MCQ Questions for Class 12 Commerce Maths Three Dimensional Geometry Quiz 10

The direction cosine of a line equally inclined to the axes are

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

Explanation

$a$ line makes angles

$\alpha ,\beta ,\gamma$ with the axes, we have

$\alpha =\beta =\gamma$

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

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

$\displaystyle \Rightarrow { l }^{ 2 }+{ l }^{ 2 }+{ l }^{ 2 }\quad$

$\Rightarrow { 3l }^{ 2 }=1$

$\displaystyle \Rightarrow { l }^{ 2 }=\frac { 1 }{ 3 } \quad$

$\Rightarrow l=\pm \dfrac { 1 }{ \sqrt { 3 } }$

$l_1, m_1, n_1$ and

$l_2, m_2, n_2$ be the DC's of two concurrent lines, the direction cosines of the line bisecting the angles between them are proportional to

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$l_1 l_2, m_1 m_2, n_1, n_2$
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$l_1m_2, l_1 n_2, l_1 n_3$
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$l_1 + l_2, m_1 + m_2, n_1 + n_2$
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None of these

Explanation

Let direction cosines of

$OA = (l_1, m_1, n_1)$ and direction cosines of

$OB = (l_2, m_2, n_2)$ .
Taking two points on

$l$ and

$m$ such that

$OA = OB = r$
Let

$C$ be the mid-point of

$AB$ . Then,

$OC$ is the bisector of the angle

$AOB$ .
Now, coordinates of

$A$ are

$(l_1r, m_1 r, n_1 r)$ and coordinates of

$B$ are

$(l_2r, m_2r, n_2r)$ .

$\therefore$ the coordinates of

$C$ are

$\displaystyle \frac{(l_1 + l_2)r}{2} , \frac{(m_1 + m_2)r}{2}, \frac{(n_1 + n_2)r}{2}$ .
Hence, the direction cosines of

$OC$ are proportional to

$l_1 + l_2, m_1 + m_2$ and

$n_1 + n_2$ .

The angle between the lines

$\displaystyle 2x=3y=-z\:and\:6x=-y=-4z$ is

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

Explanation

Angle between the lines

$2x = 3y = - z , 6x = -y = -4z$ is

$90$

Since

$a1a2 + b1b2 + c1c2 = 0.$

where

$a_1=\dfrac {1}{2},a_2=\dfrac{1}{6},b_1=\dfrac{1}{3},b_2=-1,c_1=-1,c_2=\dfrac{-1}{4}$

Thus the angle between them is

$\dfrac{\pi}{2}$

The direction cosines of a line whose equations are

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

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$\dfrac{1}{\sqrt{14}},\dfrac{-3}{\sqrt{14}},\dfrac{2}{\sqrt{14}}$
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$\dfrac{2}{\sqrt{29}},\dfrac{4}{\sqrt{29}},\dfrac{-3}{\sqrt{29}}$
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$\dfrac{1}{\sqrt{29}},\dfrac{-3}{\sqrt{29}},\dfrac{2}{\sqrt{29}}$
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$2,4,-3$

Explanation

Step 1:

The direction ratios of the line joining these points are

$(x_2−x_1),(y_2−y_1),(z_2−z_1)$

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

$\therefore$ Direction cosines are

$\dfrac{2}{\sqrt {2^2+4^2+(-3)^2}},\dfrac{4}{\sqrt {2^2+4^2+(-3)^2}},\dfrac{-3}{\sqrt {2^2+4^2+(-3)^2}}$

$\Longrightarrow \dfrac{2}{\sqrt {29}},\dfrac{4}{\sqrt {29}},\dfrac{-3}{\sqrt {29}}$

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

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

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

Explanation

Let the foot of the perpendicular be

$P(a,b,c)$ .

Then, direction ratios of

$OP$ are

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

$a,b,c$

So, the equation of the plane passing through

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

$a,b,c$ is

$a\left( x-a \right) +b\left( y-b \right) +c\left( z-c \right) =0\\ \Rightarrow ax+by+cz={ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }.$

If equation of the plane through the straight line $\displaystyle \dfrac{x -1}{2}=\dfrac{y +2}{-3}=\dfrac{z}{5}$ and perpendicular to the plane $x - y + z + 2 = 0 \:$ is $\:ax- by + cz + 4 = 0,$ then find the value of $a^2 + b^2 + c$

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$12$
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$14$
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$16$
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$18$

Explanation

Let equation of a plane containing the line be

$\ell(x- 1) + m(y + 2) + nz = 0$
then

$2\ell-3m + 5n = 0 \:$ and

$\:\ell - m + n = 0$

$\therefore \dfrac{\ell}{2}=\dfrac{m}{3}=\dfrac{n}{1}$

$\therefore$ the plane is

$\:2(x- 1) + 3 (y + 2) + z = 0$
i.e.

