

$\therefore (3\hat { i } +2\hat { j } -4\hat { k } )\lambda +(9\hat { i } +8\hat { j } -10\hat { k } )\mu +(-2\hat { i } -3\hat { j } +\hat { k } )\delta =0\\ (3\lambda +9\mu -2\delta )\hat { i } +(2\lambda +8\mu -3\delta )\hat { j } +(-4\lambda -10\mu +1\delta )\hat { k } =0\\ \therefore 3\lambda +9\mu -2\delta =0\\ 2\lambda +8\mu -3\delta =0\\ -4\lambda -10\mu +1\delta =0$

Solving by cramer's Rule,

$\left| \begin{matrix} 3 & 9 & -2 \\ 2 & 8 & -3 \\ -4 & -10 & 1 \end{matrix} \right| =0$

$\therefore$ unique solutions for

$\mu ,\lambda ,\delta$ exist and scalars are non zero.

$\therefore$

$A,B,C$ are collinear.

$(ii)P(4,5,2),Q(3,2,4),R(5,8,0)$

Similarly,

$\left| \begin{matrix} 4 & 5 & 2 \\ 3 & 2 & 4 \\ 5 & 8 & 0 \end{matrix} \right| =0$

unique solutions exists and all the scalars are non zero.

$\therefore$

$P,Q,R$ are collinear.

Let 'k' is distance of the plane through (1, 1, 1) and perpendicular to the line $\dfrac{x - 1}{3} = \dfrac{y - 1}{0} = \dfrac{z - 1}{4}$ from the origin, then value of 5 k is __

equation of plane perpendicular to line

$\displaystyle \frac{x-1}{3}=\frac{y-1}{0}=\frac{z-1}{4}$

$3x+0(y)+4z=P$

Given that plane passes through

$(1,1,1)$

$3(1)+4(1)=P$

$\Rightarrow P=7$

$\therefore$ equation of the plane

$\Rightarrow 3x+4z=7$

perpendicular distance from origin to plane

$= k$

$\Rightarrow K=\displaystyle \frac{|3(0)+4(0)-7|}{\sqrt{3^{2}+4^{2}}}=\frac{7}{\sqrt{25}}=\frac{7}{5}$

$\Rightarrow K=\dfrac{7}{5}$

$\therefore 5K=7$

$\therefore$ value of

$5K$ is

$7$

Find direction cosines of line given by $x=ay+b,\ z=cy+d.$

Line

$\rightarrow x=ay+b$

$z=cy+d$

$\dfrac{x-b}{a}=y$

$\dfrac{z-d}{c}=y$

Line

$\rightarrow \dfrac{x-b}{a}=\dfrac{y}{1}=\dfrac{z-d}{c}$

Direction cosines

$(a, 1, c)$ .

Prove that the points whose position vectors are $5\vec{a}+6\vec{b}+7\vec{c}, 7\vec{a}-8\vec{b}+9\vec{c}, 3\vec{a}+20\vec{b}+5\vec{c}$ are collinear.

Let the points be P,Q,R.

$\vec{PQ}=(7\vec{a}-8\vec{b}+9\vec{c})-(5\vec{a}+6\vec{b}+7\vec{c})=2\vec{a}-14\vec{b}+2\vec{c}$

$\vec{PR}=(3\vec{a}+2\vec{b}+5\vec{c})-(5\vec{a}+6\vec{b}+7\vec{c})=-2\vec{a}+14\vec{b}-2\vec{c}$

So,

$\vec{PQ}= -\vec{PR}$

$\therefore$ Points P,Q,R are collinear.

1104132_1111172_ans_29c584ca869147409deffaa064abf00d.jpg

If $\vec P\left( {1,5,4} \right)$ and $\vec Q\left( {4, - 1, - 2} \right)$ , find the direction ratio of ${\vec {{PQ}}}$

$\overrightarrow { p } =\hat { i } +5\hat { j } +4\hat { k } ,\overrightarrow { q } =4\hat { i } -\hat { j } -2\hat { k } \\ \therefore \overrightarrow { PQ } =\overrightarrow { q } -\overrightarrow { p } =3\hat { i } -6\hat { j } -6\hat { k }$

$\therefore$ Direction of

$\overrightarrow { PQ } =(-1,2,2)$

Find the direction cosines of a line whose direction ratios are $2, -6, 3$ .

If the direction ratios are given as

$x,y,z$ , then the direction cosines are given by

$\dfrac{x}{\sqrt{x^{2}+y + z^2}},\dfrac{y}{\sqrt{x^{2}+y^{2}+z^{2}}},\dfrac{z}{\sqrt{x^{2}+y^{2}+z^{2}}}$

So, the direction cosines for direction ratios

$2,-6,3$ are

$\dfrac{2}{\sqrt{4+36+9}},\dfrac{-6}{\sqrt{4+36+9}},\dfrac{3}{\sqrt{4+36+9}}$

$=\pm \dfrac{2}{7},\mp \dfrac{6}{7},\pm \dfrac{3}{7}$

If a line makes angles $90^{o}$ and $60^{o}$ respectively with the positive directions of $x$ and $y$ axes, find the angle which it makes with the positive direction of $z-$ axis.

We know that

$l^{2}+m^{2}+n^{2}=1$ where

$l,m,n$ are direction cosines

$\cos^{2}90^{\circ}+\cos^{2}60^{\circ}+\cos^{2}\theta =1$

$\dfrac{1}{4}+\cos^{2}\theta =1$

$\cos\theta =\pm \dfrac{\sqrt{3}}{2}$

$\theta =30^{\circ}$

It makes

$30^{\circ}$ with positive z-axis.

Find the equation of the plane which is perpendicular to $x - axis$ and that passes through the point $\left( {2, - 1,3} \right)$ .

Equation of plane passing through point

$(2,-1,3) a(x-2)+b(y+1)+c(z-3)=0$

$\because$ Plane is perpendicular to

$T-$ axis

$\therefore b=0, c=0$
Hence required equation of plane

$a(x-2)+0(y+1)+0(z-3)=0$

$\Rightarrow (x-2)=0$

$\Rightarrow x-2=0$

$(\because a \ne 0)$

If ${\cos ^4}\theta + {\cos ^2}\theta = 1 ,$ then show that : ${\sec ^4}\theta - {\sec ^2}\theta = 1$ .

We have,

${\cos ^4}\theta + {\cos ^2}\theta = 1$

$\dfrac{1}{\sec ^4\theta }+\dfrac{1} {\sec ^2\theta } = 1$

$\dfrac{1+\sec^2\theta}{\sec ^4\theta } = 1$

$1+\sec^2\theta=\sec ^4\theta$

$\sec^4\theta-\sec ^2\theta =1$

Hence, proved.

Obtain the equation of the line passing through $(1, 1, 2)$ and $(2, 1, 2)$ in the vector form.

The given line passes through

$(1,1,2)$ and

$(2,1,2)$

Therefore,

It's direction ratios are proportional to

$(2-1),\,(1-1,)\,(2-2)$ i.e,

$1,0,0$

Now,

Let,

$\vec a=\hat i+\hat j+2\hat k$ and

$\vec b=\hat i$

Therefore, vector equation of the line is

$\vec r=\vec a+\lambda +\vec b$

$=(\hat i+\hat j+2\hat k)+\lambda \hat i$

And, Cartesian equation is

$\frac{x-1}{1}=\frac{y-1}{0}=\frac{z-2}{0}$

Lines $\dfrac{x-1}{2}=\dfrac{y-1}{2}=\dfrac{z-2}{3}$ and $\dfrac{x-1}{2}=\dfrac{y-2}{2}=\dfrac{z-3}{-2}$
(Check whether the lines are parallel, mutually perpendicular or intersecting in acute angle)

Consider the

$a_1=2,\,\,b_1=2,\,\,c_1=3 \\ . \\ a_2=2,\,\,b_2=2,\,\,c_2=-2$

Let angle between the lines be

$\theta$

$\cos \theta =|\dfrac{a_1a_2+b_1b_2+c_1c_2}{\sqrt {a_1^2+b_1^2+c_1^2}\sqrt {a_2^2+b_2^2+c_2^2}}|$

$\cos \theta =|\dfrac{4+4-6}{\sqrt {4+4+9}\sqrt {4+4+4}}|$

$\cos \theta =\dfrac{1}{\sqrt {51}}$

$\theta =\cos ^{-1}(\dfrac{1}{\sqrt {51}})$

So, thses lines are intersecting in acute angle

$\cos (\dfrac{1}{\sqrt {51}})$ .

Direction ratio of $\dfrac{3-x}{1}=\dfrac{y-2}{5}=\dfrac{2z-3}{1} are.......$

$According\, to\, question, \\ \frac { { 3-x } }{ 1 } =\frac { { y-2 } }{ 5 } =\frac { { 2z-3 } }{ 1 } \\ \frac { { x-3 } }{ { -1 } } =\frac { { y-2 } }{ 5 } =\frac { { z-\frac { 3 }{ 2 } } }{ { \frac { 1 }{ 2 } } } \\ Dr's\, \, of\, \, a\, line\, is\, (-1,5,\frac { 1 }{ 2 } )$

If a line makes angles $\alpha,\ \beta,\ \gamma$ with coordinate axes, prove that
$\cos{2\alpha}+\cos{2\beta}+\cos{2\gamma}+1=0$

We Know,

$l=\cos \alpha,\,\, m=\cos \beta,\,\,n=\cos \gamma$

Also,

$l^2+m^2+n^2=1$ -----

$(1)$

$\therefore cos^2 \alpha + cos^2\beta+ cos^2\gamma=1$

Considering LHS:

$\cos 2\alpha + \cos 2\beta + \cos 2\gamma + 1 = 0$

$\begin{array} { *{ 20 }{ l } }{ =2{ { \cos }^{ 2 } }\alpha -1+2{ { \cos }^{ 2 } }\beta -1+2{ { \cos }^{ 2 } }\gamma -1+1=0 } \\ { } \\ { =2\left( { { { \cos }^{ 2 } }\alpha +{ { \cos }^{ 2 } }\beta +{ { \cos }^{ 2 } }\gamma } \right) -2 } \\ { } \\ { =2(1)-2\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad From\quad (1) } \\ { } \\ { =0 } \end{array}$

Hence Proved.

The direction $cosines\ l,m,m$ of two lines satisfy $l+m+n=0,\ { l }^{ 2 }+n^{ 2 }{ =m }^{ 2 }$ . Find the measure of the angle between the lines.

1346665_1125206_ans_041142fbb59d4c90b1a7726b15121336.jpg

Write the equation of the line passing through A $(1, 2, 3)$ and having the direction in $(1, 1, 1)$ the vector form and also in the symmetric form.

Consider,

$\begin{array}{l} \vec { a } =\hat { i } +2\hat { j } +3\hat { k } \\ \\ \vec { b } =\hat { i } +\hat { j } +\hat { k } \end{array}$

Since, the given line passes through

$(1,2,3)$ and its direction ratios are proportional to

$1,1,1$

So, its vector equation is

$\vec r=\vec a+\lambda \vec b$

$=(\hat i+2\hat j+3\hat k)+\lambda (\hat i+\hat j+\hat k)$

And its cartesian equation is

$\frac{x-1}{1}=\frac{y-2}{1}=\frac{z-3}{1}$

$x-1=y-2=z-3$

Write the equation of a line through $(1, 2, 3)$ and parallel to $\overline {r} . (\hat {i} - \hat {j} + 3\hat {k}) = 5$ .

Given, Point = (1,2,3)

$\begin{array}{l} =\, \, \, (1,2,3) \\ Vector:\, \, \, \overline { \, r. } \, (i-j+3k)=5\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left[ { \overline { r } =xi+yj+zk } \right]. \\ \, \, \, \, \, \, \, x-y+3z=5 \\ Parallel\, \, plane\, \, is\, ,\, x-y+3z+k=0 \\ \, \, \, \, \, Passes\, \, (1,2,3)\Rightarrow 1-2+9+k=0 \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \Rightarrow k=-8 \\ \, \, and, \\ x-y+3z-8=0 \\ so,\, that\, \, \overline { r } .(i-j+3k)=8.\, \end{array}$

Find the direction cosines of the line: $\displaystyle \frac { x-1 }{ 2 } =-y=\frac { z+1 }{ 2 }$

Given,

$\dfrac{x-1}{2} = -y = \dfrac{z+1}{2}$

$\dfrac{x-1}{2} = \dfrac{-y}{1} = \dfrac{z+1}{2}$

$\dfrac{x-1}{2} = \dfrac{y}{-1} = \dfrac{z+1}{2}$

Here, direction ratios are 2, -1 , 2

$\Rightarrow a =2, b = -1, c = 2$

So, the direction cosines (Dc's)are given as,

$\dfrac{2}{\sqrt{2^2+(-1)^2+2^2}} , \dfrac{-1}{\sqrt{2^2+(-1)^2+2^2}}, \dfrac{2}{\sqrt{2^2+(-1)^2+2^2}}$

$\left [\because \dfrac{a}{\sqrt{a^2+b^2+c^2}}, \dfrac{b}{a^2+b^2+c^2}, \dfrac{c}{a^2+b^2+c^2}\right]$

$\dfrac{2}{\sqrt{4+1+4}}, \dfrac{-1}{\sqrt{4+1+4}},\dfrac{2}{\sqrt{4+1+4}}$

$\Rightarrow \dfrac{2}{\sqrt{9}}, \dfrac{-1}{\sqrt{9}}, \dfrac{2}{\sqrt{9}}$

$Dc's \Rightarrow \dfrac{2}{3}, \dfrac{-1}{3}, \dfrac{2}{3}$ Answer

If $O$ is origin $OP = 3$ with direction ratios proportional to $-1, 2, -2$ then what are the coordinates of $P$ ?

Let the coordinates of point

$P$ be

$(x_1,y_1,z_1)$

Given,

$OP=3$

$\sqrt {(x_1-0)^2+(y_1-0)^2+(z_1-0)^2}=9$

$x_1^2+y_1^2+z_1^2=9$ -----

$(1)$

Direction ratio of

$OP$ are

$-1,2,-2$

Therefore,

$x_1-0=-1$

$x_1=-1$

$y_1-0=2$

$y_1=2$

$z_1-0=-2$

$z_1=-2$

Putting the values of

$x_1,y_1,z_1$ in

$(1)$ we get,

$(-1)^2+(2)^2+(-2)^2=0$

$9=9$

$x_1,y_1,z_1$ satisfies

$(1)$

Hence, coordinates of point

$P$ is

$(-1,2,-2)$

Show that the points $A(1,-2,-8),B(5,0,-2)$ and $C(11,3,7)$ are collinear, and find the ratio in which $B$ divides $AC$ .

$A= (1,-2,-8)$

$B= (5,0,-2)$

$C=(11,3,7)$

A,B,C are collinear if

$\left | \vec{AB} \right |+\left | \vec{BC} \right |=\left | \vec{AC} \right |$

$\vec{AB}= (5-1)\hat{i}+(0+2)\hat{j}+(-2+8)\hat{k}= 4\hat{i}+2\hat{j}+6\hat{k}$

$\left | \vec{AB} \right |= \sqrt{16+4+36}= 2\sqrt{14}$

$\vec{BC}= (11-5)\hat{i}+(3-0)\hat{j}+(7+2)\hat{k}= 6\hat{i}+3\hat{j}+9\hat{k}$

$\left | \vec{BC} \right |=\sqrt{36+9+81}= 3\sqrt{14}$

$\vec{AC}=(11-1)\hat{i}+(3+2)\hat{j}+(7+8)\hat{k}= 10\hat{i}+5\hat{j}+15\hat{k}$

$\left | \vec{AC} \right |= \sqrt{100+25+225}= 5\sqrt{14}$

Thus

$\left | \vec{AB} \right |+\left | \vec{BC} \right |=\left | \vec{AC} \right |$

let B divide AC in k:1

then

$(5,0,-2)= \dfrac{k(11,3,7)+1(1,-2,-8)}{k+1}$

$=\dfrac{(11k+1, 3k-2, 7k-8)}{k+1}$

$\Rightarrow k= 2/3\Rightarrow k:1\Rightarrow 2:3$

$\Rightarrow$ B divide AC in 2:3.

1147547_1142338_ans_005cfdfbd9ae4824bcc6b1662684da1f.jpg

To find the vector and the Cartesian equation in symmetric form of line passing through the points, $(2, 0, -3)$ and $(7, 3, -10)$ .

The required is passing through the points

$(2,0,-3)$ and

$(7,3,-10)$

So, it's direction ratios are proportional to

$(7-2),\,(3-0),\,(-10+3)$

i.e.

$5,3,-7$

therefore,

Vector equation of the line is

$\vec r=\vec a+\lambda \vec b$

$=(2\hat i+0\hat j+3\hat k)+\lambda (5\hat i+3\hat j-7\hat k)$

$=(2\hat i+3\hat k)+\lambda (5\hat i+3\hat j-7\hat k)$

And, Symmetric equation is

$\begin{array}{l} \dfrac { { x-2 } }{ 5 } =\dfrac { { y-0 } }{ 3 } =\dfrac { { z-\left( { -3 } \right) } }{ { -7 } } \\ \\ \dfrac { { x-2 } }{ 5 } =\dfrac { y }{ 3 } =\dfrac { { z+3 } }{ { -7 } } \end{array}$

Show that the following set of point are collinear ?
$(2,3,-4),(-1,0,5),(3,4,-7)$

Let the given points are

$A(2,3, - 4),\,B( - 1,0,5),\,\,C(3,4, - 7)$

Then,

$\begin{array}{l} \overrightarrow { AB } =\overrightarrow { OB } -\overrightarrow { OA } =-3\hat { i } -3\hat { j } +9\hat { k } \\ \\ \overrightarrow { BC } =\overrightarrow { OC } -\overrightarrow { OB } =4\hat { i } +4\hat { j } -12\hat { k } \\ \\ \overrightarrow { AC } =\overrightarrow { OC } -\overrightarrow { OA } =\hat { i } +\hat { j } -3\hat { k } \\ \\ \left| { \overrightarrow { AB } } \right| =\sqrt { { { \left( { -3 } \right) }^{ 2 } }+{ { \left( { -3 } \right) }^{ 2 } }+{ { \left( 9 \right) }^{ 2 } } } =3\sqrt { 11 } \\ \\ \left| { \overrightarrow { BC } } \right| =\sqrt { { { \left( 4 \right) }^{ 2 } }+{ { \left( 4 \right) }^{ 2 } }+{ { \left( { -12 } \right) }^{ 2 } } } =4\sqrt { 11 } \\ \\ \left| { \overrightarrow { AC } } \right| =\sqrt { { { \left( 1 \right) }^{ 2 } }+{ { \left( 1 \right) }^{ 2 } }+{ { \left( { -3 } \right) }^{ 2 } } } =\sqrt { 11 } \end{array}$

Since,

$|\vec {BC}|=|\vec {AB}|+|\vec {AC}|$

Therefore, given points are collinear.

If the vectors $x\hat{i}+\hat{j}+\hat{k},\hat{i}+y\hat{j}+\hat{k}$ and $\hat{i}+\hat{j}+z\hat{k}$ are co-planar where $x\neq 1,y\neq 1,z\neq 1$ , then prove that $\dfrac{1}{1-x}+\dfrac{1}{1-y}+\dfrac{1}{1-z}=1$ .

Since,

Three given vectors are coplanar,

Then,

$\begin{array}{l} \left| { \begin{array} { *{ 20 }{ c } }x & 1 & 1 \\ 1 & y & 1 \\ 1 & 1 & z \end{array} } \right| =0 \\ \\ x\left( { yz-1 } \right) -1\left( { z-1 } \right) +1\left( { 1-y } \right) =0 \\ \\ xyz-x-y-z+2=0\, \, \, \, \, ----\, \, \, \left( 1 \right) \end{array}$

Now,

$\begin{array}{l} \dfrac { 1 }{ { 1-x } } +\dfrac { 1 }{ { 1-y } } +\dfrac { 1 }{ { 1-z } } \\ \\ =\dfrac { { \left( { 1-y } \right) \left( { 1-z } \right) +\left( { 1-x } \right) \left( { 1-z } \right) +\left( { 1-x } \right) \left( { 1-y } \right) } }{ { \left( { 1-x } \right) \left( { 1-y } \right) \left( { 1-z } \right) } } \\ \\ =\dfrac { { 1-z-y+yz+1-z-x+zx+1-y-x+xy } }{ { \left( { 1-x } \right) \left( { 1-z-y+yz } \right) } } \\ \\ =\dfrac { { xy+yz+zx-2\left( { x+y+z } \right) +3 } }{ { 1-z-y+yz-x+zx+xy-xyz } } \\ \\ =\dfrac { { 3-2\left( { x+y+z } \right) +xy+yz+zx } }{ { 1-\left( { xyz } \right) -x-y-z+xy+yz+zx } } \\ \\ =\dfrac { { 3-2\left( { x+y+z } \right) +xy+yz+zx } }{ { 1-\left( { x+y+z } \right) -x-y-z+xy+yz+zx } } \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, from\, \left( 1 \right) \\ \\ =\dfrac { { 3-2\left( { x+y+z } \right) +xy+yz+zx } }{ { 3-\left( { x+y+z } \right) -x-y-z+xy+yz+zx } } \\ \\ =\dfrac { { 3-2\left( { x+y+z } \right) +xy+yz+zx } }{ { 3-2\left( { x+y+z } \right) +xy+yz+zx } } \\ \\ =1 \end{array}$

Hence, proved.

Find the equation of the line passing through the points $A(3, 2, -1)$ and $B (4, -1, 3)$ .

Dr’s of line

$=\left(4-3,-1-2,3+1\right)$

$=\left(1,-3,4\right)$

Equation of line

$=\dfrac{x-4}{1}=\dfrac{y+1}{-3}=\dfrac{z-3}{4}$

If the vectors $a\hat{i}+\hat{j}+\hat{k},\hat{i}+b\hat{j}+\hat{k},\hat{i}+\hat{j}+c\hat{k} \left(a\neq 1 ,b\neq 1,c\neq 1\right)$ are coplanar, then the value of $\dfrac{1}{1-a}+\dfrac{1}{1-b}+\dfrac{1}{1-c}$

Given:

$\left|\begin{matrix} a&1&1 \\ 1&b&1 \\ 1&1&c \end{matrix} \right|$ are coplanar

$\Rightarrow \left|\begin{matrix} a&1&1 \\ 1&b&1 \\ 1&1&c \end{matrix} \right|=0$

${R}_{2}\rightarrow {R}_{2}-{R}_{1},{R}_{3}\rightarrow {R}_{3}-{R}_{1}$

$\Rightarrow \left|\begin{matrix} a&1&1 \\ 1-a&b-1&0 \\ 1-a&0&c-1 \end{matrix} \right|=0$
Divide

${C}_{1}$ ,by

$\left(1-a\right)$ ,

${C}_{2}$ by

$\left(1-b\right)$ ,and

${C}_{3}$ ,by

$\left(1-c\right)$

$\Rightarrow \left|\begin{matrix} \dfrac{a}{1-a}&\dfrac{-1}{1-b}&\dfrac{-1}{1-c} \\ 1&1&0 \\ 1&0&1 \end{matrix} \right|=0$

$\Rightarrow \dfrac{a}{1-a}\left(1-0\right)-\dfrac{-1}{1-b}\left(1-0\right)+\left(\dfrac{-1}{1-c}\right)\left(0-1\right)=0$

$\Rightarrow \dfrac{a}{1-a}+\dfrac{1}{1-b}+\dfrac{1}{1-c}=0$
Add

$1$ to both sides, we get

$\Rightarrow \dfrac{a}{1-a}+1+\dfrac{1}{1-b}+\dfrac{1}{1-c}=1$

$\Rightarrow \dfrac{a+1-a}{1-a}+\dfrac{1}{1-b}+\dfrac{1}{1-c}=1$

$\Rightarrow \dfrac{1}{1-a}+\dfrac{1}{1-b}+\dfrac{1}{1-c}=1$

Show that the three lines with direction cosines
$\frac{12}{13},\frac{-3}{13},\frac{-4}{13}:\frac{4}{13},\frac{12}{13},\frac{3}{13};\frac{-4}{13},\frac{12}{13}$ are mutually perpendicular

We have,

${{l}_{1}}=\dfrac{12}{13},\,{{m}_{1}}=-\dfrac{3}{13},\,{{n}_{1}}=-\dfrac{4}{13}$

${{l}_{2}}=\dfrac{4}{13},\,{{m}_{2}}=\dfrac{12}{13},\,{{n}_{2}}=\dfrac{3}{13}$

Then, we know that,

Two lines with direction cosines ${{l}_{1}},\,{{m}_{1}},\,{{n}_{1}}\,\,\text{and}\,\,{{l}_{2}},\,{{m}_{2}},\,{{n}_{2}}$ are perpendicular to each other.

Then,

${{l}_{1}}{{l}_{2}}+{{m}_{1}}{{m}_{2}}+{{n}_{1}}{{n}_{2}}=0$

$\left( \dfrac{12}{13}\times \dfrac{4}{13} \right)+\left( \dfrac{-3}{13}\times \dfrac{12}{13} \right)+\left( \dfrac{-4}{13}\times \dfrac{3}{13} \right)$

$=\dfrac{48}{169}+\left( \dfrac{-36}{169} \right)+\left( \dfrac{-12}{169} \right)$

$=\dfrac{48-36-12}{169}$

$=\dfrac{48-48}{169}$

$=\dfrac{0}{169}$

$=0$

Hence they are perpendicular lines.

Now,

${{l}_{3}}=\dfrac{3}{13},\,{{m}_{3}}=\dfrac{-4}{13},\,{{n}_{3}}=\dfrac{12}{13}$

Then,

Two lines with direction cosines ${{l}_{1}},\,{{m}_{1}},\,{{n}_{1}}\,\,\text{and}\,\,{{l}_{2}},\,{{m}_{2}},\,{{n}_{2}}$ are perpendicular to each other.

${{l}_{2}}{{l}_{3}}+{{m}_{2}}{{m}_{3}}+{{n}_{2}}{{n}_{3}}=0$

$\left( \dfrac{4}{13}\times \dfrac{3}{13} \right)+\left( \dfrac{12}{13}\times \dfrac{-4}{13} \right)+\left( \dfrac{3}{13}\times \dfrac{12}{13} \right)$

$=\dfrac{12}{169}+\left( \dfrac{-48}{169} \right)+\left( \dfrac{36}{169} \right)$

$=\dfrac{12-48+36}{169}$

$=\dfrac{48-48}{169}$

$=\dfrac{0}{169}$

$=0$

Hence, Hence they are perpendicular lines.

Hence, all 3 lines are perpendicular to each other.

$P$ and $Q$ are the points $(-1,2,1)$ and $(4,3,5)$ respectively. Find the projection of $PQ$ on a line which makes angles of $120^{o}$ and $135^{o}$ with $y$ and $z$ axes respectively and an acute with $x-axis$ ?

$\alpha ,\beta ,\gamma$ be the angles made by the coordinate axes.

$\therefore \cos { ^{ 2 }\alpha } +\cos { ^{ 2 }\beta } +\cos { ^{ 2 }\gamma } =1$

$\Rightarrow \cos { ^{ 2 }\alpha } +\cos { ^{ 2 }120 } +\cos { ^{ 2 }135 } =1$ (given

$\beta={ 120 }^{ \circ },\gamma ={ 135 }^{ \circ }$ )

$\Rightarrow \alpha ={ 60 }^{ \circ }$

$\therefore$ Directions of the projection

$(\cos { \alpha } ,\cos { \beta } ,\cos { \gamma } )\equiv (\cfrac { 1 }{ 2 } ,\cfrac { -1 }{ 2 } ,\cfrac { -1 }{ \sqrt { 2 } } )$

$\therefore \overrightarrow { PQ } =5\hat { i } +\hat { j } -4\hat { k }$

projection of

$\overrightarrow { PQ }$ on line with directions

$(\cfrac { 1 }{ 2 } ,\cfrac { -1 }{ 2 } ,\cfrac { -1 }{ \sqrt { 2 } } )$ will be .

$\Rightarrow \overrightarrow { PQ } .(\cfrac { 1 }{ 2 } \hat { i } -\cfrac { 1 }{ 2 } \hat { j } -\cfrac { 1 }{ \sqrt { 2 } } \hat { k } )\\ =(5\hat { i } +\hat { j } -4\hat { k } ).(\cfrac { 1 }{ 2 } \hat { i } -\cfrac { 1 }{ 2 } \hat { j } -\cfrac { 1 }{ \sqrt { 2 } } \hat { k } )\\ =2+2\sqrt { 2 }$

In vector form,

$(2+2\sqrt { 2 } )(\cfrac { 1 }{ 2 } \hat { i } -\cfrac { 1 }{ 2 } \hat { j } -\cfrac { 1 }{ \sqrt { 2 } } \hat { k } )$

Find the direction cosines of the line passing through the two points $(-2,4,-5)$ and $(1,2,3)$

$P(-2,4,-5)$

$Q(1,2,3)$

So,

$x_1=-2 \,\,\, y_1=4\,\,\, z_1=-5$

$x_2=1\,\,\,y_2=2\,\,\,z_2=3$

Direction ratios

$= (x_2-x_1),(y_2-y_1),(z_2-z_1)$

$=1-(-2),2-4,3-(-5)$

$=1+2,-2,3+5$

$=3,-2,8$

Direction cosines

$\begin{array}{l} \frac { 3 }{ { \sqrt { { 3^{ 2 } }+{ { \left( { -2 } \right) }^{ 2 } }+{ 8^{ 2 } } } } } ,\frac { { -2 } }{ { \sqrt { { 3^{ 2 } }+{ { \left( { -2 } \right) }^{ 2 } }+{ 8^{ 2 } } } } } ,\frac { 8 }{ { \sqrt { { 3^{ 2 } }+{ { \left( { -2 } \right) }^{ 2 } }+{ 8^{ 2 } } } } } \\ =\frac { 3 }{ { \sqrt { 9+4+64 } } } ,\frac { { -2 } }{ { \sqrt { 9+4+64 } } } ,\frac { 8 }{ { \sqrt { 9+4+64 } } } \\ =\frac { 3 }{ { \sqrt { 77 } } } ,\frac { { -2 } }{ { \sqrt { 77 } } } ,\frac { 8 }{ { \sqrt { 77 } } } \end{array}$

A vector $\vec {r}$ is inclined at equal angles to the three axes. If the magnitude of $\vec {r}$ is $2\sqrt {3}$ units, find $\vec {r}$ .

Since

$\vec { r }$ is inclined equal angles to axes

$\therefore$

$\alpha=\beta=y$

$3\cos ^{ 2 }{ \alpha } =1$

$\therefore$

$3\cos ^{ }{ \alpha } ==\pm \cfrac { 1 }{ \sqrt { 3 } }$

Now componenets in each direction are equal

Let

$\vec { r } =a\hat{i}+a\hat{j}+a\hat{k}$

$\therefore \left| \vec { r } \right| =\sqrt { { a }^{ 2 }+{ a }^{ 2 }+{ a }^{ 2 } } =\sqrt { 3 } a$

$\sqrt { 3 } a=2\sqrt { 3 } \Rightarrow a=2$

$\vec { r } =2\hat { i } +2\hat { j } +2\hat { k }$

A triangle is formed by joining the points $(1,0,0),(0,1,0)$ and $(0,0,1)$ . Find the direction cosines of the medians.

$A \equiv (1,0,0)$

$B \equiv (0,1,0)$

$C \equiv (0,0,1)$

Middle point of side

$\displaystyle AB (d) \equiv (\frac{1}{2},\frac{1}{2},0)$

Middle point of

$\displaystyle BC (e) \equiv (0,\frac{1}{2},\frac{1}{2})$

Middle point of

$\displaystyle AC (f) \equiv (\frac{1}{2},0,\frac{1}{2})$

$\therefore$ direction ratio of cd

$\displaystyle = (\frac{1}{2},\frac{1}{2},-1)$

$\therefore$ direction ratio of bf

$\displaystyle = (\frac{1}{2},-1,\frac{1}{2})$

$\therefore$ direction ratio of ae

$\displaystyle = (-1,\frac{1}{2},\frac{11}{2})$

$\therefore$ direction cosine of cd =

$\displaystyle (\frac{1/2}{\sqrt({\frac{1}{2})^{2}+(\frac{1}{2})^{2}}+1^{2}},\frac{1/2}{\sqrt{(\frac{1}{2})^{2}+(\frac{1}{2})^{2}}+1^{2}},\frac{-1}{\sqrt{(\frac{1}{2})^{2}+(\frac{1}{2})^{2}}+1^{2}})$

$\displaystyle = (\frac{1/2}{\sqrt{3/2}},\frac{+1/2}{\sqrt{3/2}},\frac{-1}{\sqrt{3/2}})$

$\displaystyle \therefore$ direction cosines of bf

$:-$

$\displaystyle (\frac{1/2}{\sqrt{3/2}},\frac{-1}{\sqrt{3/2}},\frac{1/2}{\sqrt{3/2}})$

$\therefore$ direction cosines of ae

$\displaystyle = (\frac{-1}{\sqrt{3/2}},\frac{1}{2\sqrt{3/2}},\frac{1}{2\sqrt{3/2}})$

1118088_1163997_ans_e08094b82fbf423c9ba1319d6674a61a.jpeg

Find the direction cosines of two lines which are connected by the relations $l+m+n=0$ and $mn-2nl-2lm=0.$

$l+m+n$

$l=-m-n$

$mn-2n(-m-n)-2m(-m-n)=0$

$mn+2mn+2n^2+2m^2+2mn=0$

$2m^2+5mn+2n^2=0$

$2m^2+4mn+mn+2n^2=0$

$2m(m+2n)+n(m+2n)=0$

$(2m+n)(m+2n)=0$

$n=-2m$ ,

$m=-2n$

$n=-2m$

$l=-m-n$

$l=-m+2m$

$l=m$

$\dfrac{l}{1}=\dfrac{m}{1}=\dfrac{n}{-2}$

$m=-2n$

$l=2n-n$

$l=n$

$\dfrac{l}{1}=\dfrac{m}{-2}=\dfrac{n}{1}$

$<1,1,-2>$ and

$<1,-2,1>$

Show that the points $(2,3,4,)(-1,-,2,1),(5,8,7)$ are collinear.

