MCQ Questions for Class 12 Commerce Maths Three Dimensional Geometry Quiz 1

If a line makes an angle of

${\pi }/{4}$ with the positive direction of each of

$x$ -axis and

$y$ -axis, then the angle that the line makes with the positive direction of the

$z$ -axis is

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

Explanation

Since line makes an angle of $\displaystyle \dfrac { \pi }{ 4 }$ with positive direction of each of $x$ -axis and $y$ -axis, therefore $\displaystyle \alpha =\dfrac { \pi }{ 4 } ,\beta =\dfrac { \pi }{ 4 }$

We know that,

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

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

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

$\Rightarrow \cos ^{ 2 }{ \gamma } =0$

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

A line

$AB$ in three-dimensional space makes angles

$45^{\mathrm{o}}$ and

$120^{\mathrm{o}}$ with the positive

$\mathrm{x}$ -axis and the positive

$\mathrm{y}$ -axis respectively. lf

$AB$ makes an acute angle

$e$ with the positive

$\mathrm{z}$ -axis, then

$e$ equals

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$45^{\mathrm{o}}$
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$60^{\mathrm{o}}$
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$75^{\mathrm{o}}$
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$30^{\mathrm{o}}$

Explanation

The angle made by line

$AB$ with

$x$ axis is

$45^{0}$

The angle made by line

$AB$ with

$y$ axis is

$120^{0}$

Let the angle made by line

$AB$ with

$z$ axis is

$\theta$

We have

$\cos^{2}{(45)} +\cos^{2}{(120)}+\cos^{2}{\theta}=1$

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

Given that

$\theta$ is accute

So we get

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

$\Rightarrow \theta=60^{0}$

Therefore option

$B$ is correct

A line makes the same angle

$\Theta$ , with each of the

$\mathrm{x}$ and

$\mathrm{z}$ axis. If the angle

$\beta$ , which it makes with

$\mathrm{y}$ -axis, is such that

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

$\cos^{2}\Theta$ equals:

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$\displaystyle \frac{2}{3}$
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$\displaystyle \frac{1}{5}$
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$\displaystyle \frac{3}{5}$
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$\displaystyle \frac{2}{5}$

Explanation

A line makes angle

$\theta$ with x-axis and z axis and

$\beta$ with y-axis

$\therefore l=\cos { \theta } ,m=\cos { \beta } ,n=\cos { \theta }$
we know that

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

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

$\Rightarrow 2\cos ^{ 2 }{ \theta } =\sin ^{ 2 }{ \beta }$ ...(1)
But

$\sin ^{ 2 }{ \beta } =3\sin ^{ 2 }{ \theta }$ ...(2)
From (1) and (2)

$3\sin ^{ 2 }{ \theta } =2\cos ^{ 2 }{ \theta } \Rightarrow 3\left( 1-\cos ^{ 2 }{ \theta } \right) =2\cos ^{ 2 }{ \theta } \\ \Rightarrow 3-3\cos ^{ 2 }{ \theta } =2\cos ^{ 2 }{ \theta } \Rightarrow 3=5\cos ^{ 2 }{ \theta }$

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

A line in the 3-dimensional space makes an angle

$\theta \left(0 < \theta \leq \dfrac {\pi}{2}\right)$ with both the x and y axes, then the set of all values of

$\theta$ is the interval :

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$\left[0, \dfrac {\pi}{4}\right]$
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$\left[\dfrac {\pi}{6}, \dfrac {\pi}{3}\right]$
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$\left[\dfrac {\pi}{4}, \dfrac {\pi}{2}\right]$
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$\left[\dfrac {\pi}{3}, \dfrac {\pi}{2}\right]$

Explanation

Sum of square of direction cosines is

$1$ .

Lets assume

$\alpha$ is the angle with

$x$ ,

$\beta$ with

$y$ and

$\gamma$ with

$z$ .

$\alpha,\beta,\gamma \in \left[0,\dfrac\pi2\right]$

Now,

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

But it is given that,

$\alpha = \beta = \theta$

$\therefore \cos^2\theta + \cos^2\theta + \cos^2\gamma = 1$

$\Rightarrow 2\cos^2\theta = 1-\cos^2\gamma$

$\Rightarrow = \cos^2\theta = \dfrac{\sin^2\gamma}2$

$\Rightarrow \cos\theta = \dfrac{\sin\gamma}{\sqrt2}$

$\gamma \in \left[0,\dfrac\pi2\right]$ , then

$\theta \in \left[\dfrac\pi4,\dfrac\pi2\right]$

The points with position vectors

$60 \widehat{i} + 3 \widehat{j}$ ,

$40 \widehat{i} - 8 \widehat{j}$ ,

$a\widehat{i} -52\widehat{j}$ are collinear if

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$a = -40$
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$a = 40$
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$a = 20$
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None of these

Explanation

Three points

$A(\vec{a}), B (\vec{b}), C(\vec{c})$ are collinear if

$\vec{AB} || \vec{AC}$

$\vec{AB} = - 20 \widehat{i} - 11 \widehat{j}$

$\vec{AC} = (a - 60) \widehat{i} - 55 \widehat{j}$

$\Rightarrow \vec{AB} || \vec{AC}$

$\Rightarrow \displaystyle \dfrac{a - 60}{- 20 }$

$= \dfrac{-55}{-11}$

$\Rightarrow a = - 40$

The vector equation

$r=i-2j-k+t(6j-k)$ represents a straight line passing through the points:

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$(0, 6, -1)$ and $(1, -2, -1)$
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$(0, 6, -1)$ and $(-1, -4, -2)$
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$(1, -2, -1)$ and $(1, 4, -2)$
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$(1, -2, -1)$ and $(0, -6, 1)$

Explanation

Cartesian representation of the given line is,

$\dfrac{x-1}{0}=\dfrac{y+2}{6}=\dfrac{z+1}{-1}=t$

So any point on the given line is of the form

$(1,6t-2, -t-1)$

where

$t$ can be any real numbers

so for

$t=0$ and

$1$ the corresponding points are

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

$(1,4,-2)$

You can check other options does not satisfy above point for any

$t$ .