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

$\therefore a = 2, b = - 3, c = 1$

$\therefore a^2 + b^2 + c = 14$

if a line makes angles

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

$sin^{2}\alpha+sin^{2}\beta+sin^{2}\gamma+sin^{2}\delta$ equals

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$2$
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$\displaystyle \frac{4}{3}$
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$\displaystyle \frac{8}{3}$
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$1$

Explanation

Diagonals are

$OE=(1,1,1)$

$\vec { GB } =(1,1,0)-(0,0,1)=(1,1,-1)$

$\vec { FC } =(1,-1,1)$

$\vec { AD } =(-1,1,1)$

Let the line have direction ratio's as

$l,m,n$

Angle between

$OE$ and

$L$

$\cos { \alpha } =\dfrac { l+m+n }{ \sqrt { { l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 } } \times 2\sqrt { 3 } }$

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

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

$\cos { \delta } =\dfrac { -l+m+n }{ \sqrt { { l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 } } \times \sqrt { 3 } }$

$\sin ^{ 2 }{ \alpha } +\sin ^{ 2 }{ \beta } +\sin ^{ 2 }{ \gamma } +\sin ^{ 2 }{ \delta } =4-(\cos ^{ 2 }{ \alpha } +\cos ^{ 2 }{ \beta } +\cos ^{ 2 }{ \gamma } +\cos ^{ 2 }{ \delta } )$

$\sin ^{ 2 }{ \alpha } +\sin ^{ 2 }{ \beta } +\sin ^{ 2 }{ \gamma } +\sin ^{ 2 }{ \delta } =4-(4\{ \dfrac { { l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 } }{ { 3(l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 }) } \} )$

$=\dfrac { 8 }{ 3 }$

Three points whose position vectors are

$\overrightarrow{a}$ ,

$\overrightarrow{b}$ ,

$\overrightarrow{c}$ will be collinear if

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$\lambda \overrightarrow{a}+\mu \overrightarrow{b}=\left ( \lambda +\mu \right )\overrightarrow{c}$
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$\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{a}=\overrightarrow{0}$
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$\begin{bmatrix} \overrightarrow{a} & \overrightarrow{b} & \overrightarrow{c} \end{bmatrix}=0$
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None of these

Explanation

$\vec{a}$ ,

$\vec{b}$ and

$\vec{c}$ are collinear vectors, therefore they lie on the same line. Hence they are parallel. Hence there cross products will be 0.
Therefore

$(\vec{a}\times\vec{b})=(\vec{c}\times\vec{b})=(\vec{a}\times\vec{c})=0$
And application of the section formula gives us

$\vec{c}=\dfrac{\lambda \vec{a}+\mu\vec{b}}{\lambda+\mu}$
Or

$\vec{c}(\lambda+\mu)=\lambda \vec{a}+\mu\vec{b}$

Determine the equation of the plane on which the co - ordinates of the foot of perpendicular drawn from origin O is the point

$\displaystyle P\left ( \alpha ,\beta ,\gamma \right ).$

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$\displaystyle \alpha^{2} x+\beta^{2} y+\gamma^{2} z=\alpha ^{3}+\beta ^{3}+\gamma ^{3}$
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$\displaystyle \alpha x+\beta y+\gamma z=\alpha ^{2}+\beta ^{2}+\gamma ^{2}$
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$\displaystyle \alpha^{2} x+\beta^{2} y+\gamma^{2} z=\alpha +\beta +\gamma$
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none of these

Explanation

Given that

$OP$ is normal to plane

So the Drs of normal to the plane is

$\alpha,\beta,\gamma$

Therefore the equation of plane is

$\alpha x+\beta y+\gamma z=k$

The point

$(\alpha,\beta,\gamma)$ lies on the plane

So we have

${ \alpha }^{ 2 }+{ \beta }^{ 2 }+{ \gamma }^{ 2 }=k$

Therefore the equation of plane is

$\alpha x+\beta y+\gamma z={ \alpha }^{ 2 }+{ \beta }^{ 2 }+{ \gamma }^{ 2 }$

Therefore the correct option is

$B$

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

$\displaystyle l_{2},m_{2},n_{2}$ are D.C.'s of the two lines inclined to each other at an angle

$\displaystyle \theta$ , then the D. C.'s of the internal and external bisectors of the angle between these lines are

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$\displaystyle \frac{l_{1}+l_{2}}{2\sin \left ( \theta /2 \right )},\, \frac{m_{1}+m_{2}}{2\sin \left ( \theta /2 \right )},\, \frac{n_{1}+n_{2}}{2\sin \left ( \theta /2 \right )}$
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$\displaystyle \frac{l_{1}+l_{2}}{2\cos \left ( \theta /2 \right )},\, \frac{m_{1}+m_{2}}{2\cos \left ( \theta /2 \right )},\, \frac{n_{1}+n_{2}}{2\cos \left ( \theta /2 \right )}$
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$\displaystyle \frac{l_{1}-l_{2}}{2\sin \left ( \theta /2 \right )},\, \frac{m_{1}-m_{2}}{2\sin \left ( \theta /2 \right )},\, \frac{n_{1}-n_{2}}{2\sin \left ( \theta /2 \right )}$
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$\displaystyle \frac{l_{1}-l_{2}}{2\cos \left ( \theta /2 \right )},\, \frac{m_{1}-m_{2}}{2\cos \left ( \theta /2 \right )},\, \frac{n_{1}-n_{2}}{2\cos \left ( \theta /2 \right )}$

