MCQ Questions for Class 12 Commerce Maths Three Dimensional Geometry Quiz 5

Equation of the plane passing through a point with position vector

$\displaystyle 3\hat{i}-3\hat{j}+\hat{k}$ & normal to the line joining the points with position vectors

$\displaystyle 3\hat{i}+4\hat{j}-\hat{k}$ &

$\displaystyle 2\hat{i}-\hat{j}=5\hat{k}$ is

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$\displaystyle \overline{r}.\left ( -\hat{i}-5\hat{j}+6\hat{k} \right )+18=0$
0%
$\displaystyle \overline{r}.\left ( \hat{i}-5\hat{j}+6\hat{k} \right )=22$
0%
$\displaystyle \overline{r}.\left ( \hat{i}+5\hat{j}-6\hat{k} \right )+18=0$
0%
$\displaystyle \overline{r}.\left ( -\hat{i}+5\hat{j}+6\hat{k} \right )+12=0$

Explanation

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

$\displaystyle \Rightarrow \overline{r} \cdot \overline{n}= \overline{a} \cdot \overline{n}$ where

$\displaystyle \overline{n}=AB= -\hat{i}-5\hat{j}+6\hat{k}$

$\displaystyle \Rightarrow \overline{r} \cdot \left ( -\hat{i}-5\hat{j}+6\hat{k} \right )=\left ( 3\hat{i}-3\hat{j}+\hat{k} \right ) \cdot \left ( -\hat{i}-5\hat{j}+6\hat{k} \right )$

$\displaystyle \Rightarrow \overline{r} \cdot \left ( \hat{i}+5\hat{j}-6\hat{k} \right )+18=0$
Hence the correct option becomes c.

Vector equation of the plane passing through a point having position vector

$\displaystyle 2\hat{i}+3\hat{j}-4\hat{k}$ and perpendicular to the vector

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

0%
$\displaystyle \overrightarrow{r}\cdot \left ( 2\hat{i}-\hat{j}+2\hat{k} \right )+7=0$
0%
$\displaystyle \overrightarrow{r}\cdot \left ( 2\hat{i}-\hat{j}+2\hat{k} \right )=7$
0%
$\displaystyle \overrightarrow{r}\cdot \left ( -3\hat{i}-2\hat{j}-3\hat{k} \right )=0$
0%
$\displaystyle \overrightarrow{r}\cdot \left ( 2\hat{i}-\hat{j}+2\hat{k} \right )=9$

Explanation

Using

$\displaystyle \left ( \overrightarrow{r} - \overrightarrow{a} \right )\cdot \overrightarrow{n}=0$
i.e.

$\displaystyle \overrightarrow{r} \cdot \overrightarrow{n}= \overrightarrow{a} \cdot \overrightarrow{n}$

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

$\displaystyle \Rightarrow \overrightarrow{r} \cdot \left ( 2\hat{i}-\hat{j}+2\hat{k} \right )=4-3-8$

$\displaystyle \Rightarrow \overrightarrow{r} \cdot \left ( 2\hat{i}-\hat{j}+2\hat{k} \right )= -7$
Hence option (a) is correct.

A line makes the same angle

$\theta$ with each of the

$x$ and

$z$ axis. If the angle

$\beta$ , which it makes with

$y-$ axis is such that

$\sin ^{2}\beta =3\sin ^{2}\theta$ then

$\cos ^{2}\theta$ equals

0%
$\displaystyle \dfrac{3}{5}$
0%
$\displaystyle \frac{1}{5}$
0%
$\displaystyle \frac{2}{3}$
0%
$\displaystyle \frac{2}{5}$

Explanation

If a line makes the angle

$\alpha,\beta,\gamma$ with x,y,z axix respectively then

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

$\Rightarrow 2l^{2}+m^{2}=1 \: or \: 2n^{2}+m^{2}=1$

$\Rightarrow 2\cos^{2}\theta=1- \cos ^{2} \beta\left ( \alpha=\gamma=\theta \right )$

$2 \cos ^{2}\theta = \sin ^{2} \beta$

$\Rightarrow 2\cos ^{2} \theta = 3\sin ^{2} \theta$

$\Rightarrow 5\cos ^{2} \theta=3$

Direction cosines of the vector

$\displaystyle \bar{v}=a_{1}\hat{i}+a_{2}\hat{j}+a_{3}\hat{k}$ are

0%
$\displaystyle < a_{1},a_{2},a_{3} >$
0%
$\displaystyle <-a_{1},-a_{2},-a_{3}>$
0%
$\displaystyle <\frac {a_{1}}{\left|\bar{v}\right|},\frac {a_{2}}{\left|\bar{v}\right|,}\frac {a_{3}}{\left|\bar{v}\right|},>$
0%
none of these

Explanation

The directions cosines will be

$\dfrac{a_{1}}{\sqrt{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}},\dfrac{a_{2}}{\sqrt{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}},\dfrac{a_{3}}{\sqrt{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}}$

$\Rightarrow\dfrac{a_{1}}{|\overline{v}|},\dfrac{a_{2}}{|\overline{v}|},\dfrac{a_{3}}{|\overline{v}|}$

The Equation of the plane through a point

$\displaystyle 2\hat{i}-\hat{j}+4\hat{k}$ & parallel to the plane

$\displaystyle \overline{r}.\left ( 2\hat{i}+4\hat{j}-7\hat{k} \right )=6$ is

0%
$\displaystyle \overline{r}.\left ( 2\hat{i}+4\hat{j}-7\hat{k} \right )=21$
0%
$\displaystyle \overline{r}.\left ( 2\hat{i}+4\hat{j}-7\hat{k} \right )=14$
0%
$\displaystyle \overline{r}.\left ( 2\hat{i}+4\hat{j}-7\hat{k} \right )=42$
0%
$\displaystyle \overline{r}.\left ( 2\hat{i}+4\hat{j}-7\hat{k} \right )=28$

Explanation

Equation of the plane parallel to the given plane is

$\displaystyle \overline{r} \cdot \left ( 2\hat{i}+4\hat{j}-7\hat{k} \right )+\lambda=0$ ...........................................................

$(i)$
This plane passes through the point with P.V.

$\displaystyle 2\hat{i}-\hat{j}-4\hat{k}$

$\displaystyle \therefore \left ( 2\hat{i}-\hat{j}-4\hat{k} \right ) \cdot \left ( 2\hat{i}-\hat{j}-4\hat{k} \right )+\lambda=0$

$\displaystyle \Rightarrow 4-4+28+\lambda=0$

$\displaystyle \therefore \lambda= -28$
Putting

$\displaystyle \lambda= -28$ in

$(i)$ we get the required plane given
by

$\displaystyle \overline{r} \cdot \left ( 2\hat{i}+4\hat{j}-7\hat{k} \right )=28$
Hence (d) is the correct choice.

The line passes through the points

$\left ( 5,1,a \right )$ &

$\left ( 3,b,1 \right )$ crosses the

$yz$ plane at the point

$\displaystyle \left ( 0,\frac{17}{2},-\frac{13}{2} \right )$ ,then

0%
$a= 4, b= 6$
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$a= 6, b= 4$
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$a= 8, b= 2$
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$a= 2, b= 8$

Explanation

Equation of the line through the points

$( 5,1,a )$ &

$( 3,b,1)$ is

$\displaystyle \frac{x-5}{3-5}=\frac{y-1}{b-1}=\frac{z-a}{1-a}=\lambda$
Now it passes through

$\displaystyle \left ( 0,\frac{17}{2},\frac{-13}2{} \right )$

$\displaystyle \therefore \frac{0-5}{-2}=\frac{17/2-1}{b-1}=\frac{-13/2-a}{1-a}=\lambda \:$

$\Rightarrow \lambda =\dfrac{5}{2}$

$\displaystyle \therefore \frac{17/2-1}{b-1}=\frac{5}{2}$

$\Rightarrow b=4$
and

$\displaystyle \frac{-13/2-a}{1-a}=\frac{5}{2}$

$\Rightarrow a=6$

The scalar product form of equation of plane

$\displaystyle \overline{r}=\left ( s-2t \right )\hat{i}+\left ( 3-t \right )\hat{j}+\left ( 2s-t \right )\hat{k}$ is

