MCQ Questions for Class 12 Commerce Maths Three Dimensional Geometry Quiz 4

The point collinear with

$(4, 2, 0)$ and

$(6, 4, 6)$ among the following is

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$(0,4,6)$
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$(8,6,8)$
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$(1, -4, -6)$
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None of these

Explanation

Cartesian form of a line passing through

$(x_1,y_1,z_1)$ and

$(x_2,y_2,z_2)$ is:

$\dfrac{x-x_1}{x_2-x_1}=\dfrac{y-y_1}{y_2-y_1}=\dfrac{z-z_1}{z_2-z_1}=\lambda$

Cartesian form of the line passing through (4,2,0) and (6,4,6) is:

$\dfrac{x-4}{6-4}=\dfrac{y-2}{4-2}=\dfrac{z-0}{6-0}=\lambda$

$\dfrac{x-4}{2}=\dfrac{y-2}{2}=\dfrac{z-0}{6}=\lambda$

Therefore general point on the line is

$(2\lambda+4,2\lambda+2,6\lambda)$

Now if we compare this coordinate with given options,we get different values of

$\lambda$ .

that means none of the points given in options lie on the line(Non-Colinear).

Therefore, (D) option is correct.

Equation of the line which passes through the point with position vector

$(2, 1, 0)$ and perpendicular to the plane containing the vectors

$i + j$ and

$j + k$ is

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$\vec{r}= (2, 1, 0) + t(1, -1, 1)$
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$\vec{r} = (2, 1, 0) + t(-1, 1, 1)$
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$\vec{r}= (2, 1, 0) + t(1, 1, -1)$
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$\vec{r} = (2, 1, 0) + t(1, 1, 1)$

Explanation

$\vec{r}=position\ vector\ \vec{a}+t(normal\ vector\ \vec{n})$

normal vector of plane

Given

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

$\vec{b}=\hat{j}+\hat{k}$

$\vec{n}=\vec{a}\times\vec{b}$

$\Rightarrow\vec{a}\times\vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&1&0\\0&1&1\end{vmatrix}$

$\Rightarrow\vec{a}\times\vec{b}=\hat{i}(1-0)-\hat{j}(1-0)+\hat{k}(1-0)$

$\Rightarrow\vec{a}\times\vec{b}=\hat{i}-\hat{j}+\hat{k}$

eq become

$\Rightarrow\vec{r}=(2,1,0)+t(1,-1,1)$

If O is the origin and A is the point

$(a, b, c)$ , then the equation of the plane through A and at right angles to OA is

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$a (x - a) - b (y - b) - c (z - c) = 0$
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$a(x + a) + b (y + b) + c (z + c) = 0$
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$a(x - a) + b (y - b) + c (z - c) = 0$
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None of the above

Explanation

A plane with normal

$ai+bj+ck$ is given by

$ax+by+cz+d=0$ .

Since it passes through

$(a,b,c)$ we have

$d=-(a^2+b^2+c^2)$ .

Hence, the equation of plane is

$ax+by+cz-a^2-b^2-c^2=0$ .

By rearranging we get,

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

The values of

$a$ for which point

$(8, -7, a), (5, 2, 4)$ and

$(6, -1, 2)$ are collinear.

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$-4$
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$-2$
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$0$
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$2$

Explanation

Let the points be

$A(8,-7,a), B(5,2,4), C(6,-1,2)$
Direction ratio of

$BC =(1,-3,-2)$
If

$A,B,C$ are collinear points then direction ratio of

$BC$ and

$AB$ must be proportional.
Direction ratio of

$AB=(-3,9.4-a)$

$\therefore -3=\lambda1$

$\therefore 9=\lambda(-3)$

$\therefore 4-a=\lambda(-2)$ and

$\lambda=-3$

$\therefore \dfrac{4-a}{-3}=-2$

$\therefore a=-2$

The equation of the perpendicular from the point

$(\alpha, \beta, \gamma)$ to the plane

$ax + by + cz + d = 0$ is

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$a(x-\alpha) + b (y - \beta) + c ( z - \gamma) = 0$
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$\displaystyle \frac{x - \alpha}{a} = \frac{y - \beta}{b} = \frac{z - y}{c}$
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$a (x - \alpha) + b(y - \beta) + c (z - \gamma) = abc$
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None of these

Explanation

Equation of given plane is

$ax+by+cz+d=0$

Equation of the perpendicular plane is

$ax+by+cz+d=k$

This planes passes through

$P(\alpha,\beta,\gamma)=\dfrac{\alpha}{a}+\dfrac{\beta}{b}+\dfrac{\gamma}{c}+d=k$

$\therefore$ Equation of the required plane is

$ax+by+cz+d=\dfrac{\alpha}{a}+\dfrac{\beta}{b}+\dfrac{\gamma}{c}+d=k$

$i.e. \dfrac{x-\alpha}{a}+\dfrac{y-\beta}{b}+\dfrac{z-\gamma}{c}$

The straight line

$\displaystyle \frac{x - 3}{3} = \frac{y - 2}{1} = \frac{z - 1}{0}$ is

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parallel to x-axis
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parallel to y-axis
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parallel to z-axis
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perpendicular to z-axis

Explanation

The given line is

$\dfrac{x-3}{3}=\dfrac{y-2}{1}=\dfrac{z-1}{0}$ which passes through the points

$(3,2,1)$ and having the direction cosine's as

$a_1=(3,1,0)$

Direction cosine's of z axis are

$a_{2}=(0,0,1)$
Now,

$a_{1}\cdot a_{2}=(3,1,0)\cdot(0,0,1)=3\times 0+1\times 0+0\times 1=0$

$\therefore a_1\cdot a_2=0$
Thus,

$z-$ axis and given line are perpendicular.
Hence, option D is correct.