Explanation

If $l_1,\; m_1, n_1$ and $l_2,\; m_2, n_2$ are D.C.s of two lines then,

$l_1l_2+m_1m_2+n_1n_2= \cos \theta$

The direction ratios of the internal angle bisector are $l_1+l_2,\; m_1+m_2, \; n_1+n_2$ .

Hence, the direction cosines of the internal angle bisector are $\displaystyle \dfrac { { l }_{ 1 }+{ l }_{ 2 } }{ \sqrt { \sum { { \left( { l }_{ 1 }+{ l }_{ 2 } \right) }^{ 2 } } } } ,\dfrac { { m }_{ 1 }+{ m }_{ 2 } }{ \sqrt { \sum { { \left( { l }_{ 1 }+{ l }_{ 2 } \right) }^{ 2 } } } } ,\dfrac { { n }_{ 1 }+{ n }_{ 2 } }{ \sqrt { \sum { { \left( { l }_{ 1 }+{ l }_{ 2 } \right) }^{ 2 } } } }$

where, $\sum { { \left( { l }_{ 1 }+{ l }_{ 2 } \right) }^{ 2 } } \\ ={ \left( { l }_{ 1 }+{ l }_{ 2 } \right) }^{ 2 }+{ \left( { m }_{ 1 }+{ m }_{ 2 } \right) }^{ 2 }+{ \left( { n }_{ 1 }+{ n }_{ 2 } \right) }^{ 2 }\\ =\left( { { l }_{ 1 } }^{ 2 }+{ { m }_{ 1 } }^{ 2 }+{ { n }_{ 1 } }^{ 2 } \right) +\left( { { l }_{ 2 } }^{ 2 }+{ { m }_{ 2 } }^{ 2 }+{ { n }_{ 2 } }^{ 2 } \right) +2\left( { l }_{ 1 }{ l }_{ 2 }+{ m }_{ 1 }{ m }_{ 2 }+{ n }_{ 1 }{ n }_{ 2 } \right) \\ =1+1+2\cos { \theta } =2+2\cos { \theta } =4\cos^{2} { \dfrac { \theta }{ 2 } }$

Hence direction cosines of internal angle bisector are $\displaystyle \dfrac { { l }_{ 1 }+{ l }_{ 2 } }{ 2\cos { \dfrac { \theta }{ 2 } } } ,\dfrac { { m }_{ 1 }+{ m }_{ 2 } }{ 2\cos { \dfrac { \theta }{ 2 } } } ,\dfrac { { n }_{ 1 }+{ n }_{ 2 } }{ 2\cos { \dfrac { \theta }{ 2 } } }$

Similarly , D.R.s of external angle bisector are ${ l }_{ 1 }-{ l }_{ 2 },\; { m }_{ 1 }-{ m }_{ 2 },\; { n }_{ 1 }-{ n }_{ 2 }$

So, the direction cosines of the external angle bisector are $\displaystyle \dfrac { { l }_{ 1 }-{ l }_{ 2 } }{ \sum { { \left( { l }_{ 1 }-{ l }_{ 2 } \right) }^{ 2 } } } ,\dfrac { { m }_{ 1 }-{ m }_{ 2 } }{ \sum { { \left( { l }_{ 1 }-{ l }_{ 2 } \right) }^{ 2 } } } ,\dfrac { { n }_{ 1 }-{ n }_{ 2 } }{ \sum { { \left( { l }_{ 1 }-{ l }_{ 2 } \right) }^{ 2 } } }$

where, $\sum { { \left( { l }_{ 1 }-{ l }_{ 2 } \right) }^{ 2 } } \\ ={ \left( { l }_{ 1 }-{ l }_{ 2 } \right) }^{ 2 }+{ \left( { m }_{ 1 }-{ m }_{ 2 } \right) }^{ 2 }+{ \left( { n }_{ 1 }-{ n }_{ 2 } \right) }^{ 2 }\\ =\left( { { l }_{ 1 } }^{ 2 }+{ { m }_{ 1 } }^{ 2 }+{ { n }_{ 1 } }^{ 2 } \right) +\left( { { l }_{ 2 } }^{ 2 }+{ { m }_{ 2 } }^{ 2 }+{ { n }_{ 2 } }^{ 2 } \right) -2\left( { l }_{ 1 }{ l }_{ 2 }+{ m }_{ 1 }{ m }_{ 2 }+{ n }_{ 1 }{ n }_{ 2 } \right) \\ =1+1-2\cos { \theta } =2-2\cos { \theta } =4\sin ^{ 2 }{ \dfrac { \theta }{ 2 } }$ .

