MCQ Questions for Class 12 Commerce Maths Three Dimensional Geometry Quiz 6

$(2, -1, 2)$ and

$(K, 3, 5)$ are the triads of direction ratios of two lines and the angle between them is

$45^{\circ}$ , then a value of

$K$ is

0%
$2$
0%
$3$
0%
$4$
0%
$6$

Explanation

Applying the formula for angle between 2 lines based on direction ratios :

$\cos \theta = \dfrac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_3^2}\sqrt{a_2^2+b_2^2+c_2^3}}$

$\dfrac {1}{\sqrt {2}} = \dfrac {2K - 3 + 10}{3\times \sqrt {K^{2} + 34}}$
On verification, we get

$K = 4$

The angle between two diagonals of a cube is.

0%
$30^o$
0%
$45^o$
0%
$cos^{-1}(\frac{1}{3})$
0%
$cos^{-1}(\frac{1}{\sqrt{3}})$

Explanation

We know that

$cos\theta = \dfrac{\left | \vec{u}.\vec{v} \right |}{\left | \vec{u} \right |\left | \vec{v} \right |}$
Consider a cube with one vertex at the origin.
Then one diagonal is from

$(0, 0, 0)$ to

$(1, 1, 1)$ :

$\left | \vec{u} \right |=<1, 1, 1>$
Another diagonal is from

$(1, 0, 0)$ to

$(0, 1, 1)$ :

$\left | \vec{v} \right |=<1, -1, -1>$
Therefore,

$cos\theta = \dfrac{\left | <1, 1,1 >.<1, -1, -1> \right |}{\left | \sqrt{1^1+1^2+1^2} \right |\left | \sqrt{1^2+(-1)^2+(-1)^2} \right |}$

$\Rightarrow \dfrac{\left | 1-1-1 \right |}{\sqrt{3}\sqrt{3}}\Rightarrow \dfrac{1}{3}$
Therefore,

$\theta=cos^{-1}\left ( \frac{1}{3} \right )$

If a line segment

$OP$ makes angles of

$\dfrac {\pi}{4}$ and

$\dfrac {\pi}{3}$ with X-axis and Y-axis, respectively. Then, the direction cosines are

0%
$\dfrac {1}{\sqrt {2}}, \dfrac {\sqrt {3}}{2}, \dfrac {1}{\sqrt {2}}$
0%
$\dfrac {1}{\sqrt {2}}, \dfrac {1}{2}, \dfrac {1}{2}$
0%
$1, \sqrt {3}, 1$
0%
$1, \dfrac {1}{\sqrt {3}}, 1$

Explanation

Let

$\alpha, \beta$ and

$\gamma$ be the angles made by the line segment

$OP$ with X-axis, Y-axis and Z-axis respectively.
Given:

$\alpha = \dfrac {\pi}{4}$ and

$\beta = \dfrac {\pi}{3}$
We know that,

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

$\therefore \cos^{2}\dfrac {\pi}{4} + \cos^{2}\dfrac {\pi}{3} +\cos^{2}\gamma = 1$

$\Rightarrow \left (\dfrac {1}{\sqrt {2}}\right )^{2} + \left (\dfrac {1}{2}\right )^{2} + \cos^{2}\gamma = 1$

$\Rightarrow \dfrac {1}{2} + \dfrac {1}{4} + \cos^{2}\gamma = 1$

$\Rightarrow \cos^{2}\gamma = \dfrac {1}{4}$

$\Rightarrow \cos \gamma = \dfrac {1}{2}$

$\therefore \gamma = \dfrac {\pi}{3}$
Hence, direction cosines are

$\cos \alpha, \cos \beta, \cos \gamma$
i.e.

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

If a line makes

${ 45 }^{ o }$ ,

${ 60 }^{ o }$ with positive direction of axes

$x$ and

$y$ then the angles it makes with the

$z$ -axis is

0%
${ 30 }^{ o }$
0%
${ 90 }^{ o }$
0%
${ 45 }^{ o }$
0%
${ 60 }^{ o }$

Explanation

Fact: If

$\alpha, \beta, \gamma$ are the angle made by any line with

$x,y$ and

$z$ -axis, then

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

Here given

$\alpha=45^o, \beta = 60^o$

$\Rightarrow \cos^245^o+\cos^260^o+\cos^2\gamma=1$

$\Rightarrow \dfrac{1}{2}+\dfrac{1}{4}+\cos^2\gamma=1$

$\Rightarrow \cos^2\gamma=\dfrac{1}{4}$

$\Rightarrow \cos\gamma=\dfrac{1}{2}$

$\therefore \gamma =60^o$

The vector equation of a plane passing through a point whose. P.V. is

$\overrightarrow a$ and perpendicular to a vector

$\overrightarrow n$ , is

0%
$\overrightarrow r . \overrightarrow n = \overrightarrow a . \overrightarrow n$
0%
$\overrightarrow r \times \overrightarrow n = \overrightarrow a \times \overrightarrow n$
0%
$\overrightarrow r + \overrightarrow n = \overrightarrow a + \overrightarrow n$
0%
$\overrightarrow r - \overrightarrow n = \overrightarrow a - \overrightarrow n$

Explanation

Let

$\vec r$ be any point in the plane

So the vectors

$\vec r-\vec a$ and

$\vec n$ will be perpendicular to each other

$\Rightarrow (\vec r-\vec a)\cdot \vec n=0$

$\Rightarrow \vec r\cdot \vec n-\vec a\cdot \vec n=0$

$\Rightarrow \vec r\cdot \vec n=\vec a\cdot \vec n$

If direction cosines of a vector of magnitude

$3$ are

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

$a > 0$ , then vector is ____

0%
$2i + j + 2k$
0%
$2i - j + 2k$
0%
$i - 2j + 2k$
0%
$i + 2j + 2k$

Explanation

Unit vector with direction cosine

$\dfrac {2}{3}, -\dfrac {1}{3}, \dfrac {2}{3}$ is given by

$\hat{a}=\dfrac{2}{3}\hat i-\dfrac{1}{3}\hat j+\dfrac{2}{3}\hat k$

Thus vector with magnitude

$3$ and direction cosine as above is given by

$3\hat a =2\hat i-\hat j+2\hat k$

If a plane passing through the point

$(2, 2, 1)$ and is perpendicular to the planes

$3x + 2y + 4z + 1 = 0$ and

$2x + y + 3z + 2 = 0$ . Then, the equation of the plane is

0%
$2x - y - z - 1 = 0$
0%
$2x + 3y + z - 1 = 0$
0%
$2x + y + z + 3 = 0$
0%
$x - y + z - 1 = 0$

Explanation

Equation of plane passing through

$(2, 2, 1)$ is

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

Since, above plane is perpendicular to

$3x + 2y + 4z + 1 = 0$

and

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

$\therefore 3a + 2b + 4c = 0 .... (ii)$

and

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

$[\because$ for perpendicular,

$a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0]$

On multiplying eq. (iii) by

$2$ , we get

$4a + 2b + 6c = 0 .... (iv)$

On subtracting eq, (iv) from eq. (ii), we get

$\implies c = \dfrac {-a}{2}$

On putting

$c = \dfrac {-a}{2}$ in eq. (iii), we get

$b = \dfrac {-a}{2}$

On putting

$b = \dfrac {-a}{2}$ and

$c = \dfrac {-a}{2}$ in eq. (i),

we get

$a(x - 2) - \dfrac {a}{2}(y - 2) - \dfrac {a}{2} (z - 1) = 0$

$\implies \dfrac {a}{2} [2(x - 2) - (y - 2) - (z - 1)] = 0$

$\implies 2x - 4 - y + 2 - z + 1 = 0$

$\implies 2x - y - z - 1 = 0$ is the required equation of plane.