Three points A,B,C are collinear if direction ratios of AB and BC are proportional.

$2,3,4$ ) and B(

$-1,-2,1$ )

Direction ratios =

$-1-2,-2-3,1-4$

$=-3,-5,-3$

So,

$a_1=-3, b_1=-5, c_1=-3$

$-1,-2,1$ ) and C(

$5,8,7$ )

Direction ratios =

$5-(-1), 8-(-2), 7-1$

$=6,10,6$

so,

$a_2=6, b_2=10, c_2=6$

Now,

$\begin{array}{l} \dfrac { { { a_{ 2 } } } }{ { { a_{ 1 } } } } =\dfrac { 6 }{ { -3 } } =-2 \\ \dfrac { { { b_{ 2 } } } }{ { { b_{ 1 } } } } =\dfrac { { 10 } }{ { -5 } } =-2 \\ \dfrac { { { c_{ 2 } } } }{ { { c_{ 1 } } } } =\dfrac { 6 }{ { -3 } } =-2 \\ Since\, \dfrac { { { a_{ 2 } } } }{ { { a_{ 1 } } } } =\dfrac { { { b_{ 2 } } } }{ { { b_{ 1 } } } } =\dfrac { { { c_{ 2 } } } }{ { { c_{ 1 } } } } =-2 \end{array}$

Therefore, A,B,C are collinear.

Find the vector of magnitude (5/2) units which is parallel to the vector $3\hat i + 4\hat j$

Given Vector =

$3i+4j$

Magnitude

$= \sqrt{3^{2}+4^{2}}=5$

Unit vector in direction of resultant

$=\dfrac{3i+4j}{5}$

Vector of magnitude

$\dfrac{5}{2}$ unit in direction of resultant

$= \dfrac{5}{2}\times\left(\dfrac{3i+4j}{5}\right) = \dfrac{3i+4j}{2}$

Find the direction cosines of the sides of the triangle whose vertices are $(3, 5, -4), (-1, 1, 2)$ and $(-5, -5, -2)$ .

$AB$

$A(3,5,-4)$

$B(-1,1,2)$

Direction ratios

$=-1-3,1-5,2-(-4)=-4,-4,6$

$AB=\sqrt{(-1-3)^2+(1-5)^2+(2+4)^2}=\sqrt{68}=2\sqrt{17}$

Direction cosins

$\frac{-4}{2\sqrt{17}},\frac{-4}{2\sqrt{17}},\frac{6}{2\sqrt{17}}$

$=\dfrac{-2}{\sqrt{17}},\dfrac{-2}{\sqrt{17}},\frac{3}{\sqrt{17}}$

$BC$

$B(-1,1,2)$

$C(-5,-5,-2)$

Direction ratios

$=-5-(-1),-5-1,-2-2=-4,-6,-4$

$CA=\sqrt{(-5+1)^2+(-5-1)^2+(-2-2)^2}=\sqrt{68}=2\sqrt{17}$

Direction cosins

$\frac{-4}{2\sqrt{17}},\frac{-6}{2\sqrt{17}},\frac{-4}{2\sqrt{17}}$

$=\dfrac{-2}{\sqrt{17}},\dfrac{-3}{\sqrt{17}},\frac{-2}{\sqrt{17}}$

$CA$

$C(-5,-5,-2)$

$A(3,5,-4)$

Direction ratios

$=3-(-5),5-(-5),-4-(-2)=8,10,-2$

$CA=\sqrt{(3+5)^2+(5+5)^2+(-4+2)^2}=\sqrt{168}=2\sqrt{42}$

Direction cosins

$\frac{8}{2\sqrt{42}},\frac{10}{2\sqrt{42}},\frac{-2}{2\sqrt{42}}$

$=\dfrac{4}{\sqrt{42}},\dfrac{5}{\sqrt{42}},\frac{-1}{\sqrt{42}}$

Find the direction cosine $l,m,n$ of line which are connected by the relations $l + m + n = 0.$ , $2mn + 2ml - nl = 0$

Given:

$l+m+n=0----(1)$

$2mn+2ml+nl=0$

Now we have from

$(1), \ n=-(u+m)$

Putting

$m=-(l+m)$ in equation

$(2)$ , we get

$-2m(l+m)+2ml+(l+m)l=0$

$\Rightarrow \ -2ml-2m^2+2ml+l^2+ml$

$\Rightarrow l^2+ml-2m^2=0$

$\left (\dfrac {l}{m}\right)^2+ \left (\dfrac {l}{m}\right)-2=0$

Now g

$\dfrac {l}{m}=\dfrac {l}{m}=\dfrac {1\pm \sqrt {9}}{2}=1,\ 2$

Now when

$\dfrac {l}{m}=1\ \Rightarrow \ l=m$

From

$(1),\ 2l+n=0\ \Rightarrow \ n=-2l$

$\therefore \ l:m:n=1:1:-2$

$\therefore \$ Direction ratio of the line are

$1,1,-2$

Direction cosiner are

$I=\dfrac {1}{\sqrt {1^2+1^2+(-2)^2}},\ \dfrac {1}{\sqrt {1^2+1^2+(-2)^2}},\ \dfrac {-2}{\sqrt {1^2+1^2+(-2)^2}}$

$\Rightarrow \ \dfrac {+1}{\sqrt 6},\ \dfrac {1}{\sqrt 6},\ \dfrac {-2}{\sqrt 6}$ or

$\dfrac {-1}{\sqrt 6},\ \dfrac {-1}{\sqrt 6},\ \dfrac {2}{\sqrt 6}$

Case

$II$ when

$\dfrac {l}{m}=-2\quad \Rightarrow l=-2m$

From

$(1),\quad -2m+m+n=0\ \Rightarrow \ n=m$

$\therefore \ l:m:n=-2m:m:m$

$=-2:1:1$

$\therefore \$ Directio ratio of the line are

$-2, 1, 1$

Direction codiner are

$\dfrac {-2}{\sqrt {-2^2+1^2+1^2}},\ \dfrac {-1}{\sqrt {(-2)^2+1^2+1^2}},\ \dfrac {-1}{\sqrt {(-2)^2+1^2+1^2}}$

$\Rightarrow \ \dfrac {-2}{\sqrt 6},\ \dfrac {+1}{\sqrt 6},\ \dfrac {1}{\sqrt 6}$ or

$\dfrac {2}{\sqrt 6},\ \dfrac {-1}{\sqrt 6},\ \dfrac {-1}{\sqrt 6}$

Find the equation of the plane which contains the line of intersection of the planes $\overrightarrow r .\left( {\widehat i + 2\widehat j + 3\widehat k} \right) - 4 = 0,\overrightarrow {r.} \left( {2\widehat i + \widehat j - \widehat k} \right) + 5 = 0$ and which is perpendicular to the plane $\overrightarrow {r.} \left( {5\widehat i + 3\widehat j - 6\widehat k} \right) + 8 = 0.$

We have,

$\begin{array}{l} \overrightarrow { r } .\left( { \widehat { i } +2\widehat { j } +3\widehat { k } } \right) =4\, \, \, \, \, \, \, \, \, \, \, \, \, \overrightarrow { r } .\left( { 2\widehat { i } +\widehat { j } -\widehat { k } } \right) +5=0 \\ x+2y+3z=4\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, 2x+y-z=-5 \\ { \pi _{ 3 } }=\left( { x+2y+3z-4 } \right) +\lambda \left( { 2x+y-z+5 } \right) =0 \\ { \pi _{ 3 } }\bot { \pi _{ 4 } }\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, { \pi _{ 4 } }=\overrightarrow { r } .\left( { 5\widehat { i } +3\widehat { j } -6\widehat { k } } \right) +8=0 \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, { \pi _{ 4 } }=5x+3y-6z+8=0 \\ { \pi _{ 3 } }=x\left( { 1+2\lambda } \right) +y\left( { 2+\lambda } \right) +z\left( { 3-\lambda } \right) -4+5\lambda =0 \\ 5\left( { 1+2\lambda } \right) +3\left( { \lambda +2 } \right) -6\left( { 3-\lambda } \right) =0 \\ 11-18+10\lambda +3\lambda +6\lambda =0 \\ 19\lambda =7 \\ \lambda =\frac { 7 }{ { 19 } } \\ x\left( { 1+\frac { { 14 } }{ { 19 } } } \right) +y\left( { 2+\frac { 7 }{ { 19 } } } \right) +z\left( { 3-\frac { 7 }{ { 19 } } } \right) -4+\frac { { 35 } }{ { 19 } } =0 \end{array}$

Hence equation of plane;

$33x+45y+50z-41=0$

Find the vector equation of the line passing through the point $(3,1,2)$ and perpendicular to the plane $\vec r.\left( {2\widehat i - \widehat j + \widehat k} \right) = 8$ Also .find the point of intersection of line and plane.

Given vector equation of plane

$\vec{r}\cdot (2i - \hat{j} + \hat{k}) = 8$

$\to$ normal vector of plane

$= (2\hat{i} - \hat{j} + \hat{k})$

line passes through point

$(3, 1, 2)$

Equation of line:

$\dfrac{x-3}{2} = \dfrac{y-1}{-1} = \dfrac{z-2}{1}$

The points A(4, 5, 10), B(2, 3, 4) and C(1, 2, - 1) are three vertices of a parallelogram ABCD. Find the vector equations of the sides AB and BC and also find the coordinates of point D.

$\vec { AB } =\left( 2-4,3-5,4-10 \right) =\left( -2,-2,-6 \right)$

$\vec { AB } =-2\hat { i } -2\hat { j } -6\hat { k } \quad$

$\vec { BC } =\left( 1-2,2-3,-1-4 \right) =\left( -1,-1,-5 \right)$

$\vec { BC } =-1\hat { i } -1\hat { j } -5\hat { k }$

Equation of

$\vec { AB } =4\hat { i } +5\hat { j } +10\hat { k } +\mu \left( -2\hat { i } -2\hat { j } -6\hat { k } \right)$

Equation of

$\vec { BC } =2\hat { i } +3\hat { j } +4\hat { k } +\mu \left( \hat { i } +\hat { j } +5\hat { k } \right)$

Let the fourth point be

$(a,b,c)$

In a parallelogram diagonals bisect each other

$\therefore$ Midpoint of vector

$AC=$ midpoint of vector

$BD$

$\cfrac { 4+1 }{ 2 } ,\cfrac { 5+2 }{ 2 } ,\cfrac { 10-1 }{ 2 } =\cfrac { 2+a }{ 2 } ,\cfrac { 3+b }{ 2 } ,\cfrac { 4+c }{ 2 }$

$\Rightarrow \cfrac { 5 }{ 2 } =\cfrac { 2+a }{ 2 } \Rightarrow x=3$

$\Rightarrow \cfrac { 7 }{ 2 } =(3+b)\Rightarrow b=4$

$\Rightarrow \cfrac { 9 }{ 2 } =\cfrac { 4+c }{ 2 } \Rightarrow c=5$

$(a,b,c)=(3,4,5)$

Find the equation of line passing through points
(i) $(3\hat i + 2\hat j - 8\hat k)$ , $(5,2, - 4)$

Given two points

$\to (3,2,-8)\, \& \, (5,2,-4)$

Direction ratios

$\to (5-3), (2-2), (-4-(-8))$

$\to (2,0,4)$

So, vector equation of a line we can write as

$\vec{r} = \vec{x}_1 + \lambda \vec{d}$

$\vec{r} = (3\hat{i} + 2\hat{j} -8\hat{k}) + \lambda (2\hat{i} + 4\hat{k})$

If a variable line in two adjacent position has direction cosines $l,\ m,\ n$ and $l+\delta l,\ m+\delta m,\ n+\delta n$ show that the small angle $\delta \theta$ between two positions is given by $(\delta \theta)^{2}$ = $(\delta l)^{2}+(\delta m)^{2}+(\delta n)^{2}$ .

1807701_1179827_ans_33e1e6251d6b4fa286236ad3ff74851a.jpg

If a line makes ${90 ^\circ}$ and ${60 ^\circ}$ with x & y axis, find angle it make with z-axis.

Let say angle with x - axis is

$\alpha = 90^o$

angle with y - axis is

$\beta = 60^o$

angle with z - axis is

$\gamma = \gamma$

Now we know that sum of direction cosines's square is equal to '1'. It means -

$\cos^2\alpha + \cos^2\beta +\cos^2\gamma = 1$

$\cos^2 90^o + \cos^2 60^o + \cos^2\gamma = 1$

$0 + \dfrac{1}{4} + \cos^2\gamma = 1$

$\cos^2 \gamma = 1 - \dfrac{1}{4} = \dfrac{3}{4}$

$\cos \gamma = \pm \dfrac{\sqrt{3}}{2}$

So,

$\gamma = 30^o$

or,

$\gamma = 150^o$

Angle with z -axis is

$30^o$ or

$150^o$

If the points $(p, q) (m, n)$ and $(p-m, q-n)$ are collinear, Determine the relation between $p, q, n$ and $m$ .

$\left| \begin{matrix} p & q & 1 \\ m & n & 1 \\ p-m & q-n & 1 \end{matrix} \right|=0$

$\Rightarrow\,p\left(n-q+n\right)-q\left(m-p+m\right)+1\left(mq-mn-np+nm\right)=0$

$\Rightarrow\,p\left(2n-q\right)-q\left(2m-p\right)+1\left(mq-np\right)=0$

$\Rightarrow\,2np-pq-2qm+pq+mq-np=0$

$\Rightarrow\,np-qm=0$ or

$np=qm$ or

$\dfrac{p}{q}=\dfrac{m}{n}$ is the required relation

Find vector equation of line passing through point $\hat i + 2\hat j + 3\hat k$ and perpendicular to vector $\hat i + \hat j + \hat k$ and $2\hat i + \hat j + 3\hat k$

Equation of line passing through

$\widehat i + 2\widehat j - 3\widehat k$

Then, the equation is

$\frac{{x - 1}}{a} = \frac{{y - 2}}{b} = \frac{{z - 3}}{c}$

Line is

$\parallel$ to vector :

$\left( {\widehat i + \widehat j + \widehat k} \right) \times \left( {2\widehat i + \widehat j + 3\widehat k} \right)$

$=\left| { \begin{array} { *{ 20 }{ c } }{ \widehat { i } } & { \widehat { j } } & { \widehat { k } } \\ 1 & 1 & 1 \\ 2 & 1 & 3 \end{array} } \right|$

$= \widehat i\left( {3 - 1} \right) - \widehat j\left( {3 - 2} \right) + \widehat k\left( {1 - 2} \right)$

$= 2\widehat i - \widehat j - \widehat k$

Hence,

Equation of line :

$\frac{{x - 1}}{2} = \frac{{y - 2}}{{ - 1}} = \frac{{z - 3}}{{ - 1}}$ .

Find the direction cosines of the sides of the triangles whose vertices are $(3,5,-4),(-1,1,2)$ and $(-5,-5,-2)$ .

let

$A(3, 5, -4), B(-1, 1, 2), C(-5, -5, -2)$

Drs of

$AB=\left< -4, -4, 6 \right>$

$\therefore$ Direction cosines of

$AB=\left< \dfrac{-4}{\sqrt{68}}, \dfrac{-4}{\sqrt{68}}, \dfrac{6}{\sqrt{68}}\right>$

Drs of

$BC=<-4, -6, -4 >$

Direction cosines of

$BC=\left< \dfrac{-4}{\sqrt{68}}, \dfrac{-6}{\sqrt{68}}, \dfrac{-4}{\sqrt{68}}\right>$

Drs of

$CA=<8, 10, -2 >$

Direction cosines of

$AC=\left< \dfrac{8}{\sqrt{168}}, \dfrac{10}{\sqrt{168}}, \dfrac{-2}{\sqrt{168}}\right>$

What are the direction cosines of a line that passes through the points $P ( 6 , - 7 , - 1 )$ and $Q ( 2 , - 3,1 )$

Direction ratio of

$\overrightarrow {PQ} = - 4\widehat i + 4\widehat j + 2\widehat k$

Now,

Direction cosine

$= \frac{{ - 4\widehat i + 4\widehat j + 2\widehat k}}{{\sqrt {{{\left( { - 4} \right)}^2} + {{\left( 4 \right)}^2} + {{\left( 2 \right)}^2}} }}$

$= \frac{{ - 4\widehat i + 4\widehat j + 2\widehat k}}{6}$

$= \frac{{ - 2}}{3}\widehat i + \frac{2}{3}\widehat j + \frac{1}{3}\widehat k$

Hence, the direction cosine is

$\left( {\frac{{ - 2}}{3},\frac{2}{3},\frac{1}{3}} \right)$

1206808_1188732_ans_f29b95fb21894d439899fc0120e02e0a.JPG

The position vectors of the points A, B, C are respectively $(1,1,1) ; (1,-1,2) ; (0,2,-1)$ . Find a unit vector parallel to the plane determined by ABC and perpendicular to the vector $(1,0,1)$

Equation of plane is

$\begin{vmatrix} x-1 & y-1 & z-1 \\ 0 & -2 & 1 \\ -1 & 1 & -2 \end{vmatrix}=0$

$\Rightarrow 3(x-1)+y-1-2(z-1)=0$

$\Rightarrow 3x+y-2z-2=0$

let

$\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$

$3a_1+a_2-2a_3=0$

$a_1+0.a_2+a_3=0$

$\dfrac{a_1}{1}=\dfrac{-a_2}{3+2}=\dfrac{a_3}{-1}$

$\Rightarrow \dfrac{a_1}{1}=\dfrac{a_2}{-5}=\dfrac{a_3}{-1}=K$

$a_1=K,\ a_2=-5K,\ a_3=-K$

$a_1^2+a_2^2+a_3^2=1$

$\Rightarrow 27 K^2=1$

$\Rightarrow K=\pm \dfrac{1}{3\sqrt{3}}$

$\therefore \vec{a} = \dfrac{1}{3 \sqrt{3}} \hat{i} \dfrac{-5}{3 \sqrt{3}} \hat{j} \dfrac{-1}{3 \sqrt{3}} \hat{k}$

$=-\dfrac{1}{3\sqrt{3}}\hat{i}+\dfrac{5}{3\sqrt{3}}\hat{j}+ \dfrac{1}{3\sqrt{3}}\hat{k}$

Show that the lines whose d.c's are proportional to $(2,1,1), (4, \sqrt{3}-1,-\sqrt{3} -1)$ are inclined to one another at angle $\dfrac{\pi}{3}$ .

$d.r.1:(2,1,1)\\d.r.2:(4,\sqrt3-3,-\sqrt3-1)$

angle between lines who d.c's proportional to

$d.r 1\& d.r 2$ is equal to lines having d.r's as

$(2,1,1)\&(4,\sqrt3-1,-\sqrt3-1)\\ \therefore \cos\theta=\cfrac{2\times4+1\times(\sqrt3-1)+1\times(-\sqrt3-1)}{\sqrt{2^2+1^2+1^2}\sqrt{4^2+(\sqrt3-1)^2+(-\sqrt3-1)^2}}\\ \cos\theta=\cfrac{1}{2}\Rightarrow\theta=\cfrac{\pi}{3}$

Find the equation of a line passing through the point $(2,0,1)$ and parellel to the whose equation is $\vec { r } =(2\lambda +3)\hat i+(7\lambda -1)\hat j+(-3\lambda +2)\hat k$

$\vec { r } =(2\lambda+3) \hat i+ (7\lambda-1) \hat j+(3\lambda+L) \hat k$

$=3 \hat i=\hat j+2k+ \lambda (2 \hat i+7 \hat j-3 \hat k)$
The drs of the required line are

$(2 \hat i+7 \hat j+3 hat k)$
and line passes through (2,0,1) so the required line is

$\vec { r }=(2 \lambda+2) \hat i+(7\lambda) \hat j+(-3 \lambda+1) \hat k)$

$\vec { r }=(2 \lambda+2) i +(7\lambda ) \hat j+(-3\lambda +1) \hat k$

$\vec { r }=(2 \hat i+ \hat k)+\lambda(2\hat i+7 \hat j-3 \hat k)$

Show that the points $O(0,0), A(2,-3,3), B(-2,3,-3)$ are collinear. Find the ratio in which each point divides the segment joining the other two.

Let's find area of

$\triangle OAB$ which is given by,

$\Delta=\cfrac{1}{2}\left| OA \right| \left| OB \right| \sin { \theta }$

where

$\theta$ is angle between

$\vec { OA }$ and

$\vec { OB }$

Now,

$\cos \theta =\cfrac{\vec { OA } .\vec { OA } }{\left| \vec { OA } \right| \left| \vec { OB } \right| }$

$=\cfrac{-4-9-9}{\sqrt{4+9+9}\sqrt{4+9+9}}=\cfrac{-22}{22}$

$\Rightarrow$

$\cos \theta =-1\Rightarrow$

$\theta={180}^{o}$

$\Rightarrow$

$\Delta =0$ as

$\sin 180=0$

$\Rightarrow$

$O,A,B$ are collinear

Now, let

$A$ divides

$\vec { OB }$ in

$k:1$

$\Rightarrow$

$2=\cfrac{-2k+0}{k+1}\Rightarrow$

$2k+2=-2k\Rightarrow$

$4k=-2\Rightarrow$

$k=-\cfrac{1}{2}$

$\Rightarrow$

$A$ divides

$\vec { OB }$ in

$1:2$ externally

Now for

$B$

$\Rightarrow$

$-2=\cfrac{2k+0}{k+1}\Rightarrow$

$-2k-2=2k\Rightarrow$

$4k=-2$

$\Rightarrow$

$k=-\cfrac{1}{2}$

$\Rightarrow$

$B$ divides

$\vec { OA }$ in

$1:2$ externally for

$O$

$\Rightarrow$

$O=\cfrac{-2k+2}{k+1}\Rightarrow$

$2k=2\Rightarrow$

$k=1$

$\Rightarrow$

$O$ divides

$\vec { AB }$ in

$1:1$ internally

Show that the points $(1,2,3), (7,0,1)$ and $(-2,3,4)$ are collinear.

let

$A(1, 2, 3), B(7, 0, 1), C (-2, 3, 4)$

Drs of

$AB = (6, -2, -2)$

Drs of

$AC = (-3, 1, 1)$

Drs of

$AB = -2 \times Drs$ of AC

Hence, ABC are collinear .

The equation of the plane through the point $(4,4,0)$ and perpendicular to the plane $x+2y +2z-5 = 0$ and $3x+3y +2z -8 =0$ is

Plane passes through

$(4,4,0)$ and is perpendicular to

$x+2y+2z-5=0$ and

$3x+3y+2z-8=0$

$\therefore$ Normal to the plane is

$\begin{vmatrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & 2 & 2 \\ 3 & 3 & 2 \end{vmatrix}=\hat { i } \left( 4-6 \right) -\hat { j } \left( 2-6 \right) +\hat { k } \left( 3-6 \right)$

$=-2\hat { i } +4\hat { j } -3\hat { k }$

$\therefore$ Equation of plane is

$-2(x-4)+4(y-4)-3(z-0)=0$

$\Rightarrow -2x+8+4y-16-3z=0$

$\Rightarrow -2x+4y-3z-8=0$

$\Rightarrow 2x-4y+3z+8=0$

Hence, this is the answer.

If the direction ratios of two parallel lines $4, -3, -1$ and $p+q, 1+q, 2$ then find the values of $p$ and $q$ .

Since lines are parallel, so

$\cfrac{4}{p+q}=\cfrac{-3}{p+q}=\cfrac{-1}{2}\\ \cfrac{-3}{1+q}=\cfrac{-1}{2}\\ 6=1+q\Longrightarrow q=5$

and

$\cfrac{4}{p+q}=\cfrac{-1}{2}$

$-8=p+q\Longrightarrow p=-8-q=-8-5=-13$

If $l_{1}, m_{1}, n_{1}$ and $l_{2}, m_{2}, n_{2}$ are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are $m_{1}n_{2} - m_{2}n_{1}, n_{1}l_{2} - n_{2}l_{1},l_{1}m_{2} - l_{2}m_{1}$ .

The given lines are respectively parallel to

$\overline {a}_{1} = 1, m_{1}\overline {j} + n_{1}\overline {k}$ and

$\overline {a}_{2} = 1_{2} \overline {i} + m_{2}\overline {j} + n_{2}\overline {k}\overline {a}_{1} \times \overline {a}_{2}$ is a vector which is normal to both

$\overline {a}_{1}$ and

$\overline {a}_{2}$ and is of magnitude unity.

$\overline {a}_{1} \times \overline {a}_{2} = \begin{vmatrix} \vec {i} & \vec {j} & \vec {k}\\ l_{1} & m_{1} &n_{1} \\ l_{2} & m_{2} & n_{2}\end{vmatrix} = \vec {i} (m_{1}n_{2} - m_{2}n_{1}) - \vec {j} (l_{1}n_{2} - n_{1}I_{2}) + \vec {k} (l_{1}m_{2} - l_{2}m_{1})$

hence Dr's are

$(m_{1}n_{2} - m_{2}n_{1}, l_{1}n_{2} - l_{2}n_{1}, l_{1}m_{2} - l_{2}m_{1})$ .

Write the direction ratio of the vector $3\vec{a}+2\vec{b}$ where $\vec{a}=\hat{i}+\hat{j}-2\hat{k}$ and $\vec{b}=2\hat{i}-4\hat{j}+5\hat{k}$

1970889_1206183_ans_592df806aba84df18b73cf9a41e9ced8.PNG

If the three point $A=(2,0,1);B=(x,0,3);C=(4,0,4)$ are collineas .Find the value of x and y.

$A\equiv (2,0,1);B(x,0,3)$

$C\equiv (4,0,y)$
let B divides AC in ratio of dil.

$\frac { 4\lambda +2 }{ \lambda +1 } =x$ ............(1)

$\frac { 4\lambda +1 }{ \lambda y +1 } =3$ ................(2)
C divides AB in

$1+\lambda : 1$ externally

$\frac { 2(\lambda +1)-4 }{ \lambda } =x$ ......(3)

$frac {2\lambda -2} {\lambda}-x$

$\frac {(\lambda +1)-3} {\lambda}=y$ .............(4)
if

$A\equiv (2,0,1) ; B\equiv (x,0,3), C\equiv (4,0,y)$
are collinear

$\frac {2} {4}= \frac {1} {y} \Rightarrow y=2$

$\frac {2} {x}= \frac {1} {3}=x=-6$

A line makes angles of measure $45^{\circ}$ and $60^{\circ}$ with the positive direction of the Y and Z axes respectively. Find the angle made by line with the positive directions of X-axis.

Angle made with

$y$ axis

$=45^\circ=\beta$

Angle made with

$z$ axis

$=60^\circ=\gamma$

Angle made with

$x$ axis

$=\alpha$

Since,

$\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$

$\implies \cos^2\alpha+\cos^2 45^\circ+\cos^260^\circ=1$

$\implies cos^2\alpha+1=1$

$\implies cos^2\alpha=0$

$\implies \alpha=90^\circ$

Hence angle made with

$x$ axis

$=90^\circ$ .

Can $\dfrac { 2 }{ \sqrt { 3 } } ,\dfrac { -2 }{ \sqrt { 3 } } ,\dfrac { -1 }{ \sqrt { 3 } }$ be the direction ratios of any directed line? Justify your answer.

$\dfrac{2}{\sqrt{3}}, \dfrac{-2}{\sqrt{3}} , \dfrac{-1}{\sqrt{3}}$

$\displaystyle=\sqrt{\frac{4}{3}+\frac{4}{\sqrt{3}}+\frac{1}{3}} = \sqrt{3}$ No

Find the angle between the lines $\dfrac{-x+2}{-2}=\dfrac{y-1}{7}=\dfrac{z+3}{-3}$ and $\dfrac{x+2}{-1}=\dfrac{2y-8}{4}=\dfrac{z-5}{4}$ . Check if they are parallel or perpendicular.

$\rightarrow l_{1}: \dfrac{x-2}{2} = \dfrac{y-1}{7} = \dfrac{z+3}{-3}$

$l_{2} =\dfrac{x+2}{-1} = \dfrac{y-4}{2} = \dfrac{z-5}{4}$

$\cos\theta =\dfrac{ \vec{n_{1}}-\vec{n^{2}}}{ \left |\vec{ n_{1}} \right |\left | \vec{ n_{2}} \right |}$ where

$\theta$ is angle b/w lines

$\vec{n_{1}}$ is parallel to

$l_{1}= (2,7,-3)$

$\vec{n_{2}} = (-1,2,4)$

$\cos\theta = \dfrac{-2+14-12}{\sqrt{n_{1}}.\sqrt{n_{2}}}$

$= 0 \rightarrow \theta = 90^{\circ}$

$\Rightarrow$ perpendicular

The equation of the plane through $( 3,1 , - 3 )$ and $( 1 , - 2,2 )$ and parallel to the line with dr's $1,1 , - 2$ is

plane through

$(3, 1, -3)$ and

$(1, -2, 2)$ parallel to line with dr's

$(1, 1, -2)$

A vector in the plane is

$\vec{u} = \left(\begin{matrix} 1 \\ -2 \\ 2 \end{matrix} \right) - \left(\begin{matrix} 3 \\ 1 \\ -3 \end{matrix} \right) = \left(\begin{matrix} -2 \\ -3 \\ 5 \end{matrix} \right)=-2\hat i-3\hat j+5\hat k$

Dr's of parallel line,

$\vec{v} = \left(\begin{matrix} 1 \\ 1 \\ -2 \end{matrix} \right)=1\hat i+1\hat j-2\hat k$

normal vector to plane is

$\vec{n} = \vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & -3 & 5 \\ 1 & 1 & -2 \end{vmatrix}$

$= \hat{i} + \hat{j} + \hat{k} = (1, 1, 1)$

Equation of plane is

$\Rightarrow 1 (x - 3) + 1 (y - 1) + 1 (2 + 3) = 0$

$\Rightarrow x + y + z = 1$

Find the equation of the line passing through the point $(1 , 1 , 1)$ and intersects the line $\frac { ( x - 1 ) } { 2 } = \frac { ( y - 2 ) } { 3 } = \frac { ( z - 3 ) } { 4 }$ and $\frac { ( x + 2 ) } { 1 } = \frac { ( y - 3 ) } { 2 } = \frac { ( z + 1 ) } { 4 }$

$\begin{array}{l} Here,\, given....... \\ any\, line\, passes\, through\, point\, \, \, (1,1,1) \\ so, \\ \frac { { x-1 } }{ a } =\frac { { y-1 } }{ b } =\frac { { z-1 } }{ c } \\ and\, line\, \, interset\, through: \\ \frac { { x-1 } }{ 2 } =\frac { { x-1 } }{ 3 } =\frac { { x-1 } }{ 4 } \\ Now,\, determenent\, must\, be\, positive: \\ \, \, \, \, \, \, \, \, \, \left| \begin{array}{l} 0\, \, \, \, \, \, 1\, \, \, \, \, \, \, \, \, 2 \\ a\, \, \, \, \, \, b\, \, \, \, \, \, \, \, c \\ 2\, \, \, \, \, \, \, \, 3\, \, \, \, \, \, \, \, 4 \end{array} \right| =0\, \, \, \, or,\, \, \, \, \, \, \, \left| \begin{array}{l} a\, \, \, \, \, \, b\, \, \, \, \, \, \, \, \, c \\ 0\, \, \, \, \, \, 1\, \, \, \, \, \, \, \, \, \, 2 \\ 2\, \, \, \, \, \, \, \, 3\, \, \, \, \, \, \, \, 4 \end{array} \right| =0 \\ \Rightarrow -(4a-2c)+2(3a-2b)=0 \\ \Rightarrow -4a+2c+6a-4b=0 \\ \Rightarrow 2a-4b+2c=0 \\ \, \, \, \, \therefore \, \, \, \, a-2b+c=0-----------(i) \\ Again,{ { intersect } }\, \, another\, line........... \\ \frac { { x-(-2) } }{ 1 } =\frac { { x-3 } }{ 2 } =\frac { { x-(-1) } }{ 4 } \\ Now,\, determenent\, must\, be\, satisfiied: \\ \left| \begin{array}{l} -2-1\, \, \, \, \, \, \, \, \, \, \, 3-1\, \, \, \, \, \, \, \, \, \, -1-1 \\ a\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, b\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, c \\ 1\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, 2\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, 4 \end{array} \right| =0 \\ \Rightarrow (-3)(4b-2c)-2(4a-c)+(-2)(2a-b)=0 \\ \Rightarrow -12b+6c-8a+2c-4a+2b=0 \\ \Rightarrow -12a-10b+8c=0 \\ \Rightarrow 6a+5b-4c=0-----------(ii) \\ from\, \, \, equ\, (i)\, \& \, \, (ii),we\, get\, the\, ratio: \\ \frac { a }{ 3 } =\frac { b }{ { 10 } } =\frac { c }{ { 17 } } \\ then,\, the\, required\, line\, will:\, \frac { { x-1 } }{ 3 } =\frac { { y-1 } }{ { 10 } } =\frac { { z-1 } }{ { 17 } } \end{array}$

$A(-1,\ 2,\ 3),\ B(5,\ 0,\ -6),\ C(0,\ 4,\ -1)$ are three points. Show that the direction cosines of the bisectors of $\angle BAC$ are proportional to $(25,\ 8,\ 5)$ and $(-11,\ 20,\ 23)$ .