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

$\cos ^{ 2 }{ \theta }$ equals

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$\dfrac {2}{5}$
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$\dfrac {1}{5}`$
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$\dfrac {3}{5}$
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$\dfrac {2}{3}$

Explanation

Let

$l,m$ and

$n$ be the direction cosines

Then,

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

We have

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

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

$\Rightarrow 2\cos ^{ 2 }{ \theta } -\sin ^{ 2 }{ \beta } =0$

$\Rightarrow 2\cos ^{ 2 }{ \theta } -3\sin ^{ 2 }{ \beta } =0\left[ \because \sin ^{ 2 }{ \beta } =3\sin ^{ 2 }{ \theta } \right]$

$\Rightarrow \tan ^{ 2 }{ \theta } =\dfrac {2}{3}$

Therefore,

$\cos ^{ 2 }{ \theta } =\cfrac { 1 }{ 1+\tan ^{ 2 }{ \theta } } =\cfrac { 3 }{ 5 }$

The points with position vectors

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

$ai -52j$ are collinear if

0%
$a = -40$
0%
$a = 40$
0%
$a = 20$
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None of these

Explanation

Denoting

$a,b,c$ by the given vectors respectively
These vectors will be collinear if there is some constant

$k$ such that

$c-a=K\left( b-a \right)$

$\Rightarrow a-60=-20K$ and

$-55=-11K$

$\Rightarrow a=-100+60=-40$

Equation to a line parallel to the vector

$2\hat{i}-\hat{j}{+}\hat{k}$ and passing through the point

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

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

Explanation

Equation parallel to a vector

$\vec{a}$ and passing through point of position vector

$\vec{b}$ is given by

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

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

$x\hat{i}+y\hat{j}+z\hat{k}= \hat{i}+\hat{j}+\hat{k}+\lambda (2\hat{i}-\hat{j}+\hat{k})$

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

The direction ratios of the diagonal of the cube joining the origin to the opposite corner are (when the

$3$ concurrent edges of the cube are coordinate axes)

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

Explanation

Since, a cube is a symmetric figure, the vertex we are talking about will be at the diagonally opposite end of the origin. i.e. it will be equally inclined to the three axes.

Let the side of the cube be

$a$ , then the corner opposite to origin will have coordinates

$(a,a,a)$ .

Direction ratios of a line joining two points

$(x_1,y_1,z_1)$ and

$(x_2,y_2,z_2)$ is given by

$(x_2-x_1,y_2-y_1,z_2-z_1)$

Then, direction ratios of two point

$(0,0,0)$ and

$(a,a,a)$ will be

$(a-0,a-0,a-0)=(a,a,a)=a(1,1,1)$

Hence, the direction ratios are

$1,1,1$ .

If the direction cosines of a line are $\left(\displaystyle \dfrac{1}{c},\dfrac{1}{c},\dfrac{1}{c}\right)$ then $c=$ ______

0%
$3$
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$1/\sqrt{3}$
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$\pm\sqrt{3}$
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$\displaystyle \pm\dfrac{1}{\sqrt{3}}$

Explanation

If direction cosines are

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

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

$\Rightarrow c^{2} = 3$

$\Rightarrow c=\pm\sqrt 3$

$l = m =n = 1$ represents the direction cosines of

0%
$x-$ axis
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$y-$ axis
0%
$z-$ axis
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none of these

Explanation

Suppose,

$l,m,n$ are direction cosines

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

But

$l=m=n=1$

$\implies 3m^2=1$

$\implies l=m=n=\dfrac{1}{\sqrt{3}}$ which are not direction cosines of either of the three co-ordinate axes.

Hence, option D is correct.

List I	List II
1) d.c's of $x -$ axis	a) $(1,1,1)$
2) d.c's of $y -$ axis	b) $\left(\displaystyle \frac{]}{\sqrt{3}}\frac{]}{\sqrt{3}},\frac{]}{\sqrt{3}}\right)$
3) d.c's of $z -$ axis	c) $(1,0,0)$
4) d.c's of a line makes equal angles with axes	d) $(0,1,0)$
	e) $(0,0,1)$

The correct order for 1, 2, 3, 4 is

0%
$c,d,e,b$
0%
$a,b,c,e$
0%
$c,d,a,b$
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$b,c,a,e$

Explanation

On the

$x-$ axis, the other

$2$ coordinates

$y,z$ will be

$0$ .

So, the point is

$(1,0,0)$

Direction cosines of

$x-$ axis i.e

$(1,0,0)$ is

$(1,0,0)$

$\therefore$ Direction cosines of

$x-$ axis are

$(1,0,0)$

Similarly, direction cosines of

$y$ and

$z$ axes are

$(0,1,0)$ and

$(0,0,1)$ respectively.

$\implies 1\rightarrow c, 2\rightarrow d, 3\rightarrow e$
For the fourth part where we need to find the dc's of a line that makes equal angles with the axes.
The components of such a line will always be equal.
Moreover, the sum of the squares of the dc's of a line should be equal.

If the point is

$(a,a,a)$ , then

$a^2+a^2+a^2=3a^2$

So, the direction cosines are

$\left(\dfrac{a}{\sqrt{3a^2}},\dfrac{a}{\sqrt{3a^2}},\dfrac{a}{\sqrt{3a^2}}\right)=\left(\dfrac{1}{\sqrt3},\dfrac{1}{\sqrt3},\dfrac{1}{\sqrt3}\right)$

$\implies 4\rightarrow b$

Hence, option A is correct.

$P(x, y, z)$ moves such that

$x=0, z=0$ , then the locus of

$P$ is the line whose d.cs are

0%
$y$ -axis
0%
$1, 0, 0$
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$0, 1, 0$
0%
$0, 0, 0$

Explanation

When

$P$ moves then

$x=0,z=0$ but

$y$ is not given.