What is the sum of the squares of direction cosines of

$x$ -axis?

0%
$0$
0%
$\dfrac{1}{3}$
0%
$1$
0%
$3$

Explanation

The direction cosines of the vector a are the cosines of angles that the vector forms with the coordinate axes.

The direction cosines uniquely set the direction of vector.

$\therefore$ The sum of the squares of the direction cosines is equal to one.

$i.e. cos^2\alpha+cos^2\beta+cos^2\gamma=1$

Option C is correct answer

What is the sum of the squares of direction cosines of the line joining the points (1, 2, -3) and (-2, 3, 1) ?

0%
0
0%
1
0%
3
0%
$\frac{2}{\sqrt{26}}$

Direction ratios of the line which is perpendicular to the lines with direction ratios

$-1,2,2$ and

$0,2,1$ are

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

Explanation

Let

$l,m,n$ be the direction ratios of the required line

Given :

$l,m,n$ are perpendicular to

$-1,2,2$

So, their dot product will be zero

$\implies l(-1)+m(2)+n(2)=0$

$\implies -l+2m+2n=0$ ....

$(i)$

Also,

$l,m.n$ are perpendicular to

$0,2,1$

$\implies 2m+n=0$

$\implies m=-\dfrac{n}{2}$

Substituting

$m$ in

$(i)$ we get

$-l-n+2n=0 \implies l=n$

So, the direction ratios are in proportion to

$n:-\dfrac{n}{2}:n$

Simplifying we get

$2,-1,2$

Thus, the direction ratios are

$2,-1,2$ .

What are the direction cosines of a line which is equally inclined to the positive directions of the axes?

0%
$\left( \cfrac { 1 }{ \sqrt { 3 } } ,\cfrac { 1 }{ \sqrt { 3 } } ,\cfrac { 1 }{ \sqrt { 3 } } \right)$
0%
$\left( -\cfrac { 1 }{ \sqrt { 3 } } ,\cfrac { 1 }{ \sqrt { 3 } } ,\cfrac { 1 }{ \sqrt { 3 } } \right)$
0%
$\left( -\cfrac { 1 }{ \sqrt { 3 } } ,-\cfrac { 1 }{ \sqrt { 3 } } ,\cfrac { 1 }{ \sqrt { 3 } } \right)$
0%
$\left( \cfrac { 1 }{ 3 } ,\cfrac { 1 }{ 3 } ,\cfrac { 1 }{ 3 } \right)$

Explanation

We know sum of the squares of the direction cosines is one.

$i.e. cos^2\alpha+cos^2\beta+cos^2\gamma=1$

but its given that

$\alpha=\beta=\gamma$

$\therefore cos^2\alpha+cos^2\alpha+cos^2\alpha=1$

$3cos^2\alpha=1$

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

$\therefore$ Positive directions of the axes are

$(\dfrac{1}{\sqrt3},\dfrac{1}{\sqrt3},\dfrac{1}{\sqrt3})$

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

$(0, 2, 1)$ are direction ratios of two lines, then the direction cosines of a perpendicular to both the lines are

0%
$\left( \dfrac{1}{3}, - \dfrac{1}{3} , \dfrac{2}{3}\right )$
0%
$\left( \dfrac{2}{3}, - \dfrac{1}{3} , \dfrac{2}{3}\right )$
0%
$\left( -\dfrac{2}{3}, - \dfrac{1}{3} , \dfrac{2}{3}\right )$
0%
$\left( \dfrac{2}{\sqrt{14}}, - \dfrac{1}{\sqrt{14}} , \dfrac{3}{\sqrt{14}}\right )$

Explanation

Let a vector

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

$\vec { B } =2\hat { j } +\hat { k }$
So, vector perpendicular to

$\vec { A } \& \vec { B }$ will be

$=\vec { A } \times \vec { B }$

$=\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & -2 & -2 \\ 0 & 2 & 1 \end{matrix} \right|$

$=2\hat { i } -1\hat { j } +2\hat { k }$

So, the direction ration of the perpendicular vector or line are (2,-1,2)
Direction cosines are

$\cfrac { 2 }{ \sqrt { { 2 }^{ 2 }+{ \left( -1 \right) }^{ 2 }+{ 2 }^{ 2 } } } ,\cfrac { -1 }{ \sqrt { { 2 }^{ 2 }+{ \left( -1 \right) }^{ 2 }+{ 2 }^{ 2 } } } ,\cfrac { 2 }{ \sqrt { { 2 }^{ 2 }+{ \left( -1 \right) }^{ 2 }+{ 2 }^{ 2 } } }$
or

$\left( \cfrac { 2 }{ 3 } ,\cfrac { -1 }{ 3 } ,\cfrac { 2 }{ 3 } \right)$

The direction ratios of the line perpendicular to the lines with direction ratios

$1, -2, -2$ and

$0, 2, 1$ are

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

Explanation

Direction ratios of two lines:

$<1,-2,-2>$ and

$<0,2,1>$

Let direction ratio of required line is

$<l,m,n>$

Since, lines are perpendicular,

So,

$l-2m-2n=0$ and

$2m+n=0$

$\cfrac{l}{-2+4}=\cfrac{m}{0-1}=\cfrac{n}{2-0}$

So, Reuired direction ratio is

$<2,-1,2>$

Hence, A is the correct option.

What are the direction ratios of normal to the plane

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

0%
$(2, -1, 2)$
0%
$(1, \frac{1}{2}, 1)$
0%
$(1, -2, 1)$
0%
None of the above

Explanation

Direction ratios of normal to the plane

$ax+by+cz+d=0$ , are

$a,b,c$ .

So, here in the question the given plane is

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

Thus, the direction ratios are

$2,-1,2$

Hence, A is correct.