Hence, direction cosines of external angle bisector are $\displaystyle \dfrac { { l }_{ 1 }-{ l }_{ 2 } }{ 2\sin { \dfrac { \theta }{ 2 } } } ,\dfrac { { m }_{ 1 }-{ m }_{ 2 } }{ 2\sin { \dfrac { \theta }{ 2 } } } ,\dfrac { { n }_{ 1 }-{ n }_{ 2 } }{ 2\sin { \dfrac { \theta }{ 2 } } }$

Hence, options 'B' and 'D' are correct.

$O$ is the origin and

$A$ is the point

$\displaystyle \left ( a,b,c \right ).$ Find the direction cosines of the join of

$OA$ and deduce the equation of the plane through

$A$ at right angles to

$OA$ .

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

Explanation

D.R.'s of

$OA$ are

$\displaystyle a-0, b-0, c-0$ or a,b,c.

$\displaystyle \therefore$ D.C.'s of

$OA$ are

$\displaystyle \frac{a}{\sqrt{\left ( a^{2}+b^{2}+c^{2} \right )}},\frac{b}{\sqrt{\left ( \sum a^{2} \right )}},\frac{c}{\sqrt{\sum a^{2}}}$
Equation of any plane through

$A$ , i.e.

$\displaystyle \left ( a,b,c \right )$ is

$\displaystyle A\left ( x-a \right )+B\left ( y-b \right )+C\left ( z-c \right )=0.$
But plane is at right angles to

$OA$ which therefore is normal and whose D.R.'s are

$a, b, c$ .
Hence, the plane is

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

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

The direction cosines of the lines bisecting the internal angle

$\theta$ between the lines whose direction cosines are

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

$l_{2},m_{2},n_{2}$ are

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$< l_{1}+l_{2},m_{1} +m_{2},n_{1}+n_{2}>$
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$\displaystyle < \frac{l_{1}+l_{2}}{2\sin \frac{\theta}{2}},\frac{m_{1}+m_{2}}{2\sin \frac{\theta}{2}},\frac{n_{1}+n_{2}}{2\sin \frac{\theta}{2}}>$
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$\displaystyle < \frac{l_{1}+l_{2}}{2\cos \frac{\theta}{2}},\frac{m_{1}+m_{2}}{2\cos \frac{\theta}{2}},\frac{n_{1}+n_{2}}{2\cos \frac{\theta}{2}}>$
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none of these

Explanation

The direction cosines of the lines bisecting the internal angle

$\theta$ between the lines

direction cosines are

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

$l_{2},m_{2},n_{2}$ are

$\displaystyle < \frac{l_{1}+l_{2}}{2\cos \frac{\theta}{2}},\frac{m_{1}+m_{2}}{2\cos \frac{\theta}{2}},\frac{n_{1}+n_{2}}{2\cos \frac{\theta}{2}}>$

Option C is correct answer

The projection of a directed line segment on the co-ordinate axes are

$12, 4, 3$ , the DC's of the line are

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

Explanation

Let the line segment be represented in vector form as

$\vec a = a_1\vec i+a_2\vec j+a_3\vec k$

Vector along coordinate axes are

$\vec i,\vec j,\vec k$

Given that projection of

$\vec a$ on x axis is 12, that with y axis is 4 and that with z axis is 3.

$\Longrightarrow \vec a . \vec i = 12, \therefore \ a_1=12$

similarily,

$a_2=4,a_3=3$

$\therefore \vec a=12\vec i +4\vec j+3\vec k$

the length of the line segment=

$|\vec a| =\sqrt{144+16+9}=13$

$\therefore |PQ|=13$

Thus the direction cosines of PQ are

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

The projections of a line segment on

$x, y, z$ axes are

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

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$\displaystyle 13,< 12/13, 4/13, 3/13>$
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$\displaystyle 19,< 12/19, 4/19, 19>$
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$\displaystyle 11,< 12/11, 4/11, 3/11>$
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None of these

Explanation

Let the line segment be represented in vector form as

$\vec a =a_1\vec i+a_2\vec j+a_3\vec k$

Vector along coordinate axes are

$\vec i,\vec j,\vec k$

Vector along Given that projection of

$\vec a$ on x axis is 12, that with y axis is 4 and that with z axis is 3.