$A(-1,2,-3)\, B(5,0,-6)$

$C(0,4,-1)$

D.R's of AB are

$=(5-(-1),0-2,-6-(-3))$

$=(6,-2,-3)$

D.R's of AC rule

$=(0-(-1),4-2,-1-(-3))$

$=(1,2,2)$

DC's of AB are

$\dfrac{6}{7},\dfrac{-2}{7},\dfrac{-3}{7}$

$[\because \sqrt{6^{2}+(-2)^{2}+(-3)^{2}}=7]$

DC's of AC are

$\dfrac{1}{3},\dfrac{2}{3},\dfrac{2}{3}$

$[\because \sqrt{1^{2}+2^{2}+2^{2}}=3]$

DC's of Bisector of

$\angle$ BAC is proportional to

$\dfrac{6}{7}\pm \dfrac{1}{3},\dfrac{-2}{7}\pm \dfrac{2}{3},\dfrac{-3}{7}\pm \dfrac{2}{3}$

$\dfrac{18\pm 7}{21},\dfrac{-6\pm 14}{21},\dfrac{-9\pm 14}{21}$

$(25,8,5)$ and

$(11,-20,-23)$

i.e

$(25,8,5)$ and

$(-11,20,23)$

Find the direction cosines of a line which is perpendicular to the lines whose direction ratios are $(1,-1,2)$ and $(2,1,-1).$

${\textbf{Step - 1: Stating the direction cosines of a general vector}}{\text{.}}$

${\text{Let there be a vector }}$ $a\hat{i} + b\hat{j} + c\hat{k}$ ${\text{ ,then it's direction cosines can be written as}}$

$\cos \alpha = \pm \dfrac{a}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$

$\cos \beta = \pm \dfrac{b}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$

$\cos \gamma = \pm \dfrac{c}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$

${\text{where }}\alpha ,\beta {\text{ and }}\gamma {\text{ are the angles from x,y and z axes respectively}}{\text{.}}$

${\textbf{Step - 2: Finding the direction cosines of the vector perpendicular to the given two vectors}}{\text{.}}$

${\text{First finding the vector perpendicular to the two given vectors}}{\text{.}}$

${\text{If we have two vectors }}\overrightarrow A {\text{ and }}\overrightarrow B ,{\text{then }}\overrightarrow A \otimes \overrightarrow B {\text{ will be perpendicular to both the vectors}}{\text{.}}$

$\overrightarrow A \otimes \overrightarrow B =$ $\left| {\begin{array}{*{20}{c}}{\hat{i} }&{\hat{j} }&{\hat{k} } \\1&{ - 1}&2 \\2&1&{ - 1}\end{array}} \right|$

$\overrightarrow A \otimes \overrightarrow B = \hat{i} (( - 1)( - 1) - (2)(1)) - \hat{j} (1( - 1) - 2(2)) + \hat{k} (1 - ( - 1)(2))$

$\overrightarrow A \otimes \overrightarrow B = - \hat{i} + 5\hat{j} + 3\hat{k}$

${\text{Direction cosines of }}\overrightarrow A \otimes \overrightarrow B ,{\text{are}}$

$\cos \alpha = \pm \dfrac{1}{{\sqrt {{{( - 1)}^2} + {{(5)}^2} + {{(3)}^2}} }} = \pm \dfrac{1}{{\sqrt {35} }}$

$\cos \beta = \pm \dfrac{5}{{\sqrt {{{( - 1)}^2} + {{(5)}^2} + {{(3)}^2}} }} = \pm \dfrac{5}{{\sqrt {35} }}$

$\cos \gamma = \pm \dfrac{3}{{\sqrt {{{( - 1)}^2} + {{(5)}^2} + {{(3)}^2}} }} = \pm \dfrac{3}{{\sqrt {35} }}$

${\textbf{Hence, Direction cosines of the vector are}}\mathbf{\left( { \pm \dfrac{1}{{\sqrt {35} }}, \pm \dfrac{5}{{\sqrt {35} }}, \pm \dfrac{3}{{\sqrt {35} }}} \right).}$

If the direction ratios of two parallel lines are $4, -3, -1$ and $p+q, 1+q, 2$ then find the values of p and q.

As we know that the ratio of direction ratios of 2 parallel lines are equal

So, for 1st line it is

$4,-3,-1$

and for 2nd line it is

$p+q,1+q,2$

$\Rightarrow$

$\cfrac{4}{p+q}=\cfrac{-3}{1+q}=\cfrac{-1}{2}$

$\Rightarrow$

$\cfrac{4}{p+q}=\cfrac{-1}{2}$ and

$\cfrac{-3}{1+p}=\cfrac{-1}{2}$

$\Rightarrow$

$p+q=-8$ and

$\cfrac{-3}{1+p}=\cfrac{-1}{2}$

$\Rightarrow$

$p+q=-8....(1)$ and

$1+p=6....(2)$

from (2)

$p=5$

substitute this value in (1)

then

$q=-8-p$

$\Rightarrow$

$q=-8-5$

$q=-14$

Find the angle between the line whose direction cosines are given by $l+m+n=0$ and ${ l }^{ 2 }+{ m }^{ 2 }={ n }^{ 2 }$

$l+m+n=0 ---(1)$

$\Downarrow$

$l+m=-n$

$-(l+m)=n$

put

$m=0$ in (1)

$l=-n$

$(l,m,n)dr's=(1,0,-1)=b_1$

put

$l=0$ in (1)

$m=-n$

$(l,m,n) dr's=(0,1,-1)=b_2$

$\left| b \right| =\sqrt{1^2+0^2+(-1)^2}=\sqrt2$

$\left| b_2 \right| =\sqrt{0^2+1^2+(-1)^2}=\sqrt2$

$l^2+m^2=n^2\rightarrow l^2+m^2-n^2=0---(2)$

substitute '

$n$ '

$l^2+m^2-[-(l+m)]^2=0$

$l^2+m^2-l^2-m^2-2lm=0$

$2lm=0$

either

$l=0 \ (or) \ m=0$

i.e;

$\cos\theta=\dfrac{b_1.b_2}{\left|b_1 \right| \left| b_2 \right| }$

$=\dfrac{1}{\sqrt2.\sqrt2}$

$\cos\theta=\dfrac{1}{2}$

$\theta=\dfrac{\pi}{3}$

Show that $2x+3y+7=0$ represents a plane perpendicular to $XY$ -plane.

We have,

$2x+5y+7=0$

Let, the plane

$=2, 5, 0$

Equation of xy-plane

$z=0$

So, ratio of normal to the plane are

$0, 0, 1$

$\therefore a_1\times a_2+a_1\times a_2+a_1\times a_2$

$=2\times 0+5\times 0+0\times 7$

$=0+0+0$

$=0$

Hence, this is answer.

Find the equation of the plane through the point $(1, 1, 1)$ and perpendicular to the line $x - 2 y + z = 2,4 x + 3 y - z + 1 = 0$

Given equation of lines

$x-2y+z=2$ and

$4x+3y-z+1=0$ which are lines (1) and (2) through which the point

$(1,1,1)$ passes.

Now\quad the normal vector for line (1) is

$\overrightarrow { { n }_{ 1 } } =\widehat { i } -2\widehat { j } +\widehat { k }$

and the normal vector for line (2) is

$\overrightarrow { { n }_{ 1 } } =4\widehat { i } +3\widehat { j } -\widehat { k }$

Now we find

$\overrightarrow { n } = \overrightarrow { { n }_{ 1 } } \times \overrightarrow { { n }_{ 2 } } =\left( \widehat { i } -2\widehat { j } +\widehat { k } \right) \times \left( 4\widehat { i } +3\widehat { j } -\widehat { k } \right)$

$\Longrightarrow \overrightarrow { n } =-\widehat { i } +5\widehat { j } +11\widehat { k }$ is the required plane that passes through

$(1,1,1)$

Now the position vector for the point

$(1,1,1)$ is

$\widehat { i } +\widehat { j } +\widehat { k }$

As we know equation of plane to the given vector

$\overrightarrow { n }$ is given by

$\Longrightarrow \left( \overrightarrow { r } -\overrightarrow { a } \right) .\overrightarrow { n } =0$

$\Longrightarrow \overrightarrow { r } .\overrightarrow { n } =\overrightarrow { a } .\overrightarrow { n }$

So,

$\left( x\widehat { i } +y\widehat { j } +z\widehat { k } \right) \left( -\widehat { i } +5\widehat { j } +11\widehat { k } \right) =\left( \widehat { i } +\widehat { j } +\widehat { k } \right) \left( -\widehat { i } +5\widehat { j } +11\widehat { k } \right)$

$\Longrightarrow -x+5y+11z=-1+5+11$

$\Longrightarrow -x+5y+11z=15$

$\Longrightarrow x-5y-11z+15=0$ which is the required equation of plane.

Directions ratio of two lines are $3, -2, k$ and $-2, k, 4$ . Find k if the lines are perpendicular to each other.

Directions ratio of I-st line

$= 3, -2, k$

Directions ratio of II-nd line

$= -2, k, 4$

The given lines are perpendicular, thus,

$3(-2)-2(k)+k(4)=0$

$-6-2k+4k=0$

$2k=6$

$k=3$

Draw five two- dimensional and five three- dimensional objects your choice, and label them.

1181962_1291371_ans_a87acb0e1206489ca283ec8ab48f3f79.jpg

What is the direction cosine of angle which the vector $\sqrt {2\hat i} + \hat j + \hat k\;makes\;with\;y - axis\;?$

For a vector

$ai+bj+ck$

direction cosines are given by

$t=\dfrac{a}{\sqrt{a^2+b^2+c^2}}, m=\dfrac{b}{\sqrt{a^2+b^2+c^2}}, n=\dfrac{c}{\sqrt{a^2+b^2+c^2}}$

$L=\cos \alpha, m=\cos\beta, n=\cos\gamma$

Where

$\alpha, \beta, \gamma$

are angles made by the vectors with x-axis, y-axis and z-axis respectively for the given vector

$\sqrt{2i}+j+k$

$m=\cos\beta =\dfrac{1}{\sqrt{2+1+1}}=\dfrac{1}{\sqrt{4}}=\dfrac{1}{2}$

$\therefore \beta =\dfrac{\pi}{3}$ .

1116655_1264818_ans_513fbb3f310c4b70ba331f202a76d6f8.jpeg

Find the equations of the straight line perpendicular to the lines $\dfrac{x+1}{-3}=\dfrac{y-3}{2}=\dfrac{z+2}{1}; \dfrac{x}{1}=\dfrac{y-7}{-3}=\dfrac{z+7}{2}$

Condition for Perpendicularity

$a_1a_2+b_1b_2+c_1c_2=0$

$\implies -3a+2b+c=0\cdots(1)$

$\implies a-3b+2c=0\cdots(2)$

$\implies (a,b,c)=\pm(1,1,1)$

let the line pass through some point

$(p,q,r)$

$\implies x-p=y-q=z-r$

Find the angle between the lines joining the points $\left( 1,4,2\right ) , \left( - 2,1,2\right)$ and $\left( 1,2,3\right ) , \left( 2,5,1\right )$

Equation of the line joining

$(1, 4, 2)(-2, 1, 2)$ is

$\dfrac{x-1}{-2-1}=\dfrac{y-4}{1-4}=\dfrac{z-2}{2-2}$

( or )

$\dfrac{x-1}{-3}=\dfrac{y-4}{-3}=\dfrac{z-2}{0}$

Equation of the line joining

$(1, 2, 3) (2, 5, 1)$ is

$\dfrac{x-1}{2-1}=\dfrac{y-2}{5-2}=\dfrac{z-3}{1-3}$

(or)

$\dfrac{x-1}{1}=\dfrac{y-2}{3}=\dfrac{z-3}{-2}$

$\cos\alpha= \dfrac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}{a_2^2+b_2^2+c_2^2}}$

$\displaystyle \left.\{ where\quad a_{ 1 }=-3\\ b_{ 1 }=-3\\ c_{ 1 }=0\\ a_{ 2 }=1,\quad b_{ 2 }=3\\ c_{ 2 }=-2 \right.$

$=\dfrac{-3\times 1+(-3)\times 3+0}{\sqrt{9+9+0}\sqrt{1+9+4}}$

$=\dfrac{-3-9}{\sqrt{18}\sqrt{14}}=\dfrac{-12}{3\sqrt{2}\sqrt{14}}=\dfrac{-4\sqrt{2}}{2\sqrt{14}}$

$=-2\sqrt{\dfrac{2}{14}}=\dfrac{-2}{\sqrt{7}}$

$\alpha =\cos^{-1}\left(-2\sqrt{7}\right)$

Find the vector equation of the plane through the points $( 2,1 , - 1 )$ and $( - 1,3,4 )$ and perpendicular to the plane $x - 2 y + 4 z = 10$ .

The required plane passes through points

$P(2,1,-1)$ and

$Q(-1,3,4).$

We can write the position vectors as follows :

$a_1=2\overrightarrow{i}+\overrightarrow{j}-\overrightarrow{k}$ and

$a_2=-\overrightarrow{i}+3\overrightarrow{j}+4\overrightarrow{k}$

$\Rightarrow$

$PQ=(a_2-a_1)=-3\overrightarrow{i}+2\overrightarrow{j}+5\overrightarrow{k}$

$\overrightarrow{n_i}\,$ is the normal vector of plane

$x-2y+4z=10$

Let

$\overrightarrow{i}$ be the normal vector to the desired plane

$\overrightarrow{n}=\overrightarrow{n_i}\times \overrightarrow{PQ}=\begin{vmatrix}\overrightarrow{i}&\overrightarrow{j}&\overrightarrow{k}\\1&-2&4\\-3&2&5\end{vmatrix}$

$\Rightarrow$

$\overrightarrow{n}=\overrightarrow{i}(-10-8)-\overrightarrow{j}(5+12)+\overrightarrow{k}(2-6)$

$\Rightarrow$

$\overrightarrow{n}=-18\overrightarrow{i}-17\overrightarrow{j}-4\overrightarrow{k}$

$\\ \Rightarrow\\$

$\overrightarrow{r}.\overrightarrow{n}=\overrightarrow{a}.\overrightarrow{n}$

$\Rightarrow$

$\overrightarrow{r}.(-18\overrightarrow{i}-17\overrightarrow{j}-4\overrightarrow{k})=(2\overrightarrow{i}+\overrightarrow{j}-\overrightarrow{k})(18\overrightarrow{i}-17\overrightarrow{j}-4\overrightarrow{k})$

$\Rightarrow$

$\overrightarrow{r}.(-18\overrightarrow{i}-17\overrightarrow{j}-4\overrightarrow{k})=-36-17+4$

$\Rightarrow$

$\overrightarrow{r}.(18\overrightarrow{i}-17\overrightarrow{j}+4\overrightarrow{k})=49$

Find the direction cosine of the line passing through the points $A(-4,2,3)$ and $B(1,3,-2)$

$A\,(-4,2,3) \: B\,(1,3,-2)$

D.r's of a line passing through

$P(x_1,y_1,z_1)$ &

$Q(x_2,y_2,z_2)$

$= (x_2 - x_1), (y_2 - y_1), (z_2 - z_1)$

$D.C = \frac{x_2 - x_1}{PQ}, \frac{y_2 - y_1}{PQ} , \frac{z_2 - z_1}{PQ}$

$PQ = AB = \sqrt{(x_2 - x_1)^{2} + (y_2 - y_1)^{2} + (z_2 - z_1)^{2}}$

$A=P; \, B=Q$

$x_1 = -4; y_1 = 2; z_1 = 3$

$x_2 = 1; y_2 = 3; z_2 = -2$

$D.r's = (x_2 - x_1), (y_2 - y_1),(z_2 - z_1)$

$= 1+4; \,3-2; \, -2-3$

$= 5; \,1; \, -5$

$DC = \frac{5}{\sqrt{5^{2} + 1^{2} + (-5)^{2}}}$ ;

$\frac{1}{\sqrt{5^{2} + 1^{2} + (-5)^{2}}}$ ;

$\frac{-5}{\sqrt{5^{2} + 1^{2} + (-5)^{2}}}$

$= \frac{5}{\sqrt{25 + 1 + 25}};\, \frac{1}{\sqrt{25+1+25}};\, \frac{-5}{\sqrt{25+1+25}}$

$= \frac{5}{\sqrt{51}};\, \frac{1}{\sqrt{51}};\, \frac{-5}{\sqrt{51}}$

1210480_1284803_ans_ab38f485cdbc42a19a8746786d7c30b1.jpg

Equation of the plane passing through the three points $P(2,3,1), Q(2,1,2)$ and $R(3,6,-1)$ is

$P\left(2,3,1\right); Q\left( 2,1,2\right); R\left( 3,6,-1\right)$

$\Rightarrow \left| \begin{matrix} x-2 & y-3 & z-1 \\ 0 & -2 & 1 \\ 1 & 3 & -2 \end{matrix} \right| =0$

$\Rightarrow \left(x-2\right)\left(1\right)-\left(y-3\right)\left(-1\right)+\left(z-1\right)\left(2\right)=0$

$x-2+y-3+2z-2=0$

$x+y+2z=7$ is the equation of the plane.

Hence, the answer is

$x+y+2z=7.$

Prove that $1,1,1$ cannot be direction cosines of a straight line

$\begin{array}{l} \left( a \right) { { Re } }lation\, \, b/w\, \, the\, \, direction\, \, of\, a\, \, line \\ { l^{ 2 } }+{ m^{ 2 } }+{ n^{ 2 } }=1\to (i) \\ now\, \, given\, \, data-\alpha =1,\, \, \beta =1,\, \, \lambda =1 \\ l=\cos \alpha \\ m=\cos \beta \\ n=\cos \lambda \\ but\, \, given\, \cos \lambda =\cos { 90^{ 0 } } \\ \cos \beta =\cos { 90^{ 0 } } \\ \, \cos \lambda =\cos { 90^{ 0 } } \\ \cos \lambda =\cos \beta =\cos \lambda \, \, =\, \, \cos { 90^{ 0 } } =1 \\ both\, \, relations\, \, are\, \, not\, \, possible\, \, in\, \, 3\, \, D\, \, geometary \\ \left( b \right) \, \, because\, \, at\, \, any\, \, time\, \, one\, \, angle\, \, is\, \, zero\, \, and\, \, another\, \, two\, \, angles\, \, are\, \, { 90^{ 0 } } \\ { l^{ 2 } }+{ m^{ 2 } }+{ n^{ 2 } } \\ { \cos ^{ 2 } }\lambda +{ \cos ^{ 2 } }\beta +{ \cos ^{ 2 } }\alpha \\ 1+1+1=3\to (ii) \\ equation\, \, (i)\, \, \& \, \, (ii)\, \, are\; \, not\, \, dulfilled. \end{array}$

Do you think the points $(3,5),(6,8)$ and $(10,12)$ are straight line? Given reasons for your answer.

If the points

$\left( 3, 5 \right), \left( 6, 8 \right)$ and

$\left( 10, 12 \right)$ lie on a straight line then the slope of line joining the points

$\left( 3, 5 \right)$ and

$\left( 6, 8 \right)$ will be equal to the slope of line joining the pounts

$\left( 6, 8 \right)$ and

$\left( 10, 12 \right)$ .

Therefore,

Slope of line joining the points

$\left( 3, 5 \right)$ and

$\left( 6, 8 \right) = \cfrac{8-5}{6-3} = \cfrac{3}{3} = 1$

Slope of line joining the points

$\left( 6, 8 \right)$ and

$\left( 10, 12 \right) = \cfrac{12 - 8}{10 - 6} = \cfrac{4}{4} = 1$

Since two intersecting lines cannot have same slope, thus the given points are straight line.

Prove that the circumcenter, orthocenter, incenter and centroid of the triangle formed by the points $A(-1,11)$ $B(-9,-8)$ and $C(15,-2)$ are collinear, without actually finding any of them.

Distance of AB, BC and CA

$\begin{array}{l} \Rightarrow AB=\sqrt { { { \left( { -8 } \right) }^{ 2 } }+{ { \left( { 19 } \right) }^{ 2 } } } \\ \Rightarrow AB=\sqrt { 64+361 } =\sqrt { 425 } \, \, unit \\ \Rightarrow BC=\sqrt { { { \left( { 15+9 } \right) }^{ 2 } }+{ { \left( { -6 } \right) }^{ 2 } } } =\sqrt { { { \left( { 24 } \right) }^{ 2 } }+36 } =\sqrt { 612 } \, \, unit \\ \Rightarrow and\, \, AC=\sqrt { { { \left( { 16 } \right) }^{ 2 } }+{ { \left( { 13 } \right) }^{ 2 } } } =\sqrt { 256+109 } =\sqrt { 425 } \, \, unit \end{array}$

$\therefore \Delta ABC$ are Isosceles triangle

Hence, centroid, orthocentre, incentre and circumcentre lie on same line.

1211030_1310474_ans_5e25d8a9408a4ee38f33a666f04b1296.PNG

For what value of m , the points (3 , 5) , (m , 6) and $\left ( \dfrac{1}{2} , \dfrac{15}{2}\right )$ are collinear ?

The points are colliner if the area of triangle is 0.

$0=\dfrac{1}{2}|3 (6-\dfrac{15}{2})+m(\dfrac{15}{2}-5)+\dfrac{1}{2}(5-6)|$

$0= 3 (\dfrac{-3}{2})+m (\dfrac{5}{2})-\dfrac{1}{2}$

$0= -9 + 5 m -1$

$10= 5 m \Rightarrow m =2$

Find the value of $K$ for which the points $(7,-2) ,(5,1)$ and $(3,K)$ are collinear.

If the points are collinear then the area of triangle will be O.

$\Delta =\frac {1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

$\Rightarrow |7(1-k) +5 (k+2)+3 (-2-1)|=0$

$\Rightarrow |7 -7k+5k+10-9|=0$

$\Rightarrow |-2k +8|=0$

$\Rightarrow -2k =-8$ so

$k=4$

1194400_1309061_ans_37edc12a0b8f4ad09d8bef0f4f128d11.jpg

Find the shortest distance between the skew lines :
$l_{1} : \dfrac{x-1}{2}=\dfrac{y+1}{1}=\dfrac{z-2}{4}$

$l_{2} : \dfrac{x+2}{4}=\dfrac{y-0}{-3}=\dfrac{z+1}{1}$

Line

$l_{1}$ passes through the point

$P(1,-1,2)$ .

The equation of the plane containing line

$l_{2}$ is

$a(x+2)+b(y-0)+c(z+1)=0$ .........(1)

where,

$4a-3b+c=0$

If it is parallel to line

$l_{1}$ , then

$2a+b+4c=0$

Therefore,

$\dfrac{a}{-13}=\dfrac{b}{-14}=\dfrac{c}{10}$

Substituting values of a, b. c in equation 1 , we get,

$13x+14y-10z+16=0$

This is the equation of the plane containing line

$l_{2}$ and parallel to line

$l_{1}$ .

Shortest distance between lines

$l_{1}$ and

$l_{2}$ =

$|\dfrac{13-14-20+16}{\sqrt{13^2+14^2+(-10)^2}}|=\dfrac{5}{\sqrt{465}}$

Find the equation of the plane that passes through $x -$ axis and point $( 3,2,4 )$ .

Equation of plane passing through point

$(3,2,4) a(x-3)+b(y-2)+c(z-4)=0...(1)$

$\because$ Plane is perpendicular to

$X-$ axis

$\therefore a=0, d=0 \Rightarrow by+cz=0...(2)$
From

$(1), a=0$ so

$b(y-2)+c(z-4)=0...(3)$

$\Rightarrow by-2b+cz-4c=0$

$\Rightarrow by+cz-2b-4c\ 0$

$\Rightarrow -2b=c$

$[\because by+cz=0\ from\ (2)]$

$\Rightarrow b=0-2c$

$\therefore$ Equation of plane passing through point

$(3,2,4)$ and

$X-$ axis

$b(y-2)+c(z-4)=0$

$\Rightarrow -2c(y-2)+c(z-4)=0$

$\Rightarrow -2y+4+z-4=0$

$\Rightarrow 2y-z=0$

Find the equation of the plane passing through the point (1,-1,2) having 2,3,2 as direction ratios of normal to the plane.

The plane passes through the point having position vetor

$\overrightarrow a = \widehat{i}-\widehat{j}+2\widehat{k}$ and is normal to the vector

$\overrightarrow n = 2\widehat{i}+3\widehat{j}+2\widehat{k}$ . So, the vector equation of the plane is

$(\overrightarrow r -\overrightarrow a) . \overrightarrow n =0$

$\overrightarrow r . \overrightarrow n = \overrightarrow a . \overrightarrow n$

$\overrightarrow r. (2\widehat{i}+3\widehat{j}+2\widehat{k}) = (\widehat{i}-\widehat{j}+2\widehat{k}). (2\widehat{i}+3\widehat{j}+2\widehat{k})$

$\overrightarrow r . (2\widehat{i}+3\widehat{j}+2\widehat{k})=3$

The cartesian equation of the plane is

$2x+3y+2z=3$ .

The foot of the perpendicular drawn from the origin to the plane is $( 2,5,7 ) .$ Find the equation of the plane.

The foot of perpendicular from origin is

$(2,5,7)$

so the Drs

$(2,5,7)$

The perpendicular distance from point

$(2,5,7)\\\sqrt{2^2+5^2+7^2}\\\sqrt{4+25+49}\\\sqrt{69}$

So the equation of plane is

$2x+5y+7z-\sqrt{69}=0$

Show that the points having position vectors, $\bar a, \bar b$ and $3\bar a-2\bar b$ are colliera, where $\bar a , \bar b , \bar c$ any vectors.

Let given vectors

$\underset{A}{\bar{a}}, \underset{B}{\bar{b}}, \underset{C}{3\bar{a} - 2\bar{b}}$

$\overline{AB} = \bar{b} - \bar{a}$

$\overline{AC} = 3\bar{a} - 2\bar{b} - \bar{a} = 2\bar{a} - 2\bar{b}$

$\overline{AC} = -2\overline{AB}$

Here

$\bar{a}, \bar{b}, 3\bar{a} - 2\bar{b}$ are collinear

1210817_1398655_ans_a23fd16bd8b147c9b8bf3c4354bbb6c1.jpg

Find the directions cosines of $x,y$ and $z$ axis.

Direction cosines of

a-axis

$(cos0^{\circ},cos90^{\circ},cos90^{\circ})$

=(1,0,0)

(2)y axis

= (0,1,0)

(3)z axis

=(0,0,1)

1172963_1346896_ans_73624c674c884dea87749f636dfababd.jpg

If a line makes angles $\alpha ,\beta$ and $\gamma$ with three-dimensional coordinate axes, respectively, then find the value of $\cos 2\alpha +\cos 2\beta+\cos 2\gamma$ .

[a]l

$=cos\alpha$

$m=cos\beta$

$n=cos\gamma$

we now ,

$l^{2}+m^{2}+n^{2}=1$

$cos^{2}\alpha +cos^{2}\beta +cos^{2}\gamma =1$

$we know cos^{2}\theta =2cos^{2}\theta -1$

so,

$\frac({1-cos^{2}\alpha }{2})+(\frac{1-cos^{2}\beta }{2})+(\frac{1-cos^{2}\gamma }{2})=1$

$= (1-cos^{2}\alpha )+(1-cos^{2}\beta )+(1-cos^{2}\gamma )=2$

$=-1= cos^{2}= cos^{2}\alpha +cos^{2}\beta +cos^{2}\gamma$

Hence value is -1

1211735_1376059_ans_63db17f3505b40cf8306f35f772111bc.jpg

Find the direction cosines of the line $\dfrac{{x + 2}}{2} = \dfrac{{2y - 5}}{3};z = - 1$ .

Equation of line is

$\dfrac{x+2}{2}=\dfrac{2y-5}{3};z=-1$

$x = \dfrac{-2}{2}=\dfrac{y-5/2}{3/2}=\dfrac{z+1}{0}$

Direction ratio of line are 2,3/2,0.

$\therefore$ Direction cosines of the line are

$\dfrac{2}{\sqrt{4+(9/4)+0}},\dfrac{3/2}{\sqrt{4+9/4+0}},0$

$\Rightarrow \dfrac{2}{5/2,}\dfrac{3/2}{5/2},0$

$\Rightarrow$ 4/5,3/5,0

Show that the points $( 1,0 ) , ( 6,0 ), (0,0)$ are collinear

We can use matrix to prove that:

$\operatorname{ Since, }\left|\begin{array}{ccc}1 & 0 & 1 \\6 & 0 & 1 \\0 & 0 & 1\end{array}\right|=0\\$

Hence

$( 1,0 ) , ( 6,0 ), (0,0)$ are collinear.

Prove that points $A, B, C$ having positions vectors $\vec { a } ,\vec { b } ,\vec { c }$ are collinear, if $\left[\vec { b } \times \vec { c } +\vec { c } \times \vec { a } +\vec { a } \times \vec { b } \right]=\vec { 0 }$

If A,B,C are collinear,

$\bar{AB}$ and

$\bar{BC}$ are parallel.

$\Rightarrow \bar{AB}\times \bar{BC}=0$

$(\bar{b}-\bar{a})\times (\bar{c}-\bar{b})=0$

$\bar{b}\times \bar{c}-\bar{b}\times \bar{b}-\bar{a}\times \bar{c}+\bar{a}\times \bar{b}=0$

But,

$\bar{b}\times \bar{b}=0$

$\bar{b}\times \bar{c}-\bar{a}\bar{c}+\bar{a}\times \bar{b}=0$

$\bar{b}\times \bar{c}+\bar{c}\times \bar{a}+\bar{a}\times \bar{b}=0$

$\because (\bar{a}\times \bar{b})=-\bar{b}\times \bar{a}$

The direction ratios of $\overline { A B }$ are

$A(4,1,5)$ and

$B(3,4,5)$

The direction ratios of

$\overline {AB}$ is

$(3-4,1-4,5-5)=(-1,3,0)$

Find the vector equation of the line passing through $(1, 2, 3)$ and perpendicular to the plane $\vec {r} \cdot (\hat {i} + 2\hat {j} - 5\hat {k}) + 9 = 0$ .

DRs of line perpendicular to plane is =(1 , 2 , -5)

line equation is

$\dfrac{x-1}{1} = \dfrac{y-2}{2} = \dfrac{z-3}{-5}$

Using direction ratio show that the points $A(2,3,-4), B(1,-2,3)$ and $C(3,8,-11)$ are collinear.

Three points

$A,B,C$ are collinear if direction ratios of

$AB$ and

$BC$ are proportional.

We have

$A\left(2,3,-4\right)$ and

$B\left(1,-2,3\right)$

Direction ratios are

$=1-2,-2-3,3-\left(-4\right)$

$=-1,-5,7$

So,

${a}_{1}=-1,{b}_{1}=-5,{c}_{1}=7$

We have

$B\left(1,-2,3\right)$ and

$C\left(3,8,-11\right)$

Direction ratios are

$=3-1,8-\left(-2\right),-11-3$

$=2,10,-14$

So,

${a}_{2}=2,{b}_{2}=10,{c}_{2}=-14$

Now,

$\dfrac{{a}_{2}}{{a}_{1}}=\dfrac{2}{-1}=-2$

$\dfrac{{b}_{2}}{{b}_{1}}=\dfrac{10}{-5}=-2$

$\dfrac{{c}_{2}}{{c}_{1}}=\dfrac{-14}{7}=-2$

So,

$\dfrac{{a}_{2}}{{a}_{1}}=\dfrac{{b}_{2}}{{b}_{1}}=\dfrac{{c}_{2}}{{c}_{1}}=-2$

$\therefore\,A,B$ and

$C$ are collinear.

Find the intercepts made on the coordinates axes by the plane $2x+y-2z=3$ and find also the direction cosines of the normal to the plane.

The given equation of the plane is

$2x+y-2z=3$

$\Rightarrow\,\dfrac{x}{\dfrac{3}{2}}+\dfrac{y}{3}+\dfrac{z}{\dfrac{-3}{2}}=1$ ......

$(1)$

We know that the equation of the plane whose intercepts on the coordinate axes are

$\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$ .........

$(2)$

Comparing equations

$(1)$ and

$(2)$ , we get

$a=\dfrac{3}{2},\,b=3,\,c=-\dfrac{3}{2}$

Finding the direction cosines of the normal:

The given equation of the plane is

$2x+y-2z=8$

$\Rightarrow\,\left(x\hat{i}+y\hat{j}+z\hat{k}\right).\left(2\hat{i}+\hat{j}-2\hat{k}\right)=8$

$\Rightarrow\,\vec{r}.\left(2\hat{i}+\hat{j}-2\hat{k}\right)=8$ which is the vector equation of the plane.

$\left|\vec{n}\right|=\sqrt{4+1+4}=\sqrt{9}=3$

So, the unit vector perpendicular to

$\vec{n}$ is

$=\dfrac{\vec{n}}{\left|\vec{n}\right|}=\dfrac{2\hat{i}+\hat{j}-2\hat{k}}{3}$

So, the direction cosines of the normal to the plane are,

$\dfrac{2}{3},\dfrac{1}{3}$ and

$-\dfrac{2}{3}$

If the lines $\dfrac {x - 1}{-3} = \dfrac {y - 2}{-2k} = \dfrac {z - 3}{2}, \dfrac {x - 1}{K} = \dfrac {y - 2}{1} = \dfrac {z - 3}{5}$ are $\perp^{r}$ , find $K$ .

1700657_1587528_ans_d64a442fe68c469398db97b3347064c0.jpeg

The coordinates of the foot of the perpendicular drawn from of the origin to a plane are $(12, -4, 3)$ . Find the equation of the plane.

We have

$P\left(12,-4,3\right)$ is the foot of the perpendicular from the origin to the plane.