The equation of altitude through

$B$ to side

$AC$ is

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$r= k+t(7 i - 10 + 2k)$
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$r = k + t (- 9 i + 6 j - 2k)$
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$r = k + t (7i - 10 j - 2k)$
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$r = k + t (7i + 10 j + 2k)$

Explanation

Equation of line

$AC$ is given by,

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

$\displaystyle \dfrac{x-1}{2}=\dfrac{y-2}{1}=\dfrac{z-3}{2}=t$ (say)

So any point on line

$AC$ is

$P(2t+1, t+2, 2t+3)$ such that

$BP\perp AC$

$\Rightarrow (2t+1-0)(2)+(t+2-0)(1)+(2t+3-1)(2)=0$

$\Rightarrow 4t+2+t+2+4t+4=0\Rightarrow t=-\dfrac{8}{9}$

$\therefore\displaystyle P =\left(-\dfrac{7}{9},\dfrac{10}{9},\dfrac{11}{9} \right)$

Thus the equation altitude through

$B$ to side

$AC$ , i.e. the equation of line

$BP$ is,

$\displaystyle \dfrac{x-0}{-7/9}=\dfrac{y-0}{10/9}=\dfrac{z-1}{11/9-1}$

$\displaystyle \dfrac{x}{-7}=\dfrac{y}{10}=\dfrac{z-1}{2}$

In vector form, it is given by,

$r=k+t(-7i+10j+2k)$

The equation of the plane through

$(1, 2, 3)$ and parallel to the plane

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

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

Explanation

Equation of plane parallel to

$2x + 3y - 4z = 0$ is given by,

$2x + 3y - 4z +\lambda= 0$
Also it passes through

$(1,2,3)$

$\Rightarrow 2(1)+3(2)-4(3)+\lambda = 0\Rightarrow \lambda = 4$
Hence, required plane is

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

$M$ denotes the mid-point of the line segment joining

$A( 4\widehat{i} + 5\widehat{j}- 10 \widehat{k})$ and

$B(-\widehat{i} + 2 \widehat{j} + \widehat{k})$ , then the equation of the plane through

$M$ and perpendicular to

$AB$ , is

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$\vec{r}. ( -5 \widehat{i}- 3 \widehat{j} + 11 \widehat{k}) + \dfrac{135}{2}= 0$
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$\vec{r}.\left ( \dfrac{3}{2} \widehat{i}+\dfrac{7}{2} \widehat{j} - \dfrac{9}{2} \widehat{k}\right) + \dfrac{135}{2}= 0$
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$\vec{r}. ( 4 \widehat{i}+ 5\widehat{j} - 10 \widehat{k}) + 4= 0$
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$\vec{r}. ( - \widehat{i}+ 2 \widehat{j} + \widehat{k}) + 4= 0$

Explanation

$\vec { OM } =\dfrac { \vec { OA } +\vec { OB } }{ 2 } =\dfrac { 3i+7j-9k }{ 2 }$
Then the equation of the plane through

$M$ and perpendicular to

$AB$ is

$\left( \vec { r } -\vec { OM } \right) .\left( \vec { OB } -\vec { OA } \right) =0$

$\Rightarrow \left( \vec { r } -\dfrac { 3\vec i+7\vec j-9\vec k }{ 2 } \right) .\left( -\vec i+2\vec j+\vec k-\left( 4\vec i+5\vec j-10\vec k \right) \right) =0$

$\Rightarrow \vec { r } .(-5\vec i-3\vec j+11\vec k)+\dfrac { 135 }{ 2 } =0$

Ans: A

If the points

$a(1, 2, -1), B(2, 6, 2)$ and

$c(\lambda, -2, -4)$ are collinear then

$\lambda$ is

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

Explanation

D.R of AB are

$2 -1, 6-2,2-(-1) i.e. 1,4,3$

D.R. of AC are

$λ.−1,−2−2,−4−(−1)$

$i.e., λ−1,−4,−3$

Since A, B, C are collinear

$\therefore AB||BC$

$\therefore \dfrac {\lambda-1}{1} =\dfrac{-4}{4}=\dfrac{-3}{3}$

$\Longrightarrow \lambda-1=-1$

$\therefore \lambda=0$

Given

$A(1,-1,0)$ ;

$B(3,1,2)$ ;

$C(2,-2,4)$ and

$D(-1,1,-1)$ which of the following points neither lie on

$AB$ nor on

$CD$ ?

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$(2,2,4)$
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$(2,-2,4)$
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$(2,0,1)$
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$(0,-2,-1)$

Explanation

$A(1,-1,0) , B(3,1,2), C(2,-2,4), D(-1,1,-1)\\ \vec { AB } = <3-1, 1-(-1), 2-0>$

$= <2,2,2>$

$\therefore \quad Equation\quad of\quad line\quad AB:\\ \dfrac { x-1 }{ 2 } =\dfrac { y-(-1) }{ 2 } =\dfrac { z-0 }{ 2 } \quad \Longrightarrow \quad \dfrac { x-1 }{ 2 } =\dfrac { y+1 }{ 2 } =\dfrac { z-0 }{ 2 }$

$\vec { CD } = <-1-2, 1-(-2), -1-4>$

$= <-3,3,-5>$

$\therefore \quad Equation\quad of\quad line\quad CD:\\ \dfrac { x-2 }{ -3 } =\dfrac { y-(-2) }{ 3 } =\dfrac { z-4 }{ -5 } \quad \Longrightarrow \quad \dfrac { x-2 }{ -3 } =\dfrac { y+2 }{ 3 } =\dfrac { z-4 }{ -5 }$

By putting the values of points given in the choices in the equation of lines

$AB$ and

$CD$ ,

$(2,2,4)$ is the point which neither lie on

$AB$ nor on

$CD$ .

The projections of a directed line segment on the coordinate axes

$12, 4, 3$ . The direction cosines of the line are

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$\dfrac{12}{13}, -\dfrac{4}{13}, \dfrac{3}{13}$
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$-\dfrac{12}{13}, -\dfrac{4}{13}, \dfrac{3}{13}$
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$\dfrac{12}{13}, \dfrac{4}{13}, \dfrac{3}{13}$
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none of these

Explanation

Assuming position vector of line segment ends point are

$\vec{A}$ and

$\vec{B}$
Now given projection of line segment on the axes are

$(12, 4, 3)$
So

$\vec{AB} = 12 \hat{i} + 4\hat{j} + 3 \hat{k}$
Hence, direction cosine of line segment are

$\dfrac{12}{\sqrt{12^2+4^2+3^2}}, \dfrac{4}{\sqrt{12^2+4^2+3^2}}, \dfrac{3}{\sqrt{12^2+4^2+3^2}}$

$\Rightarrow \dfrac{12}{13}, \dfrac{4}{13}, \dfrac{3}{13}$

Given

$A(1,-1,0)$ ;

$B(3,1,2)$ ;

$C(2,-2,4)$ and

$D(-1,1,-1)$ which of the following points neither lie on

$AB$ nor on

$CD$

0%
$(2,2,4)$
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$(2,-2,4)$
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$(2,0,1)$
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$(0,-2,-1)$

Explanation

Given:-

$A(1,-1,0)$ ;

$B(3,1,2)$ ;

$C(2,-2,4)$ and

$D(-1,1,-1)$

The equation of line

$AB$ is given by

$r=i-j+t(2i+2j+2k)$ or

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

The points

$(2,0,1)$ and

$(0,-2,1)$ lies on

$AB$ .

