MCQ Questions for Class 12 Commerce Maths Three Dimensional Geometry Quiz 11

$AB=21,\ B\equiv (-2,1,-8)$ and the direction cosines of

$AB$ are

$\dfrac{6}{7},\dfrac{2}{7},\dfrac{3}{7}$ then the coordinates of

$A$ are

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$(-16,-7,-1)$
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$(-20,-5,17)$
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$(16,7,1)$
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$(20,5,17)$

Explanation

$\begin{array}{l} Let\, \, coordinate\, \, of\, \, A\left( { x,y,z } \right) \\ B\left( { -2,1,-8 } \right) \, \, given \\ \Rightarrow AB=\sqrt { (x+{ 2^{ 2 } }+{ { \left( { y-1 } \right) }^{ 2 } }{ { \left( { z+8 } \right) }^{ 2 } } } \\ \Rightarrow direction\, \, of\, \, AB=\frac { { \left( { x+2 } \right) } }{ { { z_{ 1 } } } } ,\frac { { \left( { y-1 } \right) } }{ { { z_{ 1 } } } } .\frac { { \left( { z+8 } \right) } }{ { { z_{ 1 } } } } \\ \frac { { x+2 } }{ { { z_{ 1 } } } } =\frac { 6 }{ 7 } \\ \therefore x=16 \\ \frac { { y-1 } }{ { { z_{ 1 } } } } =\frac { 2 }{ 7 } \\ \therefore y=7\, \, and\, \, { }\frac { { z+8 } }{ { { z_{ 1 } } } } =\frac { 3 }{ 7 } ,z=1 \\ Hence,\, \, \, coordinate\, \, of\left( { 16,7,1 } \right) \end{array}$

The direction ratios of the normal to the plane through

$(1,0,0)$ and

$(0,1,0)$ which makes an angle of

$\dfrac{\pi }{4}$ with the plane

$x + y = 3$ are-

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

Explanation

Let the equation of the plane through

$\left(1,0,0\right)$ and

$\left(0,1,0\right)$ be

$ax+by+cz+d=0$

where

$\vec{n}=\left(a,b,c\right)$ is unit normal vector to the plane.

$\Rightarrow \sqrt{{a}^{2}+{b}^{2}+{c}^{2}}=1$

$\Rightarrow a+d=0$ and

$b+d=0$ Subtracting these two we get

$a-b=0$ ....

$(1)$

Also since this plane makes angle

$\dfrac{\pi}{4}$ with

$x+y=3$

$\Rightarrow \dfrac{1}{\sqrt{2}}=\dfrac{\left(a,b,c\right).\left(1,1,0\right)}{1.\sqrt{2}}$

$\Rightarrow a+b=1$ ........

$(2)$

Solving

$(1)$ and

$(2)$ we get

$2a=1$ or

$a=\dfrac{1}{2}$

From

$(1)$ we have

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

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

$c=\pm\dfrac{1}{\sqrt{2}}$

$\therefore$ Direction ratios of normal is

$\left(a,b,c\right)=\left(\dfrac{1}{2},\dfrac{1}{2},\dfrac{1}{\sqrt{2}}\right)$

$=\left(1,1,\sqrt{2}\right)$ by multiplying by

$2$

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

${l}_{2},{m}_{2},{n}_{2}$ are the d.c.s of the two lines, then

${({l}_{1}{l}_{2}+{m}_{1}{m}_{2}+{n}_{1}{n}_{2})}^{2}+{({l}_{1}{m}_{2}-{l}_{2}{m}_{1})}^{2}+{({m}_{1}{n}_{2}-{m}_{2}{n}_{1})}^{2}+{({n}_{1}{l}_{2}-{n}_{2}{l}_{1})}^{2}=$

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$3$
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$2$
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$1$
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$4$

$P=(0,1,2),Q=(4,-2,1),O=(0,0,0)$ then

$\angle POQ=$

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

If a line makes angles

$\alpha, \beta, \gamma$ with positive axes, then the range of

$\sin{\alpha}\sin{\beta}+\sin{\beta} \sin{\gamma} +\sin{\gamma} \sin {\alpha}$ is