Let

$y=y$

Then the coordinates of the point will be

$(0,y,0)$

Now, direction cosines with respect to

$(0,y,0)$ is given by

$\cos \alpha = \dfrac{0}{\sqrt{0^2+y^2+0^2}}=\dfrac{0}{y}=0$

$\cos \beta = \dfrac{y}{\sqrt{0^2+y^2+0^2}}=\dfrac{y}{y} = 1$

$\cos \gamma = \dfrac{0}{\sqrt{0^2+y^2+0^2}}=\dfrac{0}{y}=0$
Hence, the direction cosines are

$0,1,0$ .

$P$ is a point on the line passing through the point

$A$ with position vector

$2\overline{i}+\overline{j}-3\overline{k}$ and parallel to

$\overline{i}+2\overline{j}+\overline{k}$ such that

$AP=2\sqrt{6}$ then the position vector of

$P$ is

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$4\overline{i}+5\overline{j}+\overline{k}$
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$3\overline{j}+5\overline{k}$
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$4\overline{i}+5\overline{j}-\overline{k}$
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$3\overline{j}-4\overline{k}$

Explanation

$P\cdot V\ of\ P= (2+t)\hat{i}+(1+2t)\hat{j}+(-3+t)\hat{k}$

$P\cdot V\ of\ A= (2,1,-3)$

$\left | \overrightarrow{AP} \right |= \sqrt{t^{2}+4t^{2}+t^{2}}$

$2\ \sqrt{6}= t\sqrt{6}$

$\Rightarrow t = 2$

$P= 4\hat{i}+5\hat{j}-\hat{k}$

If the points

$A(\overline{a}), B(\overline{b}), C(\overline{c})$ satisfy the relation

$3\mathrm{a}-8\mathrm{b}+5\mathrm{c}=0$ then the points are

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Vertices of an equilateral triangle
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Collinear
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Vertices of a right angled triangle
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Vertices of a isosceles triangle

Explanation

The points

$A(\vec{a}), B(\vec{b}), C(\vec{c})$ satisfy the relation

$3a-8b+5c=0$

$\implies 3\vec{a}-8\vec{b}+5\vec{c}=0$

$\implies 3\vec{a}+5\vec{c}=8\vec{b}$

$\implies \dfrac{3\vec{a}+5\vec{c}}{8}=\vec{b}$

$\implies \dfrac{3\vec{a}+5\vec{c}}{3+5}=\vec{b}$

Similarly,

$\vec{a}=\dfrac{8\vec{b}-5\vec{c}}{8-5}$ and

$\vec{c}=\dfrac{8\vec{b}-3\vec{a}}{8-3}$

This is a section formula.
Hence

$\vec{a},\vec{b},\vec{c}$ are collinear.

Cosine of the angle between two diagonals of a cube is equal to

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$\displaystyle \frac{2}{\sqrt 6}$
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$\displaystyle \frac{1}{3}$
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$\displaystyle \frac{1}{2}$
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None of these

Explanation

Let 0, one vertex of a cube, be the origin and three edges through O be the coordinate axes. The four diagonals are Op, AA', BB' and CC'. Let a be the length of each edge. Then the coordinates of P, A, A' are (a, a, a), (a, 0, 0), (0, a, a).
The direction ratios of OP are a, a, a.
The direction cosines of OP are

$\displaystyle \frac{a}{a \sqrt 3}, \frac{a}{a \sqrt 3}, \frac{a}{a \sqrt 3}$ i.e.,

$\displaystyle \frac{1}{\sqrt 3} , \frac{1}{\sqrt 3}, \frac{1}{\sqrt 3}$ .
Similarly direction cosines of AA'are

$\displaystyle \left ( - \frac{1}{\sqrt 3}, \frac{1}{\sqrt 3}, \frac{1}{\sqrt 3} \right )$ .
Let

$\theta$ be the angle between the diagonals OP and AA'.

$cos \theta = \displaystyle \frac{1}{\sqrt 3} \left ( - \frac{1}{\sqrt 3} \right ) + \frac{1}{\sqrt 3} \left ( \frac{1}{\sqrt 3} \right ) + \frac{1}{\sqrt 3} \left (\frac{1}{\sqrt 3} \right )$

$cos \theta = l_1 l_2 + m_1 m_2 + n_1 n_2 = - \displaystyle \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = \frac{1}{3}$

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

Area of

$\displaystyle \Delta ABC$ is

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$45$ squares units
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$55$ squares units
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$65$ squares units
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none of these

Explanation

Line PA :

$\displaystyle \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-6}{1}$

Line PB :

$\displaystyle \frac{x-1}{-1}=\frac{y-2}{2}=\frac{z-6}{1}$

Line PC :

$\displaystyle \frac{x-1}{2}=\frac{y-2}{-1}=\frac{z-6}{-2}$

Then

$\displaystyle A\left ( \frac{7}{2},-\frac{1}{2},\frac{17}{2} \right )$

$\displaystyle B\left ( \frac{17}{2},-13,-\frac{3}{2} \right )$

$\displaystyle C\left ( -14,\frac{19}{2},21 \right )$

Hence area of

$\displaystyle \Delta ABC=\frac{225\sqrt{14}}{8}$ , volume of tetrahedron PABC =

$\displaystyle \frac{1125}{8}$ cubic units

A vector is equally inclined to the

$x$ -axis,

$y$ -axis and

$z$ -axis respectively, its direction cosines are

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$< \dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}>$
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$< -\dfrac{1}{\sqrt{3}}, -\dfrac{1}{\sqrt{3}}, -\dfrac{1}{\sqrt{3}}>$
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$< \dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}>$ or $< -\dfrac{1}{\sqrt{3}}, -\dfrac{1}{\sqrt{3}}, -\dfrac{1}{\sqrt{3}}>$
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None of these

Explanation

Let the vector $v$ make an angle $\alpha$ with each of the three axes, then direction cosine of $v$ are

$<\cos\,\alpha, \cos\,\alpha, \cos\,\alpha>$

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

$\Rightarrow \cos\,\alpha=\pm \dfrac{1}{\sqrt{3}}$

Hence, direction cosine of $v$ are

$<\dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}>$ or

$<-\dfrac{1}{\sqrt{3}}, -\dfrac{1}{\sqrt{3}}, -\dfrac{1}{\sqrt{3}}>$

What are the DR's of vector parallel to

$\left( 2,-1,1 \right)$ and

$\left( 3,4,-1 \right)$ ?