The equation of line

$CD$ is given by

$r=2i-2j+4k+t(3i-3j+5k)$ or

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

$(2,-2,4)$ lie on

$CD$ .

$(2,2,4)$ does not lie on any of the lines AB or CD.

Hence, option A is correct.

The vector equation of the line

$\displaystyle \frac { x-2 }{ 2 } =\frac { 2y-5 }{ -3 } ,z=-1$ is

$\displaystyle \overrightarrow { r } =\left( 2\hat { i } +\frac { 5 }{ 2 } \hat { j } -\hat { k } \right) +\lambda \left( 2\hat { i } -\frac { 3 }{ 2 } \hat { j } +x\hat { k } \right)$ , where

$x$ is equal to

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$0$
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$1$
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$2$
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$3$

Explanation

The given line is

$\displaystyle\frac { x-2 }{ 2 } =\frac { 2y-5 }{ -3 } ,z=-1$

$\Rightarrow\displaystyle\frac { x-2 }{ 2 } =\frac { 2y-5 }{ -3 } =\frac { z+1 }{ 0 }$

This shows that the given line passes through point

$\left( 2,\displaystyle\frac { 5 }{ 2 } ,-1 \right)$ and has direction ratios

$\left( 2, \displaystyle\frac { -3 }{ 2 } ,0 \right)$

So the direction cosines are

$\displaystyle\frac { 2 }{ \sqrt { { 2 }^{ 2 }+{ \left( -\dfrac { 3 }{ 2 } \right) }^{ 2 } } } ,\frac { -\dfrac { 3 }{ 2 } }{ \sqrt { { 2 }^{ 2 }+{ \left( -\dfrac { 3 }{ 2 } \right) }^{ 2 } } } ,\frac { 0 }{ \sqrt { { 2 }^{ 2 }+{ \left( -\dfrac { 3 }{ 2 } \right) }^{ 2 } } }$ or

$\left( \displaystyle\frac { 4 }{ 5 } ,-\frac { 3 }{ 5 } ,0 \right)$

Thus the given line passes through the point having position vector

$\vec { a } =2\hat { i } +\displaystyle\frac { 5 }{ 2 } \hat { j } -\hat { k }$ and is parallel to the vector

$\vec { b } =2\hat { i } -\displaystyle\frac { 3 }{ 2 } \hat { j } +0\hat { k }$

So, its vector equations is

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

Hence,

$x=0$ .

The equation of the right bisector plane of the segment joining

$(2,3,4)$ and

$(6,7,8)$ is

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$x+y+z+15=0$
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$x+y+z-15=0$
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$x-y+z-15=0$
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None of these

Explanation

Let

$\vec { OA } =2\vec i+3\vec j+4\vec k$ &

$\vec { OB }=6\vec i+7\vec j+8\vec k$
Then

$\vec { AB }=\vec { OB } -\vec { OA } =4(\vec i+\vec j+\vec k)$ is normal of the plane and passes through midpoint(M) of

$AB$ .
Therefore,

$\vec { OM } =\dfrac { \vec { OA } +\vec { OB } }{ 2 } =4\vec i+5\vec j+6\vec k$
Equation of desired plane is

$\left( \vec { r } -\vec { OM } \right) .\vec { AB } =0$

$\Rightarrow \left( x\vec i+y\vec j+z\vec k-4\vec i-5\vec j-6\vec k \right) .\left( 4\vec i+4\vec j+4\vec k \right) =0$

$\Rightarrow x+y+z=15$

Ans: B

Equation of plane passing through the points

$(2, 2, 1)$ ,

$(9, 3, 6)$ and perpendicular to the plane

$2x+ 6y + 6z-1= 0$ is

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$3x+ 4y+ 5z= 9$
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$3x+ 4y- 5z+ 9= 0$
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$3x+ 4y- 5z- 9=0$
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None of these

Explanation

Let

$ax+by+cz=1$ be a plane
Since, it is perpendicular to

$2x+6y+6z=1$
Therefore,

$2a+6b+6c=1$ ....(1)
and passes through

$(2,2,1)$ &

$(9,3,6)$
Therefore,

$2a+2b+c=1$ ....(2)
and

$9a+3b+6c=1$ ....(3)
Solving

$(1),(2)$ and

$(3)$ simultaneoulsly, we get

$a=\dfrac {3}{9}$ ,

$b=\dfrac {4}{9}$ ,

$c=-\dfrac {5}{9}$
Therfore, desired plane is

$3x+4y-5z-9=0$

Ans: C

Vector equation of the plane

$\vec{r}=\widehat{i}-\widehat{j}+\lambda (\widehat{i}+\widehat{j}+\widehat{k})+ \mu (\widehat{i}-2\widehat{j}+3\widehat{k})$ in the scalar dot product form is

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$\vec{r}.(5\widehat{i}+2\widehat{j}-3\widehat{k})=7$
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$\vec{r}.(5\widehat{i}-2\widehat{j}+3\widehat{k})=7$
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$\vec{r}.(5\widehat{i}-2\widehat{j}-3\widehat{k})=7$
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$\vec{r}.(5\widehat{i}+2\widehat{j}+3\widehat{k})=17$

Explanation

We have been given a plane which passes through three points namely,

$P = (1, -1, 0), Q = (2, 0, 1), R = (2, -3, 3)$ substituting

$0,1$ for

$\lambda, \mu.$

$\overrightarrow{PQ} = \hat{i} - \hat{j} + \hat{k}$ and

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

Normal vector

$= \overrightarrow{PQ} \times \overrightarrow{PR}$

$= (\bar{i} - \bar{j} + \bar{k}) \times (\bar{i} -2 \bar{j} + 3\bar{k})$

$= -2\bar{k} - 3\bar{j} - \bar{k} + 3\bar{i} + \bar{j} + 2\bar{i}$

$= 5\bar{i} - 2\bar{j} - 3\bar{k}$

Now,

$5(x - 1) - 2(y + 1) - 3(z - 0) = 0$ represents the equation of plane.

$\Rightarrow 5x - 2y - 3z = 7$ or

$\bar{r}.(5\bar{i} - 2\bar{j} - 3\bar{k}) = 0$

If the points

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

$\lambda$ ,

$1)$ and (

$\mu$ ,

$7, 4$ ) are collinear, the point on the same line is

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$(5, 6, 3)$
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$(1, -1, -2)$
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$(-5, -6, -3)$
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$(0, 0, 0)$

Explanation

The direction vector using the first two points we get

$3i+(\lambda-1)j+3k$ .