$\therefore\,\vec{OP}=\vec{n}=12\hat{i}-4\hat{j}+3\hat{k}$

and

$p=\left|12\hat{i}-4\hat{j}+3\hat{k}\right|=\sqrt{144+16+9}=\sqrt{169}=13$

$\Rightarrow\,\hat{n}=\dfrac{12\hat{i}-4\hat{j}+3\hat{k}}{13}$

The vector equations of the plane is

$\vec{r}.\hat{n}=p$

$\Rightarrow\,\left(x\hat{i}+y\hat{j}+z\hat{k}\right).\dfrac{12\hat{i}-4\hat{j}+3\hat{k}}{13}=13$

$\Rightarrow\,12x-4y+3z=169$

The cartesian form is

$12x-4y+3z=169$

If a line has the direction ratios $-18, 12, -4$ , then what are its direction cosines ?

If the direction ratios of line are

$(a,b,c)$
Then direction cosine are given as

$\dfrac{a}{\sqrt{a^2+b^2+c^2}},\dfrac{b}{\sqrt{a^2+b^2+c^2}},\dfrac{c}{\sqrt{a^2+b^2+c^2}}$

Given d.r.'s of line are

$-18,12,-4$

The direction cosines will be

$\dfrac{-18}{\sqrt{(-18)^2+12^2+(-4)^2}},\dfrac{12}{\sqrt{(-18)^2+12^2+(-4)^2}},\dfrac{-4}{\sqrt{(-18)^2+12^2+(-4)^2}}\\=-\dfrac{18}{22},\dfrac{12}{22},-\dfrac{-4}{22}\\=-\dfrac{9}{11},\dfrac{6}{11},-\dfrac{2}{11}$

If $O$ be the origin and the coordinates of $P$ be $(1,2,-3)$ , then find the equation of the plane passing through $P$ and perpendicular to $OP$

Equation of plane passing through

$\left({x}_{1},{y}_{1},{z}_{1}\right)$ and perpendicular to a line with direction ratios

$A,B,C$ is

$A\left(x-{x}_{1}\right)+B\left(y-{y}_{1}\right)+C\left(z-{z}_{1}\right)=0$

The plane passes through

$P\left(1,2,-3\right)$

So,

${x}_{1}=1,\,{y}_{1}=2,\,{z}_{1}=-3$

Normal vector to plane

$=\vec{OP}$ where

$O\left(0,0,0\right)$ and

$P\left(1,2,-3\right)$

Direction ratios of

$\vec{OP}=1-0,\,2-0,\,-3-0$

$=1,2,-3$

$\therefore\,A=1,\,B=2,\,C=-3$

Equation of plane in cartesian form is

$1\left(x-1\right)+2\left(y-2\right)-3\left(z+3\right)=0$

$x-1+2y-4-3z-9=0$

$x+2y-3z-14=0$

Find the cartesian equation of a line passing through $(1, -1, 2)$ and parallel to the line whose equation are $\dfrac {x - 3}{1} = \dfrac {y - 1}{2} = \dfrac {z + 1}{-2}$ . Also, reduce the equation obtained in vector form.

1501900_1662657_ans_05f2325a200649889559aa59ef53c267.jpg

Find the equation of the plane that bisects the line segment joining points $(1,2,3)$ and $(3,4,5)$ and is at right angle to it.

The given points are

$A\left(1,2,3\right)$ and

$B\left(3,4,5\right)$

The line segement

$AB$ is given by

$\left({x}_{2}-{x}_{1}\right), \left({y}_{2}-{y}_{1}\right),\left({z}_{2}-{z}_{1}\right)$

The line segement

$AB$ is given by

$\left(3-1\right), \left(4-2\right),\left(5-3\right)=\left(2,2,2\right)$

Since the plane bisects

$AB$ at rightangles,

$\vec{AB}$ is the normal to the plane which is

$\vec{n}$

$\therefore\,\vec{n}=2\hat{i}+2\hat{j}+2\hat{k}$

Let

$C$ be the midpoint of

$AB$ .

$C=\left(\dfrac{1+3}{2},\dfrac{2+4}{2},\dfrac{3+5}{2}\right)=\left(2,3,4\right)$

Let this be

$\vec{a}=2\hat{i}+3\hat{j}+4\hat{k}$

Hence the vector equation of the plane passing through

$C$ and

$\bot AB$ is

$\left[\vec{r}-\left(2\hat{i}+3\hat{j}+4\hat{k}\right)\right].\left(2\hat{i}+2\hat{j}+2\hat{k}\right)=0$

$\Rightarrow\,\left[\left(x-2\right)\hat{i}+\left(y-3\right)\hat{j}+\left(z-4\right)\hat{k}\right].\left(2\hat{i}+2\hat{j}+2\hat{k}\right)=0$

$\Rightarrow\,2\left(x-2\right)+2\left(y-3\right)+2\left(z-4\right)=0$

$\Rightarrow\,2x+2y+2z-4-6-8=0$

$\Rightarrow\,2x+2y+2z=18$

$x+y+z=9$

This is the required equation of the plane.

Find the vector equation of plane which is at a distance of $3$ units form the origin and has $\hat{k}$ as the unit vector normal to it.

Vector equation of a plane at a distance

$d$ from the origin and normal to the vector

$\vec{n}$ is

$\vec{r}.\hat{n}=d$

Unit vector of

$\vec{n}=\hat{n}=\dfrac{1}{\left|\vec{n}\right|}\left(\vec{n}\right)$

Distance from origin

$=d=3$

$\vec{n}=\hat{k}$

Magnitude of

$\vec{n}=\sqrt{{1}^{2}}=1$

$\left|\vec{n}\right|=1$

Now,

$\hat{n}=\dfrac{1}{\left|\vec{n}\right|}\left(\vec{n}\right)$

$=1\left(\hat{k}\right)=\hat{k}$

Vector equation is

$\vec{r}.\hat{n}=d$

$\vec{r}.\hat{k}=3$

$\therefore\,$ the vector equation of the plane is

$\vec{r}.\left(\hat{k}\right)=3$

Write the normal form of the equation of the plane $2x-3y+6z+14=0$ .

Equation of the plane is

$2x-3y+6z+14=0$

$2x-3y+6z=-14$

Dividing with

$\sqrt{{2}^{2}+{\left(-3\right)}^{2}+{6}^{2}}=\sqrt{4+9+36}=\sqrt{49}=7$

$\dfrac{2}{7}x-\dfrac{3}{7}y+\dfrac{6}{7}z+\dfrac{14}{7}=0$

$\dfrac{2}{7}x-\dfrac{3}{7}y+\dfrac{6}{7}z+2=0$ which is the normal form of the plane.

Find the vector equation of the plane passing through the point $A(a,0,0), B(0, b, 0)$ and that $C(0,0,c)$ . Reduce it to normal form. If plane $ABC$ is at distance $p$ from the origin, prove that $\dfrac{1}{p^{2}}=\dfrac{1}{a^{2}}+\dfrac{1}{b^{2}}+\dfrac{1}{c^{2}}$ .

For writing equation of plane we need ,

1) Normal Vector to Plane

2) One point on plane

Let's First find normal vector to plane,

Two vectors lying in plane are ,

$b\hat{j}-a\hat{i}$

$c\hat{k}-b\hat{j}$

Thus Normal vector to plane is

$(b\hat{j}-a\hat{i})\times(c\hat{k}-b\hat{j})=(bc\hat{i}+ac\hat{j}+ab\hat{k})$

One point on plane is

$a\hat{i}$

So, Equation of plane in normal form is

$(\vec{r}).(\dfrac{bc\hat{i}+ac\hat{j}+ab\hat{k}}{\sqrt{(ab)^2+(bc)^2+(ca)^2}})=\dfrac{abc}{\sqrt{(ab)^2+(bc)^2+(ca)^2}}$

Using Normal form of plane, we can claim that,

$p=\dfrac{abc}{\sqrt{(ab)^2+(bc)^2+(ca)^2}}$

$\rightarrow \dfrac{1}{p^2}=\dfrac{(ab)^2+(bc)^2+(ca)^2}{a^2b^2c^2}=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}$

Find the vector and cartesian equation of the plane which passes through the point $(5, 2, -4)$ and perpendicular to the line with direction ratios $2, 3, -1$

Given:Plane passes through

$\left(5,2,-4\right)$

So,

$\vec{a}=5\hat{i}+2\hat{j}-4\hat{k}$

Direction ratios of line perpendicular to the plane

$=2,3,-1$

So,

$\vec{n}=2\hat{i}+3\hat{j}-\hat{k}$

Equation of the plane in vector form is

$\left(\vec{r}-\vec{a}\right).\vec{n}=0$

$\Rightarrow\,\left(\vec{r}-\left(5\hat{i}+2\hat{j}-4\hat{k}\right)\right).\left(2\hat{i}+3\hat{j}-\hat{k}\right)=0$ is the vectorial form.

$\Rightarrow\,\left(\left(x\hat{i}+y\hat{j}+z\hat{k}\right)-\left(5\hat{i}+2\hat{j}-4\hat{k}\right)\right).\left(2\hat{i}+3\hat{j}-\hat{k}\right)=0$

$\Rightarrow\,\left(\left(x-5\right)\hat{i}+\left(y-2\right)\hat{j}+\left(z+4\right)\hat{k}\right).\left(2\hat{i}+3\hat{j}-\hat{k}\right)=0$

$\Rightarrow\,2\left(x-5\right)+3\left(y-2\right)-\left(z+4\right)=0$

$\Rightarrow\,2x-10+3y-6-z-4=0$

$\Rightarrow\,2x+3y-z=20$ is the cartesian equation of the plane.

Find the vector equation of the line passing through the point $A(1, 2, -1)$ and parallel to the line $5x - 25 = 14 - 7y = 35z$ .

1501939_1662679_ans_0f2ac2d0a14b4a11acb9e6671f041495.jpg

Find the direction cosines of the line $\dfrac {4 - x}{2} = \dfrac {y}{6} = \dfrac {1 - z}{3}$ . Also, reduce it to vector form.

The given line is-

$\frac {4-x}{2} = \frac {y}{6} = \frac {1-z}{3}$

$\Rightarrow \frac {x-4}{-2} = \frac {y-0}{6} = \frac {z-1}{-3}$

$. . . . . . . . (i)$

Thus, direction ratios of the given line is-

$-2,6,-3$

Hence, the direction cosines are-

$l=\pm \frac {2}{\sqrt{(-2)^2 + (6)^2 + (-3)^2}}=\pm \frac {2}{7}$

$m=\pm \frac {6}{\sqrt{(-2)^2 + (6)^2 + (-3)^2}}=\pm \frac {6}{7}$

$n=\pm \frac {3}{\sqrt{(-2)^2 + (6)^2 + (-3)^2}}=\pm \frac {3}{7}$

From equation

$(i)$ we see that the given line passes through the point

$(4,0,1)$ and has direction ratios

$-2,6,-3$ .

Let,

$\frac {x-4}{-2} = \frac {y-0}{6} = \frac {z-1}{-3}=\lambda$ , where

$\lambda$ is some constant.

$\Rightarrow \frac{x-4}{-2}=\lambda ; \frac{y-0}{6}=\lambda ; \frac{z-1}{-3}=\lambda$

This gives us the parametric form of the line which is-

$x=-2\lambda + 4$

$y=6\lambda$

$z=-3\lambda +1$

Hence, the vector equation of the line is-

$\overrightarrow{r} = x\hat{i} + y\hat{j} + z\hat{k}$

$\Rightarrow \overrightarrow{r} = (4-2\lambda)\hat{i} + (6\lambda)\hat{j} + (1-3\lambda)\hat{k}$

The vector form of the line is thus given by-

$\overrightarrow{r} = \overrightarrow{a} + \lambda \overrightarrow{b}$ ,

where

$\overrightarrow{a} = 4\hat{i} + 0\hat{j} + \hat{k}$ and

$\overrightarrow{b} = -2\hat{i} +6\hat {j} -3\hat{k}$

$\vec {r} = (4\hat {i} + 0\hat {j} + \hat {k}) + \lambda (-2\hat {i} + 6\hat {j} - 3\hat {k})$ .

Reduce the equation $2x-3y-6z=14$ to the normal form and hence find the length of perpendicular from the origin to the plane. Also, find the direction cosines of the normal to the plane.

The given equation of the plane is

$2x-3y-6=14$ .........

$(1)$

Now,

$\sqrt{{2}^{2}+{\left(-3\right)}^{2}+{\left(-6\right)}^{2}}=\sqrt{4+9+36}=\sqrt{49}=7$

Dividing eqn

$(1)$ by

$7$ , we get

$\dfrac{2}{7}x-\dfrac{3}{7}y-\dfrac{6}{7}z=2$ .........

$(2)$

The cartesian equation of the normal form of a plane is

$lx+my+nz=p$ .........

$(3)$

where

$l,m$ and

$n$ are direction cosines of normal to the plane and

$p$ is the length of the perpendicular from the origin to the plane.

Comparing equations

$(1)$ and

$(2)$ , we get

direction cosines :

$l=\dfrac{2}{7},\,m=\dfrac{-3}{7},\,n=\dfrac{-6}{7}$

Length of the perpendicular from the origin to the plane

$=p=2$ from eqn

$(2)$

Using vectors, find the value of $\lambda$ such that the points $(\lambda , -10, 3), (1, -1, 3)$ and $(3,5,3)$ are collinear.

Given

$A\left(\lambda,−10,3\right),B\left(1,−1,3\right),C\left(3,5,3\right)$

Let

$\vec{OA}=\lambda\hat{i}-10\hat{j}+3\hat{k},\,\,\vec{OB}=\hat{i}-\hat{j}+3\hat{k}$ and

$\vec{OC}=3\hat{i}+5\hat{j}+3\hat{k}$

$\vec{AB}=\vec{OB}-\vec{OA}=\hat{i}-\hat{j}+3\hat{k}-\lambda\hat{i}+10\hat{j}-3\hat{k}=\left(1-\lambda\right)\hat{i}+9\hat{j}$

$\vec{BC}=\vec{OC}-\vec{OB}=3\hat{i}+5\hat{j}+3\hat{k}-\hat{i}+\hat{j}-3\hat{k}=2\hat{i}+6\hat{j}$

Since the points

$A,B$ and

$C$ are collinear

$\vec{AB}=\mu\vec{BC}$

$\Rightarrow\,\hat{i}\left(1-\lambda\right)+9\hat{j}=\mu\left(2\hat{i}+6\hat{j}\right)$

$\Rightarrow\,\hat{i}\left(1-\lambda\right)+9\hat{j}-\mu\left(2\hat{i}+6\hat{j}\right)=0$

But

$\hat{i}$ and

$\hat{j}$ are non-colilinear hence

$1-\lambda-2\mu=0$ and

$9-6\mu=0$

$\Rightarrow\,\mu=\dfrac{9}{6}=\dfrac{3}{2}$

Consider

$1-\lambda-2\mu=0$ we get

$1-\lambda-2\times \dfrac{3}{2}=0$

$\Rightarrow\,1-\lambda-3=0$

$\Rightarrow\,-\lambda-2=0$

$\therefore\,\lambda=-2$

Find the vector equation of the line passing through the point $(2, -1, -1)$ which is parallel to the line $6x - 2 = 3y + 1 = 2z - 2$ .

1501941_1663061_ans_918709d5e6dd4d1db1c4ee264eaca215.jpg

If the equations of a line $AB$ are $\dfrac {3 - x}{1} = \dfrac {y + 2}{-2} = \dfrac {z - 5}{4}$ , write the direction ratios of a line parallel to $AB$ .

$\dfrac{3-x}{1}=\dfrac{y+2}{-2}=\dfrac{z-5}{4}$

$\dfrac{-(x-3)}{1}=\dfrac{y+2}{-2}=\dfrac{z-5}{4}$

$\dfrac{(x-3)}{-1}=\dfrac{y+2}{-2}=\dfrac{z-5}{4}$

$(-1,-2,4)$ are the dr's of line parallel to Ars.

Cartesian equations of a line $AB$ are $\dfrac {2x - 1}{2} = \dfrac {4 - y}{7} = \dfrac {z + 1}{2}$ . Write the direction cosines of a line parallel to $AB$ .

1637295_1663256_ans_e7e6638f1a0f45de8f7f7b174a2ca4ed.jpg

Write the direction cosines of the line whose cartesian equations are $2x = 3y = -z$ .

$\dfrac{2x}{6}=\dfrac{3y}{6}=\dfrac{-z}{6}$

$\dfrac{x}{3}=\dfrac{y}{2}=\dfrac{-z}{6}$

$dr's\,(3,2,-6)$

$dr's\,\left(\dfrac{3}{\sqrt{3^2+2^2+6^2}},\dfrac{+2}{\sqrt{3^2+2^2+6^2}},\dfrac{-6}{\sqrt{3^2+2^2+6^2}}\right)$

$\left(\dfrac{3}{\sqrt{49}},\dfrac{2}{\sqrt{49}},\dfrac{-6}{\sqrt{49}}\right)$

$\left(\dfrac{3}{7},\dfrac{2}{7},\dfrac{-6}{7}\right)$

Write the angle between the lines $2x = 3y = -z$ and $6x = -y = -4z$ .

1503475_1663278_ans_8e911a906b624f47986ae67c669dd45a.jpg

Write the direction cosines of the line $\dfrac {x - 2}{2} = \dfrac {2y - 5}{-3}, z = 2$ .

1664804_1663261_ans_6da219250ca94ce293fb8e1d77e32d18.jpeg

Find the direction cosines of the line $\dfrac {x + 2}{2} = \dfrac {2y - 7}{6} = \dfrac {5 - z}{6}$ . Also, find the vector equation of the line through the point $A(-1, 2, 3)$ and parallel to the given line.

1501947_1663073_ans_e1058f77d63142bfa5ed537a2832269e.jpg

Write the direction cosines of the line, whose cartesian equations are $6x - 2 = 3y + 1 = 2z - 4$ .

$6x-2=3y+1=2z-4$

$6(x-\dfrac{2}{6})=3(y+\dfrac{1}{3})=2(z-2)$

$\dfrac{6\left(x-\dfrac{2}{6}\right)}{6}=\dfrac{3(y+\dfrac{1}{3})}{6}=\dfrac{2(z-2)}{6}$

$\dfrac{x-\dfrac{2}{6}}{1}=\dfrac{y+\dfrac{1}{3}}{2}=\dfrac{z-2}{3}$

$dr's\,\,(1,2,3)$

$dr's=\left(\dfrac{1}{\sqrt{1^2+2^2+3^2}},\dfrac{2}{\sqrt{1^2+2^2+3^2}},\dfrac{3}{\sqrt{1^2+2^2+3^3}}\right)$

$=\left(\dfrac{1}{\sqrt{14}},\dfrac{2}{\sqrt{14}},\dfrac{3}{\sqrt{14}}\right)$

Write the angle between the lines $\dfrac {x - 5}{7} = \dfrac {y + 2}{-5} = \dfrac {z - 2}{1}$ and $\dfrac {x - 1}{1} = \dfrac {y}{2} = \dfrac {z - 1}{3}$ .

1637299_1663272_ans_298ddaf9a42d4865b91d70050494b2e4.jpg

Find the value of $p$ for which the points $A(-1,3,2),B(-4,2,-2)$ and $C(5,5,p)$ are collinear.

Given points are

$A(-1,3,2),B(-4,2,-2)$ and

$C(5,5,p)$

Direction ratios of

$AB$ are

$(-4+1),(2-3),(-2-2)$

ie.,

$-3,-3,-4$ ie.,

$3,1,4$

Direction ratios of

$AC$ are

$(5+1),(5-3),(p-2)$

ie.,

$6,2,p-2$

Since

$A,B,C$ are collinear we must have

$AB\parallel AC$ ,

and Direction ratios are proportional.

$\therefore$

$\cfrac{3}{6}=\cfrac{1}{2}=\cfrac{4}{p-2}$

$\Rightarrow$

$p-2=8\Rightarrow 10$

$p=10$

Find the direction ratios and the direction cosines of the line segment joining the points:
(i) $A(1,0,0)$ and $B(0,1,1)$
(ii) $A(5,6,-3)$ and $B(1,-6,3)$
(iii) $A(-5,7,-9)$ and $B(-3,4,-6)$

1546248_1707529_ans_574cd0c5af84404a8edb487df0a3bcea.png

A passes through the point $(3,4,5)$ and is parallel to the vector $\left( 2\hat { i } +2\hat { j } -3\hat { k } \right)$ . Find the equations of the line in the vector as well as Cartesian forms.

1551459_1707637_ans_018277415b4b4707ace06bc8e3be1f02.jpg

The equations of a line are given by $\dfrac {4 - x}{3} = \dfrac {y + 3}{3} = \dfrac {z + 2}{6}$ . Write the direction cosines of a line parallel to this line.

$\dfrac{4-x}{3}=\dfrac{y+3}{3}=\dfrac{z+2}{6}$

$\dfrac{-(x-4)}{3}=\dfrac{y+3}{3}=\dfrac{z+2}{6}$

$\dfrac{(x-4)}{-3}=\dfrac{y+3}{3}=\dfrac{z+2}{6}$

$dr's\,are\,(-3,3,6)$

$dr's\,\left(\dfrac{-3}{\sqrt{3^2+3^2+6^2}},\dfrac{3}{\sqrt{3^2+3^2+6^2}},\dfrac{3}{\sqrt{3^2+3^2+6^2}}\right)$

$dr's\,\left(\dfrac{-3}{\sqrt{54}},\dfrac{3}{\sqrt{54}},\dfrac{6}{\sqrt{54}}\right)$ parallel to given line.

Write the coordinates of the foot of the perpendicular from the point $(1,2,3)$ on y-axis. Mark 1 if the answer is $(0,2,0)$ otherwise mark 0.

1414424_1667953_ans_47f649bc2e5b415fb455f6b0dea0dde0.jpg

Show that the points $A(2,3,-4),B(1,-2,3)$ and $C(3,8,-11)$ are collinear.

1551970_1707674_ans_bae122e2280240f4b5805e844562098e.jpg

Find the vector equation of a line passing through the point $A(3,-2,1)$ and parallel to the line joining the points $B(-2,4,2)$ and $C(2,3,3)$ . Also, find the Cartesian equations of the line.

1551749_1707662_ans_fcd89f0758eb4b889383409f21894f0d.jpg

Find the value of $\lambda$ for which the points $A(2,5,1),B(1,2,-1)$ and $C(3,\lambda,3)$ are collinear.

Given

$A(2,5,1),B(1,2,-1)$ and

$C(3,\lambda,3)$ are collinear

As we know that

$(x_1,y_1,z_1),(x_2,y_2,z_2)$ and

$(x_3,y_3,z_3)$ are collinear then

$\begin{vmatrix}x_1&y_1&z_1\\x_2&y_2&z_2\\x_3&y_3&z_3\end{vmatrix}=0$

$\implies \begin{vmatrix} 2 & 5 & 1 \\ 1 & 2 & -1 \\ 3 &\lambda& 3 \end{vmatrix}=0$

$\implies 2(2\times 3-(-1)(\lambda))-5((1)(3)-(-1)(3))+1((1)(\lambda)-(2)(3))=0$

$\implies 2(6+\lambda)-5(3+3)+(\lambda-6)=0$

$\implies 12+2\lambda-15-15+\lambda-6=0$

$\implies 3\lambda-24=0$

$\implies \lambda=\dfrac{24}{3}$

$\implies \lambda=8$

Show that the points $A(2,1,3), B(5,0,5)$ and $C(-4,3,-1)$ are collinear.

1551988_1707672_ans_3e7bd58b6996451289426a4116a4653d.jpg

Find the Cartesian equations of the line which passes through the point $(1,3,-2)$ and is parallel to the line given by
$\cfrac { x+1 }{ 3 } =\cfrac { y-4 }{ 5 } =\cfrac { z+3 }{ -6 }$ .
Also, find the vector form of the equations so obtained.

1551527_1707644_ans_ac29031563fd4c799b139697804a35c6.jpg

Find the equations of the line passing through the point $(1,-2,3)$ and parallel to the line $\cfrac { x-6 }{ 3 } =\cfrac { y-2 }{ -4 } =\cfrac { z+7 }{ 5 }$ .
Also find the vector form of this equation so obtained.

1551529_1707646_ans_aa08b5458979430eb93f7c6b06034909.jpg

Find the vector equation of a line passing through the point having the position vector $\left( \hat { i } +2\hat { j } -3\hat { k } \right)$ and parallel to the line joining the points with position vecgtors $\left( \hat { i } -\hat { j } +5\hat { k } \right)$ and $\left( 2\hat { i } +3\hat { j } -4\hat { k } \right)$ . Also, find the Cartesian equivalents of this equation.

1551817_1707663_ans_7ec38f6bf06a4ae6a3f8f750894429d3.jpg

Find the vector equation of the line passing through the point with position vector $\left( 2\hat { i } +\hat { j } -5\hat { k } \right)$ and parallel to the vector $\left( \hat { i } +3\hat { j } -\hat { k } \right)$ . Deduce the Cartesian equations of the line.

1551463_1707639_ans_0b9335562f844a27b892cd0f9d4396c4.jpg

Find the equations of the line passing through the point $(1,-2,3)$ and parallel to the line $\cfrac { x-3 }{ 2 } =\cfrac { y+2 }{ -5 } =\cfrac { z-6 }{ 4 }$
Also find the vector form of this equation so obtained

1551466_1707642_ans_c03c410652b94738983005a9e5c86f3d.jpg

Find the Cartesian and vector equations of a line which passes through the point $(1,2,3)$ and is parallel to the line.
$\cfrac { -x-2 }{ 1 } =\cfrac { y+3 }{ 7 } =\cfrac { 2z-6 }{ 3 }$

1551656_1707647_ans_656439d0300a4477b69c2a86831c8c7c.jpg

Find the angle between each of the following pairs of lines:
$\vec { r } =\left(3 \hat { i } +\hat { j } -2\hat { k } \right) +\lambda \left( \hat { i } -\hat { j } -2\hat { k } \right)$ and $\vec { r } =\left( 2\hat { i } -\hat { j } -5\hat { k } \right) +\mu \left( 3\hat { i } -5\hat { j } -4\hat { k } \right)$

1551366_1707733_ans_b03fd574c73148adaf34cd5370a271c5.jpg

Find the angle between each of the following pairs of lines:
$\vec { r } =\left( 3\hat { i } -4\hat { j } +2\hat { k } \right) +\mu \left( \hat { i } +3\hat { k } \right)$ and $\vec { r } =\left( 5\hat { i } \right) +\mu \left( -\hat { i } +\hat { j } +\hat { k } \right)$

1551374_1707735_ans_3bf8aceb8d48480187c7f7c253c38ed5.jpg

Find the angle between each of the following pairs of lines:
$\vec { r } =\left( \hat { i } -2\hat { j } \right) +\mu \left(2 \hat { i } -2\hat { j } +\hat { k } \right)$ and $\vec { r } =\left( 3\hat { k } \right) +\mu \left( \hat { i } +2\hat { j } -2\hat { k } \right)$

1551375_1707737_ans_f67aa5c39ca24947b840275e75564e3a.jpg

Find the values of $\lambda$ and $\mu$ so that the points $A(3,2,-4),B(9,8,-10)$ and $C(\lambda, \mu, -6)$ are collinear.

1551954_1707677_ans_7d66657854de4d0086744c9f5d822665.jpg

Find the values of $\lambda$ and $\mu$ so that the points $A(-1,4,-2), B(\lambda, \mu, 1)$ and $C(0,2,-1)$ are collinear.

1551372_1707679_ans_8e8c59373f3543669786c49afe4cfc44.jpg

Find the angle between each of the following pairs of lines:
$\cfrac{x+1}{1}=\cfrac{y-4}{1}=\cfrac{z-5}{2}$ and $\cfrac{x+3}{3}=\cfrac{y-2}{5}=\cfrac{z+5}{4}$

1551362_1707739_ans_8e240e0ff8404fed99ef08abe1bec0f7.jpg

Find the angle between each of the following pairs of lines:
$\cfrac{5-x}{3}=\cfrac{y+3}{-2},z=5$ and $\cfrac{x-1}{1}=\cfrac{1-y}{3}=\cfrac{z-5}{2}$

1551365_1707749_ans_657ea43ad8cf427190e67c97b2dd946c.jpg

Find the angle between each of the following pairs of lines:
$\cfrac{3-x}{-2}=\cfrac{y+5}{1}=\cfrac{1-z}{3}$ and $\cfrac{x}{3}=\cfrac{1-y}{-2}=\cfrac{x+2}{-1}$

1551364_1707743_ans_e1cc95f2728440ef8833b45384f469aa.jpg

Find the angle between each of the following pairs of lines:
$\cfrac{x}{1}=\cfrac{z}{-1},y=0$ and $\cfrac{x}{3}=\cfrac{y}{4}=\cfrac{z}{5}$

1551370_1707747_ans_6bb3b8aaa2744adb8116e999f45793a2.jpg

The position vectors of three points $A,B,C$ are $\left(-4 \hat { i } +2\hat { j } -3\hat { k } \right) ,\left( \hat { i } +3\hat { j } -2\hat { k } \right)$ and $\left(-9 \hat { i } +\hat { j } -4\hat { k } \right)$ . respectively. Show that the points $A,B$ and $C$ are collinear.

1551371_1707678_ans_55d33a05ad784f50888e6c8be1c45fbb.jpg

If $P(1,5,4)$ and $Q(4,1,-2)$ be two given points, find the direction ratios of $PQ$ .

Given points are

$P(1,5,4)$ and

$Q(4,1,-2)$

Direction ratios of

$PQ$ are

$(4-1),(1-5),(-2,-4)$

ie.,

$3,-4,-6$

The direction ratios of a line are $2,6,-9$ . What are its direction cosines?

We know that if

$a, b$ and

$c$ are the direction ratios, then the direction cosines

$l, m$ and

$n$ are given by

Direction cosines are

$l = \dfrac a{\sqrt{a2+b2+c2}}, m = \dfrac b{\sqrt{a2+b2+c2}}, n = \dfrac c{\sqrt{a2+b2+c2}},$

The direction ratios of a line are

$2,6,-9$ .

We have

$\sqrt { { (2) }^{ 2 }+{ (6) }^{ 2 }+{ (-9) }^{ 2 } } =\sqrt { 121 } =11$

Direction cosines of the given line are

$\cfrac{2}{11},\cfrac{6}{11},\cfrac{-9}{11}$

The Cartesian equations of a line are $\cfrac{x-1}{2}=\cfrac{y+2}{3}=\cfrac{5-z}{1}$ . Find its vetor equation.

The given equations are

$\cfrac{x-1}{2}=\cfrac{y+2}{3}=\cfrac{5-z}{1}$

$\cfrac{x-1}{2}=\cfrac{y+2}{3}=\cfrac{z-5}{-1}$

Clearly, this line passes through the point

$(1,-2,5)$ and it has d.r's

$2,3,-1$

So, its vector equation is

$\vec { r } =\left( \hat { i } -2\hat { j } +5\hat { k } \right) +\lambda \left( 2\hat { i } +3\hat { j } -\hat { k } \right)$

A line makes angles ${90}^{o},{135}^{o}$ and ${45}^{o}$ with positive directions of x-axis, y-axis and z-axis respectively. What are the direction cosines of the line?

$α, β,$ and

$γ$ are the angles made by the line segment with the coordinate axis.

$cos α, cos β$ and

$cos γ$ are called as the direction cosines and are usually denoted by

$l, m$ and

$n.$

$l = cos α, m = cos β$ and

$n = cos γ$

Direction cosines of the line are

$\cos { { 90 }^{ o } } ,\cos { { 135 }^{ o } } ,\cos { { 45 }^{ o } }$

i.e.

$\left\{ 0,\cfrac { -1 }{ \sqrt { 2 } } ,\cfrac { 1 }{ \sqrt { 2 } } \right\}$

The equations of a line are $\cfrac{4-x}{2}=\cfrac{y+3}{2}=\cfrac{z+2}{1}$ . Find the direction cosines of a line parallel to this line.

The given equations is

$\cfrac{4-x}{2}=\cfrac{y+3}{2}=\cfrac{z+2}{1}$

$\cfrac{x-4}{-2}=\cfrac{y+3}{2}=\cfrac{z+2}{1}$

The direction ratios of this line are

$-2,2,1$

The direction ratios of a line parallel to this line are

$-2,2,1$

and

$\sqrt { { (-2) }^{ 2 }+(2)^{ 2 }+{ (1) }^{ 2 } } =\sqrt { 9 } =3\quad$

The direction cosines of a line parallel to the given line are

$\cfrac{-2}{3},\cfrac{2}{3},\cfrac{1}{3}$ .

Write the vector equation of a line passing through the point $(1,-1,2)$ and parallel to the line whose equations are $\cfrac{x-3}{1}=\cfrac{y-1}{2}=\cfrac{z+1}{-2}$ .

1567268_1707835_ans_ceb757d7a5ce48c187ba19176fac88b2.jpg

Write the equations of line parallel to the line $\cfrac{x-2}{-3}=\cfrac{y+3}{2}=\cfrac{z+5}{6}$ and passing through the point $(1,-2,3)$ .

1567264_1707831_ans_2aecb02439e846a7b803c7c0b614096d.jpg

If the equations of a line are $\cfrac{3-x}{-3}=\cfrac{y+2}{-2}=\cfrac{z+2}{6}$ , find the direction cosines of a line parallel to the given line.

The given equation is

$\cfrac{3-x}{-3}=\cfrac{y+2}{-2}=\cfrac{z+2}{6}$

Direction ratios of this line are

$3,-2,6$

Direction ratios of a line parallel to this line are

$3,-2,6$

and

$\sqrt { { 3 }^{ 2 }+(-2)^{ 2 }+{ (6) }^{ 2 } } =\sqrt { 49 } =7$

Direction cosines of a line parallel to the given line are

$\cfrac{3}{7},\cfrac{-2}{7},\cfrac{6}{7}$

If $A(1,2,3),B(4,5,7),C(-4,3,-6)$ and $D(2,9,2)$ are four given points then find the angle between the lines $AB$ and $CD$ .