0%
$\left( 1,5,-2 \right)$
0%
$\left( -2,-5,2 \right)$
0%
$\left( -1,5,2 \right)$
0%
$\left( -1,-5,-2 \right)$

Explanation

Required DR's are

$\left( 3-2,4+1,-1-1 \right)$ ie,

$\left( 1,5,-2 \right)$

The direction cosines of the vectors

$2\vec {i} + \vec {j} - 2\vec {k}$ is equal to

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

Explanation

We know,

$\dfrac{a_{1}}{\left|a\right|},\dfrac{a_{2}}{\left|a\right|},\dfrac{a_{3}}{\left|a\right|}$ are the direction cosines of

$a_{1}i+a_{2}j+a_{3}k$

Given vector is

$2\hat i+\hat j-2\hat k$

$|a|=\sqrt {2^2+1^2+2^2}=\sqrt {9}=3$

Thus their direction cosines are

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

So, option A is correct.

The Cartesian equation of the plane passing through the point

$(3, -2, -1)$ and parallel to the vectors

$\overline {b} = \overline {i} - 2\overline {j} + 4\overline {k}$ and

$\overline {c} = 3\overline {i} + 2\overline {j} - 5\overline {k}$ is

0%
$2x - 17y - 8z + 63 = 0$
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$3x + 17y + 8z - 36 = 0$
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$2x + 17y + 8z + 36 = 0$
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$3x - 16y + 8z - 63 = 0$

Explanation

$\begin{vmatrix} x - x_{1}& y - y_{1} & z - z_{1}\\ a_{1} & b_{1} &c_{1} \\ a_{2} & b_{2} & c_{2} \end{vmatrix} = 0$

$\begin{vmatrix} x - 3 & y + 2 & z + 1\\ 1 & -2 & 4 \\ 3 & 2 & -5 \end{vmatrix} = 0$

$2x+ 17y +8z +36= 0$

A straight line is equally inclined to all the three coordinate axes. Then an angle made by the line with the y-axis is

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$\cos^{1}\left (\dfrac {1}{3}\right )$
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$\cos^{1}\left (\dfrac {1}{\sqrt {3}}\right )$
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$\cos^{1}\left (\dfrac {2}{\sqrt {3}}\right )$
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$\dfrac {\pi}{4}$

Explanation

Let

$l, m, n$ be the directional cosines of the line.

We know that

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

Since the line is equally inclined to all the axes,

$l = m = n$

Solving the two above equations we get,

$l = m = n = \frac{1}{\sqrt{3}}$

Hence the angle made be the line with any axis is,

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

If the foot of the perpendicular from

$(0, 0, 0)$ to a plane is

$(1, 2, 3)$ , then the equation of the plane is

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

Explanation

The normal of the plane passes through

$(1, 2, 3) and (0, 0, 0).$

Hence the normal vector is given by

$1\hat{i}+2\hat{j}+3\hat{k}$

The equation of the plane is of the form

$x+2y+3z = c$ , where c is a constant.

Since the plane passes through the point

$(1, 2, 3)$ ,

$1(1)+2(2)+3(3) = c$

$c = 14$

Hence the equation of the plane is

$x + 2y+3z = 14$

A line makes angles

$\alpha, \beta, \gamma, \delta$ with four diagonals of a cube. then

$\displaystyle \sum_{r\epsilon \{\alpha, \beta, \gamma, \delta\}} cos^2(r)=$

0%
$\dfrac{-4}{3}$
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$\dfrac{4}{3}$
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$\dfrac{3}{4}$
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$\dfrac{3}{5}$

Explanation

Let

$a$ be the length of an edge of the cube and one of the end is at origin.

Clearly,

$OP, AR, BS$ and

$CQ$ are the diagonals of the cube.

The direction ratios of

$OP, AR, BS$ and

$CQ$ are:

$a-0,a-0,a-0\ i.e. a,a,a$

$0-a,a-0,a-0\ i.e. -a,a,a$

$a-0,0-a,a-0\ i.e. a,-a,a$

and

$a-0,a-0,0-a\ i.e, a,a,-a$

Let the direction ratios of a line be proportional to

$l,m,n$

Suppose this line makes angle

$\alpha, \beta, \gamma$ and

$\delta$ with

$OP,AR,BS$ and

$CQ$ respectively.

Now

$\alpha$ is the angle between

$OP$ and the line whose direction ratios are proportional to

$l,m,n$

$\therefore\ \cos \alpha =\dfrac{a\cdot l+a\cdot m+ a\cdot n}{\sqrt{a^{2}+a^{2}+a^{2}}\sqrt{l^2 + m^2 + n^2}}$

$\implies \cos \alpha =\dfrac{l+m+n}{\sqrt{3} \sqrt{l^2 + m^2 + n^2}}$ ....... (1)

Now

$\beta$ is the angle between

$AR$ and the line whose directional ratios are proportional to

$l,m,n$

$\therefore \cos \beta =\dfrac{-a\cdot l + a\cdot m + a\cdot n}{\sqrt{(-a)^2 + a^2 + a^2}\sqrt{l^2 + m^2 + n^2}}$

$\implies$

$\cos \beta =\dfrac{-l + m + n}{\sqrt{3}\sqrt{l^2 + m^2 + n^2}}$ ..... (2)

Similarly,

$\cos \gamma =\dfrac{l - m + n}{\sqrt{3}\sqrt{l^2 + m^2 + n^2}}$ ........... (3)

and

$\cos \delta =\dfrac{l + m - n}{\sqrt{3}\sqrt{l^2 + m^2 + n^2}}$ ....... (4)