$\Longrightarrow \vec a . \vec i =12 \Longrightarrow a_1=12.$

Similarily

$a_2=4 ,a_3=3$

$\therefore \vec a =12 \vec i+4 \vec j +3 \vec k$

The length of the line segment

$= | \vec a |= \sqrt{144+16+9}=13$

and the direction cosines of the line segments are

$<\dfrac{12}{13},\dfrac{4}{13},\dfrac{3}{13}>$

The line

$\displaystyle \frac{x - 1}{2} = \frac{y}{-1} = \frac{z + 2}{2}$ cuts the plane

$\displaystyle x + y + z = 1$ at

$\displaystyle P$ . If the foot of the perpendicular from

$\displaystyle P$ to a point

$Q\displaystyle \left ( 3, \: -4, \: 1 \right )$ on the plane

$S$ then the equation of the plane

$S$ is

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$\displaystyle 3x - 2y - z = 0$
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$\displaystyle 2x - y + 2z = 12$
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$\displaystyle 2x - 10y + 5z = 51$
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none of these

Explanation

We have plne,

$x+y+z=1.......(1)$

And line,

$\dfrac{x-1}{2}=\dfrac{y}{-1}=\dfrac{z+2}{2}=\lambda......(2)$

Assume line (2) cuts plane (1) at point P,

$P=(2\lambda+1,-\lambda,2\lambda-2)$

Since line (2) cuts the plane (1) , Therefore Point P lies on plane (1).

$2\lambda+1-\lambda+2\lambda-2=1$

$3\lambda=2$

$\lambda=\dfrac{2}{3}$

Therefore,

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

direction ratios of normal of plane are,as we have two points, point P and Q(3,-4,1),

$(3-\dfrac{7}{3},-4+\dfrac{2}{3},1-\dfrac{-2}{3})$

$(\dfrac{2}{3},\dfrac{-10}{3},\dfrac{5}{3})$

Required Equation of plane,

$a(x-x_1)+b(y-y_1)+c(z-z_1)=0$

$\dfrac{2}{3}(x-3)+\dfrac{-10}{3}(y+4)+\dfrac{5}{3}(z-1)=0$

$2x-10y+5z-51=0$

$2x-10y+5z=51$

Therefore option (C) is correct.

The projection of a line segment joining the points

$P\left ( x_{1},y_{1},z_{1}, \right )$ and

$Q\left ( x_{1},y_{1},z_{1}, \right )$ on another line whose DC's are

$l, m, n$ is given by

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$l\left ( x_{1}+x_{2} \right )+m\left (y _{2}+y_{2} \right )+n\left ( z_{1}+z_{2} \right )$
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$\displaystyle 2\left [ \frac{\left ( lx_{2}+my_{2}+nz_{2} \right )}{2}-\frac{\left (lx_{1}+my_{1}+nz_{1} \right )}{2} \right ]$
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$\displaystyle l\left ( x_{2}-x_{1} \right )+ m\left ( y_{2}-y_{1} \right )+n\left ( z_{2}-z_{1} \right )$
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$\displaystyle \frac{x_{2}-x_{1}}{l}+\frac{y_{2}-y_{1}}{m}+\frac{z_{2}-z_{1}}{n}$

Explanation

Let the line be of length

$P$

D.R.'s of line are

$\dfrac { { x }_{ 2 }-{ x }_{ 1 } }{ P } ,\dfrac { { y }_{ 2 }-{ y }_{ 1 } }{ P } ,\dfrac { { z }_{ 2 }-{ z }_{ 1 } }{ P }$

D.C.'s of line on which project is taken is

$({ l },{ m },{ n })$

So, projection

$=P$ [(D.R. of line).(D.R. of line on which project is taken)]

$P[\dfrac { { x }_{ 2 }-{ x }_{ 1 } }{ P } \cdot l+\dfrac { { y }_{ 2 }-{ y }_{ 1 } }{ P } \cdot m+\dfrac { { z }_{ 2 }-{ z }_{ 1 } }{ P } \cdot n]$

$=({ x }_{ 2 }-{ x }_{ 1 })l+({ y }_{ 2 }-{ y }_{ 1 })m+({ z }_{ 2 }-{ z }_{ 1 })n$ ie option C

$={ x }_{ 2 }l+{ y }_{ 2 }m+{ z }_{ 2 }n-{ x }_{ 1 }l-{ y }_{ 1 }m-{ z }_{ 1 }n$

$={ x }_{ 2 }l+{ y }_{ 2 }m+{ z }_{ 2 }n-({ x }_{ 1 }l+{ y }_{ 1 }m+{ z }_{ 1 }n)$

$=2(\dfrac { { x }_{ 2 }l+{ y }_{ 2 }m+{ z }_{ 2 }n-({ x }_{ 1 }l+{ y }_{ 1 }m+{ z }_{ 1 }n) }{ 2 } )$

$=2(\dfrac { { x }_{ 2 }l+{ y }_{ 2 }m+{ z }_{ 2 }n }{ 2 } -\dfrac { { x }_{ 1 }l+{ y }_{ 1 }m+{ z }_{ 1 }n }{ 2 } )$

ie Option B

$A$ ,

$B$ and

$C$ are three collinear points, where

$A= i + 8 j - 5k$ ,

$B = 6i-2j$ and

$C= 9i + 4j - 3 k$ , then

$B$ divides

$AC$ in the ratio of :

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$\dfrac{5}{7}$
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$\dfrac{5}{3}$
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$\dfrac{2}{3}$
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None of these