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$\displaystyle \overrightarrow{r}\cdot \left ( 2\hat{i}-5\hat{j}-\hat{k} \right )+15=0$
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$\displaystyle \overrightarrow{r}\cdot \left ( 2\hat{i}-5\hat{j}-\hat{k} \right )=15$
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$\displaystyle \overrightarrow{r}\cdot \left ( 2\hat{i}-5\hat{j}-\hat{k} \right )=3$
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$\displaystyle \overrightarrow{r}\cdot \left ( 2\hat{i}-5\hat{j}-\hat{k} \right )= -3$

Explanation

Given

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

$\displaystyle \Rightarrow \overrightarrow{r}=3\hat{j}+s\left ( \hat{i}+2\hat{k} \right )+t\left ( -2\hat{i}+\hat{j}-\hat{k} \right )$
This equation represent the equation ofthe plane through a point having position vector

$\displaystyle \overrightarrow{a}=3\hat{j}$ parallel to the vectors

$\displaystyle \overrightarrow{b}=\hat{i}+2\hat{k}$ and

$\overrightarrow{c}=-2\hat{i}-\hat{j}+2\hat{k}$ .
Normal Vector to the plane is given by

$\displaystyle \overrightarrow{n}=\left | \overrightarrow{b} \times \overrightarrow{c} \right |=2\hat{i}-5\hat{j}-\hat{k}$
Now using scalar product form of equation of plane

$\displaystyle \overrightarrow{r} \cdot \overrightarrow{n}= \overrightarrow{a} \cdot \overrightarrow{n}$

$\displaystyle \Rightarrow \overrightarrow{r} \cdot \left ( 2\hat{i}-5\hat{j}-\hat{k} \right )= (3\hat{j})\left ( 2\hat{i}-5\hat{j}-\hat{k} \right )$

$\displaystyle \Rightarrow \overrightarrow{r} \cdot \left ( 2\hat{i}-5\hat{j}-\hat{k} \right )= -15$
Hence choice (a) is correct.

If the three points with position vectors

$\displaystyle \bar{a}-2\bar{b}+3\bar{c}, \ 2\bar{a}+\lambda \bar{b}-4\bar{c}, \ -7\bar{b}+10\bar{c}$ are collinear, then

$\displaystyle \lambda=$

0%
$1$
0%
0%
$3$
0%
none of these

Explanation

The given vectors are collinear, so

$l(\bar{a} - 2\bar{b} + 3\bar{c}) + k(2\bar{a} + \lambda\bar{b} - 4\bar{c}) = (l + k)(-7\bar{b} + 10\bar{c})$
Comparing the coefficients of

$\bar{a} \rightarrow l + 2k = 0$

$\bar{b} \rightarrow -2l + \lambda k = -7l -7k$

$\bar{c} \rightarrow 3l - 4k = 10l + 10k$

$\Rightarrow l = -2k$ and so

$\lambda = 3$

Find the equation of the plane containing the vectors

$\displaystyle \bar{\alpha}$ and

$\displaystyle\bar{ \beta}$ and passing through the point

$\displaystyle \bar{a}$

0%
$\displaystyle (\bar{r}-\bar{a})\cdot (\bar{\alpha} \times \bar{\beta}) =0$
0%
$\displaystyle (\bar{r}+\bar{a})(\bar{\alpha }\times \bar{\beta}) =0$
0%
$\displaystyle (\bar{r}-\bar{a}) (\bar {a}\cdot \bar{b}) =0$
0%
none of these

Explanation

Normal vector to the plane is,

$\bar{\alpha}\times \bar{\beta}$
Thus equation of plane passing through

$\vec{a}$ is given by,

$(\vec{r}-\vec{a})\cdot (\bar{\alpha}\times \bar{\beta})=0$

${ l }_{ 1 },{ m }_{ 1 },{ n }_{ 1 }$ and

${ l }_{ 2 },{ m }_{ 2 },{ n }_{ 2 }$ are DCs of the two lines inclined to each other at an angle

$\theta$ , then the DCs of the internal bisector of the angle between these lines are

0%
$\displaystyle \frac { { l }_{ 1 }+{ l }_{ 2 } }{ 2\sin { \frac { \theta }{ 2 } } } ,\frac { { m }_{ 1 }+{ m }_{ 2 } }{ 2\sin { \frac { \theta }{ 2 } } } ,\frac { { n }_{ 1 }+{ n }_{ 2 } }{ 2\sin { \frac { \theta }{ 2 } } }$
0%
$\displaystyle \frac { { l }_{ 1 }+{ l }_{ 2 } }{ 2\cos { \frac { \theta }{ 2 } } } ,\frac { { m }_{ 1 }+{ m }_{ 2 } }{ 2\cos { \frac { \theta }{ 2 } } } ,\frac { { n }_{ 1 }+{ n }_{ 2 } }{ 2\cos{ \frac { \theta }{ 2 } } }$
0%
$\displaystyle \frac { { l }_{ 1 }-{ l }_{ 2 } }{ 2\sin { \frac { \theta }{ 2 } } } ,\frac { { m }_{ 1 }-{ m }_{ 2 } }{ 2\sin { \frac { \theta }{ 2 } } } ,\frac { { n }_{ 1 }-{ n }_{ 2 } }{ 2\sin { \frac { \theta }{ 2 } } }$
0%
$\displaystyle \frac { { l }_{ 1 }-{ l }_{ 2 } }{ 2\cos { \frac { \theta }{ 2 } } } ,\frac { { m }_{ 1 }-{ m }_{ 2 } }{ 2\cos { \frac { \theta }{ 2 } } } ,\frac { { n }_{ 1 }-{ n }_{ 2 } }{ 2\cos{ \frac { \theta }{ 2 } } }$

Explanation

Let

$OA$ and

$OB$ be two lines with

$DC's{ l }_{ 1 },{ m }_{ 1 },{ n }_{ 1 }$ and

$\displaystyle { l }_{ 2 },{ m }_{ 2 },{ n }_{ 2 }.$

Let

$OA=OB=1.$

Then co-ordinates of

$A$ and

$B$ are

$\displaystyle \left( { l }_{ 1 },{ m }_{ 1 },{ n }_{ 1 } \right)$ and

$\displaystyle \left( { l }_{ 2 },{ m }_{ 2 },{ n }_{ 2 } \right)$ respectively.

Let

$OC Z$ be the bisector of

$\angle AOB$ such that

$C$ is the mid-point of

$AB$ and so its co-ordinates are

$\displaystyle \left( \frac { { l }_{ 1 }+{ l }_{ 2 } }{ 2 } ,\frac { { m }_{ 1 }+{ m }_{ 2 } }{ 2 } ,\frac { { n }_{ 1 }+{ n }_{ 2 } }{ 2 } \right)$

$\therefore DR's$ of

$OC$ are

$\displaystyle \frac { { l }_{ 1 }+{ l }_{ 2 } }{ 2 } ,\frac { { m }_{ 1 }+{ m }_{ 2 } }{ 2 } ,\frac { { n }_{ 2 }+{ n }_{ 2 } }{ 2 }$

$\therefore$ We have

$\displaystyle OC=\sqrt { { \left( \frac { { l }_{ 1 }+{ l }_{ 2 } }{ 2 } \right) }^{ 2 }+{ \left( \frac { { m }_{ 1 }+{ m }_{ 2 } }{ 2 } \right) }^{ 2 }+{ \left( \frac { { n }_{ 1 }+{ n }_{ 2 } }{ 2 } \right) }^{ 2 } }$

$\displaystyle =\frac { 1 }{ 2 } \sqrt { \left( { l }_{ 1 }^{ 2 }+{ m }_{ 1 }^{ 2 }+{ n }_{ 1 }^{ 2 } \right) +\left( { l }_{ 2 }^{ 2 }+{ m }_{ 2 }^{ 2 }+{ n }_{ 2 }^{ 2 } \right) +2\left( { l }_{ 1 }{ l }_{ 2 }+{ m }_{ 1 }{ m }_{ 2 }+{ n }_{ 1 }{ n }_{ 2 } \right) }$

$\displaystyle =\frac { 1 }{ 2 } \sqrt { 2+2\cos { \theta } } \quad \left[ \therefore \cos { \theta } ={ l }_{ 1 }{ l }_{ 2 }+{ m }_{ 1 }{ m }_{ 2 }+{ n }_{ 1 }{ n }_{ 2 } \right]$

$\displaystyle =\frac { 1 }{ 2 } \sqrt { 2\left( 1+\cos { \theta } \right) } =\cos { \left( \frac { \theta }{ 2 } \right) } .$

$\therefore DCs$ of

$\overrightarrow { OC }$ are

$\displaystyle \frac { { l }_{ 1 }+{ l }_{ 2 } }{ 2\left( OC \right) } ,\frac { { m }_{ 1 }+{ m }_{ 2 } }{ 2\left( OC \right) } ,\frac { { n }_{ 1 }+{ n }_{ 2 } }{ 2\left( OC \right) }$

i.e.,

$\displaystyle \frac { { l }_{ 1 }+{ l }_{ 2 } }{ \frac { 2\cos { \theta } }{ 2 } } ,\frac { { m }_{ 1 }+{ m }_{ 2 } }{ \frac { 2\cos { \theta } }{ 2 } } ,\frac { { n }_{ 1 }+{ n }_{ 2 } }{ \frac { 2\cos { \theta } }{ 2 } }$

Determine if the points

$(1,5)$

$(2,3)$ and

$(-2,-11)$ are collinear.