Using the first and third point

$\mu i+6j+6k$ .

Since they are up to multiplication by a constant we get

$\lambda=4$ and

$\mu=6$ .

Hence direction vector is

$i+j+k$ .

Any point on the line is given by

$(0,1,-2)+t(1,1,1)$ . Plugging

$t=5$ we get the point

$(5,6,3)$ .

The equation of the plane passing through the line

$\displaystyle \frac{x - 1}{2} = \frac{y + 1} {-1} = \frac{z}{3}$ and parallel to the direction whose direction numbers are

$3, 4, 2$ is

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$\displaystyle 14x - 5y - 11z = 19$
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$\displaystyle 3x + 4y + 2z + 1 = 0$
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$\displaystyle 2x - y + 3z = 3$
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none of these

Explanation

Equation of plane passing through given line is given by,

$a(x-1)+b(y+1)+cz=0$
Also direction ratio of line are

$2,-1,3$

$\Rightarrow 2a-b+3c=0......(1)$
and plane is parallel to direction with dr

$3,4,2$

$\Rightarrow 3a+4b+2c=0.....(2)$
Solving (1) and (2), we get

$a = \dfrac{-14c}{11}, b= \dfrac{5c}{11}$
Hence, required plane is

$14x-5y-11z=19$

A line makes angles

$\alpha$ ,

$\beta$ ,

$\gamma$ with the positive directions of the axes of reference. The value of

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

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

Explanation

The value of

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

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

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

$=2-3$

$=-1$

If the points

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

$B(2,6,2)$ and

$\displaystyle C\left ( \lambda,-2,-4 \right )$ are collinear, then

$\displaystyle \lambda$ is

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

Explanation

D.R. of AB are

$2−1,6−2,2−(−1)i.e.,1,4,3.$

D.R. of AC are

$λ.−1,−2−2,−4−(−1)$

$i.e., λ−1,−4,−3$

Since A, B, C are colinear,

$\therefore AB||BC$

$\therefore \dfrac{λ−1}{1}=\dfrac{−4}{4}=\dfrac{−3}{3}⇒λ−1=−1⇒λ=0$

The direction cosines of the line joining the points

$(2,3,-1)$ and

$(3,-2,1)$ are

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$-1,5,-2$
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$\dfrac{1}{\sqrt{30}}$ , $-\sqrt{\dfrac{5}{6}}$ , $\sqrt{\dfrac{2}{15}}$
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$\dfrac{-1}{30 }$ , $\dfrac{1}{6 }$ , $-\dfrac{1}{15 }$
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none of these

Explanation

Step 1:

The direction ratios of the line joining these points are

$x_2−x_1,y_2−y_1,z_2−z_1$

$(i.e) (3−2),(-2−3),(1−(-1))$

$\Longrightarrow$ we get

$(1,-5,2)$

Step 2:

Its direction cosines are :

$\dfrac {1}{\sqrt {30}},-\sqrt{\cfrac 56},\sqrt{ \cfrac {2}{15}}$

The direction cosines of the perpendicular from the origin to the plane

$\displaystyle 3x - y + 4z = 5$ are

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$\displaystyle 4, \: -1, \: 3$
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$\displaystyle 3, \: -1, \: 4$
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$\displaystyle \frac{3}{\sqrt{26}}, \: \frac{-1}{\sqrt{26}}, \: \frac{4}{\sqrt{26}}$
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$\displaystyle \frac{4}{\sqrt{26}}, \: \frac{-1}{\sqrt{26}}, \: \frac{3}{\sqrt{26}}$

Explanation

The direction ratios of the perpendicular from the origin to the plane

$\displaystyle 3x - y + 4z = 5$ are

$(3,-1,4)$
Therefore, direction cosines are

$\displaystyle \left( \frac { 3 }{ \sqrt { { 3 }^{ 2 }+{ 1 }^{ 2 }+4^{ 2 } } } ,\frac { -1 }{ \sqrt { { 3 }^{ 2 }+{ 1 }^{ 2 }+4^{ 2 } } } ,\frac { 4 }{ \sqrt { { 3 }^{ 2 }+{ 1 }^{ 2 }+{ 4 }^{ 2 } } } \right) =\left( \frac { 3 }{ \sqrt { 26 } } ,\frac { -1 }{ \sqrt { 26 } } ,\frac { 4 }{ \sqrt { 26 } } \right)$

Ans: C

The angle between the lines

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

$\displaystyle \frac{x-1}{0}=\frac{y}{3}=\frac{z}{0}$ is

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

Explanation

Given lines are

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

$\displaystyle \frac{x-1}{0}=\frac{y}{3}=\frac{z}{0}$

The direction vector of the first line is

$P=2i+2j+k$ and second line

$Q=3j$ .

Then the cosine of the angle between the line is given by

$\dfrac{|P\cdot Q|}{|P||Q|}$ .

$\Rightarrow \cos\theta=\dfrac{6}{\sqrt{4+4+1}\sqrt{3^{2}}}$

$\Rightarrow \cos\theta=\dfrac{6}{3\times3}$

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

The direction cosines of the normal to the plane

$\displaystyle 5\left ( x - 2 \right ) = 3\left ( y - z \right )$ are

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$\displaystyle 5, \: -3, \: 3$
0%
$\displaystyle \frac{5}{\sqrt{43}}, \: \frac{-3}{\sqrt{43}}, \: \frac{3}{\sqrt{43}}$
0%
$\displaystyle \frac{1}{2}, \: \frac{-3}{10}, \: \frac{3}{10}$
0%
$\displaystyle 1, \: \frac{-3}{5}, \: \frac{3}{5}$

Explanation

Given plane is,

$5(x-2) = 3(y-z)$

$\Rightarrow 5x-3y+3z-10 = 0$
So direction ratios of normal to the plane are

$5, -3$ and

$3$ .
Thus direction cosines are,

$\dfrac{5}{\sqrt{5^2+3^2+3^2}}, -\dfrac{3}{\sqrt{5^2+3^2+3^2}}, \dfrac{3}{\sqrt{5^2+3^2+3^2}}$
or

$\dfrac{5}{\sqrt{43}}, \dfrac{-3}{\sqrt{43}}, \dfrac{3}{\sqrt{43}}$

If the direction cosines of the line joining the origin and a point at unit distance from the origin are

$\displaystyle \frac{1}{\sqrt{3}}$ ,

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

$\lambda$ then value of

$\lambda$ is?