The equation of the plane perpendicular to the

$yz-$ plane and passing through the point

$(1,-2,4)$ and

$(3,-4,5)$ is

0%
$y+2z=5$
0%
$2y+z=5$
0%
$y+2z=6$
0%
$2y+z=6$

Explanation

Plane is perpendicular to yz plane

So it have x-axis as parallel to it

Find a vector lying in the plane using two given point as-

$\vec { 5 }= (3-1)\hat { i } +(-4-(-2)\hat { j } +(5-A)\hat { k }$

$\vec { 5 } = 2\hat { i } -2\hat { j } +\hat { k }$

parallel vector,

$\vec { p } =1.\hat { i } +0.\hat { j } +0.\hat { k }$

So normal to the plane,

$\vec { N } =\vec { S } \times \vec { P }$

$=\begin{vmatrix} \hat { i } & \hat { \quad j } & \hat { \quad k } \\ 2 & \quad -2 & \quad 1 \\ 1 & \quad 0 & \quad 0 \end{vmatrix}$

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

equation of plane using a point and normal,

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

$\Rightarrow y+2+2z-8=0$

$\Rightarrow y+2z=6$

Let a vector

$\vec{r}$ make angles

$60^o, 30^o$ with it and y-axes respectively.Find the angle

$\vec{r}$ make with z-axis.

0%
$30^o$
0%
$60^o$
0%
$90^o$
0%
$120^o$

Explanation

Sum of squares of direction cosines

$=1$

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

Here,

$l=\cos { 60 }^{ 0 }, m= \cos { 30 }^{ 0 }(given), n=?$

${ n }^{ 2 }=1-{ l }^{ 2 }-{ m }^{ 2 }\\ { n }^{ 2 }=1-\left(\dfrac{1}{2}\right)^{ 2 }-\left(\cfrac { \sqrt { 3 } }{ 2 } \right)^{ 2 }\\ \quad \quad =0$

Hence,

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

$\Rightarrow \theta ={ 90 }^{ 0 }$

What are the direction cosines of

$\vec{r}$ ?

0%
$\left (\frac{1}{2}, -\frac{\sqrt{3}}{2}, 0 \right )$
0%
$\left (\frac{1}{2}, \frac{\sqrt{3}}{2}, 0 \right )$
0%
$\left (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0 \right )$
0%
$\left (-\frac{1}{2}, \frac{\sqrt{3}}{2}, 0 \right )$

Explanation

Sum of squares of direction cosines

$=1$

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

Here,

$l=\cos { 60 }^{ 0 }, m= \cos { 30 }^{ 0 }, n=?$

${ n }^{ 2 }=1-{ l }^{ 2 }-{ m }^{ 2 }\\ { n }^{ 2 }=1-\left(\dfrac{1}{2}\right)^{ 2 }-\left(\cfrac { \sqrt { 3 } }{ 2 } \right)^{ 2 }\\ \quad \quad =0$

Hence,

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

$\Rightarrow \theta ={ 90 }^{ 0 }$

Direction cosines

$(l,m,n) \equiv \left(\cfrac { 1 }{ 2 } ,\cfrac { \sqrt { 3 } }{ 2 } ,0\right)$

A straight line passes through (1, -2, 3) and perpendicular to the plane

$2x + 3y - z = 7$ .Find the direction ratios of normal to plane.

0%
$< 2, 3, -1 >$
0%
$< 2, 3, 1 >$
0%
$< -1, 2, 3 >$
0%
None of the above

Explanation

$\text{concept: for any plane}\,ax+by+cz+d=0,\,\text{normal vector to this plane is}\,a\hat{i}+b\hat j+c\hat k$

the normal vector of the plane

$2x+3y-z=7$ is

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

so the direction ratios of normal to plane are

$<2,3,-1>$

What are the direction ratios of the line if it passes through the intersection of the planes

$x=3z+4$ and

$y=2z-3$ ?

0%
$(1,2,3)$
0%
$(2,1,3)$
0%
$(3,2,1)$
0%
$(1,3,2)$

Explanation

Equations of the planes are

$x=3z+4$ and

$y=2z−3$

$\therefore$ The equation of the plane passing through the line of intersection of these planes is

$x=3z+4$ and

$y=2z−3$

Thus The direction Ratios of the equation passes through intersection of the planes is

$(3,2,1)$

The direction cosines of the straight line given by the planes

$x=0$ and

$z=0$ are

0%
$1,0,0$
0%
$0,0,1$
0%
$1,1,0$
0%
$0,1,0$
0%
$0,1,1$

Explanation

Given,

$x=z=0$
It represents Z-axis

$\therefore$ Direction cosines

$=(0,1,0)$

The points, whose position vectors are

$60i + 3j, 40i - 8j$ and

$ai - 52j$ collinear, if

0%
$a = 40$
0%
$a = -40$
0%
$a = 20$
0%
$a = -20$

Explanation

Let

$60i +3j,40i-8j,ai-52j$ be the position vectors of

$A(\overline a)$ ,

$B(\overline b)$ and

$C(\overline c)$

For points A,B,C to be collinear

$\overline{AB}= \lambda \overline{BC}$ ,For some scalar

$\lambda$

$\therefore$

$\overline b - \overline a =\lambda ( \overline c - \overline b)$

$(40i - 8j) - (60i + 3j) = \lambda[(ai - 52j) - (40i - 8j)]$

$-20 = \lambda (a - 40)$ and

$-11 = -44\lambda$

$\Rightarrow a - 40 = \dfrac {-20}{\lambda}$ and

$\lambda = \dfrac {1}{4}$

$\Rightarrow a = 40 - \dfrac {20}{1} \times 4 = -40$ .

The equation of the plane perpendicular to the line

$\cfrac { x-1 }{ 1 } =\cfrac { y-2 }{ -1 } =\cfrac { z+1 }{ 2 }$ and passing through the point

$(2,3,1)$ is

0%
$r.\left( \hat { i } +\hat { j } +2\hat { k } \right) =1$
0%
$r.\left( \hat { i } -\hat { j } +2\hat { k } \right) =1$
0%
$r.\left( \hat { i } +\hat { j } +2\hat { k } \right) =7$
0%
None of these

Explanation

The given line is parallel to the vector

$n=\hat { i } -\hat { j } +2\hat { k }$ . The required plane passes through the point

$(2,3,1)$ ie.,

$2\hat { i } +3\hat { j } +\hat { k }$
so, its equation is,

$r-\left( 2\hat { i } -3\hat { j } +\hat { k } \right) \left( \hat { i } -\hat { j } +2\hat { k } \right) =0\quad$

$\Rightarrow r.\left( \hat { i } -\hat { j } +2\hat { k } \right) =1$

The point

$P(x, y, z)$ lies in the first octant and its distance from the origin is

$12$ units. If the position vector of

$P$ make

$45^{\circ}$ and

$60^{\circ}$ with the x-axis and y-axis respectively, then the coordinates of

$P$ are

0%
$(3\sqrt {3}, 6, 3\sqrt {2})$
0%
$(4\sqrt {3}, 8, 4\sqrt {2})$
0%
$(6\sqrt {2}, 6, 6,)$
0%
$(6, 6, 6\sqrt {2})$
0%
$(4\sqrt {2}, 8, 4\sqrt {3})$