0%
True
0%
False

Explanation

The given points are

$A(1,5)$ ,

$B(2,3)$ and

$C(-2,-11)$ .

Let us calculate the distance :

$AB$ ,

$BC$ and

$CA$ by using distance formula.

$AB =\sqrt { (2-1)^{ 2 }+(3-5)^{ 2 } } =\sqrt { (1)^{ 2 }+(-2)^{ 2 } }$

$=\sqrt {1+4} = \sqrt{ 5 }$ units

$BC =\sqrt { (-2-2)^{ 2 }+(-11-3)^{ 2 } }=\sqrt { (-4)^{ 2 }+(-14)^{ 2 } }$

$=\sqrt {16+196} =\sqrt {212} = 2\sqrt{53}$ units

$CA =\sqrt { (-2-1)^{ 2 }+(-11-5)^{ 2 } }$

$=\sqrt { (-3)^{ 2 }+(-16)^{ 2 } } =\sqrt {9+256} = \sqrt {265 }$

$=\sqrt {5}\times\sqrt {53}$ units

From the above we see that :

$AB+BC\neq CA$

Hence, the above stated points

$A(1,5)$ ,

$B(2,3)$ and

$C(-2,-11)$ are not collinear.

In each of the following find the value of

$k$ , for which the points are collinear.
(i)

$(7,-2)$ ,

$(5,1)$ ,

$(3,k)$
(ii)

$(8,1)$ ,

$(k,-4)$ ,

$(2,-5)$

0%

(i) $k = 4$
0%
(i) $k = 5$
0%
(ii) $k = 3$
0%
(ii) $k = 2$

Explanation

Since the given points are collinear, they do not form a triangle, which means area of the triangle is Zero.

Area of a triangle with vertices $({ x }_{ 1 },{ y }_{ 1 })$ ; $({ x }_{ 2 },{ y }_{ 2 })$ and $({ x }_{ 3 },{ y }_{ 3 })$ is $\left| \dfrac { { x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right|$

1) Substituting the points

$({ x }_{ 1 },{ y }_{ 1 }) = (7,-2)$ ;

$({ x }_{ 2 },{ y }_{ 2 }) = (5,1)$ and

$({ x }_{ 3 },{ y }_{ 3 }) = (3,k)$

In the area formula, we get

$\left| \dfrac { 7(1-k) + 5(k+2) + 3(-2-1) }{ 2 } \right| = 0$

$\left| \dfrac { 7 -7k + 5k + 10 - 9 }{ 2 } \right| = 0$

$\left| \dfrac { 8 -2k }{ 2 } \right| = 0$

$\Rightarrow 8 - 2k = 0$

$\Rightarrow k = 4$

2) Substituting the points

$({ x }_{ 1 },{ y }_{ 1 }) = (8,1)$ ;

$({ x }_{ 2 },{ y }_{ 2 }) = (k,-4)$ and

$({ x }_{ 3 },{ y }_{ 3 }) = (2,-5)$ in the area formula, we get

$\left| \dfrac { 8(-4+5) + k(-5-1) + 2(1+4) }{ 2 } \right| = 0$

$\left| \dfrac { 8 -6k +10 }{ 2 } \right| = 0$

$\left| \dfrac { 18 -6k }{ 2 } \right| = 0$

$\Rightarrow 18 - 6k = 0$

$\Rightarrow k = 3$

Equation of a plane containing

$L_{1}$ and

$L_{2}$ is

0%
$x + y + z = 0$
0%
$3x -2y -z =0$
0%
$x -3y + 2z = 0$
0%
$x + y + z = 42$

Explanation

Normal vector of the plane is given by,

$\vec{n} = \begin{vmatrix} \hat{i}&\hat{j}&\hat{k}\\-3&2&1\\1&-3&2 \end{vmatrix}=7\hat{i}+7\hat{j}+7\hat{k}=7(\hat{i}+\hat{j}+\hat{k})$
Hence equation of plane is,

$(\vec{r} - (7\hat{j}-7\hat{k}))\cdot \vec{n}=0$

$\Rightarrow 7(x-0)+7(y-7)+7(z+7)=0$

$\Rightarrow x+y+z=0$

The acute angle between the lines

$x =-2 + 2t, y = 3 -4t, z = -4 + t$ and

$x= 2 -t, y= 3 + 2t, z= -4 + 3t$ is

0%
$\displaystyle \sin^{-1}\frac{1}{\sqrt{3}}$
0%
$\displaystyle \cos^{-1}\frac{1}{\sqrt{6}}$
0%
$\displaystyle \cos^{-1}\frac{1}{\sqrt{5}}$
0%
$\displaystyle \cos^{-1}2/3$

Explanation

Given lines are

$x =-2 + 2t, y = 3 -4t, z = -4 + t$ and

$x= 2 -t, y= 3 + 2t, z= -4 + 3t$

$\Rightarrow \displaystyle\frac { x+2 }{ 2 } =\displaystyle\frac { y-3 }{ -4 } =\displaystyle\frac { z+4 }{ 1 }$ and

$\displaystyle\frac { x-2 }{ -1 } =\displaystyle\frac { y-3 }{ 2 } =\displaystyle\frac { z+4 }{ 3 }$
if

$a_{1},b_{1},c_{1}$ and

$a_{2},b_{2},c_{2}$ are direction ratios of two lines, then angle between them is

$\cos\theta = \displaystyle\frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$

$\Rightarrow \cos\theta = \displaystyle\frac{-1}{\sqrt{6}}$

$\Rightarrow \theta = \pi-\cos ^{ -1 }{ \left(\displaystyle \frac { 1 }{ \sqrt { 6 } } \right) }$
Hence, acute angle between the lines is

$\cos ^{ -1 }{ \left(\displaystyle \frac { 1 }{ \sqrt { 6 } } \right) }$

Find the value of

$p$ for which the points

$(-5,1)$ ,

$(1,p)$ and

$(4, -2)$ are collinear.

0%
$1$
0%
$0$
0%
$-1$
0%
$2$

Explanation

The given points are

$A(-5,1)$ ,

$B(1,p)$ and

$C(4,-2)$
We have

$(x_1= -5, y_1= 1), (x_2= 1, y_2= p)$ and

$(x_3= 4, y_3= -2)$
The given points

$A, B$ and

$C$ are collinear
Therefore,

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

$\Rightarrow (-5)\cdot (p+2)+1\cdot (-2-1)+4\cdot (1-p)=0$

$\Rightarrow (-5p-10-3+4-4p)=0$

$\Rightarrow -9p= -9$

$\Rightarrow p= -1$
Hence,

$p= -1$

If the foot of the perpendicular from the origin to a plane is

$\displaystyle \left ( a,b,c \right )$ , the equation of the plane is

0%
$\displaystyle \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=3$
0%
$\displaystyle ax+by+cz=3$
0%
$\displaystyle ax+by+cz=a^2+b^2+c^2$
0%
$\displaystyle ax+by+cz=a+b+c$

Explanation

Let

$\displaystyle P\left ( a,b,c \right )$ be the foot of the perpendicular from the origin to the plane, then direction ratios of

$OP$ are

$\displaystyle a-0,b-0,c-0$ i.e.