Explanation

Given points are

$A(i+8j-5k)$ ,

$B(6i-2j)$ and

$C(9i+4j-3k)$

$AB = -5i+10j-5k$ and

$CB = 3i+6j-3k$

$\Rightarrow \dfrac{|AB|}{|CB|}=\dfrac{\sqrt{150}}{\sqrt{54}}=\dfrac{5\sqrt6}{3\sqrt6}=\dfrac{5}{3}$

Therefore correct option is

$B$

If the position vectors of the points

$A$ ,

$B$ , and

$C$ be

$i + j$ ,

$i - j$ and

$ai + bj+ ck$ respective;y , then the points

$A$ ,

$B$ and

$C$ are collinear if:

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$a = b = c = 1$
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$a= 1$ , $b$ and $c$ are arbitrary scalars
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$a =b= c= 0$
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$c = 0$ , $a =1$ and $b$ is a arbitrary scalar.

Explanation

Given points are

$A(i+j)$ ,

$B(i-j)$ and

$C(ai+bj+ck)$

These three points are collinear if drs of lines

$AB$ and

$CB$ are proportional

We have

$AB=2j$ and

$CB = (a-1)i+(b+1)j+ck$

Since their drs are proportional, we get

$\dfrac{a-1}{0}=\dfrac{b+1}{2}=\dfrac{c}{0}$

$\Rightarrow c=0 , a=1$ and

$b$ can be a arbitrary scalar

Therefore option

$D$ is correct

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

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

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

Explanation

We also have

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

So that

$\displaystyle { l }^{ 2 }+{ m }^{ 2 }-{ n }^{ 2 }=0\Rightarrow 2{ n }^{ 2 }=1\Rightarrow n=\pm \frac { 1 }{ \sqrt { 2 } }$

and

$\displaystyle l+m+n=0\Rightarrow { \left( l+m \right) }^{ 2 }={ n }^{ 2 }=\frac { 1 }{ 2 } ={ l }^{ 2 }+{ m }^{ 2 }$

$\Rightarrow 2lm=0$

Either

$l=0$ or

$m=0$ , if

$\displaystyle l=0,m+n=0$

$\Rightarrow m=-n=\pm \dfrac { 1 }{ \sqrt { 2 } }$

So direction ratios of one of the lines are

$\displaystyle 0,\pm \frac { 1 }{ \sqrt { 2 } } ,\mp \frac { 1 }{ \sqrt { 2 } }$

and if

$\displaystyle m=0,l+n=0\Rightarrow l=-n=\pm \frac { 1 }{ \sqrt { 2 } }$

So the direction ratios of the other line are

$\displaystyle \pm \frac { 1 }{ \sqrt { 2 } } ,0,\mp \frac { 1 }{ \sqrt { 2 } }$

Thus the required angle is

$\displaystyle \cos ^{ -1 }{ \left[ 0\times \left( \pm \frac { 1 }{ \sqrt { 2 } } \right) +\left( \pm \frac { 1 }{ \sqrt { 2 } } \right) \left( 0 \right) +\left( \pm \frac { 1 }{ \sqrt { 2 } } \right) \left( \pm \frac { 1 }{ \sqrt { 2 } } \right) \right] }$

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

If direction cosines of two lines are proportional to

$(2,3,-6)$ and

$(3,-4,5)$ , then the acute angle between them is

0%
$\cos ^{ -1 }{ \left( \cfrac { 49 }{ 36 } \right) }$
0%
$\cos ^{ -1 }{ \left( \cfrac { 18\sqrt { 2 } }{ 35 } \right) }$
0%
${96}^{o}$
0%
$\cos ^{ -1 }{ \left( \cfrac { 18 }{ 35 } \right) }$

Explanation

We know that

$\cos { \theta } =\cfrac { \left| { a }_{ 1 }{ a }_{ 2 }+{ b }_{ 1 }{ b }_{ 2 }+{ c }_{ 1 }{ c }_{ 2 } \right| }{ \sqrt { { a }_{ 1 }^{ 2 }+{ b }_{ 1 }^{ 2 }+{ c }_{ 1 }^{ 2 } } \sqrt { { a }_{ 2 }^{ 2 }+{ b }_{ 2 }^{ 2 }+{ c }_{ 2 }^{ 2 } } }$

$\therefore \cos { \theta } =\cfrac { \left| (2)(3)+(3)(-4)+(-6)(-5) \right| }{ \sqrt { { 2 }^{ 2 }+{ 3 }^{ 2 }+{ \left( -6 \right) }^{ 2 } } \sqrt { { 3 }^{ 2 }+{ \left( -4 \right) }^{ 2 }+{ 5 }^{ 2 } } }$

$=\cfrac { \left| 6-12-30 \right| }{ \sqrt { 4+9+36 } \sqrt { 9+16+25 } }$

$\Rightarrow \cos { \theta } =\cfrac { 36 }{ 7.5\sqrt { 2 } } =\cfrac { 18\sqrt { 2 } }{ 35 }$