Explanation

Given,

$OP = 12$ units
and

$\alpha = 45^{\circ}$ and

$\beta = 60^{\circ}$
Let direction cosines of line

$OP$ is

$< l, m, n >$

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

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

$\Rightarrow \cos^{2} 45^{\circ} + \cos^{2} 60^{\circ} + \cos^{2}\gamma = 1$

$\Rightarrow \left (\dfrac {1}{\sqrt {2}}\right )^{2} + \left (\dfrac {1}{2}\right )^{2} = 1 - \cos^{2}\gamma$

$\Rightarrow \sin^{2}\gamma = \dfrac {1}{2} + \dfrac {1}{4} = \dfrac {3}{4}$

$\Rightarrow \sin \gamma = \dfrac {\sqrt {3}}{2} = \sin 60^{\circ}$
Thus

$\gamma = 60^{\circ}$
So, the coordinates of

$P\equiv (l\cdot OP, m\cdot OP, n\cdot OP)$

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

$\equiv \left (12\cdot \dfrac {1}{\sqrt {2}}, 12\cdot \dfrac {1}{2}, 12\cdot \dfrac {1}{2}\right )$

$\equiv (6\sqrt {2}, 6, 6)$

The direction cosines

$l,m,n$ of two lines are connected by the relations

$l+m+n=0$ ,

$lm=0$ , then the angle between them is

0%
$\cfrac { \pi }{ 3 }$
0%
$\cfrac { \pi }{ 4 }$
0%
$\cfrac { \pi }{ 2 }$
0%
$0$

Explanation

Given,

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

And

$lm=0$

Either,

$m=0$ or

$l=0$

$l=0$ , then put in eq (i) we get

$m=-n$

$\therefore$ Direction ratios are

$0.-n,n$ i.e.,

$0,-1,1$

$m=0$ , then put in eq.(i) we get

$l=-n$

$\therefore$ Direction ratios are

$-n,0,n$ i.e.,

$-1,0,1$

$\cos { \theta } =\cfrac { 0(-1)+(-1)\times 0+1\times 1 }{ \sqrt { { (0) }^{ 2 }+{ (-1) }^{ 2 }+{ (1) }^{ 2 } } \sqrt { { (-1) }^{ 2 }+{ (0) }^{ 2 }+{ (1) }^{ 2 } } } =\cfrac { 1 }{ 2 }$

$\Rightarrow \theta =\cfrac { \pi }{ 3 }$

The angle between the straight lines

$x - 1 = \dfrac{2y + 3}{3} = \dfrac{z + 5}{2}$ and

$x = 3r + 2;\, y = -2r - 1; \,z = 2$ , where

$r$ is a parameter, is

0%
$\dfrac{\pi}{4}$
0%
$\cos^{-1} \left( \dfrac{-3}{\sqrt{182}} \right )$
0%
$\sin^{-1} \left( \dfrac{-3}{\sqrt{182}} \right )$
0%
$\dfrac{\pi}{2}$
0%
$0$

Explanation

Given equation of lines are,

$x - 1 = \dfrac{2y + 3}{3} = \dfrac{z + 5}{2}$
and

$x = 3r +2; y = - 2r - 1; z = 2$

$\Rightarrow \dfrac{x - 1}{1} = \dfrac{y + \dfrac{3}{2}}{\dfrac{3}{2}} = \dfrac{z + 5}{2}$
and

$\dfrac{x - 2}{3} = r: \dfrac{y + 1}{-2} = r : \dfrac{z - 2}{0} = r$

$\Rightarrow \dfrac{x - 1}{1} = \dfrac{y + \dfrac{3}{2}}{\dfrac{3}{2}} = \dfrac{z + 5}{2}$
and

$\dfrac{x - 2}{3} = \dfrac{y + 1}{-2} = \dfrac{z - 2}{0}$
DR's of lines Ist and IInd lines are

$\left( 1 , \dfrac{3}{2} , 2 \right )$ and

$(3, -2, 0)$ .
The angle between two straight lines is

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

$= \dfrac{1 \times 3 - \dfrac{3}{2} \times 2 + 2 \times 0}{\sqrt{ 1^2 + \left( \dfrac{3}{2} \right )^2 + 2^2} \sqrt{3^2 + 2^2 + 0^2}}$

$= \dfrac{3 - 3 + 0}{\sqrt{1 + \dfrac{9}{4} + 4} \sqrt{9 + 4}}$

$= \dfrac{0}{\sqrt{\dfrac{29}{4}} \sqrt{13}}$

$\Rightarrow \cos \theta = 0$

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

The angle between the lines

$\dfrac {x-7}{1} = \dfrac {y+3}{-5} = \dfrac {z}{3}$ and

$\dfrac {2-x}{-7} = \dfrac {y}{2} = \dfrac {z+1}{5}$ is equal to :

0%
$\dfrac {\pi}{4}$
0%
$\dfrac {\pi}{3}$
0%
$\dfrac {\pi}{2}$
0%
$\dfrac {\pi}{6}$
0%
$0$

Explanation

Given lines are

$\dfrac {x-7}{1} = \dfrac {y+3}{-5} = \dfrac {z}{3}$ ...(i)
and

$\dfrac {2-x}{-7} = \dfrac {y}{2} = \dfrac {z+5}{1}$

$\Rightarrow \dfrac {x-2}{7} = \dfrac {y}{2} = \dfrac {z+5}{1}$ ...(ii)
DR's of line equations (i) and (ii) are

$(a_1, b_1, c_1) = (1, -5, 3)$
and

$(a_2, b_2, c_2) = (7, 2, 1)$
Therefore,

$\cos \theta = \dfrac {a_1a_2 + b_1b_2 + c_1c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2 } \sqrt{a_2^2 + b_2^2 + c_2^2} }$

$= \dfrac {1 \times 7 + (-5) \times 2 + 3 \times 1 }{\sqrt{1^2+(-5)^2+(3)^2} \sqrt{(7)^2+(2)^2+(1)^2}}$

$= \dfrac {7-10+3} {\sqrt{1+25+9} \sqrt{49+4+1}} = 0$

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

$\alpha ,\beta ,\gamma$ are the angles which a directed line makes with the positive directions of the coordinate axes, then

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

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

Explanation

The direction cosines of the line are

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

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

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

$\Rightarrow \sin ^{ 2 }{ \alpha } +\sin ^{ 2 }{ \beta } +\sin ^{ 2 }{ \gamma } =2$

If the two lines

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

$\cfrac { x-3 }{ 1 } =\cfrac { 2y-3 }{ 4 } =\cfrac { z-2 }{ 2 }$ are perpendicular, then the value of