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$\left(\dfrac{-1}{2},1\right)$
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$\left(\dfrac{1}{2},2\right)$
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$(-1,2)$
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$(-1,2]$

Explanation

If a line makes an angle $\alpha , \beta , \gamma$ with the axes, then

$\cos^{2}\alpha +\cos^{2}\beta +\cos^{2}\gamma=1$ $[\because \cos \alpha,\cos \beta,\cos \gamma$ are directional cosines $]$

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

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

Also $\sin^{2}\alpha +\sin^{2}\beta +\sin^{2}\gamma \ \ge \sin \alpha \sin \beta +\sin \beta \sin \gamma +\sin \gamma \sin \alpha$

$\Rightarrow \displaystyle \sum \sin \alpha \sin \beta \le 2$ $\dots(1)$

Also $(\displaystyle \sum \sin \alpha)^{2}=\displaystyle \sum \sin^{2}\alpha +2\displaystyle \sum \sin \alpha \sin \beta$ $[\because (a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca]$

$(\displaystyle \sum \sin \alpha)^{2} > 0$

$\displaystyle \sum \sin^{2}\alpha +2\displaystyle \sum \sin \alpha \sin \beta > 0$

$2+2\displaystyle \sum \sin \alpha \sin \beta > 0$

$2\displaystyle \sum \sin \alpha \sin \beta > -2$

$\displaystyle \sum \sin \alpha \sin \beta > \dfrac {-2}{2}$

$\displaystyle \sum \sin \alpha \sin \beta >-1$ $\dots(2)$

From $(1)\ and\ (2)$

$\displaystyle 2 \geq \sum \sin \alpha \sin \beta >-1$

Hence the range of $\displaystyle \sum \sin \alpha \sin \beta$ is $(-1, 2]$

If the angle between the lines,

$\dfrac { x }{ 2 } =\dfrac { y }{ 2 } =\dfrac { z }{ 1 }$ and

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

$\cos { ^{ -1 }\left( \dfrac { 2 }{ 3 } \right) }$ , then p is equal to:

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$-\dfrac { 4 }{ 7 }$
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$\dfrac { 7 }{ 2 }$
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$-\dfrac { 7 }{ 4 }$
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$\dfrac { 2 }{ 7 }$

The direction cosines of the line which is perpendicular to the lines whose direction cosines are proportional to ( 1, -1,2 ) and ( 2,1,-1) are:-

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$\frac { 1 }{ \sqrt { 35 } } ,-\frac { 5 }{ \sqrt { 35 } } ,\frac { 3 }{ \sqrt { 35 } }$
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$-\frac { 1 }{ \sqrt { 35 } } ,\frac { 5 }{ \sqrt { 35 } } ,\frac { 3 }{ \sqrt { 35 } }$
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$\frac { 1 }{ \sqrt { 35 } } ,\frac { 5 }{ \sqrt { 35 } } ,\frac { 3 }{ \sqrt { 35 } }$
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none of these

$(2,1,3)$ and

$(-1,2,4)$ are the extermities of a diagonal of a rhombus then the

$d.r's$ of the other diagonal are

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$(2,3,-9)$
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$(-2,3,-9)$
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$(1,-2,4)$
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$(2,3,1)$

The direction cosines of the line which is perpendicular to the lines whose direction cosines are proportional to

$(1, -1, 2)$ and

$(2, 1, -1)$ are

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$\dfrac{-1}{\sqrt{35}}, \dfrac{5}{\sqrt{35}}, \dfrac{3}{\sqrt{35}}$
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$\dfrac{13}{\sqrt{35}}, \dfrac{-1}{\sqrt{35}}, \dfrac{1}{\sqrt{35}}$
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$\dfrac{2}{\sqrt{3}}, \dfrac{5}{\sqrt{3}}, \dfrac{7}{\sqrt{3}}$
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$\dfrac{3}{\sqrt{35}}, \dfrac{5}{\sqrt{35}}, \dfrac{7}{\sqrt{35}}$