$\displaystyle a,b,c$ .
So, the equation of the plane passing through

$\displaystyle P\left ( a,b,c \right )$ the direction ratios of the normal to which
are

$\displaystyle a,b,c$ is

$\displaystyle a\left ( x-a \right )+b\left ( y-b \right )+c\left ( z-c \right )=0$

$\displaystyle \Rightarrow$

$\displaystyle ax+by+cz=a^2+b^2+c^2$

Let

$N$ be the foot of the perpendicular of length

$p$ , from the origin to a plane and

$l$ ,

$m$ ,

$n$ be the direction cosines of

$ON$ , the equation of the plane is

0%
$px\, +\, my\, +\, nz= l$
0%
$lx\, +\, py\, +\, nz= m$
0%
$lx\, +\, my\, +\, pz= n$
0%
$lx\, +\, my\, +\, nz= p$

Explanation

given direction cosines of the line normal to the plane are

$l, m, n$

$\Rightarrow \hat{n} = l\hat{i}+m\hat{j}+n\hat{n}$
we know equation of plane whose distance is p from the origin is given by,

$\vec{r} \cdot \hat{n} = p$

$\Rightarrow lx + my + nz = p$

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

For what value of

$m$ , the points

$(3,5)$ ,

$(m,6)$ and

$\begin{pmatrix} \dfrac { 1 }{ 2 },\dfrac {15 }{ 2 } \end{pmatrix}$ are collinear?

0%
$9$
0%
$5$
0%
$3$
0%
$2$

Explanation

As the points are collinear, the slope of the line joining any two points, should be same as the slope of the line joining two other points.
Slope of the line passing through points $\left( { x }_{ 1 },{ y }_{ 1 } \right)$ and $\left( { x }_{ 2 },{ y }_{ 2 } \right)$ = $\dfrac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-x_{ 1 } }$
So, slope of the line joining $(3,5) , (m,6) =$ Slope of the line joining $(3,5)$ and $\left (\dfrac {1}{2}, \dfrac {15}{2}\right )$

Therefore, $\dfrac { 6 - 5 }{ m - 3 } = \dfrac { \frac {15}{2} - 5 }{ \frac {1}{2} - 3 }$
$\Rightarrow \dfrac { 1 }{ m - 3 } = -1$

$\Rightarrow m - 3 = -1$

$\Rightarrow m = 2$

Find the direction cosines

$l,m,n$ of a line which are connected by the relation

$l+m-n=0$ and

$2ml-2mn+nl=0$

0%
$\displaystyle \frac { -2 }{ \sqrt { 6 } } ,\frac { 1 }{ \sqrt { 6 } } ,\frac { -1 }{ \sqrt { 6 } }$
0%
$\displaystyle \frac { 2 }{ \sqrt { 6 } } ,\frac { -1 }{ \sqrt { 6 } } ,\frac { 1 }{ \sqrt { 6 } }$
0%
$\displaystyle \frac { -2 }{ \sqrt { 6 } } ,\frac { -1 }{ \sqrt { 6 } } ,\frac { -1 }{ \sqrt { 6 } }$
0%
$\displaystyle \frac { 2 }{ \sqrt { 6 } } ,\frac { 1 }{ \sqrt { 6 } } ,\frac { 1 }{ \sqrt { 6 } }$

Explanation

Given

$l+m-n=0$ ...

$(1)$

$2ml-2mn+ln=0$ ...

$(2)$
From

$(1)$ ,

$n=m+l$
Putting this value of

$n$ in equation

$(2)$ , we get

$2ml-2m(m+l)+l(m+l)=0$

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

$\Rightarrow \left( l-m \right) \left( l+2m \right) =0$
Either

$l=m$ or

$l=-2m$

Case I:

$l=m$

$\Rightarrow n=m+l=2m$

$\therefore$ DR's are

$<1,1,2>.$

$\therefore$ DC's are

$\displaystyle \pm \frac { 1 }{ \sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 }+{ 2 }^{ 2 } } } ,\pm \frac { 1 }{ \sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 }+{ 2 }^{ 2 } } } ,\pm \frac { 2 }{ \sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 }+{ 2 }^{ 2 } } }$

$\therefore$ DC's are

$\displaystyle <\frac { 1 }{ \sqrt { 6 } } ,\frac { 1 }{ \sqrt { 6 } } ,\frac { 2 }{ \sqrt { 6 } } >$ or

$\displaystyle <\frac { -1 }{ \sqrt { 6 } } ,\frac { -1 }{ \sqrt { 6 } } ,\frac { -2 }{ \sqrt { 6 } } >$

Case II:

$l=-2m \Rightarrow n=l+m=-2m+m=-m$

$\therefore$ DR's are

$-2,1,-1$
DC's are

$\displaystyle \Rightarrow \pm \frac { -2 }{ \sqrt { { \left( -2 \right) }^{ 2 }+{ 1 }^{ 2 }+{ \left( -1 \right) }^{ 2 } } } ,\pm \frac { 1 }{ \sqrt { { \left( -2 \right) }^{ 2 }+{ 1 }^{ 2 }+{ \left( -1 \right) }^{ 2 } } } ,\pm \frac { -1 }{ \sqrt { { \left( -2 \right) }^{ 2 }+{ 1 }^{ 2 }+{ \left( -1 \right) }^{ 2 } } }$

$\Rightarrow$ DC's are

$\displaystyle <\frac { -2 }{ \sqrt { 6 } } ,\frac { 1 }{ \sqrt { 6 } } ,\frac { -1 }{ \sqrt { 6 } } >$ or

$\displaystyle <\frac { 2 }{ \sqrt { 6 } } ,\frac { -1 }{ \sqrt { 6 } } ,\frac { 1 }{ \sqrt { 6 } } >$

Hence, options

$A$ and

$B$ are correct.

The direction ratios of two lines are

$1,-2,-2$ and

$0,2,1$ . The direction cosines of the line perpendicular to the above lines are

0%
$\displaystyle \frac { 2 }{ 3 } ,\frac { -1 }{ 3 } ,\frac { 2 }{ 3 }$
0%
$\displaystyle \frac { -1 }{ 3 } ,\frac { 2 }{ 3 } ,\frac { 2 }{ 3 }$
0%
$\displaystyle \frac { 1 }{ 4 } ,\frac { 3 }{ 4 } ,\frac { 1 }{ 2 }$
0%
None of these

Explanation

Let

$a,b,c$ be the direction ratios of the line whose direction cosines are required.
Then as this line is perpendicular to the given lines so we have

$a(1)+b(-2)+c(-2)=0$ and

$a(0)+b(2)+c(1)=0$
Solving these simultaneously, we get

$\displaystyle \frac { a }{ \left( -2 \right) \left( 1 \right) -\left( -2 \right) \left( 2 \right) } =\frac { b }{ \left( -2 \right) \left( 0 \right) -\left( 1 \right) \left( 1 \right) } =\frac { c }{ \left( 1 \right) \left( 2 \right) -\left( 0 \right) \left( -2 \right) }$

$\displaystyle \Rightarrow \frac { a }{ 2 } =\frac { b }{ -1 } =\frac { c }{ 2 } \Rightarrow a:b:c=2:-1:2$
Therefore the required direction cosines are

$\displaystyle \frac { 2 }{ \sqrt { { 2 }^{ 2 }+{ 1 }^{ 2 }+{ 2 }^{ 2 } } } ,\frac { -1 }{ \sqrt { { 2 }^{ 2 }+{ 1 }^{ 2 }+{ 2 }^{ 2 } } } ,\frac { 2 }{ \sqrt { { 2 }^{ 2 }+{ 1 }^{ 2 }+{ 2 }^{ 2 } } } \Rightarrow \frac { 2 }{ 3 } ,\frac { -1 }{ 3 } ,\frac { 2 }{ 3 }$

If the projection of a line segment on

$x,y$ and

$z$ axes are respectively

$3,4$ and

$5$ , then the length of the line segment is

0%
$3\sqrt { 2 }$
0%
$5\sqrt { 2 }$
0%
$6\sqrt { 2 }$
0%
None of these

Explanation

Let

$l,m,n$ be the d.c's of the given line segment

$PQ.$

$\therefore l=\cos { \alpha } ,m=\cos { \beta } ,n=\cos { \gamma }$

where

$\alpha ,\beta ,\gamma$ are the angles which the line segment

$PQ$ makes with the axes.