$a$ is equal to

0%
$-4$
0%
$5$
0%
$-5$
0%
$4$
0%
$-2$

Explanation

Given equation of lines

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

$\Rightarrow \cfrac { x-1 }{ 2 } =\cfrac { y-1 }{ a } =\cfrac { z-0 }{ 4 } ...(i)$
and

$\cfrac { x-3 }{ 1 } =\cfrac { y-\cfrac { 3 }{ 2 } }{ 2 } =\cfrac { z-2 }{ 2 } ...(ii)$

Here

$\quad { a }_{ 1 }=2,{ b }_{ 1 }=a,{ c }_{ 1 }=4;{ a }_{ 2 }=1,{ b }_{ 2 }=2,{ c }_{ 2 }=2$

Since lines (i) and (ii) are perpendicular

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

$\Rightarrow 2\times 1\times +a\times 2+4\times 2=0$

$\Rightarrow 2a=-10\Rightarrow a=-5$

The number of straight lines that are equally inclined to the three-dimensional coordinate axes, is

0%
$2$
0%
$4$
0%
$6$
0%
$8$

Explanation

Since,

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

$\Rightarrow \alpha + \cos^{-1}\left (\pm \dfrac {1}{\sqrt {3}}\right )$
So, there are four line whose

$DC's$ are

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

If the direction ratios of two lines are given by

$3lm-4ln+mn=0$ and

$l+2m+2n=0$ , then, the angle between the lines is

0%
$\cfrac { \pi }{ 2 }$
0%
$\cfrac { \pi }{ 3 }$
0%
$\cfrac { \pi }{ 4 }$
0%
$\cfrac { \pi }{ 6 }$

Explanation

By solving two equations we get

$\left( { l }_{ 1 },{ m }_{ 1 },{ n }_{ 1 } \right) =\left( 2\sqrt { 2 } -3,-\sqrt { 2 } ,1 \right)$

$\left( { l }_{ 2 },{ m }_{ 2 },{ n }_{ 2 } \right) =\left( -2\sqrt { 2 } -2,\sqrt { 2 } ,1 \right)$

${ l }_{ 1 }{ l }_{ 2 }+{ m }_{ 1 }{ m }_{ 2 }+{ n }_{ 1 }{ n }_{ 2 }=0$
Therefore, the angle between them is

$\cfrac { \pi }{ 2 }$ .

The angle between two diagonals of a cube will be

0%
$\sin ^{ -1 }{ \left( \dfrac { 1 }{ 3 } \right) }$
0%
$\cos ^{ -1 }{ \left( \dfrac { 1 }{ 3 } \right) }$
0%
Variable
0%
None of these

Explanation

Let the cube be of side

$a$ . Then,

$O\left( 0,0,0 \right) , D\left( a,a,a \right) , B\left( 0,a,0 \right)$ and

$G\left( a,0,a \right)$
Now, equations of diagonals

$OD$ and

$BG$ are

$\dfrac { x }{ a } =\dfrac { y }{ a } =\dfrac { z }{ a }$ and

$\dfrac { x }{ a } =\dfrac { y-a }{ -a } =\dfrac { z }{ a }$ , respectively.
Hence, angle between

$OD$ and

$BG$ is

$\cos ^{ -1 }{ \left( \dfrac { { a }^{ 2 }-{ a }^{ 2 }+{ a }^{ 2 } }{ \sqrt { 3{ a }^{ 2 } } \cdot \sqrt { 3{ a }^{ 2 } } } \right) } =\cos ^{ -1 }{ \left( \dfrac { 1 }{ 3 } \right) }$

$A = \begin{bmatrix} l_{1}& m_{1} & n_{1}\\ l_{2} & m_{2} & n_{2}\\ l_{3} & m_{3} & n_{3}\end{bmatrix}$ and

$B = \begin{bmatrix}p_{1} & q_{1} & r_{1}\\ p_{2} & q_{2} & r_{2}\\ p_{3} & q_{3} & r_{3}\end{bmatrix}$ where

$p_{1}, q_{1}, r_{1}$ are the co-factors of the elements

$l_{i},m_{i},n_{i}$ for

$i = 1, 2, 3$ . If

$(l_{1}, m_{1}, n_{1}), (l_{2}, m_{2}, n_{2})$ and

$(l_{3},m_{3}, n_{3})$ are the direction cosines of three mutually perpendicular lines then

$(p_{1}, q_{1}, r_{1}), (p_{2}, q_{2}, r_{2})$ and

$(p_{3}, q_{3}, r_{3})$ are

0%
The direction cosines of three mutually perpendicular lines
0%
The direction ratios of three mutually perpendicular lines which are not direction cosines
0%
The direction cosines of three lines which need not be perpendicular
0%
The direction ratios but not the direction cosines of three lines which need not be perpendicular

Explanation

Given

$p_1,q_1,r_1$ are the co-factors of the elements

$l_i,m_i,n_i$

For

$i=1,2,3$

$(l_2,m_2,n_2)$ and

$(l_3,m_3,n_3)$ are the direction cosines of three mutually perpendicular lines

Thus

$(p_1,q_1,r_1),(p_2,q_2,r_2) ,(p_3,q_3,r_3)$ are also the direction cosines of three mutually perpendicular lines

Option A is Correct answer

If a line makes the angle

$\alpha ,\beta ,\gamma$ with three dimensional coordinate axes respectively, then

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

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

Explanation

We need to find value of

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

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

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

$=2(1)-3=2=-1\left( \because { l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 }=1 \right)$

If the direction cosines of a line are

$\left (\dfrac {1}{c}, \dfrac {1}{c}, \dfrac {1}{c}\right )$ , then

0%
$0 < c < 1$
0%
$c > 2$
0%
$c = \pm \sqrt {2}$
0%
None of these

Explanation

Since,

$DC's$ of a line are

$\left (\dfrac {1}{c}, \dfrac {1}{c}, \dfrac {1}{c}\right )$

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

$\therefore \left (\dfrac {1}{c}\right )^{2} + \left (\dfrac {1}{c}\right )^{2} + \left (\dfrac {1}{c}\right )^{2} = 1$

$\Rightarrow 1 + 1 + 1 = c^{2} \Rightarrow c^{2} = 3$

$\Rightarrow c = \pm \sqrt {3}$ .

A line makes the same angle

$\theta$ with each of the

$x$ and

$z$ -axes. If the angle

$\beta$ , which it makes with

$y$ -axis, is such that

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

$\cos ^{ 2 }{ \theta }$ is equal to

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

Explanation

A line makes angle

$\theta$ with

$x$ -axis and

$z$ -axis and

$\beta$ with

$y$ -axis.

$l=\cos { \theta } , m=\cos { \beta } , n=\cos { \theta }$ ....