0%
$\displaystyle \frac{1}{2}\sqrt{\frac{5}{3}}$
0%
$\displaystyle \frac{2}{\sqrt{3}}$
0%
$\displaystyle \frac{-1}{2}\sqrt{\frac{5}{3}}$
0%
none of these

Explanation

Direction cosines of the line are given as

$\dfrac{1}{\sqrt3},\dfrac{-1}{2},\lambda$
We know that

$\cos^2a+\cos^2b+\cos^2c=1$
i.e

$\dfrac{1}{\sqrt3}^2+\dfrac{-1}{2}^2+{\lambda}^2=1$

$\dfrac{1}{3}+\dfrac{1}{4}+{\lambda}^2=1$

$\dfrac{7}{12}+{\lambda}^2=1$

${\lambda}^2=\dfrac{5}{12}$
Therefore,

$\lambda=\dfrac{\sqrt5}{2\sqrt3}$

Hence, option A is correct.

$\theta$ is an angle given by

$\cos \theta$

$=$

$\displaystyle \frac{\cos ^{2}\alpha+\cos ^{2}\beta +\cos ^{2}\gamma}{\sin ^{2}\alpha +\sin ^{2}\beta +\sin ^{2}\gamma }$ where

$\alpha$ ,

$\beta$ ,

$\gamma$ are the angles made by a line with the positive directions of the axes of reference then the measure of

$\theta$ is

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

Explanation

The cosines of the angles made by a directed line segment with the coordinate axes are called as the direction cosines of that line.

$α, β, γ$ are the angles made by the line segment with the coordinate axis then these angles are termed to be the direction angles and the cosines of these angles are the direction cosines of the line.

cos α, cos β and cos γ are called as the direction cosines

hence from the above diagram it is known that

$\theta$ is equal to

$60 c^0=\dfrac{\pi}{3}$ .

If a line makes angles of

$60^{\circ}$ and

$45^{\circ}$ with the positive directions of the

$x$ -axis and

$y$ -axis respectively, then the acute angle between the line and the

$z$ -axis is

0%
$60^{\circ}$
0%
$45^{\circ}$
0%
$75^{\circ}$
0%
$15^{\circ}$

Explanation

Let the angle with

$z$ axis be

$\gamma$
And as we know that

${\cos}^{2}$

$\alpha$

$+$

${\cos}^{2}$

$\beta$

$+$

${\cos}^{2}$

$\gamma$

${=}$

$1$

${\cos}^{2}$

${60}$

$+$

${\cos}^{2}$ 45

$+$

${\cos}^{2}$

$\gamma$

${=}$

$1$

$\dfrac{1}{4}$

$+$

$\dfrac{1}{2}$

$+$

${\cos}^{2}$

$\gamma$

${=}$

$1$

${\cos}^{2}$

$\gamma$

${=}$

$1-$

$\dfrac{1}{4}$

$-$

$\dfrac{1}{2}$

${\cos}^{2}$

$\gamma$

${=}$

$\dfrac{1}{4}$
Since the angle is acute,

$\gamma$

${=}60 ^o$

$ABC$ is a triangle where

$A= \left ( 2,3,5 \right ),B= \left ( -1,3,2 \right )$ and

$C= \left (\lambda ,5,\mu \right )$ . If the median through A is equally inclined with the axes, then

0%
$\lambda = 14,\mu = 20$
0%
$\lambda = 9,\mu = 6$
0%
$\lambda = \displaystyle \frac{7}{2},\mu = 20$
0%
$\lambda = 10,\mu = 7$

Explanation

$A$ is a median, then it meets

$BC$ at their mid-point

$D$ ,
Coordinates of

$D$ are

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

$AD$ is equally inclined on the axes i.e. makes equal angle with all the three axes, which implies that, if

$l,m,n$ are the direction ratios then they should be equal.

$\therefore \dfrac{\lambda-1}{2}=4=\dfrac{2+\mu}{2}$

$\Rightarrow \lambda = 9, \mu = 6$

If the image of the point

$\displaystyle \left ( 1, \: 1, \: 1 \right )$ by a plane

$\displaystyle \left ( 3, \: -1, \: 5 \right )$ then the equation of the plane is

0%
$\displaystyle x - y + 2z = 8$
0%
$\displaystyle x - y + 2z = 16$
0%
$\displaystyle x - y + 2z = 14$
0%
None of these

Explanation

Let

$A = (1,1,1)$ and

$A'=(3,-1,5)$
So mid point of

$AA'$ is

$(2,0,3)$
and direction ratio of

$AA'$ are

$2,-2$ and

$4$ .
Since

$A'$ is the image of

$A$ so equation of plane is given by,

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

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

Which of the triplet can not represent direction cosine of a line

0%
$\displaystyle \left( \frac { 1 }{ \sqrt { 3 } } ,\frac { 1 }{ \sqrt { 3 } } ,\frac { 1 }{ \sqrt { 3 } } \right)$
0%
$\displaystyle \left( \frac { 3 }{ \sqrt { 50 } } ,\frac { 4 }{ \sqrt { 50 } } ,\frac { 5 }{ \sqrt { 50 } } \right)$
0%
$\displaystyle \left( \frac { 4 }{ \sqrt { 77 } } ,\frac { 5 }{ \sqrt { 77 } } ,\frac { 6 }{ \sqrt { 77 } } \right)$
0%
$\displaystyle \left( \frac { 2 }{ \sqrt { 25 } } ,\frac { 3 }{ \sqrt { 25 } } ,\frac { 4 }{ \sqrt { 25 } } \right)$

Explanation

In general, the direction cosine of a line is defined as the cosine of the angles between the positive directed lines and the coordinate axes.