$\Rightarrow \theta =\cos ^{ -1 }{ \left( \cfrac { 18\sqrt { 2 } }{ 35 } \right) }$

If the median through

$A$ of a

$\triangle ABC$ having vertices

$A\equiv \left( 2,3,5 \right),$

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

$C\equiv \left( \lambda ,5,\mu \right)$ is equally inclined to the axes, then

0%
$\lambda =7$
0%
$\mu=10$
0%
$\lambda =10$
0%
$\mu=7$

Find the unit vectors perpendicular to the following pair of vectors:

$2i+j+k$ ,

$i-2j+k$

0%
$\displaystyle \frac{1}{\sqrt{35}}(3i-j-5k)$
0%
$\displaystyle \frac{1}{\sqrt{35}}(3i+j-5k)$
0%
$\displaystyle \frac{1}{\sqrt{27}}(i-j-5k)$
0%
$\displaystyle \frac{1}{\sqrt{27}}(i-j+5k)$

Explanation

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

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

Vector perpendicular to Given vectors

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

$\Rightarrow\vec { a } \times \vec { b } = \left| \begin{matrix} \hat i & \hat j & \hat k \\ 2 & 1 & 1 \\ 1 & -2 & 1 \end{matrix} \right|$

$\Rightarrow\vec { a } \times \vec { b } = \hat i(1-(-2))-\hat j(2-1)+\hat k(-4-1)$

$\Rightarrow \vec { a } \times \vec { b } = 3\hat { i } -\hat { j } -\hat { 5k }$

Unit vector required is

$\dfrac { \hat { 3i } -\hat { j } -5\hat { k } }{ \left| 3\hat { i } -\hat { j } -5\hat { k } \right| }$

$=\dfrac { 1 }{ \sqrt { 35 } } (3\hat { i } -\hat { j } -\hat { 5k } )$

If the direction cosines of two lines are given by

$l+m+n=0$ and

$l^2-5m^2+n^2=0$ , then the angle between them is

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

Explanation

$l+m+n=0$

$\Rightarrow l+n=-m$ ...(i)

$l^{2}-5m^{2}+n^{2}=0$

Substituting the equation from eq (i)

$l^{2}+n^{2}-5(-l-n)^{2}=0$

$l^{2}+n^{2}-5(l^{2}+n^{2}+2ln)=0$

$-4l^{2}-4n^{2}-10ln=0$

$2l^{2}+2n^{2}+5ln=0$

$2l^{2}+5ln+2n^{2}=0$

$l=\dfrac{-5n\pm n\sqrt{25-16}}{4}$

$l=\dfrac{-5n\pm3n}{4}$

$l=-2n$ and

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

Hence

$l=\dfrac{-n}{2}, m=\dfrac{-n}{2}$
If

$l=-2n, m=n$

Hence the Dr's are

$(\dfrac{-n}{2},\dfrac{-n}{2},n)$ and

$(-2n,n,n)$

Therefore Dc's are

$(\dfrac{-1}{\sqrt{6}},\dfrac{-1}{\sqrt{6}},\dfrac{2}{\sqrt{6}})$ and

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

Hence

$\cos\theta=(\dfrac{-2}{\sqrt{6}}.\dfrac{-1}{\sqrt{6}})+(\dfrac{-1}{\sqrt{6}}.\dfrac{1}{\sqrt{6}})+(\dfrac{2}{\sqrt{6}}.\dfrac{1}{\sqrt{6}})$

$=\dfrac{2-1+2}{6}$

$=\dfrac{3}{6}$

$=\dfrac{1}{2}$

Hence

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

The vector equation of the line

$\displaystyle L_{1}$ is

$\displaystyle a+\lambda \overline{b}$ then

$\displaystyle \overline{a}$ equals

0%
$\displaystyle -\hat{i}+2\hat{j}-\hat{k}$
0%
$\displaystyle 2\hat{i}-\hat{j}+\hat{k}$
0%
$\displaystyle 2\hat{i}-2\hat{j}+3\hat{k}$
0%
$\displaystyle \hat{i}+2\hat{j}+3\hat{k}$

Explanation

The equations of the given lines in vectors form may be written as

$\displaystyle L_{1}:\overline{r}=\overline{a}+\lambda \overline{b}$ where

$\overline{a}=-\hat{i}-2\hat{j}-\hat{k},\overline{b}=3\hat{i}+\hat{j}+2\hat{k}$

$\displaystyle L_{2}:\overline{r}=\overline{c}+\mu \overline{d}$

Where

$\displaystyle \overline{c} = 2\hat{i} - 2\hat{j} + 3\hat{k}, \overline{d} = 2\hat{i} + 2\hat{j} + 3\hat{k}$

$\overline {n} = \overline {b} \times \overline {b} = \begin {vmatrix}i & j & k\\ 3 &1 & 2 \\1 & 2 & 3 \end{vmatrix} = -\hat{i}-7\hat{j}+5\hat{k}$

Equation of the line passes through the point

$\displaystyle A\left ( \overline{a} \right )$ and parallel to

$\overline{b}$ is given by

$\displaystyle \overline{r}=\overline{a}+\lambda \overline{b}$

Now line

$\displaystyle L_{1}=\dfrac{x+1}{3}=\dfrac{y+2}{1}=\dfrac{z+1}{2}$ can be written as

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

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

which passes through the point whose position vector is

$\displaystyle \overline{a}=2\hat{i}-\hat{j}+\hat{k}$ .