$\text{Since}\ { l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 }=1$

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

$\Rightarrow 2\cos ^{ 2 }{ \theta } =1-\cos ^{ 2 }{ \beta } =\sin ^{ 2 }{ \beta }$

$\Rightarrow 2\cos ^{ 2 }{ \theta } =3\sin ^{ 2 }{ \theta }$ .....(given,

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

$\Rightarrow 2\cos ^{ 2 }{ \theta } =3\left( 1-\cos ^{ 2 }{ \theta } \right)$

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

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

If a lines makes angles

$\alpha , \beta , \gamma , \delta$ with four digonals of a cube. Then

$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + \cos ^2 \delta$ will be :

0%
$4/3$
0%
$3/4$
0%
$1/4$
0%
None of these

Explanation

$DC's$ of the four diagonals of a cube on putting

$a=b=c$ are

$\left( \dfrac {1}{\sqrt3} , \dfrac {1}{\sqrt3} , \dfrac {1}{\sqrt3} \right) \left( \dfrac {-1}{\sqrt3} , \dfrac {1}{\sqrt3} , \dfrac {1}{\sqrt3} \right) \left( \dfrac {1}{\sqrt3} , \dfrac {-1}{\sqrt3} , \dfrac {1}{\sqrt3} \right) \left( \dfrac {1}{\sqrt3} , \dfrac {1}{\sqrt3} , \dfrac {-1}{\sqrt3} \right)$
Let

$l,m,n$ be the dc's of the line which is inclined at angles

$\alpha , \beta , \gamma , \delta$ resp to the four diagonals then

$\cos \alpha = \dfrac {1}{\sqrt3} \cdot l + m \cdot \dfrac{1}{\sqrt3} + n \cdot \dfrac{1}{\sqrt3} = \dfrac {l+m+n}{\sqrt3}$
Similarly

$\cos \beta = \dfrac {-l+m-n}{\sqrt3} , \cos \gamma = \dfrac {l-m+n}{\sqrt3} , \cos \delta = \dfrac {l+m-n}{\sqrt3}$

$\therefore \cos^2 \alpha , \cos^2 \beta + \cos^2 \gamma + \cos^2 \delta$

$= \dfrac {4 ( l^2 + m^2 + n^2)}{3} = \dfrac {4}{3} ( \because l^2 + m^2 + n^2 = 1)$

The acute angle between the lines whose direction ratios are given by

$l+m-n=0$ and

${ l }^{ 2 }+{ m }^{ 2 }-{ n }^{ 2 }=0$ , is

0%
$0$
0%
$\pi /6$
0%
$\pi /4$
0%
$\pi /3$

Explanation

We have

$l+m-n=0$
and

${ l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 }=0$

$\Rightarrow { l }^{ 2 }+{ m }^{ 2 }={ n }^{ 2 }$ and

$l+m=n$

$\Rightarrow { l }^{ 2 }+{ m }^{ 2 }={ (l+m) }^{ 2 }$

$\Rightarrow 2lm=0$

$\Rightarrow l=0,m=0$
If

$l=0$ , then

$m=n$

$\therefore \cfrac { l }{ 0 } =\cfrac { m }{ 1 } =\cfrac { n }{ 1 }$
If

$m=0$ , then

$l=n$

$\therefore \cfrac { l }{ 1 } =\cfrac { m }{ 0 } =\cfrac { n }{ 1 }$
thus, the dr's of the lines are proportional to

$0,1,1$ and

$1,0,1$
Therefore, angle

$\theta$ between them is given by

$\cos { \theta } =\cfrac { 0\times 1+1\times 0+\left| 1\times 1 \right| }{ \sqrt { 0+1+1 } \sqrt { 1+0+1 } } =\cfrac { 1 }{ 2 }$

$\Rightarrow \theta =\cfrac { \pi }{ 3 } \quad$

ABCD is a trapezium in which AB and CD are parallel sides. If

$l(AB) = 3 l (CD)$ and

$\bar {DC} = 2 \hat {i} - 5 \hat {k}$ . Then vector

$\bar {AB}$ is

0%
$\frac{3}{\sqrt{29} } (2 \hat{i} - 5 \hat {k})$
0%
$\frac {\sqrt{29}} {3} (5 \hat{i} - 2 \hat {k})$
0%
$-6 \hat {i} + 15 \hat {k}$
0%
A or B

Explanation

Given

$\vec{DC}=2\hat{i}-5\hat{k}$

$\vec{CD}=-2\hat{i}+5\hat{k}$

AB is parallel to CD

$\vec{AB}=\lambda\vec{CD}$

$\vec{AB}=\lambda(-2\hat{i}+5\hat{k})$

$\vec{AB}=-2\lambda\hat{i}+5\lambda\hat{k}$

$l (AB)=3 l (CD)$

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

$\sqrt{4\lambda^2+25\lambda^2}=3\sqrt{4+25}$

$\sqrt{29\lambda^2}=3\sqrt{29}$

$\sqrt{29}\lambda=3\sqrt{29}$

on comparing both sides

$\lambda=3$

putting value of

$\lambda$ in AB

$\vec{AB}=-2\times3\hat{i}+5\times3\hat{k}$

$\vec{AB}=-6\hat{i}+15\hat{k}$

Hence option C is correct

If a line makes

${ 45 }^{ o },{ 60 }^{ o }$ with positive direction of axes

$x$ and

$y$ then the angle it makes with the z-axis is:

0%
${ 30 }^{ o }\quad$
0%
${ 90 }^{ o }\quad$
0%
${ 45 }^{ o }\quad$
0%
${ 60 }^{ o }\quad$

Explanation

$\alpha, \beta, \gamma$ are the angles made by a line with the positive direction of

$x$ -axis,

$y$ -axis and

$z$ -axis then

$cos\:\alpha, cos\:\beta, cos\:\gamma$ are called the direction cosines of the line.

Also

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

Given

$\alpha=45^\circ, \beta=60^\circ$

$\Rightarrow cos^2 45^\circ+cos^2 60^\circ+cos^2\gamma=1$

$\Rightarrow \left(\dfrac{1}{\sqrt{2}}\right)^2+\left(\dfrac{1}{2}\right)^2+cos^2 \gamma=1$

$\Rightarrow \dfrac{1}{2}+\dfrac{1}{4}+cos^2 \gamma=1$

$\Rightarrow cos^2 \gamma=1-\dfrac{1}{2}-\dfrac{1}{4}$

$\Rightarrow cos^2 \gamma=1-\dfrac{3}{4}$

$\Rightarrow cos^2 \gamma=\dfrac{1}{4}$

$\Rightarrow cos \gamma=\dfrac{1}{2}$

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

$\Rightarrow \gamma=60^\circ$

The dc's

$(l,m,n)$ of two lines are connected between the relation

$l + m + n = 0, \ lm = 0$ , then the angle between the lines is

0%
$\displaystyle\frac{\pi}{3}$
0%
$\dfrac{\pi}{4}$
0%
$\dfrac{\pi}{2}$
0%
$\dfrac{\pi}{6}$

Explanation

We know that,

$\cos \theta =\dfrac{\overrightarrow{{{b}_{1}}.}\overrightarrow{{{b}_{2}}}}{\overrightarrow{\left| {{b}_{1}} \right|}\left| {{\overrightarrow{b}}_{2}} \right|}$ ...... (1)

Given that,

$l+m+n=0$

$l+m=-n$

$-\left( l+m \right)=n$

And, $lm=0$

So,

Either $l=0$ or $m=0$

When, $l=0$ , then

$m= -n$

And,

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

When, $m=0$ , then

$l= -n$

And,

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

Calculate $b_1\cdot b_2$ .