A mirror and source of light are kept at the origin and the positive x-axis respectively a ray a light from the sources strikes the mirror and is reflected. If (1,-1,1) are the Dr's of a normal to the plane. Then the D.C's of the reflected ray are

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$(\frac{-1}{3},\frac{-2}{3},\frac{2}{3})$
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$(\frac{1}{3},\frac{2}{3},\frac{2}{3})$
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$(\frac{1}{3},\frac{-2}{3},\frac{2}{3})$
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$(\frac{-1}{3},\frac{2}{3},\frac{2}{3})$

If the incident ray and normal have the directions of the vectors

$\left(1,-3,1\right),\left(1,1,1\right)$ respectively, then direction of the reflected ray is-

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$\left(4,-8,4\right)$
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$\left(5,-7,5\right)$
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$\left(6,-6,6\right)$
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$\left(-5,7,5\right)$

The direction cosines of the line which is perpendicular to the lines whose direction cosines are proportional to (1,-1,2) and (2,1,-1) are:-

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$\frac { 1 }{ \sqrt { 35 } } ,-\frac { 5 }{ \sqrt { 35 } } ,\frac { 3 }{ \sqrt { 35 } }$
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$-\frac { 1 }{ \sqrt { 35 } } ,\frac { 5 }{ \sqrt { 35 } } ,\frac { 3 }{ \sqrt { 35 } }$
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$\frac { 1 }{ \sqrt { 35 } } ,\frac { 5 }{ \sqrt { 35 } } ,\frac { 3 }{ \sqrt { 35 } }$
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None of these

If from the point

$P ( f , g , h )$ perpendiculars

$PL , PM$ be drawn to

$y z$ and

$zx$ planes then the equation to the plane

$OLM$ is -

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$\frac { x } { f } + \frac { y } { g } + \frac { z } { h } = 0$
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$\frac { x } { f } + \frac { y } { g } - \frac { z } { h } = 0$
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$- \frac { x } { f } + \frac { y } { g } + \frac { z } { h } = 0$
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$\frac { x } { f } - \frac { y } { g } + \frac { z } { h } = 0$

The equation of the plane passing through

$(1, 1, 1)$ and

$(1, -1, -1)$ and perpendicular to

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

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$2x + 5y + z - 8 = 0$
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$x + y - z - 1 = 0$
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$2x + 5y + z + 4 = 0$
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$x - y + z - 1 = 0$

The direction ratios of two lines are

$(4,3,5)$ and

$(\lambda, -1, 2)$ . If the angle between them is

$45^{o}$ , a value of

$\lambda$ is

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

The st lines whose direction cosines satisfy:

$al+bm+cn=0$ and

$fmn+gnl+hlm=0$ are perpendicular if:

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$\dfrac {f}{a}+\dfrac {g}{b}+\dfrac {h}{c}=0$
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$\dfrac {a^{2}}{f}+\dfrac {b^{2}}{g}+\dfrac {c^{2}}{h}=0$
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$\sqrt {af}+\sqrt {bg}+\sqrt {ch}=0$
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$a^{2}f+b^{2}g+c^{2}h=0$ .

$l_1$ ,

$m_1$ ,

$n_1$ and

$l_2$ ,

$m_2$ ,

$n_2$ are the direction cosines of two perpendicular lines, then the direction cosine of the line which is perpendicular to both the lines , will be

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( $m_1$ $n_2$ - $m_2$ $n_1$ ), ( $n_1$ $l_2$ - $n_2$ $l_1$ ), ( $l_1$ $m_2$ - $l_2$ $m_1$ )
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( $l_1$ $l_2$ - $m_1$ $m_2$ ), ( $m_1$ $m_2$ - $n_1$ $n_2$ ), ( $n_1$ $n_2$ - $l_1$ $l_2$ )
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$\dfrac{1} {\sqrt {l^{2}_1+m^{2}_1+n^{2}_1}}$ , $\dfrac{1} {\sqrt {l^{2}_2+m^{2}_2+n^{2}_2}}$ , $\dfrac{1} {\sqrt3}$
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$\dfrac{1} {\sqrt3}$ , $\dfrac{1} {\sqrt3}$ , $\dfrac{1} {\sqrt3}$