$\therefore$ from (1),(2),(3) and (4) we have

$\displaystyle \sum_{r\epsilon \{\alpha, \beta, \gamma, \delta\}} cos^2(r)$

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

$=\dfrac{(l+m+n)^2}{3(l^2 + m^2 + n^2)} + \dfrac{(-l+m+n)^2}{3 (l^2+m^2+n^2)} + \dfrac{(l-m+n)^2}{3(l^2+m^2+n^2)} + \dfrac{(l+m-n)^2}{{3}(l^2+m^2+n^2)}$

$=\dfrac{(l+m+n)^2 + (-l+m+n)^2 + (l-m+n)^2 + (l+m-n)^2}{3(l+m+n)^2}$

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

Find vector equation for the line passing through the points

$3\overline i+4\overline j-7\overline k,\overline i-\overline j+6\overline k$ .

0%
$\overline r=(3-2\lambda)\overline i+(4-5\lambda)\overline j+(-7+13\lambda)\overline k$
0%
$\overline r=(2\lambda)\overline i+(4+5\lambda)\overline j+(-7-13\lambda)\overline k$
0%
$\overline r=(3-2\lambda)\overline i-(4-5\lambda)\overline j+(-7+13\lambda)\overline k$
0%
$\overline r=(3-2\lambda)\overline i+(4-5\lambda)\overline j-(-7+13\lambda)\overline k$

Explanation

Given points

$\vec{a}=3\hat{i}+4\hat{j}-7\hat{k}$

$\vec{b}=\hat{i}-\hat{j}+6\hat{k}$

normal vector of line be

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

$\vec{n}=\hat{i}-\hat{j}+6\hat{k}-(3\hat{i}+4\hat{j}-7\hat{k})$

$\vec{n}=\hat{i}-\hat{j}+6\hat{k}-3\hat{i}-4\hat{j}+7\hat{k}$

$\vec{n}=-2\hat{i}-5\hat{j}+13\hat{k}$

vector eq of line

$\vec{r}=\vec{a}+\lambda(\vec{n})$

$\vec{r}=3\hat{i}+4\hat{j}-7\hat{k}+\lambda(-2\hat{i}-5\hat{j}+13\hat{k})$

$\vec{r}=3\hat{i}+4\hat{j}-7\hat{k}-2\lambda\hat{i}-5\lambda\hat{j}+13\lambda\hat{k})$

$\vec{r}=(3-2\lambda)\hat{i}+(4-5\lambda)\hat{j}+(-7+13\lambda)\hat{k}$

Find vector equation of line passing through

$(1,4,1),(1,2,2)$ .

0%
$\overline r=\overline i+(4-2\lambda)\overline j+(1+\lambda)\overline k$
0%
$\overline r=\overline i+(4+2\lambda)\overline j+(1+\lambda)\overline k$
0%
$\overline r=\overline i-(4-2\lambda)\overline j+(1+\lambda)\overline k$
0%
$\overline r=\overline i+(4-2\lambda)\overline j+(1-\lambda)\overline k$

Explanation

Given points

$A(1,4,1)$ and

$B(1,2,2)$

position vector of point A

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

position vector of point B

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

normal vector of line be

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

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

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

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

vector eq of line

$\vec{r}=\vec{a}+\lambda(\vec{n})$

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

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

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

The projections of a directed line segment on the coordinate axes are

$12, 4, 3$ respectively.

What are the direction cosines of the line segment?

0%
$(12/13, 4/13, 3/13)$
0%
$(12/13, -4/13, 3/13)$
0%
$(12/13, -4/13, -3/13)$
0%
$(-12/13, -4/13, 3/13)$

Explanation

Let the direction cosines of the line segment in x,y,z are 12k, 4k, 3k respectively

But

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

$\Rightarrow 144{ k }^{ 2 }+16{ k }^{ 2 }+9{ k }^{ 2 }=1\\ \Rightarrow { k }^{ 2 }=\dfrac { 1 }{ 169 } \\ \Rightarrow k=\dfrac { 1 }{ 13 }$

$\therefore$ The direction cosines of the line segment are

$\left(\dfrac { 12 }{ 13 } ,\dfrac { 4 }{ 13 } ,\dfrac { 3 }{ 13 } \right)$

From the point

$P(3, -1, 11)$ , a perpendicular is drawn on the line

$L$ given by the equation

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

$Q$ be the foot of the perpendicular.

What are the direction ratios of the line segment

$PQ$ ?

0%
$(1, 6, 4)$
0%
$(-1, 6, -4)$
0%
$(-1, -6, 4)$
0%
$(2, -6, 4)$

Explanation

The general point on the line

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

$(x,y,z)=(2\lambda ,3\lambda +2,4\lambda +3)$

Let

$Q(x,y,z)=(2\lambda ,3\lambda +2,4\lambda +3)$

The directional cosines of

$PQ= 2\lambda -3,3\lambda +3,4\lambda -8$

Since the line

$L$ and

$PQ$ are perpendicular

$(2\lambda -3)(2)+(3\lambda +3)(3)+(4\lambda -8)4=0\\ \Longrightarrow \lambda =1\\ \Longrightarrow Q=(2,5,7)$

The direction ratios of

$PQ=(-1,6,-4)$

Find the vector equation of line joining the points

$(2,1,3)$ and

$(-4,3,-1)$

0%
$\overline r=2(1-3\lambda)\overline i-(1+2\lambda)\overline j-(3-4\lambda)\overline k$
0%
$\overline r=2(1-3\lambda)\overline i-(1+2\lambda)\overline j+(3-4\lambda)\overline k$
0%
$\overline r=2(1-3\lambda)\overline i+(1+2\lambda)\overline j+(3-4\lambda)\overline k$
0%
$\overline r=2(1+3\lambda)\overline i+(1+2\lambda)\overline j+(3+4\lambda)\overline k$