Suppose length of line segment

$PQ=r.$

Thus, projection of line segment

$PQ$ on x-axis

$=PQ \cos { \alpha } ,=rl.$

Also the projection of line segment

$PQ$ on x-axis

$=3$ (given).

$\therefore lr=3$

Similarly,

$mr=4, nr=5.$

Now squaring and adding there equations, we get

${ \left( lr \right) }^{ 2 }+{ \left( mr \right) }^{ 2 }+{ \left( nr \right) }^{ 2 }=r^2[{ 3 }^{ 2 }+{ 4 }^{ 2 }+{ 5 }^{ 3 }]$

$\Rightarrow { r }^{ 2 }\left( { l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 } \right) =9+16+25$

$\Rightarrow { r }^{ 2 }=50\quad \quad \quad \quad \left[ \because { l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 }=1 \right]$

$\Rightarrow r=\sqrt { 50 } =5\sqrt { 2 }$

Lines

$OA,OB$ are drawn from

$O$ with direction cosines proportional to

$(1,-2,-1),(3,-2,3).$ Find the direction cosines of the normal to the plane

$AOB$

0%
$\displaystyle \left< \pm \frac { 4 }{ \sqrt { 29 } } \pm \frac { 3 }{ \sqrt { 29 } } \pm \frac { -2 }{ \sqrt { 29 } } \right>$
0%
$\displaystyle \left< \pm \frac { 2 }{ \sqrt { 29 } } \pm \frac { 3 }{ \sqrt { 29 } } \pm \frac { -2 }{ \sqrt { 29 } } \right>$
0%
$\displaystyle \left< \pm \frac { 8 }{ \sqrt { 29 } } \pm \frac { 6 }{ \sqrt { 29 } } \pm \frac { -2 }{ \sqrt { 29 } } \right>$
0%
$\displaystyle \left< \pm \frac { 8 }{ \sqrt { 29 } } \pm \frac { 3 }{ \sqrt { 29 } } \pm \frac { -2 }{ \sqrt { 29 } } \right>$

Explanation

Let

$ax+by+cz+d=0$ be the plane, then

$O(0,0,0);A(1,-2,-1);B(3,-2,3)$

$\Rightarrow d=0$ and

$a-2b-c=0$ also

$3a-2b+3c=0$
Putting

$c=(a-2b),$ we get

$6a=8b$

$\displaystyle a=\frac { 4 }{ 3 } b\therefore a=\frac { 4 }{ 3 } b,b=b$ and

$\displaystyle c=-\frac { 2b }{ 3 }$
And the plane is

$\displaystyle b\left( \frac { 4 }{ 3 } x+y-\frac { 2 }{ 3 } z \right) =0$

$4x+3y-2z=0$
The normal

$\pm \left( \overrightarrow { n } =4\hat { i } +3\hat { j } -2\hat { k } \right)$

$\displaystyle \hat { n } =\pm \left( \frac { 4 }{ \sqrt { 29 } } \hat { i } +\frac { 3 }{ \sqrt { 29 } } \hat { j } +\frac { 2 }{ \sqrt { 29 } } \hat { k } \right)$
D.C' s of normal vector

$\displaystyle \left< \pm \frac { 4 }{ \sqrt { 29 } } \pm \frac { 3 }{ \sqrt { 29 } } \pm \frac { -2 }{ \sqrt { 29 } } \right>$

If the points

$(p,0)$ ,

$(0,q)$ and

$(1,1)$ are collinear, then

$\dfrac { 1 }{ p }+\dfrac { 1 }{ q }$ is equal to:

0%
$-1$
0%
$1$
0%
$2$
0%
$0$

Explanation

As the points are collinear, the slope of the line joining

any two points, should be same as the slope of the line joining two other

points.

Slope of the line passing through points $\left( { x }_{ 1 },{ y }_{ 1 } \right)$ and $\left( { x }_{ 2 },{ y }_{ 2 } \right)$ $=$ $\dfrac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-x_{ 1 } }$

So, slope of the line joining $(p,0) , (0,q) =$ Slope of the line joining

$(0,q)$ and $(1,1)$

$\dfrac { q - 0 }{ 0 - p } = \dfrac { 1 - q }{ 1 - 0 }$

$- \dfrac { q }{ p } = 1 - q$

Dividing both sides by $q$ ,
$- \dfrac { 1 }{ p } = \dfrac { 1 }{ q } - 1$

$=> \dfrac { 1 }{ p } + \dfrac { 1 }{ q } = 1$

The angle between the straight lines whose direction cosines are given by

$2l+2m-n=0,mn+nl+lm=0$ , is

0%
$\displaystyle \frac { \pi }{ 2 }$
0%
$\displaystyle \frac { \pi }{ 3 }$
0%
$\displaystyle \frac { \pi }{ 4 }$
0%
None of these

Explanation

Relations are

$2l-2m-n=0\Rightarrow n=2l+2m$ and

$mn+nl+lm=0$ .
Eliminating

$n$ , we have

$(m+l)(2l+2m)+lm=0$

$\Rightarrow2{ l }^{ 2 }+5lm+2{ m }^{ 2 }=0$

$\Rightarrow \left( 2l+m \right) \left( l+2m \right) =0$
When

$2l+m=0$ , we have from

$2l+2m-n=0,n=m.$
Thus

$\displaystyle \frac { l }{ -1 } =\frac { m }{ 2 } =\frac { n }{ 2 } ,$

$\Rightarrow$ d.c's of one line are proportional to

$[-1,2,2]$ .
Again, when

$l+2m=0$ , we have from

$2l+2m-n=0,n=l$

$\displaystyle \therefore \frac { l }{ 2 } =\frac { m }{ -1 } =\frac { n }{ 2 } ,$

$\Rightarrow$ d.c's of the other line are proportional to

$[2,-1,2]$ .
Now, if

$\theta$ be the angle between the two lines, then

$\displaystyle \cos { \theta } =\frac { -1.2+2.-1+2.2 }{ \sqrt { \left( { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 2 }^{ 2 } \right) } \sqrt { \left( { 2 }^{ 2 }+{ 1 }^{ 2 }+{ 2 }^{ 2 } \right) } } =0,$

$\Rightarrow \displaystyle \theta =\frac { \pi }{ 2 }$ or the two lines are at right angles to each other.

Equation of the line

$L$ is -

0%
$\vec{r} = 2\hat{k} + \lambda(\hat{i} + \hat{k})$
0%
$\vec{r} = 2\hat{k} + \lambda(2\hat{j} + \hat{k})$
0%
$\vec{r} = 2\hat{k} + \lambda(\hat{j} + \hat{k})$
0%
none of these

Explanation

$\textbf{Step-1: Find the normal vector of the plane using given information}$

$\text{Given four points A(2,1,0), B(1,0,1), C(3,0,1) and D(0,0,2)}$

$\text{From the given points, we have}$

$\vec{BA}=(2-1)\hat i+(1-0)\hat j+(0-1)\hat k$ ,

$\vec{BC}=(3-1)\hat i+(0-0) \hat j+(1-1)\hat k$

$\vec{BA}=\hat i+\hat j- \hat k$ ,

$\vec{BC}=2\hat i$

$\text{So, the normal vector of the plane containing}$

$ABC$

$\text{is}$

$\vec{BA}\times \vec{BC}=(2i)\times (\hat i+\hat j-\hat k)$

$=2(\hat j+\hat k)$

$\text{Since the line is orthogonal to the plane, hence its directional vector is}$

$\text{ parallel to the normal vector of the plane.}$

$\text{Therefore the direction vector of the line is}$

$(j+k)$ .