If two straight lines having directions cosines

$\lambda, m, n$ and

$f, g, h$ satisfy

$\lambda+m+n=0$ and

$fmn+gn\lambda+h\lambda m=0$ and are perpendicular then

$f+g+h$ is equal to

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

The Dr's of two lines are 1, -2, -2 and 0, 2, 1 the Dc's of the line perpendicular to the above lines are :-

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$\frac { 2 }{ 3 } ,-\frac { 1 }{ 3 } ,\frac { 2 }{ 3 }$
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$-\frac { 1 }{ 3 } ,\frac { 2 }{ 3 } ,\frac { 2 }{ 3 }$
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$\frac { 1 }{ 14 } ,\frac { 3 }{ 4 } ,\frac { 2 }{ 3 }$
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None of these

In a plane there are 10 points, no three are in same straight line except 4 points which are collinear, then the number of straight lines are

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39
0%
41
0%
45
0%
40

The point collinder with (1,-2,-3) and (2,0,0) amoung the following is

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(0,4,6)
0%
(0, -4, -5)
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(0, -4, -6)
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non of these

A line with direction ratio 2,7,-5 is drawn to intersect the lines

$\frac { x-y }{ 3 } =\frac { y-7 }{ -1 } =\frac { z+2 }{ 1 }$ and

$\frac { x+3 }{ -3 } =\frac { y-3 }{ 2 } =\frac { z-6 }{ 4 }$ at P and Q respectively, then length of PQ is-

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$\sqrt { 78 }$
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$\sqrt { 77 }$
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$\sqrt { 54 }$
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$\sqrt { 74 }$

${ L }_{ 1 }$ and

${ L }_{ 2 }$ are two lines whose vector equations are

${ L }_{ 1 }:\vec { r } =\lambda \left[ \left( cos\theta +\sqrt { 3 } \right) \hat { i } +\left( \sqrt { 2 } sin\theta \right) \hat { j } +\left( cos\theta -\sqrt { 3 } \right) \hat { k } \right]$

${ L }_{ 2 }:\vec { r } =\mu \left( a\hat { i } +b\hat { j } +c\hat { k } \right) ,$ where

$\lambda$ and

$\mu$ are scalars and

$\alpha$ is the acute angle between

${ L }_{ 1 }$ and

${ L }_{ 2 }$ If the angle

$'\alpha '$ is independent of

$\theta$ then the value of

$\alpha$ is

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

Direction ratio of two lines are

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

$l_{2},m_{2},n_{2}$ then direction ratios of the line perpendicular to both the lines are

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$l_{1}-l_{2}, m_{1}-m_{2}, n_{1}-n_{2}$
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$l_{1}+l_{2}, m_{1}+m_{2}, n_{1}+n_{2}$
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$m_{1}n_{2}-n_{1}m_{2}, n_{1}l_{2}-n_{2}l_{1}, l_{1}m_{2}-m_{1}l_{2}$
0%
$m_{1}n_{2}-n_{1}m_{2}, n_{1}l_{2}-n_{1}l_{1}, l_{1}m_{2}-m_{1}l_{2}$

Explanation

Let

${u}_{1}=\left({l}_{1},{m}_{1},{n}_{1}\right)$ be a unit vector along one line and

${u}_{2}=\left({l}_{2},{m}_{2},{n}_{2}\right)$ be a unit vector along other line.