1551633_1707758_ans_ad4500ade40b4586a3c09d9d30fb8f08.jpg

Find the vector equation of a line passing through the point $(1,2,3)$ and parallel to the vector $\left( 3\hat { i } +2\hat { j } -2\hat { k } \right)$

1567272_1707841_ans_2df0107587c14dd6ac74a5aa768d3835.jpg

What are the direction cosines of the y-axis?

$α, β,$ and

$γ$ are the angles made by the line segment with the coordinate axis.

$cos α, cos β$ and

$cos γ$ are called as the direction cosines and are usually denoted by

$l, m$ and

$n.$

$l = cos α, m = cos β$ and

$n = cos γ$

Clearly the y-axis makes an angle of

${ 90 }^{ o },{ 0 }^{ o },{ 90 }^{ o }$ ie., with the x-axis, y-axis, z-axis respectively

so its direction cosines are

$\cos { { 90 }^{ o } } ,\cos { { 0 }^{ o } } ,\cos { { 90 }^{ o } }$ ie.,

$0,1,0$

Find the Cartesian and vector equations of a plane passing through the point $(1, 2, 3)$ and perpendicular to the line with direction ratios $2, 3, -4$ .

1543419_1707949_ans_258c6ce757444491a004d0391ffebdf3.png

Find the equation of the plane passing through group of points.
$A(2, 2, -1), B(3, 4, 2)$ and $C(7, 0, 6)$ .

1542528_1707916_ans_f509627f5fb64f9a9c048fa8b90e777c.jpg

If O be the origin and $P(1, 2, -3)$ be a given point, then find the equation of the plane passing through P and perpendicular to OP.

1543420_1707950_ans_c510f8408b804c19918ac6747edd8bcc.jpg

Find the equation of the plane passing through group of points.
$A(-2, 6, -6), B(-3, 10, -9)$ and $C(-5, 0, -6)$ .

1542530_1707935_ans_5df0443b0d4f4df1a0fd811ac89d9c4a.jpg

Find the equation of the plane passing through group of points.
$A(0, -1, -1), B(4, 5, 1)$ and $C(3, 9, 4)$ .

1542529_1707928_ans_d453a7d5581c4ef193ec6af3ae5d07c7.jpg

For the following planes, find the direction cosines of the normal to the plane and the distance of the plane from the origin.
$z=3$ .

1543432_1708483_ans_63e1be36ff714f0caa76f9d28261ed14.png

Find the vector and Cartesian equations of a plane which is at a distance of $\dfrac{6}{\sqrt{29}}$ from the origin and whose normal vector from the origin is $(2\hat{i}-3\hat{j}+4\hat{k})$ .

1543423_1708472_ans_1e3f3b399dc0455f922bd096a187b07d.png

Find the vector and Cartesian equations of a plane which is at a distance of $5$ units from the origin and which has $3\hat{i}-2\hat{j}+6\hat{k}$ as the unit vector normal to it.

2016116_1708469_ans_c26f582968644bef9dfbba3b416ceb67.png

Find the vector and Cartesian equations of a plane which is at a distance of $7$ units from the origin and whose normal vector from the origin is $(3\hat{i}+5\hat{j}-6\hat{k})$ .

1543422_1708471_ans_abf2248be04c41d09bddfa7197a88716.png

Find the vector equation of a plane whose Cartesian equation is $5x-7y+2z+4=0$ .

1543428_1708478_ans_14a6a1aadbf04244b07fe584104349a9.jpg

Find the direction cosines of the normal to the plane $3x-6y+2z=7$ .

Given equation in vector form is

$\vec{r}\cdot (3\hat{i}-6\hat{j}+2\hat{k})=7$

$|3\hat{i}-6\hat{j}+2\hat{k}|=\sqrt{3^2+(-6)^2+2^2}=\sqrt{49}=7$

$\therefore \vec{r}\cdot \left(\dfrac{3}{7}\hat{i}-\dfrac{6}{7}\hat{j}+\dfrac{2}{7}\hat{k}\right)=1$ .

$\therefore$ direction cosines of the normal to the plane are

$\dfrac{3}{7}, \dfrac{-6}{7}, \dfrac{2}{7}$ .

Find a unit vector normal to the plane $x-2y+2z=6$ .

Given equation in vector form is

$\vec{r}\cdot (\hat{i}-2\hat{j}+2\hat{k})=6$

Here,

$|\hat{i}-2\hat{j}+2\hat{k}|=\sqrt{1^2+(-2)^2+2^2}=\sqrt{9}=3$

$\vec{r}\cdot\left(\dfrac{1}{3}\hat{i}-\dfrac{2}{3}\hat{j}+\dfrac{2}{3}\hat{k}\right)=\dfrac{6}{3}=2$

Required unit vector

$=\left(\dfrac{1}{3}\hat{i}-\dfrac{2}{3}\hat{j}+\dfrac{2}{3}\hat{k}\right)$ which is normal to the given plane.

Find the vector and cartesian equations of the planes that passes through the point $(1, 4, 6)$ and the normal vector to the plane is $\hat {i} - 2\hat {j} + \hat {k}$ .

1543425_1708474_ans_970abcb7659349d7adfd351df1076a2f.png

For the following planes, find the direction cosines of the normal to the plane and the distance of the plane from the origin.
$2x+3y-z=5$ .

Given equation in vector form is

$\vec{r}\cdot (2\hat{i}+3\hat{j}-\hat{k})=5$

$|2\hat{i}+3\hat{j}-\hat{k}|=\sqrt{2^2+3^2+(-1)^2}=\sqrt{14}$

$\therefore\vec{r}\cdot\left(\dfrac{2}{\sqrt{14}}\hat{i}+\dfrac{3}{\sqrt{14}}\hat{j}-\dfrac{1}{\sqrt{14}}\hat{k}\right)=\dfrac{5}{\sqrt{14}}$

Direction cosines of

$\hat{n}$ (normal) are

$\dfrac{2}{\sqrt{14}}, \dfrac{3}{\sqrt{14}}, \dfrac{-1}{\sqrt{14}}$ and

$p=\dfrac{5}{\sqrt{14}}$ .

Find the equation of a plane passing through the points $A(a, 0, 0), B(0, b, 0)$ and $C(0, 0, c)$ .

1546885_1708850_ans_4c791115523942f7a29c13471625b841.png

Find the direction cosines of the normal to the plane $y=3$ .

1546789_1708831_ans_0869367600b24ec480bb2f5c284fdf2a.jpg

Find the direction ratios of the normal to the plane $x+2y-3z=5$ .

1546707_1708828_ans_3ce5c52038d0499fa62065f9ef073e6a.png

Find the direction cosines of the perpendicular from the origin to the plane $\vec{r}\cdot(6\hat{i}-3\hat{j}-2\hat{k})+1=0$ .

1546992_1708880_ans_2a4c41dbd4614a82af1483fb575111df.jpg

For the following planes, find the direction cosines of the normal to the plane and the distance of the plane from the origin.
$3y+5=0$ .

Given equation in vector form is

$\vec{r}\cdot (0\hat{i}-1\cdot \hat{j}+0\cdot\hat{k})=\dfrac{5}{3}$

$\left[\because -y=\dfrac{5}{3}\right]$ .

Clearly,

$|0\hat{i}-1\cdot \hat{j}+0\cdot \hat{k}|=\sqrt{0^2+(-1)^2+0^2}=1$

$\therefore \vec{r}\cdot \hat{n}=p$ , where

$\hat{n}=(0\hat{i}-1\cdot \hat{j}+0\hat{k})$ and

$p=\dfrac{5}{3}$ .

Direction cosines of

$\hat{n}$ are

$0, -1, 0$ and

$p=\dfrac{5}{3}$ .

Find the direction cosines of the normal to the plane $3x+4=0$ .

Given plane is

$3x+4=0\Rightarrow -x=\dfrac{4}{3}$ .

Direction ratios of the normal to this plane are

$-1, 0, 0$

and

$\sqrt{(-1)^2+0^2+0^2}=1$ .

We know that if

$a, b$ and

$c$ are the direction ratios, then the direction cosines

$l, m$ and

$n$ are given by

Direction cosines are

$l = \dfrac a{\sqrt{a^2+b^2+c^2}}, m = \dfrac b{\sqrt{a^2+b^2+c^2}}, n = \dfrac c{\sqrt{a^2+b^2+c^2}},$

Hence, the required direction cosines are

$-1, 0, 0$ .

A vector $\vec{n}$ of magnitude $8$ units is inclined to the x-axis is $45^o$ , y-axis at $60^o$ and an acute angle with the z-axis. If a plane passes through a point $(\sqrt{2}, -1, 1)$ and is normal to $\vec{n}$ , find its equation in vector form.

$α, β,$ and

$γ$ are the angles made by the line segment with the coordinate axis.

$cos α, cos β$ and

$cos γ$ are called as the direction cosines and are usually denoted by

$l, m$ and

$n.$

$l = cos α, m = cos β$ and

$n = cos γ$

Here

$l=\cos 45^o=\dfrac{1}{\sqrt{2}}$ ,

$m=\cos 60^o=\dfrac{1}{2}$ and let

$n=\cos\gamma$ . Then,

$l^2+m^2+n^2=1\\$

$\Rightarrow \dfrac{1}{2}+\dfrac{1}{4}+\cos^2\gamma =1$

$\Rightarrow \cos^2\gamma =\dfrac{1}{4}$

$\Rightarrow \cos\gamma =\dfrac{1}{2}$

We know that

$\hat{n}=\dfrac{\vec{n}}{|\vec{n}|}=(l\hat{i}+m\hat{j}+n\hat{k})$ .

$\therefore \vec{n}=|\vec{n}|(l\hat{i}+m\hat{j}+n\hat{k})\Rightarrow \vec{n}=8\left(\dfrac{1}{\sqrt{2}}\hat{i}+\dfrac{1}{2}\hat{j}+\dfrac{1}{2}\hat{k}\right)=(4\sqrt{2}\hat{i}+4\hat{j}+4\hat{k})$ .

So, the equation of the plane is

$\vec{r}\cdot \vec{n}=\vec{a}\cdot \vec{n}\Rightarrow \vec{r}\cdot (4\sqrt{2}\hat{i}+4\hat{j}+4\hat{k})=(\sqrt{2}\hat{i}-\hat{j}+\hat{k})\cdot (4\sqrt{2}\hat{i}+4\hat{j}+4\hat{k})\\$

$\Rightarrow \vec{r}\cdot (4\sqrt{2}\hat{i}+4\hat{j}+4\hat{k})=(8-4+4)=8\\$

$\Rightarrow \vec{r}\cdot (\sqrt{2}\hat{i}+\hat{j}+\hat{k})=2$ .

Prove that the normals to the planes $4x+11y+2z+3=0$ and $3x-2y+5z=8$ are perpendicular to each other.

Given planes are

$4x+11y+2z+3=0$ and

$3x-2y+5z=8$
Direction ratios of normals are <4,11,2> and <3,-2,5>
Finding the dot product of DRs of two normals = 4(3)+11(-2)+2(5) = 12-22+10 =0
Since the dot product is zero, the two normals are perpendicular to each other.

Find the direction cosines of the normal to the plane $2x+3y-z=4$ .

The given plane is

$2x+3y-z=4$ .

Direction ratios of the normal to the given plane are

$a=2, b=3, c=-1$

and

$\sqrt{2^2+3^2+(-1)^2}=\sqrt{14}$ .

We know that if

$a, b$ and

$c$ are the direction ratios, then the direction cosines

$l, m$ and

$n$ are given by

Direction cosines are

$l = \dfrac a{\sqrt{a^2+b^2+c^2}}, m = \dfrac b{\sqrt{a^2+b^2+c^2}}, n = \dfrac c{\sqrt{a^2+b^2+c^2}},$

Hence, the required direction cosines are

$\dfrac{2}{\sqrt{14}}, \dfrac{3}{\sqrt{14}}, \dfrac{-1}{\sqrt{14}}$ .

Show that points $A(a, b+c), B(b, c+a)$ and $C(c, a+b)$ are collinear.

${\textbf{Step -1: Placing the values using formula of triangle}}{\textbf{.}}$

${\text{For collinear points, area of triangle = 0}}{\text{.}}$

${\text{Area of triangle = }}\dfrac{1}{2}\left| {\begin{array}{*{20}{c}}{{x_1}}&{{y_1}}&1 \\ {{x_2}}&{{y_2}}&1 \\ {{x_3}}&{{y_3}}&1 \end{array}} \right|$

${\text{Here, }}{x_1} = a,$

${y_1} = b + c$

${x_2} = b,{\text{ }}{y_2} = c + a$

${x_3} = c,{\text{ }}{y_3} = a + b$

${\text{Putting the values,}}$

$\Delta = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}} a&{b + c}&1 \\ b&{c + a}&1 \\ c&{a + b}&1 \end{array}} \right|$

${\textbf{Step -2: Applying }}$

${\mathbf{{C_1} \to {C_1} + {C_2}}}$

${\textbf{ to solve further}}{\text{.}}$

$\Delta = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}{a + b + c}&{b + c}&1 \\ {b + c + a}&{c + a}&1 \\ {c + a + b}&{a + b}&1 \end{array}} \right|$

${\text{Taking }}\left( {a + b + c} \right){\text{ common from }}{C_1}{\text{. }}$

$\Delta = \dfrac{1}{2}\left( {a + b + c} \right)\left| {\begin{array}{*{20}{c}}1&{b + c}&1 \\ 1&{c + a}&1 \\ 1&{a + b}&1 \end{array}} \right|$

${\text{As, }}{1^{st}}{\text{ and }}{{\text{3}}^{rd}}{\text{ column are identical value of determinant will be zero}}{\text{.}}$

$\Delta = \dfrac{1}{2}\left( {a + b + c} \right) \times 0$

$\Delta = 0$

${\textbf{Hence, all three points are collinear}}{\textbf{.}}$

Show that the following points are collinear:
$A(3, -5, 1), B(-1, 0, 8)$ and $C(7, -10, -6)$ .

1863066_1711080_ans_65a6a0bbd49d4e61aa10e9b6424e019e.jpg

If the equation of a line AB is $\dfrac{x - 3}{1} = \dfrac{y + 2}{-2} = \dfrac{z - 5}{4}$ . Then find the direction ratios of line parallel to AB.

The direction ratios of line parallel to AB is

$1, -2$ and

$4$ .

Write the cartesian equation of a plane, bisecting the line segment joining the points $A (2, 3, 5)$ and $B(4, 5, 7)$ at right angles.

One point on the required plane = mid point of given line segment.

$= \left ( \dfrac{2 + 4}{2}, \dfrac{3 + 5}{2}, \dfrac{5 + 7}{2} \right ) = (3, 4, 6)$

And direction ratios of normal to the plane

$= (4 -2), (5 -3), (7 -5) = (2, 2, 2)$

$\therefore$ Required equation of plane is

$2(x - 3) + 2(y - 4) + 2(z - 6) = 0$

$\Rightarrow2x + 2y + 2z = 26$

$\Rightarrow x + y + z = 13$

Show that the points $A(2, 3, -4), B(1, -2, 3)$ and $C(3, 8, -11)$ are collinear.

Direction ratios of the line joining A and B are

$(1, -2), (-2, -3), (3 + 4) \,i.e., -1, -5, 7$
Direction ratios of the line joining A and C are

$(3, -2), (8, -3), (-11 + 4) \,i.e., 1, 5, -7$
It is clear that direction ratios of line AB and AC are proportional.

$\Rightarrow$ Parallel vectors of both lines are parallel to each other.

$\Rightarrow$ Both given lines are parallel.

$\Rightarrow$ But point A is common. Hence points ABC are collinear.

Find the angle between the lines $2x = 3y = -z$ and $6x = -y = -4z.$

Given lines are

$2x = 3y = -z$ and

$6x = -y = -4z$ which may be written as

$\dfrac{x - 0}{1/2} = \dfrac{y - 0}{1/2} = \dfrac{z - 0}{-1}\qquad...(i)$ and

$\dfrac{x - 0}{1/6} = \dfrac{y - 0}{-1} = \dfrac{z - 0}{-1/4}\qquad...(ii)$

Parallel vectors

$\vec{b_1}$ and

$\vec{b_2}$ of

$(i)$ and

$(ii)$ respectively are

$\vec{b_1} = \dfrac{1}{2}\hat{i} + \dfrac{1}{3}\hat{j} - \hat{k}$ and

$\vec{b_2} = \dfrac{1}{6}\hat{i} - \hat{j} - \dfrac{1}{4}\hat{k}$

$\therefore$ Angle between lines (i) and (ii) = Angle between

$\vec{b_1}$ and

$\vec{b_2}$

$= cos^{-1}\left ( \left | \dfrac{\vec{b_1}.\vec{b_2}}{|\vec{b_1}| |\vec{b_2}|} \right | \right ) = cos^{-1}\left ( \left | \dfrac{\left ( \dfrac{1}{2}\hat{i} + \dfrac{1}{3}\hat{j} - \hat{k} \right ). \left ( \dfrac{1}{6}\hat{i} - \hat{j} - \dfrac{1}{4}\hat{k} \right )}{\sqrt{\left ( \dfrac{1}{2} \right )^2 + \left ( \dfrac{1}{3} \right )^2 + (-1)^2} \sqrt{\left ( \dfrac{1}{6} \right )^2 + (-1)^2 + \left ( -\dfrac{1}{4} \right )^2}} \right | \right )$

$= cos^{-1}\left ( \left | \dfrac{\dfrac{1}{12} - \dfrac{1}{3} + \dfrac{1}{4}}{\sqrt{\dfrac{1}{4} + \dfrac{1}{9} + 1} \sqrt{\dfrac{1}{36} + 1 + \dfrac{1}{16}}} \right | \right ) = cos^{-1}0 = \dfrac{\pi}{2}$

Write the direction ratios of a line parallel to $AB:\dfrac{2x - 1}{2} = \dfrac{4 - y}{7} = \dfrac{x - 1}{2}$

We have equations of line

$\dfrac{2x - 1}{2} = \dfrac{4 - y}{7} = \dfrac{x - 1}{2} \\\Rightarrow \dfrac{x - 1/2}{1} = \dfrac{y - 4}{-7} = \dfrac{x - (-1)}{2}$

Direction ratios of given line are

$1, -7, 2.$

Hence, direction ratios of any line parallel to

$AB$ are

$\lambda, -7\lambda, 2\lambda$ where

$\lambda$ is any real number.

Find the direction cosines of the line passing through two points $(-2, 4, -5)$ and $(1, 2, 3).$

We know that direction cosines of the line passing through two points

$P (x_1, y_1, z_1)$ and

$Q(x_2, y_2, z_2)$ are given by:

$\dfrac{x_2 - x_1}{PQ}, \dfrac{y_2 - y_1}{PQ}, \dfrac{z_2 - z_1}{PQ},$

where

$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

Here P is

$(-2, 4, -5)$ and Q is

$(1, 2, 3).$

$PQ = \sqrt{(1 - (-2))^2 + (2 - 4)^2 + (3 - (-5))^2} = \sqrt{77}$

Thus, the direction cosines of the line joining two points are

$\dfrac{1 + 2}{\sqrt{77}}, \dfrac{2 - 4}{\sqrt{77}}, \dfrac{3 + 5}{\sqrt{77}}$
or

$\dfrac{3}{\sqrt{77}}, \dfrac{-2}{\sqrt{77}}, \dfrac{8}{\sqrt{77}}$

Find the angle between the pair of lines given by
$\vec{r} = 3i + 2j - 4k + \lambda (i + 2j + 2k)$ and $\vec{r} = 5i - 2j + \mu(3 i + 2 j + 6 k)$

Here,

$\vec{b_1} = \hat{i} + 2 \hat{j} + 2\hat{k}$ and

$\vec{b_2} = 3\hat{i} + 2\hat{j} + 6\hat{k}$

The angle between the two lines is given by

$cos\,\theta = \left | \dfrac{\vec{b_1}.\vec{b_2}}{|\vec{b_1}| |\vec{b_2|}} \right | = \left | \dfrac{(\hat{i} + 2\hat{j} + 2\hat{k}) . (3\hat{i} + 2\hat{j} + 6\hat{k})}{\sqrt{1 + 4 + 4}\sqrt{9 + 4 + 36}} \right | = \left | \dfrac{3 + 4 + 12}{3 \times 7} \right | = \dfrac{19}{21}$

Hence,

$\theta = cos^{1}\left ( \dfrac{19}{21} \right )$

If a line makes an angle $30^\circ, 60^\circ, 90^\circ$ with the positive direction of $x, y, z$ axes respectively, then find its direction cosines.

Since, the direction cosine of the line which makes an angle of a, b, g with x, y and z are cos a, cos b and cos g.

$\therefore$ Direction cosines of given line are

$cos \,30^\circ, cos \,60^\circ, cos\,90^\circ$

$i.e.,\ \ \dfrac{\sqrt 3}{2}, \dfrac{1}{2}, 0$

Find the value of p, so that the lines $l_1 = \dfrac{1 - x}{3} = \dfrac{7y - 14}{p} = \dfrac{z - 3}{2}$ and $l_2: \dfrac{7 - 7x}{3p} = \dfrac{y - 5}{1} = \dfrac{6 - z}{5}$ are perpendicular to each other. Also find the equations of a line passing through a point $(3, 2, -4)$ and parallel to line $l_1.$

Given line

$l_1$ and

$l_2$ are

$l_1 \equiv \dfrac{1 - x}{3} = \dfrac{7y - 14}{p} = \dfrac{z - 3}{2}$

$\Rightarrow \dfrac{x - 1}{-3} = \dfrac{y - 2}{\dfrac{p}{7}} = \dfrac{z - 3}{2}$

$l_2 \equiv \dfrac{7 - 7x}{3p} = \dfrac{y - 5}{1} = \dfrac{6 - z}{5}$

$\Rightarrow \dfrac{x - 1}{\dfrac{-3p}{7}} = \dfrac{y - 5}{1} = \dfrac{z - 6}{-5}$
Since

$l_1 \perp l_2$

$\Rightarrow (-3) \left ( -\dfrac{3p}{7} \right ) + \dfrac{p}{7} \times 1 + 2 \times (-5) = 0 \Rightarrow \dfrac{9p}{7} + \dfrac{p}{7} - 10 = 0 \Rightarrow \dfrac{10p}{7} = 10$

$\Rightarrow p = \dfrac{7 \times 10}{10} \Rightarrow p = 7$
The equation of line passing through

$(3, 2, -4)$ and parallel to

$l_1$ is given by

$\dfrac{x - 3}{-3} = \dfrac{y - 2}{\dfrac{p}{7}} = \dfrac{z + 4}{2}$

$i.e., \dfrac{x - 3}{-3} = \dfrac{y - 2}{1} = \dfrac{z + 4}{2} (\because p = 7)$

What are the direction consines of a line , which makes equal angles with the co-ordinates axes?

Let

$\alpha$ be the angle made by line with cordinates axes.

$\Rightarrow$ direction cosines of line are

$cos \alpha , cos \alpha , cos \alpha$

$\Rightarrow cos^2 \alpha + cos^2 \alpha + cos^2 \alpha = 1$

$3 cos^2 \alpha = 1 \Rightarrow cos^2 \alpha = \frac {1}{3}$

$\Rightarrow cos \alpha = \pm \frac {1}{ \sqrt {3} }$
Hence the direction cosines, of the line equally inclined to the coordinate axes are

$\pm \frac {1}{ \sqrt {3} }, \pm \frac {1}{ \sqrt {3} } , \pm \frac {1}{ \sqrt {3} }$
[ Note : If

$I, m, n$ are direction cosine of line , then

$I^2 + m^2 +n^2 = 1]$

Find the vector and cartesian equations of the plane which bisects the line joining the points $(3, -2, 1)$ and $(1, 4, -3)$ at right angles.

Let

$P(3, -2, 1); Q(1, 4, -3)$ be two points such that

$R$ (a point on the plane) is midpoint of

$\vec{PQ}$ and

$\vec{PQ}$ is perpendicular to the required plane.

Now, coordinate of

$R = \left ( \dfrac{3 + 1}{2}, \dfrac{4 - 2}{2}, \dfrac{-3 + 1}{2} \right ) = (2, 1, -1)$

Also,

$\vec{PQ} = (1 - 3)\hat{i} + (4 + 2)\hat{j} + (-3 - 1)\hat{k} = -2\hat{i} + 6\hat{j} - 4\hat{k}$

Now, we have a normal vector

$\vec{PQ}$ and a point

$R(2, 1, -1)$ of required plane.

Therefore, vector equation of required plane is

$(\vec{r} - (2\hat{i} + \hat{j} - \hat{k})). (-2\hat{i} + 6\hat{j} - 4\hat{k}) = 0$

$\left \{ \vec{r} - (2\hat{i} + \hat{j} - \hat{k}) \right \}. (\hat{i} - 3\hat{j} + 2\hat{k}) = 0$

$\Rightarrow \vec{r}. (\hat{i} - 3\hat{j} + 2\hat{k}) - (2 - 3 - 2) = 0$

$\Rightarrow \vec{r}. (\hat{i} - 3\hat{j} + 2\hat{k}) + 3 = 0$

Also, cartesian equation of required plane is

$x - 3y + 2z + 3 = 0$

Find the vector equation of the line passing through the point $A(1, 2, -1)$ and parallel to the line $5x - 25 = 14 - 7y = 35z.$

Given line is

$5x - 25 = 14 - 7y = 35z$

$\Rightarrow \dfrac{x - 5}{1/5} = \dfrac{2 - y}{1/7} = \dfrac{z - 0}{1/35}$

$\Rightarrow \dfrac{x - 5}{1/5} = \dfrac{y - 2}{-1/7} = \dfrac{z - 0}{1/35}$

$= \dfrac{x - 5}{7} = \dfrac{y - 2}{-5} = \dfrac{z - 0}{1}$ ...(i)
Hence, parallel vector of given line i.e.,

$\vec{b} = 7\hat{i} - 5\hat{j} + \hat{k}$
Since required line is parallel to given line (i)

$\Rightarrow \vec{b} = 7\hat{i} - 5\hat{j} + \hat{k}$ will also be parallel vector of required line which passes through

$A(1, 2, -1).$
Therefore, required vector equation of line is

$\vec{r} = (\hat{i} + 2\hat{j} - \hat{k}) + \lambda (7\hat{i} - 5\hat{j} + \hat{k})$

The direction cosines of the vector $(2 \hat{i}+2 \hat{j}-\hat{k})$ are ____.

Direction cosines of

$(2 \hat{i}+2 \hat{j}-\hat{k})$ are

$\dfrac{2}{\sqrt{4+4+1}}, \dfrac{2}{\sqrt{4+4+1}}, \dfrac{-1}{\sqrt{4+4+1}}$

i.e.,

$\dfrac{2}{3}, \dfrac{2}{3}, \dfrac{-1}{3}$

Plot the following points and check whether they are collinear or not :
$(i) \left(1,3\right), \left(-1 ,-1\right), \left(-2, -3\right)$
$(ii) \left(1,1\right), \left(2 ,-3\right), \left(-1, -2\right)$
$(iii) \left(0,0\right), \left(2 ,2\right), \left(5, 5\right)$

(i) From the graph, we find that all the three points lie on the same straight line. Hence, the given points are collinear.

(ii) From the graph, we find that all the three points do not lie on the same straight line. Hence, the given points are not collinear.

(iii) From the graph, we find that all the three points lie on the same straight line. Hence, the given points are collinear.

1659827_1794439_ans_0c543e08cbd7458591aff66527096435.png

If the points $(-1, -1, 2), (2,m,5)$ and $(3,11,6)$ are collinear, find the value of $m$ .

Let the given points be

$A(-1, -1,2), B(2,m,5)$ and

$C(3,11,6)$ .
Then

$\bar{AB}=(2+1)\hat{i}+(m+1)\hat{j}+(5-2)\hat{k}=3\hat{i}+(m+1)\hat{j}+3\hat{k}$

And

$\bar{AC}=(3+1)\hat{i}+(11+1)\hat{j}+(6-2)\hat{k}-4\hat{i}+12\hat{j}+4\hat{k}$

Since

$A, B, C$ are collinear, we have

$\bar{AB}=\lambda \bar{AC}$
i.e.,

$(3\hat{i}+(m+1)\hat{j}+3\hat{k})=\lambda (4\hat{i}+12\hat{j}+4\hat{k})$

$\Rightarrow 3=4\lambda$

$\lambda=\dfrac{3}{4}$

and

$m+1=12\lambda$

$m=12\times\dfrac{3}{4}-1$
(solving equations)
Therefore,

$m=8$ .

The plane $ax+b y=0$ is rotated about its line of intersection with the plane $z=0$ through an angle $\alpha .$ Prove that the equation of the plane in its new position is $a x+b y \pm(\sqrt{a^{2}+b^{2}} \tan \alpha) z=0$

Equation of the plane is

$\alpha x+b y=0 \ldots\ldots$ (i)

$\therefore$ Equation of the plane after new position is

$\dfrac{a x \cos \alpha}{\sqrt{a^{2}+b^{2}}}+\dfrac{b y \cos \alpha}{\sqrt{b^{2}+a^{2}}} \pm z \sin \alpha=0$

$\Rightarrow \dfrac{a x}{\sqrt{a^{2}+b^{2}}}+\dfrac{b y}{\sqrt{b^{2}+a^{2}}} \pm z \tan \alpha=0[\text { on dividing by } \cos \alpha]$

$\Rightarrow a x+b y \pm z \tan \alpha \sqrt{\alpha^{2}+b^{2}}=0[\text { on multiplying with } \sqrt{a^{2}+b^{2}}]$

Alternate Method

Given, planes are

$ax+b y=0 \ldots\ldots$ (i)

And

$z=0 \ldots(i i)$

Therefore, the equation of any plane passing through the line of intersection of planes (i) and(ii) may be taken as

$\alpha x+b y+k=0 \ldots\ldots$ (iii)

Then, direction cosines of a normal to the plane(iii) are

$\dfrac{a}{\sqrt{a^{2}+b^{2}+k^{2}}}, \dfrac{b}{\sqrt{a^{2}+b^{2}+k^{2}}}$

$\dfrac{c}{\sqrt{a^{2}+b^{2}+k^{2}}}$ and direction cosines of the normal to the plane (i) are

$\dfrac{a}{\sqrt{a^{2}+b^{2}}}, \dfrac{b}{\sqrt{a^{2}+b^{2}}}, 0$

since, the angle between the planes(i) and(ii) is

$\alpha$

$\therefore \cos \alpha=\dfrac{a \cdot a+b \cdot b+k \cdot 0}{\sqrt{a^{2}+b^{2}+k^{2} \sqrt{a^{2}+b^{2}}}}$

$=\sqrt{\dfrac{a^{2}+b^{2}}{a^{2}+b^{2}+k^{2}}}$

$\Rightarrow k^{2} \cos ^{2} \alpha=a^{2}\left(1-\cos ^{2} \alpha\right)+b^{2}\left(1-\cos ^{2} \alpha\right)$

$\Rightarrow k^{2}=\dfrac{\left(a^{2}+b^{2}\right) \sin ^{2} \alpha}{\cos ^{2} \alpha}$

$k=\pm \sqrt{a^{2}+b^{2}} \tan \alpha$

On putting this value in plane (iii), we get the equation of the plane as

$a x+b y+z \sqrt{a^{2}+b^{2}} \tan \alpha=0$

Show that the straight lines whose direction cosines are given by the equations $a l+b m+c n=0$ and $u l^{2}+v m^{2}+w n^{2}=0$ are parallel or perpendicular as $\dfrac{a^{2}}{u}+\dfrac{b^{2}}{v}+\dfrac{c^{2}}{w}=0$ or $a^{2}(v+w)+b^{2}(w+u)+c^{2}(u+v)=0$

Here,

$l=-\dfrac{(b m+c n)}{a}$ and

$u l^{2}+m^{2} v+w n^{2}=0$

Eliminating

$l$ , we get

$\dfrac{u(b m+c n)^{2}}{a^{2}}+v m^{2}+w n^{2}=0$

$u\left(b^{2} m^{2}+2 b c m n+c^{2} n^{2}\right)+v a^{2} m^{2}+w a^{2} n^{2}=0$

$\left(b^{2} u+a^{2} v\right) m^{2}+(2 b c u) m n+\left(c^{2} u+a^{2} w\right) n^{2}=0$

$\Rightarrow\left(b^{2} u+a^{2} v\right)\left(\dfrac{m}{n}\right)^{2}+(2 b c u)\left(\dfrac{m}{n}\right)+\left(c^{2} u+a^{2} w\right)=0,$

which is quadratic in

$(m / n)$ having roots

$m_{1} / n_{1}$ and

$m_{2} / n_{2}$

a. If the straight lines are parallel, the quadratic in

$m / n$ has equal roots, i.e., discriminant

$=0$

$\Rightarrow(2 b c u)^{2}-4\left(b^{2} u+a^{2} v\right)\left(c^{2} u+a^{2} w\right)=0$

$\Rightarrow b^{2} c^{2} u^{2}=\left(b^{2} u+a^{2} v\right)\left(c^{2} u+a^{2} w\right)$

$\Rightarrow a^{2} v w+b^{2} u w+c^{2} u v=0$

$\Rightarrow \dfrac{a^{2}}{u}+\dfrac{b^{2}}{v}+\dfrac{c^{2}}{w}=0$

b. If the straight lines are perpendicular

$\Rightarrow \dfrac{m_{1}}{n_{1}}. \dfrac{m_{2}}{n_{2}}=\dfrac{c^{2} u+a^{2} w}{b^{2} u+a^{2} v} \quad(\text { product of roots })$

$\Rightarrow \dfrac{m_{1} m_{2}}{c^{2} u+a^{2} w}=\dfrac{n_{1} n_{2}}{b^{2} u+a^{2} v}$

Similarly, by eliminating

$n$ , we get

$\dfrac{l_{1} l_{2}}{b^{2} w+c^{2} v}=\dfrac{m_{1} m_{2}}{c^{2} u+a^{2} w}$

From (i) and (ii)

$\dfrac{l_{1} l_{2}}{b^{2} w+c^{2} v}=\dfrac{m_{1} m_{2}}{c^{2} u+a^{2} w}=\dfrac{n_{1} n_{2}}{b^{2} u+a^{2} v}=\lambda$

since they are perpendicular,

$l_{1} l_{2}+m_{1} m_{2}+n_{1} n_{2}=0$

$\Rightarrow \lambda\left(b^{2} w+c^{2} v\right)+\lambda\left(c^{2} u+a^{2} w\right)+\lambda\left(b^{2} u+a^{2} v\right)=0$

$\Rightarrow a^{2}(v+w)+b^{2}(w+u)+c^{2}(u+v)=0$

Find the vector equation of the line which is parallel to the vector $3 \hat{i}-2 \hat{j}+6 \hat{k}$ and which passes through the point $(1,-2,3)$ .