$\text{Now, it passes through}$

$D=2k$

$\text{Therefore the required equation of the line is}$

$\vec{r}=2k+\lambda(j+k)$

$\textbf{Hence, option - C is the answer}$

If direction ratios of the normal of the plane which contains the lines

$\displaystyle\frac{x-2}{3}=\displaystyle\frac{y-4}{2}=\displaystyle\frac{z-1}{1}\;\&\;\displaystyle\frac{x-6}{3}=\displaystyle\frac{y+2}{2}=\displaystyle\frac{z-2}{1}$ are

$(a,\,1,\,-26)$ , then

$a$ is equal to

0%
$5$
0%
$6$
0%
$7$
0%
$8$

Explanation

$\dfrac { x-2 }{ 3 } =\dfrac { y-4 }{ 2 } =\dfrac { z-1 }{ 1 }$

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

$(a,1,-26) \rightarrow$ DR's of normal

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

$\Rightarrow 3(a)+(1)+1(-26) = 0$

$\Rightarrow 3a=24$

$\Rightarrow$

$a = 8$

Hence the answer is

$8.$

The equation of the plane which contains the lines

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

$\vec{r}\, =\, \hat{i}\, +\, 2 \hat{j}\, -\, \hat{k}\, +\, \mu\, (\hat{i}\, +\, \hat{j}\, +\, 3 \hat{k})$ must be

0%
$\vec{r}.\, (7 \hat{i}\, -\, 4\, \hat{j}\, -\, \hat{k})\, =\, 0$
0%
$7(x\, -\, 1)\, -\, 4(y\, -\, 2)\, -\, (z\, +\, 1)\, =\, 0$
0%
$\vec{r}.\, (\hat{i}\, +\, 2\, \hat{j}\, -\, \hat{k})\, =\, 0$
0%
$\vec{r}.\, (\hat{i}\, +\, \hat{j}\, +\, 3\, \hat{k})\, =\, 0$

Explanation

Clearly lines are parallel to

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

$\hat{i}+\hat{j}+3\hat{k}$
Thus normal vector of required plane is,

$\vec{n}=\begin{vmatrix} \hat{i}&\hat{j}&\hat{k}\\1&2&-1\\1&1&3\end{vmatrix}=7\hat{i}-4\hat{j}-\hat{k}$
Hence, required plane is

$[\vec{r}-(\hat{i}+2\hat{j}-\hat{k})]\cdot \vec{n}=0$

$\Rightarrow [\vec{r}-(\hat{i}+2\hat{j}-\hat{k})]\cdot (7\hat{i}-4\hat{j}-\hat{k})=0$

$\Rightarrow 7(x-1)-4(y-2)-(z+1)=0$

If the foot of the perpendicular from the origin to a plane is

$\left( a,b,c \right) ,$ the equation of the plane is

0%
$\displaystyle \frac { x }{ a } +\frac { y }{ b } +\frac { z }{ c } =3$
0%
$ax+by+cz=3$
0%
$ax+by+cz={a}^{2}+{b}^{2}+{c}^{2}$
0%
$ax+by+cz=a+b+c$

Explanation

Let

$P\left( a,b,c \right)$ be the foot of the perpendicular from the origin to the plane, then direction ratios of

$OP$ are

$a-0,b-0,c-0$ i.e.

$a,b,c.$

So the equation of the plane passing through

$P\left( a,b,c \right)$ , the direction ratios of the normal to which are

$a,b,c$ is

$a\left( x-a \right) +b\left( y-b \right) +c\left( z-c \right) =0$

$\Rightarrow ax+by+cz={a}^{2}+{b}^{2}+{c}^{2}$

Are the points (1, 1), (2, 3) and (8, 11) collinear ?

0%
collinear
0%
Non collinear
0%
coplaner
0%
None of above

Explanation

Area of triangle formed by these vertices is

$\displaystyle \Delta =\frac { 1 }{ 2 } \begin{vmatrix} 1 & 1 & 1 \\ 2 & 3 & 1 \\ 8 & 11 & 1 \end{vmatrix}$
Applying

${ R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 },{ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 1 }$

$\displaystyle \Delta =\frac { 1 }{ 2 } \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 0 \\ 7 & 10 & 0 \end{vmatrix}=\frac { 1 }{ 2 } \left( 10-14 \right) =2$
Hence points are non collinear

The planes

$3x-y+z+1=0,5x+y+3z=0$ intersect in the line

$PQ$ . The equation of the plane through the point

$(2,1,4)$ and perpendicular to

$PQ$ is

0%
$x+y-2z=5$
0%
$x+y-2z=-5$
0%
$x+y+2z=5$
0%
$x+y+2z=-5$

Explanation

Let

$[l,m,n]$ be the direction cosines of

$PQ$ , then

$3l-m+n=0$ and

$5l+m+3n=0$

$\displaystyle \therefore \frac { l }{ -3-1 } =\frac { m }{ 5-9 } =\frac { n }{ 3+5 } \Rightarrow \frac { l }{ 1 } =\frac { m }{ 1 } =\frac { n }{ -2 }$
Now, a plane

$\bot$ to

$PQ$ will have

$l,m,n$ as the coefficient of

$x,y$ and

$z$ .
Hence, the plane

$\bot$ to

$PQ$ is

$x+y-2z=\lambda$
It passes through

$(2,1,4)$

$\therefore 2+1-2.4=\lambda \Rightarrow \lambda =-5$
Hence, the required plane is

$x+y-2z=-5$

The direction ratios of a line followed by the insect during its journey from A to G along the shortest path are

0%
$(3, 4, 5)$
0%
$(0, 4, 5)$
0%
$(7, 0, 5)$
0%
$(0, 0, 1)$

Explanation

Paths teken can be three:

$i)AF$ &

$FG=5+5=10$

$ii) AE$ &

$EG=3+\sqrt { 14 }$

$iii)AB$ &

$BG=4+\sqrt { 34 }$

shortest one is

$ii)$

$\therefore A\longrightarrow E\longrightarrow G$

$\therefore$ direction ratio

$\therefore A\longrightarrow E=(0,0,3)\equiv (0,0,1)$

1042622_332937_ans_1237baee71264fc0b728e51d057d0a45.PNG

The direction cosines of a vector

$\displaystyle \hat {i} + \hat {j} + \sqrt {2} \hat {k}$ are

0%
$\displaystyle \frac {1}{2} , \frac {1} {2} , 1$
0%
$\displaystyle \frac {1}{\sqrt 2} , \frac {1} {\sqrt 2} , \frac {1} {2}$
0%
$\displaystyle \frac {1}{2} , \frac {1} {2} , \frac {1} {\sqrt 2}$
0%
$\displaystyle \frac {1}{\sqrt 2} , \frac {1} {\sqrt 2} , \frac {1} {\sqrt 2}$

Explanation

Direction cosine of

$a\hat{i}+b\hat{j}+c\hat{k}$ can be given as

$\dfrac{a}{\sqrt{a^2+b^2+c^2}}, \dfrac{b}{\sqrt{a^2+b^2+c^2}}\dfrac{c}{\sqrt{a^2+b^2+c^2}}$
Thus, here we have directional cosines as

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

$\displaystyle \frac {1}{2} , \frac {1} {2} , \frac {1} {\sqrt 2}$

In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

$3y+4z-6=0$

0%
$\left( 0,\cfrac { 24 }{ 25 } ,\cfrac { 18 }{ 25 } \right)$
0%
$\left( 0,\cfrac { 24 }{ 25 } ,\cfrac { 24 }{ 25 } \right)$
0%
$\left( 0,\cfrac { 18 }{ 25 } ,\cfrac { 24 }{ 25 } \right)$
0%
None of these

Explanation

Let the coordinates of the foot of perpendicular

$P$ from the origin to the plane be

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

$3y+4z-6=0$

$\Rightarrow$

$0x+3y+4z=6=6....(1)$
The direction ratios of the normal are

$0,3$ and

$4$

$\therefore$

$\sqrt { 0+{ \left( 3 \right) }^{ 2 }+{ \left( 4 \right) }^{ 2 } } =5$
Dividing both sides of equation (1) by

$5$ , we obtain

$0x+\cfrac{3}{5}y+\cfrac{4}{5}z=\cfrac{6}{5}$
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

$\left( 0,\cfrac { 3 }{ 5 } .\cfrac { 6 }{ 5 } ,\cfrac { 4 }{ 5 } .\cfrac { 6 }{ 5 } \right)$ i.e.,

$\left( 0,\cfrac { 18 }{ 25 } ,\cfrac { 24 }{ 25 } \right)$

If a line makes angles

$\alpha, \beta, \gamma$ and

$\delta$ with the diagonals of a cube, Then,

$\cos^2\alpha +\cos^2\beta +\cos^2\gamma +\cos^2\delta =\dfrac {a}{b}$ , where

$a$ and

$b$ are in lowest form, find

$a+b$

0%
$7$
0%
$6$
0%
$8$
0%
None of these

Explanation

A cube is a rectangular parallelopiped having equal length, breadth and height. Let OADBFEGC be the cube with each side of length a units.

The four diagonals are OE, AF, BG and CD.