Explanation

given Points

A(2,1,3) and B(-4,3,-1)

$\vec{b}=-4\hat{i}+3\hat{j}-\hat{k}$

Position vector of line

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

normal vector can be calculated by

$\vec{b}-\vec{a}=\vec{n}=-4\hat{i}+3\hat{j}-\hat{k}-(2\hat{i}+\hat{j}+3\hat{k})$

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

$\vec{n}=-6\hat{i}+2\hat{j}-4\hat{k}$

eq of line

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

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

$\vec{r}=2\hat{i}+\hat{j}+3\hat{k}-6\lambda\hat{i}+2\lambda\hat{j}-4\lambda\hat{k}$

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

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

The vector equation of line passing through two points

$A(x_1,y_1,z_1),B(x_2,y_2,z_2)$ is

0%
$\vec r=\vec a-\vec b$
0%
$\vec r=\vec a +\lambda(\vec b-\vec a)$
0%
$\vec r=\vec a+\lambda\vec b$
0%
$\vec r=\vec a+\vec b$

Explanation

Given points

$A(x_{1},y_{1},z_{1})$ and

$B(x_{2},y_{2},z_{2})$

position vector of point A

$\vec{a}=x_{1}\hat{i}+y_{1}\hat{j}+z_{1}\hat{k}$

position vector of point B

$\vec{b}=x_{2}\hat{i}+y_{2}\hat{j}+z_{2}\hat{k}$

normal vector of line be

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

eq of line

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

Which of the following represents direction cosines of the line:

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

Explanation

If direction cosine of a line is

$l,m,n$ , then

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

${ 0 }^{ 2 }+{ \left( \cfrac { \sqrt { 3 } }{ 2 } \right) }^{ 2 }+{ \left( \cfrac { 1 }{ 2 } \right) }^{ 2 }=1$
Hence, the correct answer from the given alternative is (c)

$0,\cfrac { \sqrt { 3 } }{ 2 } ,\cfrac { 1 }{ \sqrt { 2 } }$ .

A line passes through the points (6, -7, -1) and (2, -3, 1). What are the direction ratios of the line ?

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

Explanation

Direction ratios of a line passing through points

$(x_1,y_1,z_1)$ and

$(x_2,y_2,z_2)$ are represented by

$\pm (x_1-x_2, \ y_1-y_2, \ z_1-z_2)$

Hence for the given line, direction ratios are

$(6-2,-7-(-3),-1-1)$

$\Rightarrow \pm (4,-4,-2)$

$\Rightarrow (-4,4,2)$ or

$(4,-4,-2)$

The length of perpendicular from the origin to the plane which makes intercepts

$\dfrac{1}{3},\dfrac{1}{4},\dfrac{1}{5}$ respectively on the coordinate axes is

0%
$\dfrac{1}{\sqrt[5]{2}}$
0%
$\dfrac{1}{10}$
0%
$\displaystyle\sqrt[5]{2}$
0%
5

Explanation

Equation of plane

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

$3x+4y+5z-1=0$

diatance from origin

$\dfrac{1}{\sqrt{50}}=\dfrac{1}{\sqrt[5]{2}}$

ABCD is a trapezium in which AB and CD are parallel.that the mid points of the sides AB ,CD and the intersection of the diagonals are collinear.

0%
True
0%
False

If points (1,2), (3 , 5) and (0 , b ) are collinear the value of b is

0%
$\dfrac{1}{2}$
0%
$\dfrac{7}{2}$
0%
2
0%
-1

Explanation

$Area=\dfrac{1}{2}| 1(5-b)+3(b-2)+0(2-5)|$
As points are collinear , so area =0

$\therefore \dfrac{1}{2}| 1(5-b)+3(b-2)+0(2-5)|=0$

$\Rightarrow 5-b+3b-6=0$

$\Rightarrow=1=2b$

$\therefore b=\dfrac{1}{2}$

The direction angles of the line

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

$\alpha, \beta$ and

$\gamma$ , then

$\cos \alpha + \cos \beta + \cos \gamma =$ ________.

0%
$\dfrac {2}{\sqrt {26}}$
0%
$\dfrac {8}{\sqrt {26}}$
0%
$1$
0%
$2$

Explanation

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

$\cos\alpha$ is given by

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

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

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

$\cos\alpha=\dfrac{4}{\sqrt{9+16+1}}=\dfrac{4}{\sqrt {26}}$

$\cos\beta=\dfrac{4}{\sqrt{9+16+1}}=\dfrac{-3}{\sqrt {26}}$

$\cos\gamma=\dfrac{1}{\sqrt{9+16+1}}=\dfrac{1}{\sqrt {26}}$

$\cos \alpha = \dfrac {4}{\sqrt {26}} ,\cos \beta = \dfrac {-3}{\sqrt {26}} ,\cos \gamma = \dfrac {1}{\sqrt {26}}$

$\therefore \cos \alpha + \cos \beta + \cos \gamma = \dfrac {2}{\sqrt {26}}$ .

The angle between the lines whose direction cosines are

$\left( \dfrac {\sqrt{3}}{4}, \dfrac {1}{4}, \dfrac {\sqrt{3}}{2} \right)$ and

$\left( \dfrac {\sqrt{3}}{4}, \dfrac {1}{4}, -\dfrac {\sqrt{3}}{2} \right)$ is :

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

Explanation

Let

$(l_{1},m_{1},n_{1})=\left( \dfrac {\sqrt{3}}{4}, \dfrac {1}{4}, \dfrac {\sqrt{3}}{2} \right)$ and

$(l_{2},m_{2},n_{2})=\left( \dfrac {\sqrt{3}}{4}, \dfrac {1}{4}, -\dfrac {\sqrt{3}}{2} \right)$ are the direction cosines of two lines

Angle between two lines with direction cosines

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

$(l_{2},m_{2},n_{2})$ is given by

$\cos \theta = |l_1 l_2 + m_1m_2 + n_1n_2|$

$= \begin{vmatrix} \dfrac{\sqrt{3}}{4} \times \dfrac{\sqrt{3}}{4} + \dfrac{1}{4}\times \dfrac{1}{4} + \dfrac{\sqrt{3}}{2} \times \left( \dfrac {- \sqrt {3}}{2}\right) \end{vmatrix}$

$= \begin{vmatrix} \dfrac{3}{16} + \dfrac{1}{16} - \dfrac{3}{4} \end{vmatrix} = \begin{vmatrix} \dfrac {-2}{4} \end{vmatrix} = \dfrac {1}{2}$

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

If a line makes the angles

$\alpha , \beta$ and

$\gamma$ with the axes, then what is the value of

$1+\cos 2\alpha +\cos 2\beta+\cos 2\gamma$ equal to ?