Let

$\vec{a}=3 \hat{i}-2 \hat{j}+6 \hat{k}$ and

$\vec{b}=\hat{i}-2 \hat{j}+3 \hat{k}$

So, vector equation of the line, which is parallel to the vector

$\vec{a}=3 \hat{i}-2 \hat{j}+6 \hat{k}$ and passes through the vector

$\vec{b}=\hat{i}-2 \hat{j}+3 \hat{k}$ is

$\vec{r}=\vec{b}+\lambda \vec{a}$

$\therefore \vec{r}=(\hat{i}-2 \hat{j}+3 \hat{k})+\lambda(3 \hat{i}-2 \hat{j}+6 \hat{k})$

$\Rightarrow(x \hat{i}+y \hat{j}+z \hat{k})-(\hat{i}-2 \hat{j}+3 \hat{k})=\lambda(3 \hat{i}-2 \hat{j}+6 \hat{k})$

$\Rightarrow(x-1) \hat{i}+(y+2) \hat{j}+(z-3) \hat{k}=\lambda(3 \hat{i}-2 \hat{j}+6 \hat{k})$

If a line makes angles $90^{o}, 135^{o}, 45^{o}$ with the $X-, Y-$ and $Z-$ axes respectively, then find its direction cosines.

Let

$l, m,n$ be the direction cosines of the line

Then

$l=\cos \alpha, m=\cos \beta, n=\cos \gamma$

here,

$\alpha=90^o, \beta=135^o=\gamma =45^o$

$\therefore l=\cos 90^o=0$

$m=\cos 135^o=\cos (180^o-45^o)=-\cos 45^o$

$=-\dfrac{1}{\sqrt{2}}$ and

$N=\cos 45^o =\dfrac{1}{\sqrt{2}}$

$\therefore$ the direction cosines of the line are

$0, -\dfrac{1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}}$

If $\bar {AB}=2\hat i-4\hat j+7\hat k$ and initial $A(1,5,0)$ . Find the terminal point $B$ .

Give:

$A(1,5,0)$

$\therefore \bar a=\hat i+5\hat j$

Now,

$\bar{AB}=2\hat i-4\hat j+7\hat k$

$\therefore \bar b-\bar a=2\hat i-4\hat j+7\hat k$

$\therefore \bar b=(2\hat i-4\hat j+7\hat k)+\bar a$

$=(2\hat i-4\hat j+7\hat k)+(\hat i+5\hat j)$

$=3\hat i+\hat j+\hat k$

Hence, the terminal point

$B(3,1,7).$

If a vector has direction angles $45^{o}$ and $60^{o}$ , find the third direction angle.

Let

$\alpha=45^o, \beta=60^o$

We have to find

$\gamma$

$\because \cos^2 \alpha+\cos^2 \beta+\cos^2 \gamma=1$

$\therefore \cos^2 45^o+\cos^2 60^o +\cos^2 r=1$

$\therefore \left(\dfrac{1}{\sqrt{2}}\right)^{2}+\left(\dfrac{1}{2}\right)+\cos^{2}\gamma=1$

$\therefore \cos^{2}\gamma =1-\dfrac{1}{2}-\dfrac{1}{4}=\dfrac{1}{4}$

$\therefore \cos\gamma=\pm \dfrac{1}{2}$

$\therefore \cos \gamma=\dfrac{1}{2}$ or

$\cos^{2}\gamma=-\dfrac{1}{2}$

$\therefore \cos \gamma =\cos \dfrac{\pi}{3}$ or

$\cos\gamma =-\cos \dfrac{\pi}{3}$

$\therefore \cos \left(\pi-\dfrac{\pi}{3}\right)=\cos \dfrac{2\pi}{3}$

$\therefore \gamma =\dfrac{\pi}{3}$ or

$\gamma =\dfrac{2\pi}{3}$

Hence, the third direction angle is

$\dfrac{\pi}{3}$ or

$\dfrac{2\pi}{3}$

If a line has the direction ratios $4, -12, 18$ , then find its direction cosines.

The direction ratios of the line are

$a=4, b=-12, c=18$ .

Let

$l, m, n$ be the direction cosines of the line.

Then

$l=\dfrac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}=\dfrac{4}{\sqrt{4^{2}+(-12)^{2}+(18)^{2}}}$

$=\dfrac{4}{\sqrt{16+144+324}}=\dfrac{4}{22}=\dfrac{2}{11}$

$m=\dfrac{b}{\sqrt{a^{2}+b^{2}+c^{2}}}=\dfrac{-12}{\sqrt{4^{2}+(-12)^{2}+(18)^{2}}}$

$=\dfrac{-12}{\sqrt{16+144+324}}=\dfrac{-12}{22}=\dfrac{-6}{11}$

and

$n=\dfrac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}$

$=\dfrac{18}{\sqrt{4^{2}+(-12)^{2}+(18)^{2}}}$

$=\dfrac{18}{16+144+324}=\dfrac{18}{22}=\dfrac{9}{11}$

Hence, the direction cosines of the line are

$\dfrac{2}{11}, \dfrac{-6}{11}, \dfrac{9}{11}$ .

Show that the following points are collinear:
$P=(4,5,2),Q=(3,2,4),R=(5,8,0)$ .

Let

$\overline {a},\overline {b},\overline {c}$ be the position vector of the points.

$P=(4,5,2),B=(3,2,4)$ and

$c=(5,8,0)$ respectively.

Then,

$\overline {a}=4\hat i+5\hat j-2\hat k,\overline {b}=3\hat i+2\hat j-4\hat k,\overline {c}=5\hat i-8\hat j+0\hat k$

$\overline {AB}=\overline {b}-\overline {a}$

$=(3\hat i+2\hat j+4\hat k)-(4\hat i+5\hat j+2\hat k)$

$=-(\hat i+3\hat j-2\hat k$ ......(i)

and

$\overline {BC}=\overline {c}-\overline {b}$

$=(5\hat i+8\hat j+0\hat k)-(3\hat i+2\hat j+4\hat k)$

$=2\hat i+6\hat j-4\hat k$

$=2(\hat i+3\hat j-2\hat l)$

$=2.\overline {AB}$ ........[BY(i)]

$\therefore \overline {BC}$ is a non-zero scalar multiple of

$\overline {AB}$

$\therefore$ they are parallel to each other.

But they have point

$B$ in common.

$\therefore \overline {BC}$ and

$\overline {AB}$ are collinear vector.

Hence, points

$A,B$ and

$C$ are collinear.

If the points $A(3,0,p), B(-1,q, 3)$ and $C(-3,3,0)$ are collinear, then find
the values of $p$ and $q$ .

Let

$\bar{a}, \bar{b}, \bar{c}$ be the position vectors of

$A, B$ and

$C$ respectively.

Then

$\bar{a}=3\hat{i}+0.\hat{j}+p\hat{k}$ ,

$\bar{b}=-\hat{i}+q\hat{j}+3\hat{k}$ and

$\bar{c}=-3\hat{i}+3\hat{j}+0.\hat{k}$

Putting

$\lambda =-3$ in equation

$(2)$ , we get

$3(-3+1)=-3q$

$\therefore -6=-3q$

$\therefore q=2$

Also, putting

$\lambda=-3$ in equation

$(3)$ , we get

$0=-9+p$

$\therefore p=9$

Hence,

$p=9$ and

$q=2$ .

Find the value of $\lambda$ so thta the lines $\dfrac{1-x}{3}=\dfrac{7y-14}{\lambda }=\dfrac{z-3}{2} and \dfrac{7-7x}{3\lambda }=\dfrac{y-5}{1}=\dfrac{6-z}{5}$ are at riht angles.

1988280_1847323_ans_f872ac5f053a46a29f3adc8102cca00e.jpg

Find the vector equation of the line which passes through the point ( 3 , 2 , 1 ) and is parallel to the vector $2\hat{i}+2\hat{j}-3\hat{k}$

The vector equation of the line passing through

$A(\bar{a})$ and parallel to the vector

$\bar{b}$ is

$\bar{r}=\bar{a}+\lambda \bar{b}$ ,
where

$\lambda$ is a scalar

$\therefore$ the vector equation of the line passing through the point having position vector

$3\hat{i}+2\hat{j}+\hat{k}$ and parallel to the vector

$2\hat{i}+2\hat{j}-3\hat{k}$ is

$\bar{r}=(3\hat{i}+2\hat{j}+\hat{k})+\lambda (2\hat{i}+2\hat{j}-3\hat{k})$

Find the vector equation of the line passing through point through point through points having position vector $3\hat{i}+4\hat{j}-7\hat{k} and 6\hat{i} - \hat{j}-+\hat{k}$

The vector equation of the line passing through the

$A(\bar{a})$ and

$B(\bar{b})$ is

$\bar{r}=\bar{a}+\lambda (\bar{b}-\bar{a}),\lambda$ is a scalar

$\therefore$ the vector equation of the line passing through the point having position vector

$3\hat{i}+4\hat{j}-7\hat{k}$ and

$6\hat{i} - \hat{j}-+\hat{k}$

$\bar{r}=(3\hat{i}+4\hat{j}-7\hat{k})+\lambda \left [ \left ( 6\hat{i}-\hat{j}+\hat{k} \right )-\left ( 3\hat{i}+4\hat{j}-7\hat{k} \right ) \right ]$

$i.e.\bar{r}=(3\hat{i}+4\hat{j}-7)+\lambda (3\hat{i}-5\hat{j}+8\hat{k})$

Find the vector equation of the plane passing through the point having position vector $\hat{i}+\hat{j}+\hat{k}$ and perpendicular to the vector $4\hat{i}+5\hat{j}+6\hat{k}$

1988343_1847211_ans_15f7497cd81e4d0f94646ea5c1041652.jpg

Find the vector equation of the line passing through the point having position vector
$-\hat{i}-\hat{j}+2\hat{k}$ and parallel to the line $\bar{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda (3\hat{i}+2\hat{j}+\hat{k})$

Let A be point having position vector

$\bar{a}=\hat{i}-\hat{j}+2\hat{k}$
The required line is parallel to the line

$\bar{r}=(\hat{i}+2\hat{j}+3\hat{k})\lambda (3\hat{i}+2\hat{j}+\hat{k})$

$\therefore$ it is parallel to the vector

$\bar{b}=3\hat{i}+2\hat{j}+\hat{k}$
The vector equation of the line passing through

$A(\bar{a})$ and parallel to

$\bar{b}$ is

$\bar{r}=\bar{a}+\lambda \bar{b}$ where

$\lambda$ is a scalar

$\therefore$ the required vector equation of the line is

$\bar{r}=(-\hat{i}-\hat{j}+2\hat{k})+\lambda (3\hat{i}+2\hat{j}+\hat{k})$

Find the vector equation of line passing through the point having vector $5\hat{i}+4\hat{j}+3\hat{k}$ and having direction ratios $-3,4,2$

Let A be the point whose position vector is

$=5\hat{i}+4\hat{j}+3\hat{k}$
The vector equation of the line passing through

$\bar{b}=-3\hat{i}+4\hat{j}+2\hat{k}$
The vector equation of the line passing through

$A(\bar{a})$ and parallel to

$\bar{b}$ is

$\bar{r}=\bar{a}+\lambda \bar{b}$ , where

$\lambda$ is a scalar .

$\therefore$ the required vector equation of the line is

$\bar{r}=5\hat{i}+4\hat{j}+3\hat{k}\lambda \left ( -3\hat{i}+4\hat{j}+2\hat{k} \right )$

Find the acute between the lines x = y , z = 0 and x = 0 , z = 0

1988274_1847330_ans_100d8188b8e740fc8863509008676a3a.jpg

Reduce the vector equation $\bar{r} \cdot (3\hat{i}+4\hat{j}+12\hat{k})$ to normal form and hence find the length of the perpendicular from the origin to the plane.

The normal form of equation of a plane is

$\bar{r}.\hat{n}=p$ where

$.\hat{n}$ is unit vector along the normal and

$p$ is the length of perpendicular drawn from origin to the plane.
Here, given pane is

$\bar{r}.\left ( 3\hat{i}+4\hat{j}+12\hat{k} \right )=78.....(1)$ or

$\vec{r}.\vec{n}=78$
where

$\vec{n}=3\hat{i}+4\hat{j}+12\hat{k}$ .
Since,

$|\vec{n}|=\sqrt{3^2+4^2+12^2}=13\neq 1$

Therefore, the given equation is not in the normal form.
To reduceit to the normal form, we divide both sides by

$\|\vec{n}\|$ , that is,

$\vec{r}.\dfrac{\vec{n}}{|\vec{n}|}=\dfrac{78}{|\vec{n}|}=6$

or

$\vec{r}.\left(\dfrac{3}{6}\hat{i}+\dfrac{4}{6}\hat{j}+\dfrac{12}{6}\hat{k}\right)=6$

$\Rightarrow \vec{r}.\left(\dfrac{1}{2}\hat{i}+\dfrac{2}{3}\hat{j}+2\hat{k}\right)=6$

This is the normal form of the equation of the given plane.
Hence, the length of the perpendicular from the origin is

$6$

Find the direction cosines of the lines $\bar{r}=(-2\hat{i}+\frac{5}{2}\hat{j}-\hat{k})+\lambda (2\hat{i}+3\hat{j})$

1973982_1847369_ans_7b2bb29de5bf48f58cc9da83c2d6feca.jpg

Show that the points $(2, 3, 4), (-1, -2, 1), (5, 8, 7)$ are collinear.

1959457_1859979_ans_e73ee55cfa374bba8044d5ecacb977fc.png

Find the vector equation of the plane which is at a distance of 5 units from the origin and which is normal to the vector $2\hat{i}+\hat{j}+2\hat{k}$

1973854_1847621_ans_77fbe467c0c8420e8ce5c3974e38299e.jpg

Show that the line through the points $(4, 7, 8), (2, 3, 4)$ is parallel to the line through the points $(-1, -2, 1), (1, 2, 5)$ .

1959460_1860007_ans_b337b7b86b5c4b589f02e54980f31988.png

Find the direction cosines of the sides of the triangle whose vertices are $(3, 5, -4), (-1, 1, 2)$ and $(-5, -5, -2)$ .

1959458_1859981_ans_da63f86ab26c43ca9806bdb5f6d0eee9.png

Find the cartesian equation of the line which passes through the point $(-2, 4, -5)$ are parallel to the line given by $\dfrac {x + 3}{3} = \dfrac {y - 4}{5} = \dfrac {z + 8}{6}$ .

1959461_1860013_ans_5bd7fbe396ef44c4a101610a88389dfe.png

Find the vector equations of the plane passing through the point $A (1 , 92 , 1 ) , B ( 2 , 91 , 93 ) and ( C ( 0 , 1 , 5)$

1954175_1847694_ans_c20fdea7e75947c297a1768098bb2016.png

Write the dc's of the line joining the points $(1, 0, 0)$ and $(0, 1, 1)$ .

1965517_1860654_ans_dc02d27e74544be68f9c80922b6ce583.png

Find the vector and cartesian equations of the planes that passes through the point $(1, 0, -2)$ and the normal to the plane is $\hat {i} + \hat {j} - \hat {k}$ .

1965364_1860500_ans_2e905fcfc4c0427e9421b3e03568d1f7.gif

Find the direction cosines of the vctor joining the points $A(1, 2 , -3)$ and $B ( -1, -2, 1)$ directed from $A$ to $B$ .

$\bar{AB}=+2\overrightarrow{i}+4\overrightarrow{j}-4\bar{k} \quad \therefore |\bar{AB}|=\sqrt{4+16+16}=6$

direction cosines are

$\left( \frac { -2 }{ 6 } ,\frac { -4 }{ 6 } ,\frac { 4 }{ 6 } \right) =\left( \frac { -1 }{ 3 } ,\frac { -2 }{ 3 } ,\frac { 2 }{ 3 } \right)$

Find the vector equation of a plane which is at a distance of $7$ units from the origin and normal to the vector $6i+3j-2k$

Unit vector along

$6i+3j-2k$

$\hat n=\dfrac{6i+3j-2k}{\sqrt{(6)^2+(3)^2+(-2)^2}}$

$=\dfrac{6i+3j-2k}{\sqrt{36+9+4}}=\dfrac{6i+3j-2k}{\sqrt{49}}$

$\therefore$ Vector equation of plane

$\vec r. \hat n=d$

$\therefore \vec r \left( \dfrac{6i+3j-2k}{\sqrt{49}}\right)=7$

$\Rightarrow \vec r.(6i+3j-2k)=7 \sqrt{49}$

$\Rightarrow \vec r.(6i+3j-2k)=49$

Find the vector equation of a plane which is at a distance of $7$ unit from the origin and has $\hat i$ as the unit vector normal to it.

Given unit vector along normal

$\hat n=i$
and distance from origin

$(0,0,0)=7\ units$

$\therefore$ from vector equation of plane

$\vec r. \hat n=d$

$\vec r. \hat i=7$

If the coordinates of the points $A, B, C, D$ be $(1, 2, 3), (4, 5, 7), (-4, 3, -6)$ and $(2, 9, 2)$ respectively, then find the angle between the lines $AB$ and $CD$ .

1965421_1860564_ans_e93dcc55f5e9455b930ce60d7a15c1ea.png

Find the equations of the planes that passes through three points.
$(1, 1, 0), (1, 2, 1), (-2, 2, -1)$ .

1965390_1860513_ans_564ec2b535fc441b82bd17cd77e28c42.gif

Find the direction of the line joining two points $(4,2,3)$ and $(4,5,7)$

Let dc's of line joining two points

$P(x_1, y_1, z_1)$ and

$Q(x_2, y_2, z_2)$ are

$\dfrac{x_2-x_1}{PQ},\dfrac{y_2-y_1}{PQ},\dfrac{z_2-z_1}{PQ}$

where

$PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

According to equation

$x_1=4, y_1=2, z_1=3$

$x_2=4, y_2=5, z_2=7$

$\therefore PQ=\sqrt{(4-4)^2+(5-2)^2+(7-3)^2}$

$=\sqrt{0+9+16}$

$=\sqrt{25}$

$=5$

$\therefore$ dC's of line joining

$(4,2,3)$ and

$(4,5,7)$ are

$\dfrac{4-4}{5},\dfrac{5-2}{5},\dfrac{7-3}{5}$ or

$0,\dfrac{3}{5},\dfrac{4}{5}$

Find the equations of the planes that passes through three points.
$(1, 1, -1), (6, 4, -5), (-4, -2, 3)$ .

1965370_1860511_ans_2de69cfbddb84dcc861518e9d1e33c46.gif

If O be the origin and the coordinates of P be (1, 2, -3), then find the equation of the plane passing through P and perpendicular to OP.

1966808_1860611_ans_d4bded9d9cb64402ae5f623b1df13e5a.png

Find the equation of the line which passes through the points $(5,7,9)$ and is parallel to the following axis:
$Y-$ axis

Let the line parallel to

$Y$ axis passe through point

$(0,1,0)$

$\therefore$ Position vector of point

$(0,1,0)$

$\therefore$ required vector equation of line

$\vec r=\vec {r_1}+\lambda \vec {r_2}$

$\vec r=(5\hat i+7\hat j+9\hat k)+\lambda (0.\hat i+1.\hat j+0.\hat k)$

$\vec r=5\hat i+(7+\lambda)\hat j+9\hat k$
Cartesian equatrion: Let

$x\hat i+y\hat j+z\hat k$

$x\hat i+y\hat j+z\hat k=5\hat i+(7+\lambda)\hat j+9\hat k$
On comparing

$x=5, y=7+\lambda , z=9$

$\Rightarrow \dfrac {x-5}{0}=\dfrac {y-7}{1}=\dfrac {z-9}{0}=\lambda$

$\therefore$ Equation of line

$\dfrac {x-5}{0}=\dfrac {y-7}{1}=\dfrac {z-9}{0}$

Find the equation of plane perpendicular of the origin from the plant is $13$ and direction ratios of this perpendicular are $4,-3,12$

Given DR's of normal on plane are

$4,-3,12$

$\therefore$ DC's o fnormal are

$\dfrac{4}{\sqrt{(4)^2+(-3)^2+(12)^2}},\dfrac{-3}{\sqrt{(4)^2+(-3)^2+(12)^2}}, \dfrac{12}{\sqrt{(4)^2+(-3)^2+(12)^2}}$

$\Rightarrow \dfrac{4}{\sqrt{169}},\dfrac{-3}{\sqrt{169}},\dfrac{12}{\sqrt{169}}$

$\Rightarrow \dfrac{4}{13},\dfrac{-3}{13},\dfrac{12}{13}$
So, the equation of the plane is

$ax+by+cz=d$
where

$d=13$

$\dfrac{4}{13}x-\dfrac{-3}{13}y+\dfrac{12}{13}z=13$

Projection of a line on axis are $-3,4,-12$ . Find length of line segment and direction cosines.

Projection of a line coordinate axis are the direction ratios of a line.
If direction cosines are

$l,m,n$ then

$\dfrac{l}{-3}=\dfrac{m}{4}=\dfrac{n}{-12}=k$

$\therefore l=-3k,m=4k,n=-12k$
But

$l^2+m^2+n^2=1$

$\therefore (-3k)^2+(4k)^2+(-12k)^2=1$

$\Rightarrow 9k^2+16k^2+144k^2=1$

$\Rightarrow 169k^2=1$

$\Rightarrow k=\pm\dfrac{1}{13}$

$\therefore l=\dfrac{-3}{13},m=\dfrac{4}{13},n=\dfrac{-12}{13}$
Length of

$OP=$ Distance between

$O(0,0,0)$ and point

$P(-3,4,-12)$

$\therefore OP=\sqrt{(-3-0)^2+(4-0)^2+(-12-0)^2}$

$=\sqrt{9+16+144}=\sqrt{169}$

$=13$

$\therefore$ Direction cosines are

$-\dfrac{3}{13},\dfrac{4}{13},\dfrac{-12}{13}$

Find the equation of the line which passes through the points $(5,7,9)$ and is parallel to the following axis:
$X-$ axis

Let position vector of point

$(5,7,9)$

$\vec {r_1}=5\hat i+7\hat j+9\hat k$
The line parallel to

$X-$ axis passes through point

$(1,0,0)$ whose position vector

$\vec {r_2}=(\hat i+0\hat j+0\hat k)$

$\therefore$ Vector equation of required line.

$\vec r=\vec {r_1}+\lambda \vec {r_2}$

$\vec r=(5\hat i+7\hat j+9\hat k)+\lambda (\hat i+0\hat j+0\hat k)$

$\vec {r}=(5+\lambda)\hat i+7\hat j+9\hat k$
Cartesian equation line:
Let

$\vec r =i+yj +zk$

$x\hat i+y\hat j+z\hat k=(5+\lambda)\hat i+7\hat j+9\hat k$
On comparing

$\dfrac {x-5}{1}=\dfrac {y-7}{0}=\dfrac {z-9}{0}\lambda$

$\therefore$ Required equation of line is

$\dfrac {x-5}{1}=\dfrac {y-7}{0}=\dfrac {z-9}{0}$

Find the vector equation of the line that passes through point $(2, 3, 2)$ and is parallel to the line $\vec r=( -2\hat i +3\hat j)+\mu (2\hat i -3\hat j +6\hat k)$ .

Line Passes through point

$(2, 3, 2)$

$\therefore$ Position vector of the point is

$\vec a=2i+3j+2k$
Again vector equation

$\vec r=-2i+3j+\mu (2i-3j+6k)$ and of

$\vec b=2i-3j+6k$ are equal.

$\therefore$ Vector equation of line is

$\vec r =\vec a+\lambda \vec b$

$=(2i+3j+2k)+\lambda (2i-3j+6k)$
The equation of distance line between the given lines can be written as

$\vec r=(-2i+3j)+\mu (2i-3j+6k)$

$=\vec a_1+\mu \vec b_1$
and

$\vec r =(2i+3j+2k)+\lambda (2i-3j+6k)$

$=\vec a_2 +\lambda \vec b_2$
But

$\vec b_1 =\vec b_2=\vec b$ (

$\because$ lines are parallel )

$\therefore \vec b \times ( \vec a_2-\vec a_1)=(2i-3j+6k)\times (2i+3j+2k)-(-2i+3j)$

$=(2i-3j+6k) \times (4i+2k)$

$\therefore \vec b \times ( \vec a_2-\vec a_1)=\begin{vmatrix} i & j & k \\ 2 & -3 & 6 \\ 4 & 0 & 2 \end{vmatrix}$

$=i(-6-0)-j(4-24)+k(0+12)$

$=-6i+20j+12k$

$\therefore$ Distance between lines

$d=\left| \dfrac {\vec b \times ( \vec a_2 -\vec a_1)}{| \vec b|}\right|$

$=\dfrac {|-6i+20j+12k|}{|2i-3j+6k|}$

$=\dfrac {\sqrt{(-6)^2+(20)^2+(12)^2}}{\sqrt{(2)^2+(-3)^2+(6)^2}}$

$=\dfrac {\sqrt{36+400+144}}{\sqrt {4+9+36}}$

$=\dfrac {\sqrt{580}}{\sqrt{49}}=\dfrac {\sqrt{580}}{7}$

Find the vector equation of a plane which is at a direction of $4$ units from the origin and direction cosines of the normal to the plane are $2,-1,2.$

Given Dc's of the normal to the plane are

$2,-1,2$ .

$\therefore$ Dc's of the plane are

$\dfrac{2}{3},\dfrac{-1}{3},\dfrac{2}{3}$

$\sqrt{(2)^2+(-1)^2+(2)^2}=\sqrt 9=3$

$\vec n=\dfrac{2}{3}i-\dfrac{1}{3}j+\dfrac{2}{3}k$
and

$d=4\ unit$

$\therefore$ from equation of plane

$\vec r. \vec n=d$

$\vec r \left( \dfrac{2}{3}i-\dfrac{1}{3}j+\dfrac{2}{3}k \right)=4$

Find a unit normal vector to the plane $x+y+z-3=0$

Let given

$x+y+z-3=0$

$\dfrac{1}{\sqrt{1^2+1^2+1^2}},\dfrac{1}{\sqrt{1^2+1^2+1^2}},\dfrac{1}{\sqrt{1^2+1^2+1^2}}$ or

$\dfrac{1}{\sqrt 3},\dfrac{1}{\sqrt 3},\dfrac{1}{\sqrt 3}$
So, from

$lx+my+nz=d$

$(l \hat i+m \hat j+n \hat k) (x \hat i+ y\hat j+ z \hat k)=d$

$\left( \dfrac{1}{\sqrt 3}\hat i+\dfrac{1}{\sqrt 3}\hat j+\dfrac{1}{\sqrt 3}\hat k\right) (x \hat i+ y\hat j+ z \hat k)=3$
From

$\hat n. \vec r=3$
Unit vector

$\hat n=\dfrac{1}{\sqrt 3}i+\dfrac{1}{\sqrt 3}j+\dfrac{1}{\sqrt 3}k$

$\Rightarrow \hat n =\dfrac{1}{\sqrt 3}(i+j+k)$

Find the vector equation of the line which passes through the point $(3, 4, 5)$ and is parallel to the vector $2\hat{ i} + 2\hat{ j} – 3\hat{k}$ .

Vector equation of a line is given by

$\vec { r} =\vec { a } +\vec { \lambda b }$ ; where

$\vec{a}$ = position vector of a point on the line.

$\vec{b}$ = vector parallel to the line

Here,

$\vec {a } =3\hat{i}+4\hat{j}+5\hat{k}$

$\vec {b } =2\hat{i}+2\hat{j}-3\hat{k}$

$\therefore \boxed{\vec{r}=(3\hat{i}+4\hat{j}+5\hat{k} ) +\lambda(2\hat{i}+2\hat{j}-3\hat{k})}$

Find the equation of the line parallel to the vector
$2\hat i-\hat j+3\hat k$
and passes through the point $(5,-2,4)$ .

Line passes through

$(5,-2,4)$

$\therefore$ Position vector of

$(5,-2,4)$

$\vec a =5\hat i-2\hat j+4\hat k$
Given vector

$\vec b=2\hat i-\hat j+3\hat k$

$\therefore$ Vector equation of line parallel to

$\vec b=2\hat i-\hat j+4\hat k$
passing through point

$(5,-2,4)$

$\vec r=\vec a+\lambda \vec b$

$\vec r=(5\hat i-2\hat j+4\hat k)+\lambda (2\hat i-\hat j+3\hat k)$

Find the equation of a line in vector and Cartesian form that passes through the point with position vector
$2\hat i-\hat j+4\hat k$
and in the direction
$\hat i+2\hat j-\hat k$ .

Position vector

$\vec{a}=2\hat{i}-\hat{j}+4\hat{k}$ and

$\vec{b}=\hat{i}+2\hat{j}-\hat{k}$

$\therefore$ Vector equation of line parallel to

$\vec{b}$ and passing through

$\vec{a}$

$\vec{r}=\vec{a}+\lambda \vec{b}$

$\Rightarrow \vec{r}=(2\hat{i}-\hat{j}+4\hat{k})+\lambda (\hat{i}+2\hat{j}-\hat{k})$

$...(1)$

$\lambda$ is a parametric.
In cartesian form,

$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$

$x\hat{i}+y\hat{j}+z\hat{k}=(2\hat{i}-\hat{j}+4\hat{k})+\lambda (\hat{i}+2\hat{j}-\hat{k})$

$\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}$

$=(2+\lambda)\hat{i}+(-1+2\lambda) \hat{j}+(4-\lambda) \hat{k}$
Comparing coefficient of

$\hat{i}, \hat{j}$ and

$\hat{k}$

$x=2+\lambda, y=-1+2\lambda, z=4-\lambda$

$\Rightarrow x-2=\lambda, \dfrac{y+1}{2}=\lambda, \dfrac{z-4}{-1}=\lambda$

$\Rightarrow \dfrac{x-2}{1}=\dfrac{y+1}{2}=\dfrac{z-4}{-1}=\lambda$

$\therefore$ Required equation of line is

$\dfrac{x-2}{1}=\dfrac{y+1}{2}=\dfrac{z-4}{-1}$

Find the equation of the line which passes through the points $(5,7,9)$ and is parallel to the following axis:
$Z-$ axis

The line parallel to

$Z-$ axis pass through

$(0,0,1)$

$\therefore r_2=0.\hat i+0.\hat j+1.\hat k$

$\therefore$ Required vector equation of line

$\vec =\vec {r_1}+\lambda \vec {r_2}$

$\vec r=(5\hat i+7\hat j+9\hat k)+\lambda (0.\hat i+0.\hat j+1.\hat k)$

$\vec r= =5\hat i+7\hat j+(9+\lambda )\hat k$
Cartesian equation : Let

$x\hat i+y\hat j+z\hat k$

$x\hat i+y\hat j+z\hat k=5\hat i +7\hat j+(9+\lambda)\hat k$
On comparing :

$x=5, y=7, z=9+\lambda$

$\Rightarrow \dfrac {x-5}{0}=\dfrac {y-7}{0}=\dfrac {z-9}{1}\lambda$

$\therefore$ Required equation

$\dfrac {x-5}{0}=\dfrac {y-7}{0}=\dfrac {z-9}{1}$

Find an equation passes through $(1,2,3)$ and is parallel to the vector.
$(3\hat i+2\hat j-2\hat k)$ .

$\because$ Line is parallel to

$3\hat{i}+2\hat{j}-2\hat{k}$

$\therefore$ dc's of line are

$3,2,-2$ .

$\because$ Line is passing through

$1,2,3$ .

$\therefore$ Cartesian equation of line

$\dfrac{x-x_{1}}{a}=\dfrac{y-y_{1}}{b}=\dfrac{z-z_{1}}{c}$

$\Rightarrow \dfrac{x-1}{3}=\dfrac{y-2}{2}=\dfrac{z-3}{-2}$
Vector equation of line

$\vec{r}=\vec{a}+\lambda \vec{b}$

$\Rightarrow \vec{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda (3\hat{i}+2\hat{j}-2\hat{k})$

Find the equation of the line cartesian from which passes through $(1,2,3)$ and is parallel to $\dfrac {-x-2}{1}=\dfrac {y+3}{7}=\dfrac {2z-6}{3}$ .

Let

$a, b, c$ are dc's of line passing through

$(x_1, y_1 , z_1)$ then equation of the line

$\dfrac {x-x_1}{a}=\dfrac {y-y_1}{b}=\dfrac {z-z_1}{c}$
Here line passes through point

$(1,2,3)$ and parallel to the line

$\dfrac {x-2}{-1}=\dfrac {y+3}{7}=\dfrac {2z-6}{3}$

$\therefore$ dc's of line are

$-1,7,3/2$ or

$-2,14,3$ form line

$\dfrac {x-2}{-1}=\dfrac {y+3}{7}=\dfrac {z-3}{3/2}$

$\therefore$ Required equation of line

$\dfrac {x-1}{2}=\dfrac {y-2}{14}=\dfrac {z-3}{3}$

Find the equation of the line passes through the point $(2,-1,1)$ and is parallel to the line $\dfrac {x-3}{2}=\dfrac {y+1}{7}=\dfrac {z-2}{-3}$ .