The direction \cos ines of the diagonal OE which is the line joining two points O and E are

$\dfrac {a-0}{\sqrt {a^2+a^2+a^2}}, \dfrac {a-0}{\sqrt {a^2+a^2+a^2}}, \dfrac {a-0}{\sqrt {a^2+a^2+a^2}}$

i.e.

$\dfrac {1}{\sqrt 3}, \dfrac {1}{\sqrt 3}, \dfrac {1}{\sqrt 3}$

Similarly, the direction \cos ines of AF, BG and CD are

$(-\dfrac {1}{\sqrt 3}, \dfrac {1}{\sqrt 3}, \dfrac {1}{\sqrt 3}); (\dfrac {1}{\sqrt 3}-\dfrac {1}{\sqrt 3}, \dfrac {1}{\sqrt 3}); (\dfrac {1}{\sqrt 3},\dfrac {1}{\sqrt 3},-\dfrac {1}{\sqrt 3})$ , respectively.

Let l, m, n be the direction \cos ines of the given line which makes angles

$\alpha, \beta, \gamma, \delta$ with OE, AF, BG, CD, respectively. Then

$\cos \alpha =\dfrac {1}{\sqrt 3}(l+m+n); \cos \beta =\dfrac {1}{\sqrt 3}(-l+m+n)$ ;

$\cos \gamma =\dfrac {1}{\sqrt 3}(l-m+n); \cos \delta =\dfrac {1}{\sqrt 3}(l+m+n)$ ;

Squaring and adding, we get

$cso^2\alpha +\cos ^2\beta +\cos ^2\gamma +\cos ^2\delta \\=\dfrac {1}{3}[(1+m+n)^2+(-1+m+n)^2+(l-m+n)^2+(l+m+n)^2]\\=\dfrac {1}{3}[4(l^2+m^2+n^2)]=\dfrac {4}{3}$

(as

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

In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

$2x+3y+4z-12=0$

0%
$\left( \cfrac { 24 }{ 29 } ,\cfrac { 36 }{ 49 } ,\cfrac { 48 }{ 29 } \right)$
0%
$\left( \cfrac { 24 }{ 49 } ,\cfrac { 36 }{ 49 } ,\cfrac { 48 }{ 49 } \right)$
0%
$\left( \cfrac { 24 }{ 29 } ,\cfrac { 36 }{ 29 } ,\cfrac { 48 }{ 29 } \right)$
0%
$\left( \cfrac { 24 }{ 49 } ,\cfrac { 36 }{29 } ,\cfrac { 48 }{ 49 } \right)$

Explanation

Let the coordinates of the foot of perpendicular

$O$ from the origin to the plane be

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

$2x+3y+4z-12=0$

$\Rightarrow$

$2x+3y+4z=12......(1)$
The direction ratios of normal are

$2,3$ and

$4$

$\therefore$

$\sqrt { { \left( 2 \right) }^{ 2 }+{ \left( 3 \right) }^{ 2 }+{ \left( 4 \right) }^{ 2 } } =\sqrt { 29 }$
Dividing both sides of equation (1) by

$\sqrt {29}$ , we obtain

$\cfrac { 2 }{ \sqrt { 29 } } x+\cfrac { 3 }{ \sqrt { 29 } } y+\cfrac { 4 }{ \sqrt { 29 } } z=\cfrac { 12 }{ \sqrt { 29 } }$
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 coordinates of the foot of the perpendicular are

$\left( \cfrac { 2 }{ \sqrt { 29 } } \cdot\cfrac { 12 }{ \sqrt { 29 } } ,\cfrac { 3 }{ \sqrt { 29 } } .\cfrac { 12 }{ \sqrt { 29 } } ,\cfrac { 4 }{ \sqrt { 29 } } \cdot\cfrac { 12 }{ \sqrt { 29 } } \right)$ i.e.,

$\left( \cfrac { 24 }{ 29 } ,\cfrac { 36 }{ 29 } ,\cfrac { 48 }{ 29 } \right)$

$\alpha, \beta$ and

$\gamma$ are the angles which a half ray makes with the positive direction of the axes, then

$\sin^2\alpha+\sin^2\beta+\sin^2\gamma$ is equal to

0%
$1$
0%
$2$
0%
$0$
0%
$-1$

Explanation

Given expression,

$\sin^2\alpha+\sin^2\beta+\sin^2\gamma$

$=(1-\cos^2\alpha)+(1-\cos^2\beta)+(1-\cos^2\gamma)$

$=3-\cos^2\alpha+\cos^2\beta+\cos^2\gamma=3-1=2$

$(\because \cos^2\alpha+\cos^2\beta+\cos^2\gamma=1)$

The equation of the plane containing the line

$\dfrac { x+1 }{ -3 } =\dfrac { y-3 }{ 2 } =\dfrac { z+2 }{ 1 }$ and the point

$\left( 0,7,-7 \right)$ is

0%
$x+y+z=1$
0%
$x+y+z=2$
0%
$x+y+z=0$
0%
None of these

Explanation

The equation of plane containing the line

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

$a\left( x+1 \right) +b\left( y-3 \right) +c\left( z+2 \right) =0$ .....(i)

where

$-3a+2b+c=0$ ......(ii)

This passes through

$\left( 0,7,-7 \right)$ .

$\therefore a+4b-5c=0$ .....(iii)

From equations (ii) and (iii),

$\dfrac { a }{ -14 } =\dfrac { b }{ -14 } =\dfrac { c }{ -14 }$

$\dfrac { a }{ 1 } =\dfrac { b }{ 1 } =\dfrac { c }{ 1 }$

Thus, the required plane is

$x + y + z = 0$

If the three points

$A(1,6), B(3,-4)$ and

$C(x,y)$ are collinear, then the equation satisfying by

$x$ and

$y$ is

0%
$5x+y-11=0$
0%
$5x+13y+5=0$
0%
$5x-13y+5=0$
0%
$13x-5y+5=0$

Explanation

Since, the points

$A(1,6), B(3,-4)$ and

$C(x,y)$ are colinear

$\therefore$

$\begin{vmatrix} 1 & 6 & 1 \\ 3 & -4 & 1 \\ x & y & 1 \end{vmatrix}=0$

$\Rightarrow$

$1(-4-y)-6(3-x)+1(3y+4x)=0$

$\Rightarrow$

$10x+2y-22=0$

$\Rightarrow$

$5x+y-11=0$

If the line

$\vec{OR}$ makes angles

$\theta_1, \theta_2, \theta_3$ with the planes

$XOY, YOZ, ZOX$ respectively, then

$\cos^2\theta_1+\cos^2\theta_2+\cos^2\theta_3$ is equal to

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

If the line

$\vec{OR}$ makes angles

$\theta_1, \theta_2, \theta_3$ with the planes

$XOY, YOZ, ZOX$ respectively, then

$\cos^2\theta_1+\cos^2\theta_2+\cos^2\theta_3$

$=1$

The direction cosines of the line segment joining points

$(-3, 1, 2)$ and

$(1, 4, -10)$ is.

0%
$\displaystyle\frac{4}{13}, \frac{3}{13},- \frac{12}{13}$
0%
$\displaystyle\frac{-4}{13}, \frac{3}{13}, -\frac{12}{13}$
0%
$\displaystyle\frac{-4}{13}, \frac{-3}{13}, \frac{12}{13}$
0%
None of these

Explanation

The direction ratios of the line segment joining

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

$B(1, 4, -10)$ are proportional to

$1-(-3), 4-1, -10-2,$ i.e.,

$4, 3, -12$ .
Now, its direction cosines are

$\displaystyle\frac{4}{\sqrt{4^2+3^2+(-12)^2}}, \displaystyle\frac{3}{\sqrt{4^2+3^2+(-12)^2}}, \displaystyle\frac{-12}{\sqrt{4^2, 3^2+(-12)^2}}$
or

$\displaystyle\frac{4}{13}, \frac{3}{13}, -\frac{12}{13}$

The direction cosine of a line which is perpendicular to both the lines whose direction ratios are

$1, 2, 2$ and

$0, 2, 1$ are

0%
$\displaystyle \frac { -2 }{ 3 } ,\frac { 1 }{ 3 } ,\frac { 2 }{ 3 }$
0%
$\displaystyle \frac { 2 }{ 3 } ,\frac { -1 }{ 3 } ,\frac { 2 }{ 3 }$
0%
$\displaystyle \frac { 2 }{ 3 } ,\frac { 1 }{ 3 } ,\frac { -2 }{ 3 }$
0%
$\displaystyle \frac { 2 }{ 3 } ,\frac { -1 }{ 3 } ,\frac { -2 }{ 3 }$

Explanation

Since we need the direction cosines of a line perpendicular to both the lines, having direction ratios

$1,2,2$ and

$0,2,1$ ; we first find the direction ratios of the required line by finding the cross product of the given ratios.