So the direction cosine of a line

$(\dfrac{2}{\sqrt 25},\dfrac{3}{\sqrt 25},\dfrac{4}{\sqrt 25})$ cannot be represented in triplet form

Option D is right answer

The direction ratios of a normal to the plane through

$\left( 1,0,0 \right) ,\left( 0,1,0 \right) ,$ which makes an angle of

$\displaystyle \frac { \pi }{ 4 }$ with the plane

$x+y=3$ are

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

Explanation

Any plane through $\left( 1,0,0 \right)$ is

$A\left( x-1 \right) +B\left( y-0 \right) +C\left( z-0 \right) =0$ ... $(1)$

It contains $\left( 0,1,0 \right)$ if $-A+B=0$ .... $(2)$

Also, $(1)$ makes an angle of $\displaystyle \dfrac { \pi }{ 4 }$ with the plane $x+y=3$

Therefore, $\displaystyle \cos { \dfrac { \pi }{ 4 } } =\dfrac { \left| A+B \right| }{ \sqrt { { A }^{ 2 }+{ B }^{ 2 }+{ C }^{ 2 } } \sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 } } }$

$\Rightarrow { \left( A+B \right) }^{ 2 }={ A }^{ 2 }+{ B }^{ 2 }+{ C }^{ 2 }\Rightarrow 2AB={ C }^{ 2 }$ .... $(3)$

From (2) and (3), ${ C }^{ 2 }=2{ A }^{ 2 }\Rightarrow C=\pm \sqrt { 2 } A$

Hence, $A:B:C::A:A:\pm \sqrt { 2 } A$

$\therefore$ direction ratios are $1:1:\pm \sqrt { 2 }$

let

$P(4,1,\lambda)$ and

$Q(2,-1,\lambda)$ be two points. A line having direction ratios

$1,-1,6$ is perpendicular to the plane passing through the origin,

$P$ and

$Q$ , then

$\lambda$ equals

0%
$-\dfrac{1}{2}$
0%
$\dfrac{1}{2}$
0%
$1$
0%
none of these

Explanation

By given point

$P\left( 4,1,\lambda \right)$ and

$Q\left( 2,-1,\lambda \right)$

$\begin{vmatrix} 1 & \lambda \\ -1 & \lambda \end{vmatrix},\begin{vmatrix} \lambda & 4 \\ \lambda & 2 \end{vmatrix},\begin{vmatrix} 4 & 1 \\ 2 & -1 \end{vmatrix}\\ \Rightarrow \left( \lambda +\lambda \right) ,\left( 2\lambda -4\lambda \right) ,\left( -4-2 \right) \\ \Rightarrow -2\lambda ,2\lambda ,6\\ \Rightarrow -2\lambda =1,2\lambda =-1$

$\displaystyle \Rightarrow \lambda =-\frac { 1 }{ 2 }$

The position vectors of three points are

$2\vec{a}-\vec{b}+3\vec{c}$ ,

$\vec{a}-2\vec{b}+\lambda \vec{c}$ and

$\mu \vec{a}-5\vec{b}$ where

$\vec{a}, \vec{b}, \vec{c}$ are non coplanar vectors, then the points are collinear when

0%
$\displaystyle \lambda =-2, \mu =\dfrac{9}{4}$
0%
$\displaystyle \lambda =-\dfrac{9}{4}, \mu =2$
0%
$\displaystyle \lambda =\dfrac{9}{4}, \mu =-2$
0%
None of these

Explanation

When points

$x, y, z$ are collinear, we have

$\alpha x + \beta y = (\alpha + \beta)z$
Similarly,

$x(2\vec{a} - \vec{b} + 3\vec{c}) + y(\vec{a} - 2\vec{b} + \lambda \vec{c}) = (x + y)(\mu \vec{a} - 5\vec{b})$

$\Rightarrow$ comparing the coefficients of

$\vec{a} \rightarrow 2x + y = x\mu + y\mu$

$\vec{b} \rightarrow - x - 2y = - 5x - 5y$

$\vec{c} \rightarrow 3x + \lambda y = 0$

$\Rightarrow 4x = -3y$ and so

$\lambda = \dfrac{9}{4}$
Also,

$\mu = -2$

The angle between two lines whose direction cosines satisfy the equations

$n= l+m$ and

$m= 2l+3n$ is

0%
$180^{\circ}$
0%
$90^{\circ}$
0%
$60^{\circ}$
0%
none of these

Explanation

Eliminating $l$ from equations $n=l+m, \; m=2l+3n$ , we get

$\dfrac{m}{n}= \dfrac{5}{3}$ ... $(1)$

Similarly, eliminating $m$ from the two equations we get

$\dfrac{n}{l}=\dfrac{3}{-2}$ ... $(2)$

From $(1)$ and $(2)$ , $m:n:l = 5:3:-2$

Let $m=5t,\; n=3t,\; l=-2t$

Now, use $l^2+m^2+n^2=1$

$\Rightarrow (25+9+4)t^2=1$

$\Rightarrow t^2= \dfrac{1}{38}$

$\Rightarrow t= \pm \dfrac{1}{\sqrt{38}}$

Positive value of $t$ will give direction cosines of one line and negative value will give direction cosines of another line.

Hence, they will be parallel to each other with angle between them $180^{o}$

Let the direction - cosines of the line which is equally inclined to the axis be

$\displaystyle \pm \frac{1}{\sqrt{k}}$ . Find

$k$ ?

0%
$2$
0%
$3$
0%
$5$
0%
$6$

Explanation

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

But it is given that

$\alpha=\beta=\gamma$

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

$3cos^2\alpha=1$

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

Thus

$\cos\alpha=\dfrac{1}{\sqrt3}$

$\therefore \dfrac{1}{\sqrt k}=\dfrac{1}{\sqrt 3}$

Taking squares on both sides we get

$\therefore k=3$

If the direction ratios of a line are

$1+\lambda ,1-\lambda ,2,$ and it makes an angle of

$60^{o}$ with the y-axis then

$\lambda$ is

0%
$1+\sqrt{3}$
0%
$2+\sqrt{5}$
0%
$1-\sqrt{3}$
0%
$2-\sqrt{5}$

Explanation

$a,b,c$ are direction ratios and

$\theta$ is the angle between y axis and line.

Then

$\cos\theta=\dfrac{b}{\sqrt{a^2+b^2+c^2}}$ .