${u}_{1}\times {u}_{2}$ is a unit vector perpendicular to both

${u}_{1}$ and

${u}_{2}$

${u}_{1}\times {u}_{2}=\left|\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ {l}_{1} & {m}_{1} & {n}_{1} \\ {l}_{2} & {m}_{2} & {n}_{2}\end{matrix}\right|$

$=\hat{i}\left({m}_{1}{n}_{2}-{n}_{1}{m}_{2}\right)-\hat{j}\left({n}_{2}{l}_{1}-{n}_{1}{l}_{2}\right)+\hat{k}\left({l}_{1}{m}_{2}-{m}_{1}{l}_{2}\right)$

$=\hat{i}\left({m}_{1}{n}_{2}-{n}_{1}{m}_{2}\right)+\hat{j}\left({n}_{1}{l}_{2}-{n}_{2}{l}_{1}\right)+k\left({l}_{1}{m}_{2}-{m}_{1}{l}_{2}\right)$

So, the required direction cosines are:

${m}_{1}{n}_{2}-{n}_{1}{m}_{2}$ ,

${n}_{1}{l}_{2}-{n}_{2}{l}_{1}$ and

${l}_{1}{m}_{2}-{m}_{1}{l}_{2}$

The equation to the altitude of the triangle formed by

$(1, 1, 1)$ ,

$(1, 2, 3)$ ,

$(2, -1, 1)$ through

$(1, 1, 1)$ .

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$\bar{r}=(\bar{i}+\bar{j}+\bar{k})+t(\bar{i}-3\bar{j}-2\bar{k})$
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$\bar{r}=(\bar{i}+\bar{j}+\bar{k})+t(3\bar{i}+\bar{j}+2\bar{k})$
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$\bar{r}=(\bar{i}+\bar{j}+\bar{k})+t(\bar{i}-\bar{j}+2\bar{k})$
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$|\bar{r}|=5$

Explanation

$\text { Solve: }\\$

$\overrightarrow{B D}=(a-1) \hat{i}+(b-2) \hat{j}+(c-3) \hat{k} \\$

$\text { Points } B_{1} D \text { and } c \text { are three collinear } \\$

$\text { point } \\$

$\qquad \Rightarrow \lambda \overrightarrow{B C}=(\overrightarrow{B D}) \\$

$\Rightarrow \quad(a-1)=\lambda \quad \Rightarrow \quad a=\lambda+1$

$(b-2) =-3 \lambda \quad \Rightarrow b=2-3 \lambda \\$

$c-3 =-2 \lambda \quad \Rightarrow c=3-2 \lambda$

$\overrightarrow{A D}$ is perpendicular to

$\overrightarrow{B C}$

$\Rightarrow \quad(\overrightarrow{A D}) \cdot(\overrightarrow{B C})=0$

$\overrightarrow{A D}=\lambda \hat{i}+(1-3 \lambda) \hat{j}+(2-2 \lambda) \hat{k}$

$\Rightarrow \quad \overrightarrow{A D} \cdot \overrightarrow{B C}=\lambda-3+9 \lambda-4+4 \lambda=0$

$\Rightarrow 14 \lambda=7 \\$

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

$\text { SO, } a=\lambda+1= \frac{3}{2} \\$

$b=2-3 \lambda= \frac{1}{2} \text { and } c=3-2 \lambda$

So, equation of its altitude i.e.

$\overrightarrow{A D}$

$\left.\overrightarrow{A D}=\hat{i}+\hat{\jmath}+k+{{t[}}({a}-1) \hat{i}+(b-1) \hat{\jmath}+(c-1) \hat{k}\right]$

$\Rightarrow \overrightarrow{A D}=\hat{i}+\hat{j}+\hat{k}+t\left(\frac{1}{2} \hat{i}+\left(-\frac{1}{2}\right) \hat{\jmath}+\hat{k}\right)$

$\Rightarrow \overrightarrow{A D}=\hat{i}+\hat{\jmath}+\hat{k}+t(\hat{\imath}-\hat{\jmath} +2 \hat{k})$

where

$t=\frac{t}{2}$

The plane passing through

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

$(-7, -3, -5)$ is parallel to

0%
$X-axis$ .
0%
$Y-axis$ .
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$Z-axis$ .
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None of these

Each group from the alternatives represents lengths of sides of a triangleStare which group does not represent a right-angled triangle.