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

Explanation

Since, the line makes the angles

$\alpha , \beta$ and

$\gamma$ with the axes

So,

$\cos\alpha, \cos\beta, \cos\gamma$ will be the direction cosines of the line.

And we know that the sum of squares of the direction cosines of any line is

$1$ .

Now,

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

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

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

$=1+2-3=0$

Hence, B is correct.

Direction cosines of the line

$\cfrac { x+2 }{ 2 } =\cfrac { 2y-5 }{ 3 } ,z=-1$ are ____

0%
$\cfrac { 4 }{ 5 } ,\cfrac { 3 }{ 5 } ,0$
0%
$\cfrac { 3 }{ 5 } ,\cfrac { 4 }{ 5 } ,\cfrac { 1 }{ 5 }$
0%
$-\cfrac { 3 }{ 5 } ,\cfrac { 4 }{ 5 } ,0$
0%
$\cfrac { 4 }{ 5 } ,-\cfrac { 2 }{ 5 } ,\cfrac { 1 }{ 5 }$

Explanation

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

$\Rightarrow$

$\dfrac { x+2 }{ 4 } =\dfrac { y-2.5 }{ 3 } , z=-1$

$\therefore$ direction cosine in

$x$

$($

$\alpha$

$) = 4k$ , direction cosine in

$y$

$($

$\beta$

$)$

$= 3k$ , direction cosine in

$z ($

$\gamma$

$)=0$

But

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

Therefore,

$16{ k }^{ 2 }+9{ k }^{ 2 }=1$

$\Rightarrow 25{ k }^{ 2 }=1$

$\Rightarrow k=\dfrac { 1 }{ 5 }$

Therefore, direction cosines of the line are

$\left (\dfrac { 4 }{ 5 } ,\dfrac { 3 }{ 5 } ,0\right)$ .

The following lines are

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

0%
collinear
0%
skew-lines
0%
co-planar lines
0%
parallel lines

Explanation

Condition for three lines

$\vec { { r }_{ 1 } }$ ,

$\vec { { r }_{ 2 } }$ , and

$\vec { { r }_{ 3 } }$ to be collinear is:

$\vec { { r }_{ 1 } } +\lambda \vec { { r }_{ 2 } } +\vec { { \mu r }_{ 3 } } =0$

where

$\vec { { r }_{ 1 } } =\left( \vec { i } +\vec { j } \right)$

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

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

and

$\lambda$ and

$\mu$ are scalars

Hence, the answer is collinear.

$L$ and

$M$ are two points with position vectors

$2\overline { a } -\overline { b }$ and

$a+2\overline { b }$ respectively. The position vector of the point

$N$ which divides the line segment

$LM$ in the ratio

$2:1$ externally is

0%
$3\overline { b }$
0%
$4\overline { b }$
0%
$5\overline { b }$
0%
$3\overline { a } +4\overline { b }$

Explanation

Given points are

$L$ and

$M$ are

$2\bar { a } -\bar { b }$ and

$\bar { a } -2\bar { b }$ ,

$N$ divides in the ratio

$2:1$ externally,

$\therefore M$ is the mid point of

$L$ and

$N$ ,

$\Longrightarrow N=2M-L$ ,

$\Longrightarrow N=2\bar { a } +4\bar { b } -2\bar { a } +\bar { b }$ ,

$N=5\bar { b }$

If the normal of the plane makes an angles

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

$\dfrac {\pi}{2}$ with positive X-axis, Y-axis and Z-axis respectively and the length of the perpendicular line segment from origin to the plane is

$\sqrt {2}$ , then the equation of the plane is ________.

0%
$x + y + z = \sqrt {2}$
0%
$x + y + z = 1$
0%
$x + y = 2$
0%
$x = \sqrt {2}$

Explanation

Normal of the plane makes an angle of

$\dfrac{\pi}{4},\dfrac{\pi}{4},\dfrac{\pi}{2}$ with X-axix,Y-axis, and Z-axis respectively,

Direction ratio of normal to the plane is

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

Direction ratio of normal to the plane is

$\left (\cos \dfrac {\pi}{4}, \cos \dfrac {\pi}{4}, \cos \dfrac {\pi}{2}\right )$

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

Equation of the plane is given by

$x\cos\alpha+y\cos\beta+z\cos\gamma=\perp\ distance$

$\therefore \dfrac {x}{\sqrt {2}} + \dfrac {y}{\sqrt {2}} = \sqrt {2}$

$x + y = 2$ .

Direction cosines of ray from

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

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

0%
$-2, 3, -6$
0%
$2, -3, 6$
0%
$2, 3, 6$
0%
$\dfrac{-2}{7}, \dfrac{3}{7}, \dfrac{-6}{7}$

Explanation

Given the points are

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

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

Now the direction ratios of the ray

$PQ$ are

$(-1-1,1+2,-2-4)=(-2,3,-6)$ .

The direction cosines of the line

$PQ$ will be

$\left(\dfrac{2}{\sqrt{2^2+3^2+6^2}},\dfrac{3}{\sqrt{2^2+3^2+6^2}},\dfrac{-6}{\sqrt{2^2+3^2+6^2}}\right)=\left(\dfrac{-2}{7},\dfrac{3}{7},\dfrac{-6}{7}\right)$ .