Hence by the given informations we have

$\dfrac{1}{4}=\cos^2(60)=\dfrac{(1-\lambda)^2}{(1+\lambda)^2+(1-\lambda)^2+4}$

$\Rightarrow \lambda^2-4\lambda-1$

$\Rightarrow \lambda=2 \pm \sqrt{5}$

Let

$A= \left ( 1,2,2 \right )$ ,

$B=\left ( 2,3,6 \right )$ and

$C= \left ( 3,4,12 \right )$ . The direction cosines of a line equally inclined with

$OA,OB$ and

$OC$ , where

$O$ is the origin, are

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

Explanation

Let the direction cosines of the line be

$(a,b,c)$
Since it's equally inclined to

$OA,\; OB,\; OC$

$\dfrac{a+2b+2c}{3}=\dfrac{2a+3b+6c}{7}=\dfrac{3a+4b+12c}{13}=k$

$\Rightarrow a+2b+2c=3k...(1) \\ 2a+3b+6c=7k ....(2)\\ 3a+4b+12c=13k....(3)$
Eliminating

$a$ from

$(1)$ and

$(2)$ , we get

$b-2c=-k....(3)$
Similarly, eliminating

$a$ from

$(1)$ and

$(3)$ , we get

$2b-6c=-4k...(4)$
From

$(3)$ and

$(4)$ , we get

$b=c=k...(5)$
Using the above value in

$(1)$ we get

$a=-k....(6)$
From

$(5)$ and

$(6)$ ,

$a:b:c=-1:1:1$
Also,

$a^2+b^2+c^2=1 \Rightarrow k= \pm \dfrac{1}{\sqrt{3}}$
Hence, direction cosines are

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

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

A st. line which makes angle of

$\displaystyle 60^{\circ}$ with each of y - and z - axes, is inclined with x - axis at an angle

0%
$\displaystyle 45^{\circ}$
0%
$\displaystyle 30^{\circ}$
0%
$\displaystyle 75^{\circ}$
0%
$\displaystyle 60^{\circ}$

Explanation

As we know in

$3D$ the relation between angles made by a vector with

$x$ axis

$(x),y$ axis

$(y), z$ axis

$(z)$ is:-

${\cos}^{2}{x}+{\cos}^{2}{y}+{\cos}^{2}{z}=1$

$\Rightarrow\,{\cos}^{2}{{60}^{\circ}}+{\cos}^{2}{{60}^{\circ}}+{\cos}^{2}{z}=1$

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

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

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

$\Rightarrow\,\dfrac{1}{2}+{\cos}^{2}{z}=1$

$\Rightarrow\,{\cos}^{2}{z}=1-\dfrac{1}{2}=\dfrac{1}{2}$

$\Rightarrow\,\cos{z}=\dfrac{1}{\sqrt{2}}$

$\therefore\,z={45}^{\circ}$

The acute angle between two lines whose direction ratios are

$2,3,6$ and

$1,2,2$ is

0%
$\displaystyle \cos ^{ -1 }{ \left( \frac { 20 }{ 21 } \right) }$
0%
$\displaystyle \cos ^{ -1 }{ \left( \frac { 18 }{ 21 } \right) }$
0%
$\displaystyle \cos ^{ -1 }{ \left( \frac { 8 }{ 21 } \right) }$
0%
None of these

Explanation

We have

${ a }_{ 1 }=2,{ b }_{ 1 }=3,{ c }_{ 1 }=6,{ a }_{ 2 }=1,{ b }_{ 2 }=2,{ c }_{ 2 }=2$

$\theta$ is the angle between two lines whose direction ratios are given, then

$\cos { \theta } =\displaystyle\frac { { a }_{ 1 }{ a }_{ 2 }+{ b }_{ 1 }{ b }_{ 2 }+{ c }_{ 1 }{ c }_{ 2 } }{ \sqrt { { \left( { a }_{ 1 } \right) }^{ 2 }+{ \left( { b }_{ 1 } \right) }^{ 2 }+{ \left( { c }_{ 1 } \right) }^{ 2 } } \sqrt { { \left( { a }_{ 2 } \right) }^{ 2 }+{ \left( { b }_{ 2 } \right) }^{ 2 }+{ \left( { c }_{ 2 } \right) }^{ 2 } } }$

$\Rightarrow \cos { \theta } =\displaystyle \frac { 2\times 1+3\times 2+6\times 2 }{ \sqrt { { \left( 2 \right) }^{ 2 }+{ \left( 3 \right) }^{ 2 }+{ \left( 6 \right) }^{ 2 } } \sqrt { { \left( 1 \right) }^{ 2 }+{ \left( 2 \right) }^{ 2 }+{ \left( 2 \right) }^{ 2 } } } =\frac { 20 }{ 21 }$

$\Rightarrow \theta =\cos ^{ -1 }{ \dfrac { 20 }{ 21 } }$

Find the equations of the plane through the point

$\displaystyle \left ( x_{1},y_{1},z_{1} \right )$ and perpendicular to the straight line

$\displaystyle \frac{x-\alpha }{l}=\frac{y-\beta }{m}=\frac{z-\gamma }{n}$

0%
$\displaystyle l\left ( x-x_{1} \right )+m\left ( y-y_{1} \right )+n\left ( z-z_{1} \right )=0.$
0%
$\displaystyle l\left ( x-x_{1} \right )+m\left ( y-y_{1} \right )-n\left ( z-z_{1} \right )=0.$
0%
$\displaystyle l\left ( x-x_{1} \right )-m\left ( y-y_{1} \right )+n\left ( z-z_{1} \right )=0.$
0%
none of these

Explanation

Since the line is perpendicular to the plane, hence the normal vector of the plane is parallel to the direction vector of the line.

Hence the equation of the plane will be

$l(x)+m(y)+n(z)=d$

It is given that the plane passes through

$x_{1},x_{2},x_{3}$

Hence

$lx_{1}+my_{1}+nz_{1}=d$

Substituting in the equation of the plane, we get

$lx+my+nz=lx_{1}+my_{1}+nz_{1}$

$l(x-x_{1})+m(y-y_{1})+n(z-z_{1})=0$

Which of the following triplets give the direction cosines of a line ?

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

Explanation

$l,m,n$ are the directions cosine of a line then

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

Thus we get

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

Option D is correct answer

If a point

$P$ in the space such that

$\overline{OP}$ is inclined to

$OX$ at

$45$ and

$OZ$ to

$60$ then

$\overline{OP}$ inclined to

$OY$ is

0%
$75^{\circ}$
0%
$75^{\circ} \text{or} 105^{\circ}$
0%
$60^{\circ} \text{or} 120^{\circ}$
0%
None of these

Explanation

$\displaystyle l=\cos { { 45 }^{ O } } =\frac { 1 }{ \sqrt { 2 } }$ and

$\displaystyle m=\cos { { 60 }^{ O } } =\frac { 1 }{ 2 }$

$\displaystyle \Rightarrow { n }^{ 2 }=1-\frac { 1 }{ 2 } -\frac { 1 }{ 4 } =\frac { 1 }{ 4 } \Rightarrow n=\pm \frac { 1 }{ 2 }$
So