0%
$( 8,40,41 )$
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$( 20,25,30 )$
0%
$( 8,15,17 )$
0%
$( 6,8,10 )$

there are 20 points in the plane on three of which are collinear. the number of straight lines by joining them is

0%
190
0%
200
0%
40
0%
500

The lines

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

$\vec{r}=(2i-j)+\mu(i+j-k)$ intersect for

0%
$\lambda=1, \mu =1$
0%
$\lambda=2, \mu =1$
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All values of $\lambda$ and $\mu$
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No value of $\lambda$ and $\mu$

The value of p so that the lines

$\frac { 1-x }{ 3 } =\frac { 7y-14 }{ 2p } =\frac { z-3 }{ 2 }$ and

$\frac { 7-7x }{ 3p } =\frac { y-5 }{ 1 } =\frac { 6-z }{ 5 }$ are at right angles are

0%
70/11
0%
7/11
0%
10/7
0%
17/11

What is the area of the triangle with vertices

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

$(3,4,0)$ ?

0%
$\frac{{15}}{2}sq$ unit
0%
$15sq$ unit
0%
$\frac{{7}}{2}sq$ unit
0%
$7sq$ unit

In the three points with position vectors (1, a. b) : (a, b, 3) are collinear in space, then the value of a + b is

0%
3
0%
4
0%
5
0%
none

The number of straight lines that can be drawn through any two points out of

$10$ points, of which

$7$ are collinear.

0%
$25$
0%
$30$
0%
$35$
0%
$45$

If the vectors

$2\hat{i} + 3\hat{j} , ~5\hat{i} + 6\hat{j} ,$ and

$8\hat{i} +\lambda{\hat{j}}$ have their initial points at

$(1 , 1)$ , then the value of

$\lambda$ so that the vectors terminate on one straight line is

0%
$0$
0%
$3$
0%
$6$
0%
$9$

$( 0,0 ) , ( a , 0 )$ and

$( 0 , b )$ are collinear, then

0%
$a b = 0$
0%
$a = b$
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$a = - b$
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$a - b = c$

If points

$(a - 2, a - 4); (a, a + 1)$ and

$(a + 4, 16)$ are collinear, then

$a$ is equal to

0%
$5$
0%
$-5$
0%
$7$
0%
$-7$

The direction cosines of the normal to the plane

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

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

The plane passing through the point

$\left ( 5,1,2 \right )$ perpendicular to the line

$2\left ( x - 2 \right ) = y - 4 = z - 5$ will meet the line in the point

0%
$\left ( 1,2,3 \right )$
0%
$\left ( 2,3,1 \right )$
0%
$\left ( 1,3,2 \right )$
0%
$\left ( 3,2,1 \right )$

The equation of the plane passing through the points

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

$\left ( 7,0,6 \right )$ is

$5x + 3y - 2z =\lambda$ where

$\lambda$ is

0%
$23$
0%
$21$
0%
$19$
0%
$27$

The vector equation of line 2x - 1 = 3 y + 2 = z - 2 is

0%
$\bar{r}=\left ( \dfrac{1}{2}\hat{i}-\dfrac{2}{3}\hat{j}+2\hat{k} \right )+\lambda \left ( 3\hat{i}+2\hat{j}+6\hat{k} \right )$
0%
$\bar{r}=\hat{i}-> j+(2\hat{i}+\hat{j}+\hat{k})$
0%
$\bar{r}=\left ( \dfrac{1}{2}\hat{i}-\hat{j}\right )+\lambda \left ( \hat{i}+2\hat{j}+6\hat{k} \right )$
0%
$\bar{r}=\left ( \hat{i}+\hat{j} \right )+\lambda \left ( \hat{i}-2\hat{j}+6\hat{k} \right )$

If P, Q R are collinear points such that P( 7, 7) Q( 3, 4) and PR = 10 then R is

0%
(1, 1)
0%
( 1, -1)
0%
( -1, 1)
0%
( -1, -1)

CBSE Questions for Class 12 Commerce Maths Three Dimensional Geometry Quiz 11 - 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 11 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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