If the lines

$x=1+a,y=-3-\lambda a,z=1+\lambda a$ and

$x=\cfrac { b }{ 2 } ,y=1+b,z=2-b$ are coplanar, then

$\lambda$ is equal to

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

Explanation

The given lines are

$\cfrac { x-1 }{ 1 } =\cfrac { y+3 }{ -\lambda } =\cfrac { z-1 }{ \lambda } =\left( a \right)$ and

$\cfrac { x-0 }{ 1/2 } =\cfrac { y-1 }{ 1 } =\cfrac { z-2 }{ -1 } =(b)$

$\therefore$ coplanarity, we must have

$\begin{vmatrix} -1 & 4 & 1 \\ 1 & -\lambda & \lambda \\ 1/2 & 1 & -1 \end{vmatrix}=0$

$\Rightarrow -1\left( \lambda -\lambda \right) -4\left( -1-\cfrac { \lambda }{ 2 } \right) +\left( 1+\cfrac { \lambda }{ 2 } \right) =0$

$4+2\lambda +1+\cfrac { \lambda }{ 2 } =0\quad \Rightarrow 5+\cfrac { 5\lambda }{ 2 } =0\quad \therefore \lambda =-2$

Vector equation of the line

$6x - 2 = 3y + 1 = 2z - 2$ is

0%
$\vec r = \hat i + \hat j + 3\hat k + \lambda (\hat i - 2\hat j + 3\hat k)$
0%
$\vec r = \hat i - 2\hat j + 3\hat k + \lambda \left( {\frac{1}{3}\hat i + \frac{1}{3}\hat j + \hat k} \right)$
0%
$\vec r = \frac{1}{3}\hat i - \frac{1}{3}\hat j\, + \hat k + \lambda (\hat i + 2\hat j + 3\hat k)$
0%
$\vec r = - 2\hat i + \hat j\, - 2\hat k + \lambda (6\hat i + 3\hat j + 2\hat k)$

Explanation

$6x - 2 = 3y + 1 = 2z - 2$

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

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

$a = \left( {{1 \over 3}, - {1 \over 3},1} \right),b = \left( {{1 \over 6},{1 \over 3},{1 \over 2}} \right)$

$velue\,e{q^n}\,is\, = \left( {1,2,3} \right)$

$\mathop {{\text{ }}r}\limits^ \to = \dfrac{1}{3}\widehat i - \dfrac{1}{3}\widehat j + \widehat k + \lambda \left( {\dfrac{1}{6}\widehat i + \dfrac{1}{3}\widehat j + \dfrac{1}{2}\widehat k} \right)$

$\mathop {{\text{ }}r}\limits^ \to = \dfrac{1}{3}\widehat i - \dfrac{1}{3}\widehat j + \widehat k + \lambda \left( {\widehat i + 2\widehat j + 3\widehat k} \right)$

The points with position vectors

$60\hat{i}+3\hat{j}$ ,

$40\hat{i}-8\hat{j}$ ,

$a\hat{i}-52\hat{j}$ are collinear if

0%
$a=-40$
0%
$a=40$
0%
$a=20$
0%
$None\ of\ these$

Explanation

suppose

${60i + 3j}$ ,

${40i - 8j}$ and

${ai - 52j}$ is the three position of vector

$A,B,C$

$\begin{array}{l} \overrightarrow { AB } =\left( { 40i-8j } \right) -\left( { 60i+3j } \right) \\ \overrightarrow { AB } =-20i-11j \\ \overrightarrow { BC } =\left( { ai-52j } \right) -\left( { 40i-8j } \right) \\ \overrightarrow { BC } =\left( { a-40 } \right) i-44j \\ \left( { a-40 } \right) i-44j=m\left( { -20i-11j } \right) \\ \left( { a-40 } \right) i-44j=-20im-11jm \\ -44=-11m \\ m=\frac { { -44 } }{ { -11 } } \\ m=4 \\ a-40=-20m \\ a-40=-20\left( 4 \right) \\ a=-80+40 \\ a=-40 \end{array}$

Direction cosines

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

$l-5m+3n=0$ and

$7l^{2}+5m^{2}-3n^{2}=0$ . The direction cosines of one of the lines are

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

Explanation

Consider the problem

Given

$l-5m+3n=0$

$l=5m-3m$ ----

$(i)$

And

$7l^2+5m^2-3n^2=0$ ----

$(ii)$

From

$(i)$ put the value of

$l$ in equation

$(ii)$

Then,

$7(5m-3n)^2+5m^2-3n^2=0$

Hence

$30(2m-n)(3m-2n)=0$

thus,

$2m=n$ and

$3m=2n$

therefore,

$\dfrac m 1=\dfrac n 2=\dfrac{(5m-3n)}{5-2.3}=\dfrac{1}{\sqrt 6}$

And

$\dfrac{m}{2}=\dfrac n 3=\dfrac{(5m-3n)}{5.2-3.3}=\dfrac{1}{\sqrt {14}}$

Hence the required Directions cosines of line are

$- \dfrac{1}{\sqrt 6 },\dfrac{1}{{\sqrt 6 }},\dfrac{2}{{\sqrt 6 }}$

The direction ratios of the line joining the points

$A(4,-3,7)$ and

$B(1,3,5)$ are:

0%
$(5,0,12)$
0%
$(3,-6,2)$
0%
$(5,6,2)$
0%
$\left(\dfrac{5}{2},0,6\right)$

Can

$\dfrac{1}{\sqrt{3}}, \dfrac{2}{\sqrt{3}}, \dfrac{-2}{\sqrt{3}}$ be the direction cosines of any directed line?

0%
Yes
0%
No
0%
Cannot say
0%
None of these

Explanation

No, they can not be the direction cosines of any directed line.

As the sum of square of them is not

$1$ .

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

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

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

$=\dfrac{1+4+4}{3}$

$=3$ .

CBSE Questions for Class 12 Commerce Maths Three Dimensional Geometry Quiz 1 - 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 1 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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