$\alpha ={ 60 }^{ O }$ or

${ 120 }^{ O }$

The equation of the plane which bisect the join of point

$(7, 2, 3)$ and

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

0%
$x + 2y + 3= 0$
0%
$2x -y + 3 = 0$
0%
$x + y +2z + 7= 0$
0%
$4x + 3y =9$

Explanation

Lets assume

$P = (7, 2, 3)$ and

$Q = (-1, -4, 3)$
So mid point of

$PQ$ is

$(3, -1, 3) = M$ (say), since plane bisect

$PQ$ so

$M$ will lie in required plane.
Also direction ratios of

$PQ$ are

$8, 6, 0$ .
Also given

$PQ$ is perpendicular to the plane, thus equation of plane is given by,

$8(x-3)+6(y+1)+0(z-3)=0$

$\Rightarrow 4x+3y-9=0$

If a line makes angles

$\displaystyle \alpha ,\beta ,\gamma$ with axes of co-ordinates, then

$\displaystyle \cos 2\alpha +\cos 2\beta +\cos 2\gamma$ is equla to

0%
$\displaystyle -2$
0%
$\displaystyle -1$
0%
$1$
0%
$2$

Explanation

We know that direction cosines are

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

Then sum of the squares of the direction cosine is 1

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

we know Trignometric identity that

$\sin^2A+\cos^2A=1$

From the above identity we can write as

$\cos^2A=1-\sin^2A$

Now replacing we get

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

By adding all we get

$3-(\sin^2\alpha+\sin^2\beta+\sin^2\gamma)=1$

$\therefore \cos2\alpha+\cos2\beta+\cos2\gamma = 2-3 = -1$

If a line makes an angle

$\displaystyle \theta_1, \theta_2, \theta_3$ which the axis respectively, then

$\displaystyle cos 2\theta_1 + cos 2 \theta_2 + cos 2 \theta_3 = ?$

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

Explanation

Let

$P(x,y,z)$ be any point on the line OP, D-Ratio of OP are x,y,z. D.R. of normal plane

$i.e.,z = 0$ are

$0, 0, 1$

$\sin\theta_1=\dfrac{z}{\sqrt{x^2+y^2+z^2}}$

similarily for

$sin\theta_2$ and

$sin \theta_3$

$\therefore sin^2\theta_1 + sin^2\theta_2 + sin^2\theta_3$

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

$\Longrightarrow 1-cos^2\theta_1+ 1-cos^2\theta_2+ 1-cos^2\theta_3=1$

$\therefore cos^\theta_1+cos^\theta_2+cos^\theta_3=3-1=2$

$\bar a,\bar b,\bar c$ are three non-zero vectors such that any two of them are non-collinear. If

$\bar a+\bar b$ is collinear with

$\bar c$ and

$\bar b+\bar c$ is collinear with

$\bar a$ , then what is their sum?

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

Explanation

We have

$\bar a+\bar b =t\bar c$ ----

$(1)$

$\bar b+\bar c =s\bar a$ ----

$(2)$
From

$(1)$ and

$(2)$

$\bar a+\bar b=t(s\bar a-\bar b)$
Since no two of them are collinear, comparing coeffficients gives

$st=1$ and

$t=-1$

$\Rightarrow s=-1$ and

$t=-1$
From

$(1)$

$\therefore \bar a+\bar b+\bar c=0$
Hence, option

$B$ .

The equation of plane through

$(1, 2, 3)$ and parallel to the plane

$\bar{r}.\left ( \hat{3i}+\hat{4j}+\hat{5k} \right )=0$

0%
$\bar{r}.\left ( \hat{3i}+\hat{4j}+\hat{5k} \right )=26$
0%
$\bar{r}.\left ( \hat{3i}-\hat{4j}+\hat{5k} \right )+4$
0%
$\bar{r}.\left ( \hat{3i}+\hat{4j}-\hat{5k} \right )+4=0$
0%
None of these

Explanation

Equation of plane passing through

$(1,2,3)$ and parallel to plane

$\vec{r} \cdot (3\hat{i}+4\hat{j}+5\hat{k} = 0$ is given by,

$(\vec{r}-(\hat{i}+2\hat{j}+3\hat{k}) )\cdot (3\hat{i}+4\hat{j}+5\hat{k}) = 0$

$\Rightarrow \vec{r} \cdot (3\hat{i}+4\hat{j}+5\hat{k}) -(\hat{i}+2\hat{j}+3\hat{k}) \cdot (3\hat{i}+4\hat{j}+5\hat{k} ) = 0$

$\Rightarrow \vec{r}\cdot (3\hat{i}+4\hat{j}+5\hat{k} )- 26 = 0$
Hence, option 'A' is correct.

0%
Both Assertion & Reason are individually true & Reason is correct explanation of Assertion
0%
Both Assertion & Reason are individually true but Reason is not the ,correct (proper) explanation of Assertion
0%
Assertion is true but Reason is false
0%
Assertion is false but Reason is true

Explanation

Points A(4,0,4),B(1,2,3),C(−2,4,2) are collinear

formula to check whether these three points are collinear or not

AB+BC=AC

to find AB the equation is

$\sqrt {(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$ ---(1)

$x_1 = 4 , y_1 =0 and z_1 = 4$

$x_2= 1 , y_2 =2 and z_2 = 3$

by substituting the values in (1) we will get

AB = 3.7

similarly for BC and AC

BC=3.7

AC=7.4

hence finally its is known that these points are collinear

the answer will be Both Assertion & Reason are individually true & Reason is correct explanation of Assertion.

Equation of the plane which passes through the point (-1, 3, 2) & is

$\perp$ to each of the planes

$p_{1}$ &

$p_{2}$ is

0%
$2x+y+z+1= 0$
0%
$3x+8y-2z-17= 0$
0%
$2x+y-z+1= 0$
0%
$x-2y+z+1= 0$

Explanation

Normal of

$p1 = [ 2 -3 1 ]$

Normal of

$p2 = [ 1 -1 1 ]$

If planes are perpendiculars then dot products of their normals must be 0,

So consider normal of desired plane is

$[ a b c]$

Hence,

$2a - 3b + c = 0$ (Dot product with normal of p1)

$a - b + c = 0$ (Dot product with normal of p2)

Solving above 2 we get ,

$a - 2b = 0$

So if we use normals of planes in option b & d,

we notice that they don't satisfy the above equation

We observe that option a does not pass through (-1, 3, 2)

Since on substitution ,

$2.(-1) + 3 + 2 + 1 \neq 0$

The correct answer is

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

CBSE Questions for Class 12 Commerce Maths Three Dimensional Geometry Quiz 4 - 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 4 - MCQExams.com

0:0:2

Practice Class 12 Commerce Maths Quiz Questions and Answers

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