The equation of line passing through point

$(2,-1,1)$ and parallel to the given line

$\dfrac {x-3}{2}=\dfrac {y+1}{7}=\dfrac {z-2}{-3}$ because dc's of parallel line are same

$\therefore \vec m=2\hat +7\hat j-3\hat k$
For vector equation of line passing through

$(2,-1,1)$ and parallel to

$\vec m$ .
The position vector of

$(2,-1,1)$

$\vec a=2\hat i-\hat j+\hat k$

$\therefore$ Required equation of line

$\vec r=\vec a +\lambda \vec m$

$=(2\hat i-\hat j+\hat k)+\lambda (2\hat i+7\hat j-3\hat k)$

Find the Cartesian equation of line passes through the point $(-2,4,-5)$ and is parallel to
$\dfrac {x+3}{3}=\dfrac {y-4}{5}=\dfrac {z+8}{6}$ .

Let line passes through point

$(x_{1}, y_{1}, z_{1})$ whose dc's are

$a,b,c$ then equation of line is

$\dfrac{x-x_{1}}{a}=\dfrac{y-y_{1}}{b}=\dfrac{z-z_{1}}{c}$
Here line passes through

$(-2,4,-5)$ and parallel to

$\dfrac{x+3}{3}=\dfrac{y-4}{5}=\dfrac{z+8}{6}$

$\therefore dc's$ of the required line are

$3,5,6$ .
Hence equation of line is

$\dfrac{x+2}{3}=\dfrac{y-4}{5}=\dfrac{z+5}{6}$

Find the equation of the line vector and in certesian form that passes through the point with vector
$2\hat i-3\hat j-4\hat k$
and
$3\hat i+4\hat j-5\hat k$
is parallel to the vector.

Position vector of the point is

$\therefore \vec a=2\hat i-3\hat j+4\hat k$
given vector

$\vec b=3\hat i+4\hat j-5\hat k$

$\therefore$ Vector equation of line parallel

$\vec b=3\hat i+4\hat j-5\hat k$
passing through

$2\hat i-3\hat j+4\hat k$

$\vec r=\vec a+\lambda \vec b$

$\Rightarrow \vec r=(2\hat i-3\hat j+4\hat k)+\lambda (3\hat i+4\hat j-5\hat k)$

$=(2+3\lambda)\hat i+(-3+4\lambda)\hat j+(4-5\lambda)\hat k$
Let Cartesian equation is

$xi+yj+zk$

$x\hat i+y\hat j+z\hat k=(2+3\lambda)\hat i+(-3+4\lambda) \hat j+(4-5\lambda)\hat k$
On comparing

$x=2+3\lambda, =-3+4\lambda, z=4-5\lambda$

$\Rightarrow \dfrac {x-2}{3}=\dfrac {y+3}{4}=\dfrac {z-4}{-5}=\lambda$

$\Rightarrow \dfrac {x-2}{3}=\dfrac {y+3}{4} =\dfrac {z-4}{-5}$

Let the equation of the plane containing the straight line $\displaystyle \frac{x - 1}{2} = \frac{y + 2}{-3} = \frac{z}{5}$ are perpendicular to the plane $x - y + z + 2 = 0$ be $kx + ny + z + 4 = m$ . Find $k+n$ ?

Line :

$\displaystyle \frac{x - 1}{2} = \frac{y + 2}{-3} = \frac{z}{5}$
Plane :

$x - y + z + 2 = 0$
The vector perpendicular to required plane is

$\begin{vmatrix} i & j & k \\ 2 &-3 &5 \\ 1 &-1 &1 \end{vmatrix} = 2i +3j + k$
Now equation of plane passing through

$(1, -2, 0)$ and perpendicular to

$2\hat{i} + 3\hat{j} + \hat{k}$ is given by,

$(x - 1) 2 +(y + 2) 3 +(z - 0)1 = 0$

$\Rightarrow 2x + 3y + z + 4 = 0$

On comparing with the equation

$kx+ny+z+4$ , we have

$k=2, n=3$ .

Therefore,

$k+n=2+3=5$

Ans: 5

Find the direction cosine of the line $\dfrac{x}{4} = \dfrac{y}{7} = \dfrac{z}{4}$

Looking at the given equation of the line, the line is parallel to

$4\hat{i} + 7\hat{j} + 4\hat{k}$

The direction cosines have to be such that they should be components of a unit vector.

Dividing the direction ratio by its modulus, we have

$\cfrac{4 \hat{i} + 7\hat{j} + 4\hat{k}}{\sqrt{4^2 + 7^2 + 4^2}}$

$= \cfrac{4 \hat{i} + 7\hat{j} + 4\hat{k}}{\sqrt{81}}$

$= \cfrac{4 \hat{i} + 7\hat{j} + 4\hat{k}}{9}$

The direction cosines thus can be written as

$\cfrac{4}{9}, \cfrac{7}{9}, \cfrac{4}{9}$

Find the acute angle between the lines whose direction ratios are $(5,12,-13)$ and $(3,-4,5)$ .

${ a }_{ 1 }=5,{ b }_{ 1 }=12,{ c }_{ 1 }=-13$

${ a }_{ 2 }=3,{ b }_{ 2 }=-4,{ c }_{ 2 }=5$
Acute angle

$\cos { \theta } =\left| \cfrac { { a }_{ 1 }{ a }_{ 2 }+{ b }_{ 1 }{ b }_{ 2 }+{ c }_{ 1 }{ c }_{ 2 } }{ \sqrt { { { a }_{ 1 } }^{ 2 }+{ { b }_{ 1 } }^{ 2 }+{ { c }_{ 1 } }^{ 2 } } \sqrt { { { a }_{ 2 } }^{ 2 }+{ { b }_{ 2 } }^{ 2 }+{ { { c }_{ 2 } }^{ 2 } } } } \right|$

$\quad \Rightarrow \cos { \theta } =\left| \cfrac { 15-48-65 }{ \sqrt { 25+144+169 } \sqrt { 9+16+25 } } \right|$

$\Rightarrow \cos { \theta } =\left| \cfrac { -98 }{ 13\sqrt { 2 } \times 5\times \sqrt { 2 } } \right|$

$\Rightarrow \cos { \theta } =\cfrac { -98 }{ 13\times 5\times 2 }$

$\Rightarrow \cos { \theta } =\left| -\cfrac { 49 }{ 65 } \right|$

$\Rightarrow \cos { \theta } =\cfrac { 49 }{ 65 } \quad \Rightarrow \theta =\cos ^{ -1 }{ \left( \cfrac { 49 }{ 65 } \right) }$

If a.b.c are noncoplanar find the point of intersection of the line passing through the points $2a+3b-c , \,3a+4b-2c$ with the line joining the points $a-2b+3c,\, d-6b+bc$

Let

$\vec{OA}=2a+3b-c$ ,

$\vec{OB}=3a+4b-2c$ ,

$\vec{OC}=a-2b+3c$ ,

$\vec{OD}=a-6b+6c$

The vector equation of the line joining the points

$\vec{OA}, \vec{OB}$ is

$\vec{r}=\vec{OA}+t(\vec{OB}-\vec{OA}), t\in R$

$=2a+3b-c+t(3a+4b-2c-2a-3b+c)$

$=2a+3b-c+t(a+b-c)$ .........

$(1)$

The vector equation of the line joining the points

$\vec{OC}, \vec{OD}$ is

$\vec{r}=\vec{OC}+s(\vec{OD}-\vec{OC}), s\in R$

$=a-2b+3c+s(a-6b+6c-a+2b-3c)$

$=a-2b+3c+s(-4b+3c)$ ........

$(2)$

from

$(1)$ , we have

$\vec{r}=(2+t)\vec{a}+(3+t)\vec{b}+(-1-t)\vec{c}$

and from

$(2)$ , we have

$\vec{r}=(1-a)a+(-2-4s)b+(3+3s)c$

$=a+(-2-4s)b+(3+3s)+c$

comparing

$\vec{a}, \vec{b}, \vec{c}$ coefficients on both sides, we get

$2+t=1\Rightarrow t=-1$

$3+t=-2-4s\Rightarrow 3-1=-2-4s$

$\Rightarrow s=-1$

Put

$t=-1$ in

$(1)$

$\vec{r}=2a+3b-c-1(a+b-c)$

$=2a+3b-c-a-b+c$

$=a+2b$ .

Find the direction cosines of the line which is perpendicular to the lines which direction cosines proportional to 1, -2, -2 and 0, 2, 1.

Let

$\left(a,b,c\right)$ be the direction cosine perpendicular to the lines

$\left(1,-2,-2\right)$ and

$\left(0,2,-1\right)$

$\Rightarrow {a}_{1}{a}_{2}+{b}_{1}{b}_{2}+{c}_{1}{c}_{2}=0$

$\Rightarrow a-2b-2c=0$ ...

$(1)$

and

$2b-c=0$ .....

$(2)$

$\dfrac{a}{-2+4}=\dfrac{-b}{1-0}=\dfrac{c}{2}$

$\Rightarrow \dfrac{a}{2}=\dfrac{b}{-1}=\dfrac{c}{2}$

Thus,

$\left(2,-1,2\right)$ is the direction cosine of the line perpendicular to the above lines.

Find the value of $k$ , if the points $(k, 3), (6, -2)$ and $(-3, 4)$ are collinear.

Points

$A(K,3)$ ;

$B(6,-2)$ ;

$C(-3,4)$ are coplanar.

Hence, area

$= 0$

$\Rightarrow \cfrac { 1 }{ 2 } \left| \begin{matrix} { x }_{ 1 }-{ x }_{ 2 } & { y }_{ 1 }-{ y }_{ 2 } \\ { x }_{ 1 }-{ x }_{ 3 } & { y }_{ 1 }-{ y }_{ 3 } \end{matrix} \right| =0$

$\Rightarrow \left| \begin{matrix} k-6 & 3+2 \\ k+3 & 3-4 \end{matrix} \right| = 0$

$\Rightarrow (k-6)(-1) - (5)(k+3) = 0$

$\Rightarrow -k+6-5k-15 = 0$

$\Rightarrow 6k = -6$

$\Rightarrow k =- \cfrac {3 }{ 2 }$

If l, m, n are the direction cosines of a straight line, then prove that $l^2+m^2+n^2=1$ .

Let

$OP$ be any line through the origin

$O$ which has direction cosines

$l,m,n$ .

Let

$P$ be the point having coordinates

$(x,y,z)$ and

$OP=r$ .

Then

$OP^2=x^2+y^2+z^2=r^2 \quad ...(1)$

From

$P$ draw

$PA,PB,PC$ perpendicular on the coordinate axes, so that

$OA=x,OB=y,OC=z$

Also,

$\angle POA=\alpha, \angle POB= \beta, \angle POC= \gamma$

From triangle

$AOP, l=cos\:\alpha=\dfrac{x}{r}\Rightarrow x=lr$

Similarly,

$y=mr, z=nr$

Adding all we get

$x^2+y^2+z^2=r^2(l^2+m^2+n^2)$

$\Rightarrow r^2=r^2(l^2+m^2+n^2)$

$\Rightarrow l^2+m^2+n^2=1$

Find the orthocentre of the triangle whose vertices are $A\left(a,0,0\right),\,B\left(0,b,0\right),\,C\left(0,0,c\right)$

The plane of the triangle is

$x/a+y/b+z/c=1$

Let

$O(\alpha,\beta,y)$ be the orthocenter

drs of OA are

$\alpha-a,\beta,y$

drs of BC are

$o,-b,c$

$OA \bot BC \Rightarrow \beta b=yc$

$OB \bot CA \Rightarrow \alpha a=yc$

$\therefore \alpha a=\beta b=yc=k^2(say)$

$o(\alpha ,\beta,y)$ lies on the plane

$\therefore \dfrac{\alpha}{a}+\dfrac{\beta}{b}+\dfrac{y}{c}=1\Rightarrow\dfrac{1}{k^2}=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}$

The orthocentre is

$\left(\dfrac{{k}^{2}}{a},\dfrac{{k}^{2}}{b},\dfrac{{k}^{2}}{c}\right)$
1351973_878298_ans_19cd7829b7ba49a6af3380af28388b88.png

Find the vector equation of the line which passes through the point with position vector $4\hat {i} - \hat {j} + 2\hat {k}$ and is in the direction of $-2\hat {i} + \hat {j} + \hat {k}$ .

Let

$\vec a=4\hat i-\hat j+2\hat k, \vec b=-2\hat i+\hat j+\hat k$

Equation of the line passing through point

$\vec a$ and having direction

$\vec{b}$ is

$\vec r=\vec a+\lambda \vec b$

Thus the vector equation of the line is

$\vec r=4\hat i-\hat j+2\hat k+\lambda(-2\hat i+\hat j+\hat k)$

The direction ratios of $\vec{ AB }$ are $(-2, 2, 1)$ If $A=(4, 1, 5)$ and $l$ (AB) $=6$ units, find the coordinates of $B.$

Let the coordinates of point B be

$\left( {{x_1},{y_1},{z_1}} \right)$

direction ratios of AB are

$-2,2,1$

So,

$x_1-4=-2r$

$y_1-1=2r$

$z_1-5=r$

But lenght of AB

$=6$

$\Rightarrow \sqrt {{{\left( {{x_1} - 4} \right)}^2} + {{\left( {{y_1} - 1} \right)}^2} + {{\left( {{z_1} - 5} \right)}^2}} = 6$

$\Rightarrow {\left( {{x_1} - 4} \right)^2} + {\left( {{y_1} - 1} \right)^2} + {\left( {{z_1} - 5} \right)^2} = 36$

$\Rightarrow 4{r^2} + 4{r^2} + {r^2} = 36$

$\Rightarrow 9{r^2} = 36$

$\Rightarrow r = 2$ or

$r = -2$

$r=2$

$x_1=0,$

$y_1=5,$

$z_1=7$

and if

$r=-2$

$x_1=8,$

$y_1=-3,$

$z_1=3$

Therefore the point B is

$\left( {0,5,7} \right)$ or

$\left( {8, - 3,3} \right)$

Find the vector equation of the passing through the point $2i+3j+k$ and parallel to the vector $4i-2j+3k$ .

The line passes through the point,

$\overrightarrow a = 2\hat i + 3\hat j + \hat k$

And parallel to the line,

$\overrightarrow b = 4\hat i - 2\hat j + 3\hat k$

The vector equation of the line passing through a point and parallel to the given line is,

$\overrightarrow r = \overrightarrow a + \lambda \overrightarrow b$

Substituting the values, we get,

$\overrightarrow r = \left( {2\hat i + 3\hat j + \hat k} \right) + \lambda \left( {4\hat i - 2\hat j + 3\hat k} \right)$

$\overrightarrow r = \left( {2 + 4\lambda } \right)\hat i + \left( {3 - 2\lambda } \right)\hat j + \left( {1 + 3\lambda } \right)\hat k$

The points with the co-ordinates $(2a, 3a)$ , $(3b, 2b)$ & $(c, c)$ are collinear.

If the area of triangle is

$0$ , then there will be no triangle, the points will be collinear,

$=4ab+3bc-2ac-9ab-2bc-2ac$

$=4ab-3bc-2b-9ab$

$=4ab-9ab+bc$

$=-5ab+bc$

$5ab=bc$

$5a=c$ .

1152251_1087080_ans_67838756f5f94fb5b61155efb0b25016.png

Find the equation of the plane which passes through the point $(3, 2, 0)$ and the line $\dfrac{x-4}{1}=\dfrac{y-6}{5}=\dfrac{z-4}{4}$ is?

The equation of a plane passing through a point

$\left(3,2,0\right)$ is

$a\left(x-3\right)+b\left(y-2\right)+c\left(z-0\right)=0$ .........

$(1)$

Since the above plane contains the line

$\dfrac{x-3}{1}=\dfrac{y-6}{5}=\dfrac{z-4}{4}$

So, the point

$\left(3,6,4\right)$ satisfies the equation of plane

$(1)$

$\therefore a\left(3-3\right)+b\left(6-2\right)+c\left(4-0\right)=0$

$\Rightarrow 4b+4c=0$

$\Rightarrow b+c=0$ or

$b=-c$ .........

$(2)$

As the plane contains the line, therefore normal to the plane is perpendicular to the line

${a}_{1}{a}_{2}+{b}_{1}{b}_{2}+{c}_{1}{c}_{2}=0$ where

${a}_{1}=a,\,{b}_{1}=b,\,{c}_{1}=c$ and

${a}_{2}=1,\,{b}_{2}=5,\,{c}_{2}=4$

$\therefore a\times 1+b\times 5+c\times 4=0$

$a+5b+4c=0$

$a-5b+4c=0$ from

$(2)$

$\Rightarrow a=c$

On putting the values of

$a,\,b$ and

$c$ in eqn

$(1)$ , we get

$c\left(x-3\right)-c\left(y-2\right)+c\left(z-0\right)=0$

$c\left(x-3-y+2+z-0\right)=0$

$x-y+z=1$ is the required equation of the plane.

The direction cosine of the two lines are determined by the relations $l-5m+3n=0$ and $7{l}^{2}+5{m}^{2}-3{n}^{2}=0$ .Find the solution.

It is given that,

$l - 5m + 3n = 0$

$l = 5m - 3n$

And,

$7{l^2} + 5{m^2} - 3{n^2} = 0$

$7{\left( {5m - 3n} \right)^2} + 5{m^2} - 3{n^2} = 0$

$7\left( {25{m^2} + 9{n^2} - 30mn} \right) + 5{m^2} - 3{n^2} = 0$

$180{m^2} + 60{n^2} - 210mn = 0$

$6{m^2} + 2{n^2} - 7mn = 0$

$6{m^2} - 4mn - 3mn + 2{n^2} = 0$

$2m\left( {3m - 2n} \right) - n\left( {3m - 2n} \right) = 0$

$\left( {2m - n} \right)\left( {3m - 2n} \right) = 0$

$2m - n = 0$

$m = \dfrac{n}{2}$

And,

$3m - 2n = 0$

$m = \dfrac{2}{3}n$

So, when $m = \dfrac{n}{2}$ , then, $n = 2m$ ,

$l = 5m - 3\left( {2m} \right)$

$l = 5m - 6m$

$l = - m$

$\dfrac{l}{{ - 1}} = \dfrac{m}{1}$

And since $\dfrac{m}{1} = \dfrac{n}{2}$ , so,

$\dfrac{l}{{ - 1}} = \dfrac{m}{1} = \dfrac{n}{2}$

The direction cosines will be,

$\dfrac{l}{{\sqrt {{l^2} + {m^2} + {n^2}} }} = \dfrac{{ - 1}}{{\sqrt {{{\left( { - 1} \right)}^2} + {{\left( 1 \right)}^2} + {{\left( 2 \right)}^2}} }}$

$= - \dfrac{1}{{\sqrt 6 }}$

$\dfrac{m}{{\sqrt {{l^2} + {m^2} + {n^2}} }} = \dfrac{1}{{\sqrt {{{\left( { - 1} \right)}^2} + {{\left( 1 \right)}^2} + {{\left( 2 \right)}^2}} }}$

$= \dfrac{1}{{\sqrt 6 }}$

$\dfrac{n}{{\sqrt {{l^2} + {m^2} + {n^2}} }} = \dfrac{2}{{\sqrt {{{\left( { - 1} \right)}^2} + {{\left( 1 \right)}^2} + {{\left( 2 \right)}^2}} }}$

$= \dfrac{2}{{\sqrt 6 }}$

Thus, direction cosines is $- \dfrac{1}{{\sqrt 6 }},\dfrac{1}{{\sqrt 6 }},\dfrac{2}{{\sqrt 6 }}$ .

When, $m = \dfrac{2}{3}n$ , then, $n = \dfrac{3}{2}m$ ,

$l = 5m - 3\left( {\dfrac{3}{2}m} \right)$

$l = 5m - \dfrac{9}{2}m$

$l = \dfrac{m}{2}$

$\dfrac{l}{1} = \dfrac{m}{2}$

And since $\dfrac{m}{1} = \dfrac{{2n}}{3}$ , then $\dfrac{m}{2} = \dfrac{n}{3}$ , so,

$\dfrac{l}{1} = \dfrac{m}{2} = \dfrac{n}{3}$

The direction cosines will be,

$\dfrac{l}{{\sqrt {{l^2} + {m^2} + {n^2}} }} = \dfrac{1}{{\sqrt {{{\left( 1 \right)}^2} + {{\left( 2 \right)}^2} + {{\left( 3 \right)}^2}} }}$

$= \dfrac{1}{{\sqrt {14} }}$

$\dfrac{m}{{\sqrt {{l^2} + {m^2} + {n^2}} }} = \dfrac{2}{{\sqrt {{{\left( 1 \right)}^2} + {{\left( 2 \right)}^2} + {{\left( 3 \right)}^2}} }}$

$= \dfrac{2}{{\sqrt {14} }}$

$\dfrac{n}{{\sqrt {{l^2} + {m^2} + {n^2}} }} = \dfrac{3}{{\sqrt {{{\left( 1 \right)}^2} + {{\left( 2 \right)}^2} + {{\left( 3 \right)}^2}} }}$

$= \dfrac{3}{{\sqrt {14} }}$

Thus, direction cosines is $\dfrac{1}{{\sqrt {14} }},\dfrac{2}{{\sqrt {14} }},\dfrac{3}{{\sqrt {14} }}$ .

If the line passing through the origin makes angles $\theta_{1},\theta_{2},\theta_{3}$ with the planes $XOY,XOZ$ and $ZOX$ respectively, then prove that $\cos^{2}\theta_{1}+\cos^{2}\theta_{2}+\cos^{2}\theta_{3}=2.$

Direction cosine are given as

$\cos \theta_1, \cos\theta_2, \cos\theta_3$

Here

$P(x, y, z)$ &

$0(0, 0, 0)$

$OP=\sqrt{(x-0)^2+(y-0)^2+(z-0)^2}$

$OP=\sqrt{x^2+y^2+z^2}$

$r=x^2+y^2+z^2$

According to diagonal

$\cos\theta_1=\dfrac{x}{r}$

$\cos\theta_2=\dfrac{y}{r}$

$\cos\theta_3=\dfrac{z}{r}$

$x=r\cos\theta_1$

$y=r\cos\theta_2$

$z=r\cos\theta_3$

Square & add

$x^2+y^2+z^2=r^2(\cos^2\theta_1+\cos^2\theta_2+\cos^2\theta_3)$

$r^2=r^2\cos^2\theta_1+\cos^2\theta_2+\cos^2\theta_3$

$\boxed{\cos^2\theta_1+\cos^2\theta_2+\cos^2\theta_3=1}$ .

1259790_1055615_ans_adec55e4e23c4cb58845d1dc57db6285.png

If $\vec {n}$ is a vector of magnitude $\sqrt{3}$ and is equally inclined with an acute angle with the coordinate axes. Find the vector and cartesian forms of equation of a plane which passes through $(2,1,-1)$ and is normal to $\vec {n}$ .

Let

$\alpha,\beta, \gamma$ angles of

$\bar{n}$

Given angels are equal So:

$\alpha=\beta=\gamma$

ie:

$\cos\alpha=\cos\beta=\cos\gamma$

So direction cosines

$\Rightarrow l=m=n\Rightarrow l=l=l$

As we know

$l^{2}+m^{2}+n^{2}=n\Rightarrow 8l^{2}=1$

$l=1/\sqrt{3}$

Given magnitude

$=\sqrt{3}$

$\bar{n}=\sqrt{3}\left(\dfrac{1}{\sqrt{3}}\hat{i}+\dfrac{1}{\sqrt{3}}\hat{j}+\dfrac{1}{\sqrt{3}}\hat{k}\right)=\hat{i}+\hat{j}+\hat{k}$

Given

$(2, 1, -1)\Rightarrow \bar{a}=2\hat{i}+\hat{j}+\hat{k}$ &

$\bar{n}=\hat{i}+\hat{j}+\hat{k}$

get plane equation

$\bar{r}.n=\bar{a}.\bar{n}$

$\bar{r}(\hat{i}+\hat{j}+\hat{k})=(2\hat{i}+\hat{j}-\hat{k})(\hat{i}+\hat{j}+\hat{k})$

$=2+1-1$

$\bar{r}(\hat{i}+\hat{j}+\hat{k})=2$

$(x\hat{i}+y\hat{j}+z\hat{k})(\hat{i}+\hat{j}+\hat{k})=2$

$[\bar{r}=x\hat{i}+y\hat{j}+z\hat{k}]$

$x+y+z=2$ As we want Question form

Find the equation of the plane passing through $(-2,1,3)$ and having $(3,-5,4)$ as $d.r's$ of its normal.

equation of plane having

$(3,-5,4)$ as d.r's is

$3x-5y+4z=c$

This plane pass through

$(-2,1,3)$ so the point will satisfy the equation

$3(-2)-5(1)+4(3)=c$

$c=1$

So, equation of plane is

$3x-5y+4z=1$

Find the direction cosines of two lines if they satisfy $l+3n=5m$ and $7l^2+5m^2=3n^2$ .

Given equations are

$l - 5m + 3n = 0 \, .....(1)$

$7 l^2 + 5m^2 - 3n^2 = 0 \, ......(2)$

From (1),

$l = 5m - 3n$

Substituting in (2)

$\Rightarrow 7 (5m - 3n)^2 + 5m^2 - 3n^2 = 0$

$\Rightarrow 7 (25 m^2 + 9n^2 - 30 mn) + 5m^2 - 3n^2 = 0$

$\Rightarrow 6m^2 - 7mn + 2n^2 = 0$

$\Rightarrow (3m - 2n) (2m - n) = 0$

$\Rightarrow m = \dfrac{2}{3} n$ or

$m = \dfrac{n}{2}$

$l = 5 \left(\dfrac{2}{3}\right) n - 3n = \dfrac{1}{3} n$ or

$l = \dfrac{5n}{2} - 3n$

$= -\dfrac{n}{2}$

So, the direction cosines are

$\dfrac{1}{3} n, \dfrac{2}{3} n, n$

$\dfrac{-n}{2} , \dfrac{n}{2} , n$

$\Rightarrow$ direction ratios are

$1, 2, 3$ or

$-1, 1, 2$

$\therefore$ The direction cosines are

$\pm \dfrac{1}{\sqrt{14}} , \pm \dfrac{2}{\sqrt{14}} , \pm \dfrac{3}{\sqrt{14}}$

$\mp \dfrac{1}{\sqrt{6}} , \pm \dfrac{1}{\sqrt{6}} , \pm \dfrac{2}{\sqrt{6}}$

Show that the points are collinear
(1) $A(3,2,-4), B(9,8,-10), C(-2,-3,1)$

Hence, these points are collinear.
1350587_1121921_ans_4a6ec5cfe3f441a3b954b159139e480c.jpg

The direction cosines l, m, n of two lines satisfy $l+m+n=0$ , $l^2+m^2=n^2$ . Find the measure of the angle between the lines.

1358420_1135334_ans_a7b55be977bb4008837721e569c69591.jpg

Find the direction cosines of perpendicular from the origin to the plane $\bar{r}(2\hat{i}+3\hat{j}+6\hat{k})+7=0$

$\bar{r} (2i+3j+6k)+7=0$ .

$\bar{r}(-2i-3j-6k)=7$

now

$n=-2i-3j-6k$

$|n|=\sqrt{4+9+36}=\sqrt{49}=7$

$\bar{r}\left(\dfrac{-2}{7}i-\dfrac{3}{7}j-\dfrac{6}{7}k\right)=\dfrac{7}{7}$

Now

dc's are

$\left(\dfrac{-2}{7}, \dfrac{-3}{7}, \dfrac{-6}{7}\right)$ .

The direction cosines of two lines are given by $al+bm+cn=0,\,\,u{l}^{2}+v{m}^{2}+w{n}^{2}=0$ .Find the conition for the lines to be parallel.

1351956_1245145_ans_d32881f3e3f44c1d80f2915507b44bf2.PNG

Find the direction cosines and direction ratios of the line joining the points
$A(5,6,-3), B(1,-6,3)$

Given points are

$A=\left( { 5,6,-3 } \right) \, \, \, \& \, \, B=\left( { 1,-6,3 } \right)$

Direction ratios are

$a={ x_{ 2 } }-{ x_{ 1 } }=1-5=-4, \\ b={ y_{ 2 } }-{ y_{ 1 } }=-6-6=-12, \\ c={ z_{ 2 } }-{ z_{ 1 } }=3+3=6,$

Now,

$\sqrt { { a^{ 2 } }+{ b^{ 2 } }+{ c^{ 2 } } } =\sqrt { { { \left( { -4 } \right) }^{ 2 } }+{ { \left( { -12 } \right) }^{ 2 } }+{ 6^{ 2 } } } =\sqrt { 16+144+36 }=14$

Direction cosines are

$l=\dfrac { a }{ { \sqrt { { a^{ 2 } }+{ b^{ 2 } }+{ c^{ 2 } } } } } \, \, \, \, =\dfrac { { -4 } }{ { 14 } } \, \, \, \, \, =-\dfrac { 2 }{ 7 } \\ m=\dfrac { b }{ { \sqrt { { a^{ 2 } }+{ b^{ 2 } }+{ c^{ 2 } } } } } \, \, =\dfrac { { -12 } }{ { 14 } } \, \, \, =-\dfrac { 6 }{ 7 } \\ n=\dfrac { c }{ { \sqrt { { a^{ 2 } }+{ b^{ 2 } }+{ c^{ 2 } } } } } \, \, =\dfrac { 6 }{ { 14 } } \, \, \, =\dfrac { 3 }{ 7 }$

Direction ratios are

$\left( { -4,-12,6 } \right)$ and direction cosines are

$\left( { -\dfrac { 2 }{ 7 } ,\dfrac { { -6 } }{ 7 } ,\dfrac { 3 }{ 7 } } \right)$

Find the direction cosines and direction ratios of the line joining the points
$A(0,0,0,), B(4,8,-8)$

Given the points

${ { A } }\left( { 0,0,0 } \right) \, \, \& \, \, B\left( { 4,8,-8 } \right)$

Direction ratios are

$a=4-0\, \, \, =4 \\ b=8-0\, \, \, \, =8 \\ c=-8-0\, \, =-8$

Now,

$\sqrt { { a^{ 2 } }+{ b^{ 2 } }+{ c^{ 2 } } } =\sqrt { { 4^{ 2 } }+{ 8^{ 2 } }+{ 8^{ 2 } } } \\ \, =\sqrt { 16+64+64 } \, \, \, \, =\sqrt { 144 } \, \, \, \, =12$

Direction cosines are

$l=\dfrac { a }{ { \sqrt { { a^{ 2 } }+{ b^{ 2 } }+{ c^{ 2 } } } } } \, \, \, \, =\dfrac { 4 }{ { 12 } } \, \, \, =\dfrac { 1 }{ 3 } \\ m=\dfrac { b }{ { \sqrt { { a^{ 2 } }+{ b^{ 2 } }+{ c^{ 2 } } } } } \, \, =\dfrac { 8 }{ { 12 } } \, \, \, =\dfrac { 2 }{ 3 } \\ n=\dfrac { c }{ { \sqrt { { a^{ 2 } }+{ b^{ 2 } }+{ c^{ 2 } } } } } \, \, \, =\dfrac { { -8 } }{ { 12 } } \, \, \, =-\dfrac { 2 }{ 3 }$

Find the direction cosines of a line that pass through the point $P\left(1,4,6\right)$ and $Q\left(5,1,11\right)$ and is so directed that it make an acute angle with the positive direction of $y-axis$

$\begin{array}{l} P\left( { 1,4,6 } \right) \\ Q=\left( { 5,1,11 } \right) \\ Direction\, \, \cos ine=\frac { { { x_{ 2 } }-{ x_{ 1 } } } }{ { PQ } } ,\frac { { { y_{ 2 } }-{ y_{ 1 } } } }{ { PQ } } ,\frac { { { z_{ 2 } }-{ z_{ 1 } } } }{ { PQ } } \\ Direction \\ \left( { 5-1 } \right) \left( { 1-4 } \right) \left( { 11-6 } \right) \\ 4,-3,5 \\ OC=\frac { 4 }{ { \sqrt { { 4^{ 2 } }+{ { \left( { -3 } \right) }^{ 2 } }+{ 5^{ 2 } } } } } ,\frac { { -3 } }{ { \sqrt { { 4^{ 2 } }+{ { \left( { -3 } \right) }^{ 2 } } } +{ 5^{ 2 } } } } ,\frac { 5 }{ { \sqrt { { 4^{ 2 } }+{ { \left( { -3 } \right) }^{ 2 } }+{ 5^{ 2 } } } } } \\ =\frac { 4 }{ { 5\sqrt { 2 } } } ,\frac { { -3 } }{ { 5\sqrt { 2 } } } ,\frac { 5 }{ { 5\sqrt { 2 } } } \end{array}$

Find the direction cosines and direction ratios of the line joining the points
$A(4,2,-6), B(-2,1,3)$

$\begin{array}{l} direction\, \, ratio\, \, of\, \, line\, \, AB=\left( { -2,0 } \right) ,\left( { 1-2 } \right) ,\left( { 3+6 } \right) =-6,-1,9 \\ \Rightarrow and\, \, direction\, \, { { cosine } }\, \, is \\ \frac { { -6 } }{ { \sqrt { { { \left( { -6 } \right) }^{ 2 } }+{ { \left( { -1 } \right) }^{ 2 } }+{ 9^{ 2 } } } } } ,\, \, \, \frac { 1 }{ { \sqrt { { { \left( { -6 } \right) }^{ 2 } }+{ { \left( { -1 } \right) }^{ 2 } }+{ 9^{ 2 } } } } } ,\, \, \frac { 9 }{ { \sqrt { { { \left( { -6 } \right) }^{ 2 } }+{ { \left( { -1 } \right) }^{ 2 } }+{ 9^{ 2 } } } } } \\ \frac { { -6 } }{ { \sqrt { 118 } } } ,\, \, \frac { { -1 } }{ { \sqrt { 118 } } } ,\, \, \frac { 9 }{ { \sqrt { 118 } } } \, \, \, Ans. \end{array}$

Find the direction cosines and direction ratios of the line joining the points
$A(1,3,5), B(-1,0,-1)$

Given the points

${ { A } }\left( { 1,3,5 } \right) \, \, \& \, \, B\left( { -1,0,-1 } \right)$

Direction ratios are

$a=-1-1\, \, \, \, =-2 \\ b=0-3\, \, =-3 \\ c=-1-5\, \, =-6,\, \, \, \, \, \, \, \, \, \,$

Now,

$\sqrt { { a^{ 2 } }+{ b^{ 2 } }+{ c^{ 2 } } } =\sqrt { { { \left( { -2 } \right) }^{ 2 } }+{ { \left( { -3 } \right) }^{ 2 } }+{ { \left( { -6 } \right) }^{ 2 } } } =\sqrt { 4+9=36 } \, \, \, \, =\sqrt { 49 } \, \, \, \, =7$

Direction cosines are

$l=\dfrac { a }{ { \sqrt { { a^{ 2 } }+{ b^{ 2 } }+{ c^{ 2 } } } } } \, \, \, \, =\dfrac { { -2 } }{ 7 } \\ m=\dfrac { b }{ { \sqrt { { a^{ 2 } }+{ b^{ 2 } }+{ c^{ 2 } } } } } \, \, =\dfrac { { -3 } }{ 7 } \\ n=\dfrac { c }{ { \sqrt { { a^{ 2 } }+{ b^{ 2 } }+{ c^{ 2 } } } } } \, \, \, =\dfrac { { -6 } }{ 7 }$ .