It would result in

$(\hat{i} + 2\hat{j} + 2\hat{k}) \times (2\hat{j} + \hat{k})$ , i.e.

$2\hat{k} - \hat{j} + 0 - 2\hat{i} + 4\hat{i} + 0$

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

Thus, the direction ratios become

$2, -1, 2$

And the corresponding direction cosines become

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

Three district points A, B and C with p.v.s. and

$\displaystyle \vec { a } ,\vec { b }$ and

$\displaystyle \vec { c }$ respectively are collinear if there exist non-zero scalars x, y, z such that

0%
$\displaystyle x\vec { a } +y\vec { b } +z\vec { c } =0$ and $\displaystyle x+y+z=0$
0%
$\displaystyle x\vec { a } +y\vec { b } +z\vec { c }$ and $\displaystyle x+y+z\neq 0$
0%
$\displaystyle x\vec { a } +y\vec { b } +z\vec { c } \neq 0$ and $\displaystyle x+y+z=0$
0%
$\displaystyle x\vec { a } +y\vec { b } +z\vec { c } =3$ and $\displaystyle x+y+z\neq 0$

Explanation

If three points are collinear, we can see that as one point dividing the line joining the other two points either internally or externally in some ratio.

So,

$x\vec {a} + y\vec {b} = (x + y)\vec{c}$ when

$\vec{c}$ divides the line joining

$\overline{a}$ and

$\overline{b}$ in the ratio

$y : x$

This can also be written as

$x\vec {a} + y\vec {b} - (x + y)\vec {c} = 0$

$z = - (x + y)$ , we get

$x + y + z = 0$ and

$x\vec {a} + y\vec {b} + z\vec {c} = 0$

If a line makes angles

$\alpha, \beta, \gamma$ with the coordinate axes, then the value of

$\cos 2\alpha + \cos 2\beta + \cos 2\gamma$ is

0%
$3$
0%
$-2$
0%
$2$
0%
$-1$

Explanation

$\cos 2\alpha + \cos 2\beta + \cos 2\gamma$

$= 2(\cos^{2} \alpha + \cos^{2}\beta + \cos^{2}\gamma) - 3$

$= 2 - 3 = -1$

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

The points

$(k -1, \ k +2), (k, \ k +1), (k +1, \ k)$ are collinear for

0%
any value of $k$
0%
$k=-\dfrac{1}{2}$ only
0%
no value of $k$
0%
integral values of $k$ only

Explanation

For given points to be collinear

$\begin{vmatrix}k-1&k+2&1\\k&k+1&1\\k+1&k&1 \end{vmatrix}=0$

$\Rightarrow \begin{vmatrix}-1&1&0\\-1&1&0\\k+1&k&1\end{vmatrix}=0$ ,

$R_1\rightarrow R_1-R_2$ and

$R_2\rightarrow R_2-R_3$

$\Rightarrow 0=0$ , Since 1st and 2nd row are same.

Hence given points are collinear for all real values of

$k$

A plane passing through

$(-1, 2, 3)$ and whose normal makes equal angle with the coordinate axes is

0%
$x + y + z + 4 = 0$
0%
$x - y + z + 4 = 0$
0%
$x + y + z - 4 = 0$
0%
$x + y + z = 0$

Explanation

Since normal makes equal angles with coordinate axis.

So, it intercept with all the axis will be same. So equation of plane will be

$\dfrac{x}{a}+\dfrac{y}{a}+\dfrac{z}{a}=1$

$\Rightarrow x+y+z=a$

Now, it passes through

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

$-1+2+3=a$

$\Rightarrow a=4$

$\Rightarrow x+y+z-4=0$

$\cos { \alpha } ,\cos { \beta } ,\cos { \gamma }$ are the direction cosines of a vector

$\vec { a }$ , then

$\cos { 2\alpha } +\cos { 2\beta } +\cos { 2\gamma }$ is equal to

0%
$2$
0%
$3$
0%
$-1$
0%
$0$

Explanation

We know that

$\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$ ....(1)

Hence

$\cos2\alpha+\cos2\beta+\cos2\gamma=(2\cos^2\alpha-1)+(2\cos^2\beta-1)+(2\cos^2\gamma-1)$

$=2(\cos^2\alpha+\cos^2\beta+\cos^2\gamma)-3=2\cdot 1-3=-1$ , using (1)

The vector equation of the plane which is at a distance of

$\cfrac { 3 }{ \sqrt { 14 } }$ from the origin and the normal from the origin is

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

0%
$\vec { r } .(2\hat { i } -3\hat { j } +\hat { k } )=3$
0%
$\vec { r } .(\hat { i } +\hat { j } +\hat { k } )=9$
0%
$\vec { r } .(\hat { i } +2\hat { j } )=3$
0%
$\vec { r } .(2\hat { i } +\hat { k } )=3$

Explanation

The unit vector along the normal of the plane is given by

$\vec{n}=\dfrac{2i-3j+k}{\sqrt{4+9+1}}=\dfrac{2i-3j+k}{\sqrt{14}}$
Therefore, the vector equation of the plane with the above normal is

$\vec{r}.(\dfrac{2i-3j+k}{\sqrt{14}})=d$ where 'd' is the distance of the plane from the origin.
Now it is given that

$d=\dfrac{3}{\sqrt{14}}$ , hence the vector equation of the plane becomes

$\vec{r}.(\dfrac{2i-3j+k}{\sqrt{14}})=\dfrac{3}{\sqrt{14}}$ or

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

The direction ratios of two lines AB, AC are 1, -1, -1 and 2, -1,The direction ratios of the normal to the plane ABC are

0%
$2, 3, -1$
0%
$2, 2, 1$
0%
$3, 2, -1$
0%
$-1, 2, 3$

Explanation

Given : Direction ratios of

$AB$ and

$AC$ are

$1,-1,-1$ and

$2,-1,1$

i.e.

$\vec{AB}=\hat{i}-\hat{j}-\hat{k}$ and

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

$\therefore$ Normal vector to the plane ABC is given by

$=\vec{AC}\times \vec{AB}$

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

$=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 1 & -1 & -1 \end{vmatrix}$

$=2\hat{i}+3\hat{j}-\hat{k}$
Hence, the direction ratios of normal are

$2,3,-1$

$A(3, 4, 5), B(4, 6, 3), C(-1, 2, 4)$ and

$D(1, 0, 5)$ are such that the angle between the lines

$\overline{DC}$ and

$\overline{AB}$ is

$\theta$ then

$cos\,\theta =$

0%
$\dfrac{7}{9}$
0%
$\dfrac{2}{9}$
0%
$\dfrac{4}{9}$
0%
$\dfrac{5}{9}$

Explanation

Given,

$A(3,4,5), B(4,6,3), C(-1,2,4), D(1,0,5)$

$\vec{DC}=-2i+2j-k$ ...(i)

$\vec{AB}=i+2j-2k$ ...(ii)

Hence,

$DC.AB\cos\theta=\vec{DC}.\vec{AB}$

$9\cos\theta=-2+4+2$

$\cos\theta=\dfrac{4}{9}$

If the angles made by a straight line with the coordinate axes are

$\alpha, \dfrac{\pi}{2}-\alpha, \beta$ then

$\beta=$

0%
$0$
0%
$\dfrac{\pi}{6}$
0%
$\dfrac{\pi}{2}$
0%
$\pi$

Explanation

Given angles are

$\alpha, 90^0-\alpha, \beta$

The summations of cosines of these angles has to be

$1$ .

So,

$\cos^2\alpha + \cos^2(90^0-\alpha)+\cos^2\beta=1$

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

$\cos^2\beta=0$

$\beta=\dfrac{\pi}{2}$

CBSE Questions for Class 12 Commerce Maths Three Dimensional Geometry Quiz 5 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Three Dimensional Geometry Quiz 5 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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