${{b}_{1}}.{{b}_{2}}=(0,1,-1).(1,0,-1)$

$=0+0+1$

$=1$

Therefore,

$\left| {{b}_{1}} \right|=\sqrt{{{0}^{2}}+{{1}^{2}}+{{(-1)}^{2}}}=\sqrt{2}$

$\left| {{b}_{2}} \right|=\sqrt{{{0}^{2}}+{{1}^{2}}+{{(-1)}^{2}}}=\sqrt{2}$

Now, substitute the values in equation (1).

$\cos \theta =\dfrac{\overrightarrow{{{b}_{1}}}.\overrightarrow{{{b}_{2}}}}{\left| {{\overrightarrow{b}}_{1}} \right|\left| \overrightarrow{{{b}_{2}}} \right|}$

$\Rightarrow \cos \theta =\dfrac{1}{\sqrt{2}.\sqrt{2}}=\dfrac{1}{2}$

$\Rightarrow \theta =\dfrac{\pi }{3}$

Hence, the angle between the lines is $\dfrac{\pi}{3}$ .

The equation of the plane passing through the point

$(1, 1, 1)$ and perpendicular to the planes

$2x+y-2z=5$ and

$3x-6y-2z=7$ is?

0%
$14x+2y-15z=1$
0%
$-14x+2y+15z=3$
0%
$14x-2y+15z=27$
0%
$14x+2y+15z=31$

Explanation

NOTE:Equation if the second plane should be

$3x-6y-2z=7$

Hence option (D) is the correct anwer.
1446001_770925_ans_67043567b1664eb9be38537606129b2a.jpeg

ABC is a triangle where

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

$C(\lambda , 5, \mu)$ . Let the median through

$A$ is equally inclined to the axes.
The value of

$\mu - \lambda$ is equal to:

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

Explanation

Now, the mid point of the side

$BC$ be

$D\left(\dfrac{\lambda-1}{2},4,\dfrac{\mu+2}{2}\right)$ .

Let

$\alpha$ ,

$\beta$ and

$\gamma$ be the angles made by the median

$AD$ of

$\Delta$ ABC.

Then direction cosines of the side

$AD$ are

$\cos \alpha=\dfrac{\dfrac{\lambda-1}{2}-2}{\vec{AD}}$ ,

$\cos \beta=\dfrac{4-3}{\vec{AD}}$ ,

$\cos \gamma=\dfrac{\dfrac{\mu+2}{2}-5}{\vec{AD}}$ .

Since

$\alpha=\beta=\gamma$

$\Rightarrow \cos \alpha=\cos \beta=\cos \gamma$ ......

$\because$ (equally inclined to the axes)

$\Rightarrow \dfrac{\lambda-5}{2}=1=\dfrac{\mu-8}{2}$

$\Rightarrow \lambda=7$ and

$\mu=10$ .

$\therefore \mu-\lambda=3$

The point of intersection of the line joining the points

$(-3, 4, -8)$ and

$(5, -6, 4)$ with the

$XY$ -plane is

0%
$\left (\dfrac {7}{3}, -\dfrac {8}{3}, 0\right )$
0%
$\left (-\dfrac {7}{3}, -\dfrac {8}{3}, 0\right )$
0%
$\left (-\dfrac {7}{3}, \dfrac {8}{3}, 0\right )$
0%
$\left (\dfrac {7}{3}, \dfrac {8}{3}, 0\right )$

Explanation

$\overrightarrow { r } =\overrightarrow { a } +x(\overrightarrow { b } -\overrightarrow { a } )\\ \overrightarrow { a } =-3\hat { i } +4\hat { j } +(-8)\hat { k } \\ \overrightarrow { b } =5\hat { i } +(-6)\hat { j } +4\hat { k } \\ \overrightarrow { r } =-3\hat { i } +4\hat { j } +(-8)\hat { k } +x(8\hat { i } +(-10)\hat { j } +12\hat { k } )=(8x-3)\hat { i } +(4-10x)\hat { j } +(12x-8)\hat { k }$

On $XY$ plane $\hat { k }$ will be $0$

So $12x-8=0$ it means $x=\dfrac { 2 }{ 3 }$

Putting the value of $x$ in component of $\hat { i }$ and $\hat { j }$ we get

$8\times \dfrac { 2 }{ 3 } -3$ and $4-10\times \dfrac { 2 }{ 3 }$

$\dfrac { 7 }{ 3 } \ and\ \dfrac { -8 }{ 3 }$

So correct optiopn will be option A.

Find the direction cosines of the line which is perpendicular to the lines with direction cosines proportional to 1, -2, -2 and 0, 2, 1.

0%
$\dfrac{1}{3}, \dfrac{2}{3}, -\dfrac{1}{3}$
0%
$\dfrac{2}{3}, -\dfrac{1}{3}, \dfrac{2}{3}$
0%
$1, -\dfrac{1}{3}, \dfrac{2}{3}$
0%
None of these

Explanation

Let

$l,m,n$ be the direction cosines of the required line. Since it is perpendicular to the lines whose direction cosines are proportional to

$1, -2, -2$ and

$0, 2, 1$ respectively.

Thus,

$l-2m-2n=0$

$0l+2m+n=0$

On solving these equations, we get,

$\dfrac{l}{2}=\dfrac{m}{-1}=\dfrac{n}{2}$

Thus, the direction ratios of the required line are proportional to

$2,-1,2$ . Hence, its direction cosines are :

$\dfrac{2}{\sqrt{2^2+(-1)^2+2^2}}, \dfrac{-1}{\sqrt{2^2+(-1)^2+2^2}}, \dfrac{2}{\sqrt{2^2+(-1)^2+2^2}}=\dfrac{2}{3}, -\dfrac{1}{3}, \dfrac{2}{3}$ .

The measure of acute angle between the lines whose direction ratios are

$3, 2, 6$ and

$-2, 1, 2$ is __________.