Find the direction ratio of a line perpendicular to the two lines whose direction ratios are
$-2,1,-1$ and $-3,-4,1$

Given direction ratios are :

$-2,1,-1$ and

$-3,-4,1$

Let

$a,b$ and

$c$ be the direction ratios of the line perpendicular to the given lines.

Thus, we have,

$-2a+b-c=0$

$-3a-4b+c=0$

Cross multiplying, we get

$\dfrac{a}{1\times 1-\left(-4\right)\times\left(-1\right)}=\dfrac{b}{\left(-3\right)\times\left(-1\right)-\left(-2\right)\times 1}=\dfrac{c}{-2\times -4-\left(-3\right)\times 1}$

$\Rightarrow \dfrac{a}{1-4}=\dfrac{b}{3+2}=\dfrac{c}{8+3}$

$\Rightarrow \dfrac{a}{-3}=\dfrac{b}{5}=\dfrac{c}{11}$

Let us find

$\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}=\sqrt{{\left(-3\right)}^{2}+{5}^{2}+{11}^{2}}=\sqrt{9+25+121}=\sqrt{155}$

Thus, the direction ratios of the required line are

$-3,5,11$

The direction cosines are

$:\dfrac{-3}{\sqrt{155}},\dfrac{5}{\sqrt{155}},\dfrac{11}{\sqrt{155}}$

Find the vector equation for the line passing through the points $(-1,0,2)$ and $(3,4,6)$ .

$\textbf{Step -1: Defining Vector equation of a line . }$

$\implies \vec r=-\hat i+2\hat k+\lambda(4\hat i+4\hat j+4\hat k)$

$\textbf{Step -2: Finding vector .}$

$\text{Given points are (-1, 0, 2) and (3, 4, 6)}$

$\text{Direction cosines:}(3-(-1),4-0,6-2)=(4,4,4)$

$\text{Vector equation of a line}=\vec r=\vec a+\lambda (\vec b-\vec a)$

$\implies \vec r=-\hat i+2\hat k+\lambda(4\hat i+4\hat j+4\hat k)$

$\textbf{Hence, the vector equation of the line passing through the given points is}$

$\mathbf{\vec r=-\hat i+2\hat k+\lambda(4\hat i+4\hat j+4\hat k) .}$

Find the equation of the plane passing through $(1,2-4)$ and perpendicular to the line $\cfrac{x+1}{3}=\cfrac{y}{-1}=\cfrac{z-8}{4}$ . Also find the distance of the point $(3,-1,4)$ from the plane.

2057020_1443890_ans_50c00eccfd974f40af565af61db689cb.jpg

Find the vector and cartesian equations of the plane passing through the point $(2, -1, 1)$ and perpendicular to the line having direction ratios $4, 2, -3$ .

1683707_1708485_ans_8d5354ee279a46729184e581c15cabeb.jpeg

A line passes through the points (3,1, 2) and (5 , - 1, 1), find the direction ratios and direction cosines of the line.

2043926_1539766_ans_2e984df9b82244148b79bf2e05ce2a1c.jpeg

Show that the following points are collinear:
$P(3, -2, 4), Q(1, 1, 1)$ and $R(-1, 4, 2)$ .

1969611_1711081_ans_c2efa5ad40b34a71a399ca18ad81fb0a.jpg

Find the equation of the plane passing through the point $(-1, 2, 1)$ and perpendicular to the line joining the points $(-3, 1, 2)$ and $(2, 3, 4)$ . Also, find the perpendicular distance of the origin from this plane.

Let the equation of plane passing through

$(-1, 2, 1)$ be

$a(x + 1) + b(y - 2) + c(z - 1) = 0 ...(i)$

Now the equation of line joining point

$(-3, 1, 2)$ and

$(2, 3, 4)$ is

$\dfrac{x + 3}{2 + 3} = \dfrac{y - 1}{3 - 1} = \dfrac{z - 2}{4 - 2} \Rightarrow \dfrac{x + 3}{5} = \dfrac{y - 1}{2} = \dfrac{z - 2}{2} ...(ii)$

$\because$ Plane (i) is perpendicular to line (ii)

$\Rightarrow \dfrac{a}{5} = \dfrac{b}{2} = \dfrac{c}{2} = \lambda (say)$ [

$\because$ Normal vector of (i) is parallel to line (ii) ]

$\Rightarrow a = 5\lambda, b = 2\lambda, c = 2\lambda$

Hence, equation of required plane is

$5\lambda (x + 1) + 2\lambda (y - 2) + 2\lambda (z - 1) = 0$

$\Rightarrow 5x + 5 + 2y - 4 + 2z = 0$

$\Rightarrow 5x + 2y + 2z - 1 = 0$

If d is the distance from origin to plane then

$d = \left | \dfrac{0.x + 0.y + 0.z - 1}{\sqrt{5^2 + 2^2 + 2^2}} \right | =\dfrac{1}{\sqrt{33}}$

Three Dimensional Geometry - Class 12 Commerce Maths - Extra Questions

(-2, -1), (-1, -4) and (-4,-1)

Find the values of pp so the line 1−x3=7y−142p=z−32\cfrac{1-x}{3}=\cfrac{7y-14}{2p}=\cfrac{z-3}{2} and 7−7x3p=y−51=6−z5\cfrac{7-7x}{3p}=\cfrac{y-5}{1}=\cfrac{6-z}{5} are at right angles.

Find the direction cosines of the vector joining the points A(1,2,−3)A (1, 2, -3) and B(−1,−2,1)B (-1, -2, 1) directed from AA to BB.

If OO be the origin and the coordinates of PP be (1,2,-3)(1,2,-3), then find the equation of the plane passing through PP and perpendicular to OPOP.

If a line makes angles 90^{\circ}, 135^{\circ}, 45^{\circ}90^{\circ}, 135^{\circ}, 45^{\circ} with the x, yx, y and Z-\text{axes}Z-\text{axes} respectively, find its direction cosines.

Find the equation plane passing (a, b, c)(a, b, c) and parallel to the plane \vec {r} \cdot (\hat {i} + \hat {j} + \hat {k}) = 2\vec {r} \cdot (\hat {i} + \hat {j} + \hat {k}) = 2.

Find the direction cosines of the vector \hat {i}+2\hat {j}+3\hat {k}\hat {i}+2\hat {j}+3\hat {k}.

Find the angle between the lines whose direction ratios are a,b,ca,b,c and b-c, c-a, a-bb-c, c-a, a-b

Find the vector and the Cartesian equations of the lines that pass through the origin and (5, -2, 3)(5, -2, 3)

Show that the points A (1, 2, 7), B (2, 6, 3)A (1, 2, 7), B (2, 6, 3) and C (3, 10, -1)C (3, 10, -1) are collinear .

The equation of a line is 5x-3=15y+7=3-10z5x-3=15y+7=3-10z. Write the direction cosines of the line.

The acute angle between the planes 2x-y+z=62x-y+z=6 and x+y+2z=3x+y+2z=3 is _____.

Write the direction cosines of xx-axis.

Determine the direction cosines of the line through the points (2,0,3)(2,0,3) and (4,1,1)(4,1,1)

The Cartesian equation of a line is: 2x-3=3y+1=5-6z2x-3=3y+1=5-6z. Find the vector equation of a line passing through (7, -5, 0)(7, -5, 0) and parallel to the given line.

Find the acute angle between the lines passing through (-3 , -1 , 0), (2, -3 , 1) and (1 , 2 , 3),(-1 ,4 , -2) respectively.

Find direction cosine line x=3z+2,y=2-5zx=3z+2,y=2-5z.

Find the direction collinear of a line which makes equal angles with the positive co-ordinate axes.

Find the length of the perpendicular drawn from the origin to the plane 2x-3y+6z+21=02x-3y+6z+21=0.

Find the value of \beta\beta so that the line \dfrac {x - 2}{6} = \dfrac {y - 1}{\beta} = \dfrac {z + 5}{-4}\dfrac {x - 2}{6} = \dfrac {y - 1}{\beta} = \dfrac {z + 5}{-4} is perpendicular to the plane 3x - y - 2z = 73x - y - 2z = 7.

Find the direction cosines of the line \dfrac{{x + 2}}{2} = \dfrac{{2y - 5}}{3};z = - 1\dfrac{{x + 2}}{2} = \dfrac{{2y - 5}}{3};z = - 1. Also, find the vector equation of the line.

Find the vector equation of the line through A(3,4,-7)A(3,4,-7) and B(6,-1,1)B(6,-1,1). Also find the cartesian form.

Show that the lines whose d.c.s are given by l + m + n = 0 , 2mn + 3ln - 5lm = 0l + m + n = 0 , 2mn + 3ln - 5lm = 0 are perpendicular to each other.

If a line is making same angle \theta\theta with the positive direction of X-,Y-X-,Y- and Z-Z-axes, then show that \theta=\cos^{-1}(\dfrac {1}{\sqrt {3}}\theta=\cos^{-1}(\dfrac {1}{\sqrt {3}}.

Find the equation of the line in vector and in cartesion form that passes through the point with position vector 2\hat {i}-\hat{j}+4\ \hat {k}2\hat {i}-\hat{j}+4\ \hat {k} and is in the direction \hat {i}+2\hat {j}-\hat {k}\hat {i}+2\hat {j}-\hat {k}.

Find the direction cosines of the lines , connected by the relations: l + m + n = 0 and 2lm + 2ln - mn = 0.

If the lines through the points (4, 1, 2)(4, 1, 2) and (5, k, 0)(5, k, 0) is parallel to the line through the points (2, 1, 1)(2, 1, 1) and (3, 3, 1)(3, 3, 1), find kk.

If the d.c's of a line are \left( {\dfrac{1}{c}\,,\,\dfrac{1}{c}\,,\,\dfrac{1}{c}} \right)\left( {\dfrac{1}{c}\,,\,\dfrac{1}{c}\,,\,\dfrac{1}{c}} \right) , find c.

find the angle between the lines whose direction cosines satisfy the equation \ell + m + n = 0,\,\,{\ell ^2} + {m^2} - {n^2} = 0\ell + m + n = 0,\,\,{\ell ^2} + {m^2} - {n^2} = 0

Find the co-ordinate of the pt. pp, if \ell \left( PQ \right) =21 \ell \left( PQ \right) =21, Q\equiv \left( -2,1,-8 \right) Q\equiv \left( -2,1,-8 \right) and directional cosines of PQPQ are (6/7), (2/7), (3/7)(6/7), (2/7), (3/7)?

Find the vector equation of the line joining (1,2,3)(1,2,3) and (-3,4,3)(-3,4,3) and show pependicular to the z-axis

The cartesian equation of a line is \dfrac{{x + 3}}{2} = \dfrac{{y - 5}}{4} = \dfrac{{z + 6}}{2}\dfrac{{x + 3}}{2} = \dfrac{{y - 5}}{4} = \dfrac{{z + 6}}{2} find the vector equation of the line?

If the d.c's of a line are \left( {\frac{1}{c},\,\frac{1}{c},\,\frac{1}{c}} \right) \left( {\frac{1}{c},\,\frac{1}{c},\,\frac{1}{c}} \right) , find c.

If a line makes angles \alpha,\beta,\gamma\alpha,\beta,\gamma with positive directions of X,Y,Z-axes, what is the value of {sin}^{2}{\alpha}\ +\ {sin}^{2}{\beta} +\ {sin}^{2}{\gamma}{sin}^{2}{\alpha}\ +\ {sin}^{2}{\beta} +\ {sin}^{2}{\gamma}?

Find the direction ratio of the line \dfrac{x-1}{2}=3y=\dfrac{2z+3}{4}\dfrac{x-1}{2}=3y=\dfrac{2z+3}{4}

If a line makes angle \left( {\alpha ,\,\beta ,\,\gamma } \right)\left( {\alpha ,\,\beta ,\,\gamma } \right) with the co-ordinates axes. Prove that {\sin ^2}\alpha \, + \,{\sin ^2}\beta \, - \,{\cos ^2}\gamma \, = \,1{\sin ^2}\alpha \, + \,{\sin ^2}\beta \, - \,{\cos ^2}\gamma \, = \,1

Find the equation of line passing through (5, 0, 5)(5, 0, 5) & (2, 1, 3)(2, 1, 3). Also show that (5, 0, 5), (2, 1, 3)(5, 0, 5), (2, 1, 3) & (-4, 3, -1)(-4, 3, -1) are collinear.

Find the vector equation of the line joining points 2\hat i + \hat j + 3\hat k2\hat i + \hat j + 3\hat k and - 4\hat i + 3\hat j - \hat k - 4\hat i + 3\hat j - \hat k .

Find the measure of the angle between two lines if their direction cosines \ell, m, n\ell, m, n satisfy \ell+m-n=0, \ell^{2}+m^{2}-n^{2}=0\ell+m-n=0, \ell^{2}+m^{2}-n^{2}=0.

The angle between the lines whose direction cosines satisfy the equations l+m+n=0l+m+n=0 and l^2=m^2+n^2l^2=m^2+n^2 is?

Find the direction cosines of the vector \vec{r}=(6\hat{i}+2\hat{j}-3\hat{k})\vec{r}=(6\hat{i}+2\hat{j}-3\hat{k}).

Given three points whose position are vectors x\hat i + y\hat j + z\hat kx\hat i + y\hat j + z\hat k and - \hat i - \hat j - \hat i - \hat j. Find the condition for the points to be collinear.

Using vectors, find the value of \lambda\lambda such that the points (\lambda, -10, 3), (1, -1, 3)(\lambda, -10, 3), (1, -1, 3) and (3, 5, 3)(3, 5, 3) are collinear.

Find the vector equation of the line passing through the points 2\hat i + 3\hat j + \hat k2\hat i + 3\hat j + \hat k and parallel to the vector 4\hat i - 2\hat j + 3\hat k4\hat i - 2\hat j + 3\hat k .

Show that points A\left( a,b+c \right), B\left( b,c+a \right), C\left( c,a+b \right)A\left( a,b+c \right), B\left( b,c+a \right), C\left( c,a+b \right) are collinear.

The number of straight line that are equally inclined to the three dimensional co- ordinate axes , is

Find the equation of plane passing through the point (3,2,4)(3,2,4) and perpendicular to the X-X-axis.

The direction cosines of a line are \dfrac{1}{2}, \dfrac{1}{2}, k\dfrac{1}{2}, \dfrac{1}{2}, k and k > 0k > 0, then k=?k=?

The cartasian equation of a line is \dfrac {x-5}{3}=\dfrac {y+4}{7}=\dfrac {z-6}{2}\dfrac {x-5}{3}=\dfrac {y+4}{7}=\dfrac {z-6}{2}. Find its vector equation.

Show that the points A(-2\hat{i}+3\hat{j}+5\hat{k}), B(\hat{i}+2\hat{j}+3\hat{k})A(-2\hat{i}+3\hat{j}+5\hat{k}), B(\hat{i}+2\hat{j}+3\hat{k}) and C(7\hat{i}-\hat{k})C(7\hat{i}-\hat{k}) are collinear.

Find the equation of plane passing through the point (2-1,3)(2-1,3) and perpendicular to the X-X-axis.

Show that A(1,-2,-8),B(5,0,-2)A(1,-2,-8),B(5,0,-2) and C(11,3,7)C(11,3,7) are collinear.

If the direction cosines of a line are \left(\dfrac{1}{2},0,n\right)\left(\dfrac{1}{2},0,n\right) and n<0n<0. Then find nn.

Determine whether the points are collinear.P(-2,3),B(1,2),C(4,1)P(-2,3),B(1,2),C(4,1)

The position vectors \vec {a},\vec {b},\vec {c}\vec {a},\vec {b},\vec {c} of three points satisfy the relation 2\vec {a}+7\vec {b}+5\vec {c}=\vec {0}2\vec {a}+7\vec {b}+5\vec {c}=\vec {0}. Are these points collinear?

If a line makes angles \alpha ,\beta\alpha ,\beta & \gamma\gamma with OX,OYOX,OY & OZOZ respectively Prove that: (i) \sin ^{ 2 }{ \alpha } +\sin ^{ 2 }{ \beta } +\sin ^{ 2 }{ \gamma } =2\sin ^{ 2 }{ \alpha } +\sin ^{ 2 }{ \beta } +\sin ^{ 2 }{ \gamma } =2

The direction ratios of a vector are 2, -3, 42, -3, 4 . Find its direction cosines

Obtain the equation of the plane which passes through (3, 4, -5) and (1, 2, 3) and parallel to Z axis.

Prove that the points A(4,-3,-1), B(5,-7,6)A(4,-3,-1), B(5,-7,6) and C(3,1,-8)C(3,1,-8) are collinear.

Find the value of p so the line \dfrac{1-x}{3}=\dfrac{7y-14}{2p}=\dfrac{z-3}{2}\dfrac{1-x}{3}=\dfrac{7y-14}{2p}=\dfrac{z-3}{2} and \dfrac{7-7x}{3p}=\dfrac{y-5}{1}=\dfrac{6-z}{5}\dfrac{7-7x}{3p}=\dfrac{y-5}{1}=\dfrac{6-z}{5} are at right angles.

Find the vector equation of the line through A(3,4,-7) and B(6,-1,1).

If a line makes angles 90^{o}, 135^{o}, 45^{o}90^{o}, 135^{o}, 45^{o} with x, y and z-axis respectively, find its direction cosines.

Find the angles at which the normal vector to the plane 4x+8y+z=54x+8y+z=5 is inclined to the coordinate axes.

Write the equation of the plane passing through the point \left({x}_{1},{y}_{1},{z}_{1}\right)\left({x}_{1},{y}_{1},{z}_{1}\right) and having a, b, ca, b, c, as direction ratios of its normal.

If \vec{PO}+\vec{OQ}=\vec{QO}+\vec{OR}\vec{PO}+\vec{OQ}=\vec{QO}+\vec{OR}, prove that the points P, Q, RP, Q, R are collinear.

A ray makes angles \dfrac{\pi}{3},\dfrac{\pi}{3}\dfrac{\pi}{3},\dfrac{\pi}{3} with \overrightarrow { OX }\overrightarrow { OX } and \overrightarrow { OY }\overrightarrow { OY } respectively. Find the angle made by it with \overrightarrow { OZ }\overrightarrow { OZ }.

Find the equation of the line passing through the point (-1, 2, 1)(-1, 2, 1) and parallel to the line \dfrac {2x - 1}{4} = \dfrac {3y + 5}{2} = \dfrac {2 - z}{3}\dfrac {2x - 1}{4} = \dfrac {3y + 5}{2} = \dfrac {2 - z}{3}.

Find the vector equation of the coordinates planes.

Find the equation of the line joining the points (-1,3)(-1,3) and (4,-2)(4,-2).

The foot of the perpendicular drawn from the origin to the plane is (4,2,6)(4,2,6). Find the equation of the plane.

Find a normal to the plane x+2y+3z-6=0x+2y+3z-6=0

Find the direction cosines if the passing through two points (-2,4,-5)(-2,4,-5) and (1,2,3).(1,2,3).

Find the equation of the plane passing through the point (2,3,1)(2,3,1) given that the direction ratios of normal to the plane are proportional to 5, 3,25, 3,2

If a line has direction ratio 2,-1,-2,2,-1,-2, determine its direction cosines.

Find the vector equation of the line passing through the point (1, 2, -4)(1, 2, -4) and parallel to the line \dfrac {x - 3}{4} = \dfrac {y - 5}{2} = \dfrac {z + 1}{3}\dfrac {x - 3}{4} = \dfrac {y - 5}{2} = \dfrac {z + 1}{3}.

Find the vector equation of a plane which is at 42 unit distance from the origin and which is normal to the vector 2\hat{i}+\hat{j}-2\hat{k}2\hat{i}+\hat{j}-2\hat{k}

Write the vector equation of a line whose Cartesian equations are \cfrac{x-5}{3}=\cfrac{y+4}{7}=\cfrac{6-z}{2}\cfrac{x-5}{3}=\cfrac{y+4}{7}=\cfrac{6-z}{2}

Find the vector equation of the line passing through the point vector \hat{i}+\hat{j}+\hat{k}\hat{i}+\hat{j}+\hat{k} and 2\hat{i}-\hat{j}+\hat{k}2\hat{i}-\hat{j}+\hat{k}

Show that the points A(2,3,4),B(-1,-2,1)A(2,3,4),B(-1,-2,1) and C(5,8,7)C(5,8,7) are collinear.

Find the values of $p$ so the line $\cfrac{1-x}{3}=\cfrac{7y-14}{2p}=\cfrac{z-3}{2}$ and $\cfrac{7-7x}{3p}=\cfrac{y-5}{1}=\cfrac{6-z}{5}$ are at right angles.

Find the direction cosines of the vector joining the points $A (1, 2, -3)$ and $B (-1, -2, 1)$ directed from $A$ to $B$ .

If $O$ be the origin and the coordinates of $P$ be $(1,2,-3)$ , then find the equation of the plane passing through $P$ and perpendicular to $OP$ .

If a line makes angles $90^{\circ}, 135^{\circ}, 45^{\circ}$ with the $x, y$ and $Z-\text{axes}$ respectively, find its direction cosines.

Find the equation plane passing $(a, b, c)$ and parallel to the plane $\vec {r} \cdot (\hat {i} + \hat {j} + \hat {k}) = 2$ .

Find the direction cosines of the vector $\hat {i}+2\hat {j}+3\hat {k}$ .

Find the angle between the lines whose direction ratios are $a,b,c$ and $b-c, c-a, a-b$

Find the vector and the Cartesian equations of the lines that pass through the origin and $(5, -2, 3)$

Show that the points $A (1, 2, 7), B (2, 6, 3)$ and $C (3, 10, -1)$ are collinear .

The equation of a line is $5x-3=15y+7=3-10z$ . Write the direction cosines of the line.

The acute angle between the planes $2x-y+z=6$ and $x+y+2z=3$ is _____.

Write the direction cosines of $x$ -axis.

Determine the direction cosines of the line through the points $(2,0,3)$ and $(4,1,1)$

The Cartesian equation of a line is: $2x-3=3y+1=5-6z$ . Find the vector equation of a line passing through $(7, -5, 0)$ and parallel to the given line.

Find direction cosine line $x=3z+2,y=2-5z$ .

Find the length of the perpendicular drawn from the origin to the plane $2x-3y+6z+21=0$ .

Find the value of $\beta$ so that the line $\dfrac {x - 2}{6} = \dfrac {y - 1}{\beta} = \dfrac {z + 5}{-4}$ is perpendicular to the plane $3x - y - 2z = 7$ .

Find the direction cosines of the line $\dfrac{{x + 2}}{2} = \dfrac{{2y - 5}}{3};z = - 1$ . Also, find the vector equation of the line.

Find the vector equation of the line through $A(3,4,-7)$ and $B(6,-1,1)$ . Also find the cartesian form.

Show that the lines whose d.c.s are given by $l + m + n = 0 , 2mn + 3ln - 5lm = 0$ are perpendicular to each other.

If a line is making same angle $\theta$ with the positive direction of $X-,Y-$ and $Z-$ axes, then show that $\theta=\cos^{-1}(\dfrac {1}{\sqrt {3}}$ .

Find the equation of the line in vector and in cartesion form that passes through the point with position vector $2\hat {i}-\hat{j}+4\ \hat {k}$ and is in the direction $\hat {i}+2\hat {j}-\hat {k}$ .

If the lines through the points $(4, 1, 2)$ and $(5, k, 0)$ is parallel to the line through the points $(2, 1, 1)$ and $(3, 3, 1)$ , find $k$ .

If the d.c's of a line are $\left( {\dfrac{1}{c}\,,\,\dfrac{1}{c}\,,\,\dfrac{1}{c}} \right)$ , find c.

find the angle between the lines whose direction cosines satisfy the equation $\ell + m + n = 0,\,\,{\ell ^2} + {m^2} - {n^2} = 0$

Find the co-ordinate of the pt. $p$ , if $\ell \left( PQ \right) =21$ , $Q\equiv \left( -2,1,-8 \right)$ and directional cosines of $PQ$ are $(6/7), (2/7), (3/7)$ ?

Find the vector equation of the line joining $(1,2,3)$ and $(-3,4,3)$ and show pependicular to the z-axis

The cartesian equation of a line is $\dfrac{{x + 3}}{2} = \dfrac{{y - 5}}{4} = \dfrac{{z + 6}}{2}$ find the vector equation of the line?

If the d.c's of a line are $\left( {\frac{1}{c},\,\frac{1}{c},\,\frac{1}{c}} \right)$ , find c.

If a line makes angles $\alpha,\beta,\gamma$ with positive directions of X,Y,Z-axes, what is the value of ${sin}^{2}{\alpha}\ +\ {sin}^{2}{\beta} +\ {sin}^{2}{\gamma}$ ?

Find the direction ratio of the line $\dfrac{x-1}{2}=3y=\dfrac{2z+3}{4}$

If a line makes angle $\left( {\alpha ,\,\beta ,\,\gamma } \right)$ with the co-ordinates axes. Prove that ${\sin ^2}\alpha \, + \,{\sin ^2}\beta \, - \,{\cos ^2}\gamma \, = \,1$

Find the equation of line passing through $(5, 0, 5)$ & $(2, 1, 3)$ . Also show that $(5, 0, 5), (2, 1, 3)$ & $(-4, 3, -1)$ are collinear.

Find the vector equation of the line joining points $2\hat i + \hat j + 3\hat k$ and $- 4\hat i + 3\hat j - \hat k$ .

Find the measure of the angle between two lines if their direction cosines $\ell, m, n$ satisfy $\ell+m-n=0, \ell^{2}+m^{2}-n^{2}=0$ .

The angle between the lines whose direction cosines satisfy the equations $l+m+n=0$ and $l^2=m^2+n^2$ is?

Find the direction cosines of the vector $\vec{r}=(6\hat{i}+2\hat{j}-3\hat{k})$ .

Given three points whose position are vectors $x\hat i + y\hat j + z\hat k$ and $- \hat i - \hat j$ . Find the condition for the points to be collinear.

Using vectors, find the value of $\lambda$ such that the points $(\lambda, -10, 3), (1, -1, 3)$ and $(3, 5, 3)$ are collinear.

Find the vector equation of the line passing through the points $2\hat i + 3\hat j + \hat k$ and parallel to the vector $4\hat i - 2\hat j + 3\hat k$ .

Show that points $A\left( a,b+c \right), B\left( b,c+a \right), C\left( c,a+b \right)$ are collinear.

Find the equation of plane passing through the point $(3,2,4)$ and perpendicular to the $X-$ axis.

The direction cosines of a line are $\dfrac{1}{2}, \dfrac{1}{2}, k$ and $k > 0$ , then $k=?$

The cartasian equation of a line is $\dfrac {x-5}{3}=\dfrac {y+4}{7}=\dfrac {z-6}{2}$ . Find its vector equation.

Show that the points $A(-2\hat{i}+3\hat{j}+5\hat{k}), B(\hat{i}+2\hat{j}+3\hat{k})$ and $C(7\hat{i}-\hat{k})$ are collinear.

Find the equation of plane passing through the point $(2-1,3)$ and perpendicular to the $X-$ axis.

Show that
$A(1,-2,-8),B(5,0,-2)$ and $C(11,3,7)$ are collinear.

If the direction cosines of a line are $\left(\dfrac{1}{2},0,n\right)$ and $n<0$ . Then find $n$ .

Determine whether the points are collinear.
$P(-2,3),B(1,2),C(4,1)$

The position vectors $\vec {a},\vec {b},\vec {c}$ of three points satisfy the relation $2\vec {a}+7\vec {b}+5\vec {c}=\vec {0}$ . Are these points collinear?

If a line makes angles $\alpha ,\beta$ & $\gamma$ with $OX,OY$ & $OZ$ respectively Prove that: (i) $\sin ^{ 2 }{ \alpha } +\sin ^{ 2 }{ \beta } +\sin ^{ 2 }{ \gamma } =2$

The direction ratios of a vector are $2, -3, 4$ . Find its direction cosines

Prove that the points $A(4,-3,-1), B(5,-7,6)$ and $C(3,1,-8)$ are collinear.

Find the value of p so the line $\dfrac{1-x}{3}=\dfrac{7y-14}{2p}=\dfrac{z-3}{2}$ and $\dfrac{7-7x}{3p}=\dfrac{y-5}{1}=\dfrac{6-z}{5}$ are at right angles.

If a line makes angles $90^{o}, 135^{o}, 45^{o}$ with x, y and z-axis respectively, find its direction cosines.

Find the angles at which the normal vector to the plane $4x+8y+z=5$ is inclined to the coordinate axes.

Write the equation of the plane passing through the point $\left({x}_{1},{y}_{1},{z}_{1}\right)$ and having $a, b, c$ , as direction ratios of its normal.

If $\vec{PO}+\vec{OQ}=\vec{QO}+\vec{OR}$ , prove that the points $P, Q, R$ are collinear.

A ray makes angles $\dfrac{\pi}{3},\dfrac{\pi}{3}$ with $\overrightarrow { OX }$ and $\overrightarrow { OY }$ respectively. Find the angle made by it with $\overrightarrow { OZ }$ .

Find the equation of the line passing through the point $(-1, 2, 1)$ and parallel to the line $\dfrac {2x - 1}{4} = \dfrac {3y + 5}{2} = \dfrac {2 - z}{3}$ .

Find the equation of the line joining the points $(-1,3)$ and $(4,-2)$ .

The foot of the perpendicular drawn from the origin to the plane is $(4,2,6)$ . Find the equation of the plane.

Find a normal to the plane $x+2y+3z-6=0$

Find the direction cosines if the passing through two points $(-2,4,-5)$ and $(1,2,3).$

Find the equation of the plane passing through the point $(2,3,1)$ given that the direction ratios of normal to the plane are proportional to $5, 3,2$

If a line has direction ratio $2,-1,-2,$ determine its direction cosines.

Find the vector equation of the line passing through the point $(1, 2, -4)$ and parallel to the line $\dfrac {x - 3}{4} = \dfrac {y - 5}{2} = \dfrac {z + 1}{3}$ .

Find the vector equation of a plane which is at 42 unit distance from the origin and which is normal to the vector $2\hat{i}+\hat{j}-2\hat{k}$

Write the vector equation of a line whose Cartesian equations are $\cfrac{x-5}{3}=\cfrac{y+4}{7}=\cfrac{6-z}{2}$

Find the vector equation of the line passing through the point vector $\hat{i}+\hat{j}+\hat{k}$ and $2\hat{i}-\hat{j}+\hat{k}$

Show that the points $A(2,3,4),B(-1,-2,1)$ and $C(5,8,7)$ are collinear.

Find the equation of the line passing through the point $(2, -1, 3)$ and parallel to the line $\vec {r} = (\hat {i} - 2\hat {j} + \hat {k}) + \lambda (2\hat {i} + 3\hat {j} - 5\hat {k})$ .

Find the vector equation of the line passing through the point having position vector $-2\hat{i} + \hat{j}+\hat{k}$ and parallel to vector $4\hat{i}-\hat{j}+2\hat{k}$

State True or False.
The following position vectors are collinear.
$a-2b+3c$ ,
$2a+3b-4c$ ,
$-7b+10c$ .

Test the co-linearity of the set of points with the following position vectors: If they are co-linear then write 1 otherwise write 0.
$5a+4b+2c, \ 6a+2b-c, \ 7a+b-c.$

If $A(1, 2, 1), B(2, 6, 2)$ and $C(\lambda, 14, 4)$ are collinear, then value of $\lambda$ is

The cartesian equation of a line is $\dfrac {x - 5}{3} = \dfrac {y + 4}{7} = \dfrac {z - 6}{2}$ write its vector form.

If a line has the direction ratios $-18, 12,- 4$ , then what are its direction cosines?

The angle between the following pair of lines:
$\cfrac{x}{2}=\cfrac{y}{2}=\cfrac{z}{1}$ and $\cfrac{x-5}{4}=\cfrac{y-2}{1}=\cfrac{z-3}{8}$ is $\theta = \cos^{-1}\dfrac{a}{3}$ then $a=$

Find the equation of the line in vector and in cartesian form that passes through the point with position vector $2\hat {i} - \hat {j} + 4\hat {k}$ and is in the direction $\hat {i} + 2\hat {j} - \hat {k}$ .