0%
$\cos^{-1}\left (\dfrac {1}{7}\right )$
0%
$\cos^{-1}\left (\dfrac {8}{15}\right )$
0%
$\cos^{-1}\left (\dfrac {1}{3}\right )$
0%
$\cos^{-1}\left (\dfrac {8}{21}\right )$

Explanation

Angle between the lines with direction ratios

$a_1,b_1,c_1$ and

$a_2,b_2,c_2$ is given by

$cos\: \theta= \left| \dfrac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \cdot \sqrt{a_2^2+b_2^2+c_2^2}}\right|$

We have

$a_1=3,b_1=2,c_1=6,a_2=-2,b_2=1,c_2=2$

$\cos\: \theta= \left| \dfrac{3(-2)+2(1)+6(2)}{\sqrt{3^2+2^2+6^2} \cdot \sqrt{-2^2+1^2+2^2}}\right|$

$=\left| \dfrac{-6+2+12}{\sqrt{9+4+36}\cdot \sqrt{4+1+4}} \right|$

$=\left| \dfrac{8}{\sqrt{49}\cdot \sqrt{9}} \right|$

$= \dfrac{8}{7\cdot 3}$

$\cos\: \theta=\left( \dfrac{8}{21} \right)$

$\theta=\cos^{-1}\left( \dfrac{8}{21} \right) \approx 67.60$

The perpendicular distance from the point

$(3,1,1)$ on the plane passing through the point

$(1,2,3)$ and containing the line,

$\vec { r } =\hat { i } +\hat { j } +\lambda \left( 2\hat { i } +\hat { j } +4\hat { k } \right)$ , is:

0%
$\cfrac { 1 }{ \sqrt { 11 } }$
0%
$\cfrac { 4 }{ \sqrt { 41 } }$
0%
$0$
0%
$\cfrac { 3 }{ \sqrt { 11 } }$

Explanation

Let equation of plane through

$P, Q$ &

$R$ be

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

$a\hat{i}+b\hat{j}+c\hat{k}=\begin{vmatrix} \hat{i}&\hat{j}&\hat{k} \\0&1 &3 \\2&2&7\end{vmatrix} = \hat{i}+6{j}-2\hat{k}$

$\therefore (a, b, c) = (1, 6, -2)$

$\therefore$ equation of plane will be

$x+6-2z=0$

$\therefore$ distance of

$(3,1,1)$ from the plane,

$d=\left|\dfrac{3+6-2-7}{\sqrt{1+36+4}}\right|=0$

If the angle between the lines,

$\displaystyle\frac{x}{2}=\frac{y}{2}=\frac{z}{1}$ and

$\displaystyle\frac{5-x}{-2}=\frac{7y-14}{P}=\frac{z-3}{4}$ is

$\cos^{-1}\displaystyle \left(\frac{2}{3}\right)$ ,

then p is equal to?

0%
$-\displaystyle\frac{7}{4}$
0%
$\displaystyle\frac{2}{7}$
0%
$-\displaystyle\frac{4}{7}$
0%
$\displaystyle\frac{7}{2}$

Explanation

Above formula is used to find angle

$\theta$ between two lines having direction ratios

$a_1,b_1,c_1$ and

$a_2,b_2,c_2$

Now let us find the direction cosines of the given lines

$\dfrac{x}{2}=\dfrac{y}{2}=\dfrac{z}{1}$ direction cosines are

$2,2,1$

next line can also be written as

$\dfrac{x-5}{2}=\dfrac{y-2}{\dfrac{P}{7}}=\dfrac{z-3}{4}$
hence direction cosines are

$2,\dfrac{P}7,4$

and we know that

$cos\theta=\dfrac23$

Keeping the value of these cosines in above formula we get

$\dfrac{2}{3}=\left|\dfrac{2\times 2+2\times \dfrac{P}{7}+1\times 4}{\sqrt{2^2+2^2+1^2}\sqrt{2^2+\dfrac{P^2}{49}+4^2}}\right|=\dfrac{8+\dfrac{2P}{7}+4}{3\times \sqrt{2^2+\dfrac{P^2}{49}+4^2}}$

after cross multiplication we get

$\Rightarrow \left(4+\dfrac{P}{7}\right)^2=20+\dfrac{P^2}{49}$

$\Rightarrow 16+\dfrac{8P}{7}+\dfrac{P^2}{49}=20+\dfrac{P^2}{49}$

$\Rightarrow \dfrac{8P}{7}=4$

$\Rightarrow P=\dfrac{7}{2}$

Therefore Answer is

$D$

The acute angle between two lines such that the direction cosines

$l,\ m,\ n$ of each of them satisfy the equation

$l+m+n=0$ and

$l^2+m^2-n^2=0$ is

0%
$30^\circ$
0%
$45^\circ$
0%
$60^\circ$
0%
$15^\circ$

Explanation

Lines are

$l+m+n=0\implies -l=(m+n)$ and

$l^2-m^2+n^2=0\implies l^2=m^2-n^2$ Solving them gives,

$(-(m+n))^2=m^2+n^2+2mn=m^2-n^2\implies 2n(n+m)=0$ Now, for

$n=0$

$m=-l$ , so d.c's

$\left(\dfrac 1{\sqrt 2}, -\dfrac 1{\sqrt 2}, 0\right)$ And for

$n=-m$

$l=0$ , so d.c's

$\left(0, \dfrac 1{\sqrt 2}, -\dfrac 1{\sqrt 2}\right)$ Angles between the lines

$|\cos \theta|=\left|\dfrac 12\right|\implies \theta =\dfrac \pi 3$

The direction cosines of the ray

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

$Q(-1,1,-2)$ are

0%
$\left(-2,-3,-6\right)$
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$\left(2,-3,-6\right)$
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$\left(\dfrac{2}{7},\dfrac{3}{7},\dfrac{6}{7}\right)$
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$\left(-\dfrac{2}{7},\dfrac{3}{7},-\dfrac{6}{7}\right)$

Explanation

$P(1, -2, 4)\ Q(-1, 1, -2)$

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

$= \sqrt{4 + 9 + 36}$

$= \sqrt{49}$

$= 7$

$DC = (\dfrac{-1 - 1}{7}, \dfrac{1 - (-2)}{7}, \dfrac{-2 - 4}{7})$

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

The measure of the angle between the lines, whose direction numbers are

$l,m,n$ and

$m-n,n-l,l-m$ is ______

0%
$\cfrac { \pi }{ 4 }$
0%
$\cfrac { \pi }{ 6 }$
0%
$\cfrac { \pi }{ 2 }$
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$\cfrac { \pi }{ 3 }$

Explanation

Here,

$(l,m,n).(m-n,n-l,l-m)$

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

$=lm-nl+mn-ml+nl-nm$

$=0$
So, Lines are mutuall perpendicular
So, Angle between them is

$\cfrac { \pi }{ 2 }$

CBSE Questions for Class 12 Commerce Maths Three Dimensional Geometry Quiz 6 - 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 6 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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