The, position vector of the foot of the

$\perp$ er drawn from origin to the plane is

$\displaystyle 4\hat{i}-2\hat{j}-5\hat{k}$ then equation of the plane is

0%
$\displaystyle \overline{r}.\left ( 4\hat{i}+2\hat{j}+5\hat{k} \right )=45$
0%
$\displaystyle \overline{r}.\left ( 4\hat{i}-2\hat{j}-5\hat{k} \right )=45$
0%
$\displaystyle \overline{r}.\left ( 4\hat{i}-2\hat{j}-5\hat{k} \right )+45=0$
0%
$\displaystyle \overline{r}.\left ( 4\hat{i}+2\hat{j}-5\hat{k} \right )=37$

Explanation

Equation of the plane passes through the point

$B$ with position vector

$\displaystyle 4\hat{i}-2\hat{j}-5\hat{k} \ and \ \perp \ to \ \overline{OB}=\overline{n}$

$\displaystyle \therefore \overline{n}=\overline{OB}=4\hat{i}-2\hat{j}-5\hat{k}$

$\displaystyle \therefore$ Equation of the plane is

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

$\displaystyle \Rightarrow \overline{r} \cdot \left ( 4\hat{i}-2\hat{j}-5\hat{k} \right )=\left ( 4\hat{i}-2\hat{j}-5\hat{k} \right )\cdot \left ( 4\hat{i}-2\hat{j}-5\hat{k} \right )$

$\displaystyle \Rightarrow \overline{r} \cdot \left ( 4\hat{i}-2\hat{j}-5\hat{k} \right )=45$
Hence (B) is correct choice.

If the points

$a(cos \alpha + i sin \alpha)$ ,

$b(cos \beta + i sin \beta)$ and

$c(cos \gamma + isin \gamma)$ are collinear then the value of

$|z|$ is:
( where

${z = bc \ sin(\beta-\gamma) + ca \ sin(\gamma-\alpha) + ab \ sin(\alpha - \beta) + 3i -4k}$ )

0%
$2$
0%
$5$
0%
$1$
0%
None of these.

Explanation

Given

$a=cos\alpha+i\sin\alpha=e^{i\alpha}$ ,

$b=cos\beta+isin\beta=e^{i\beta}$ and

$c=cos\gamma+isin\gamma=e^{i\gamma}$

Consider

$bcsin(\beta-\gamma)=e^{i(\beta+\gamma)}sin(\beta-\gamma)=\frac{1}{2i}e^{i(\beta+\gamma)}(e^{i(\beta-\gamma)}-e^{-i(\beta-\gamma)}) = \frac{1}{2i}(e^{i(2\beta)}-e^{i(2\gamma)})$

Similarly we get

$casin(\gamma-\alpha) = \frac{1}{2i}(e^{i(2\gamma)}-e^{i(2\alpha)})$ and

$absin(\alpha-\beta) = \frac{1}{2i}(e^{i(2\alpha)}-e^{i(2\beta)})$

Therefore we get

$bcsin(\beta-\gamma)+casin(\gamma-\alpha)+absin(\alpha-\beta)=0$

So we get

$z=3i-4i$

$\Rightarrow |z|=5$

The angle between two diagonals of a cube is

0%
$\cos^{-1}\left (\dfrac {1}{\sqrt {3}}\right )$
0%
$\cos^{-1}\left (\dfrac {1}{3}\right )$
0%
$30^{\circ}$
0%
$45^{\circ}$

Explanation

Let

$OABCDEFG$ be a cube with vertices as below

$O(0,0,0), A(a,0,0), B(a,a,0), C(0,a,0), D(0,a,a), E(0,0,a), F(a,0,a), G(a,a,a)$

There are four diagonals

$OG,CF,AD$ and

$BE$ for the cube.

Let us consider any two say

$OG$ and

$AD$

We know that if

$A(x_1,y_1,z_1)$ and

$B(x_2,y_2,z_2)$ are two points in space then

$\vec{AB}=(x_2-x_1)i + (y_2-y_1)j + (z_2-z_1)k$

$\Rightarrow \vec{OG}=(a-0)i+(a-0)j+(a-0)k=ai+aj+ak$

and

$\vec{AD} = (0-a)i+(a-0)j+(a-0)k=-ai+aj+ak$ .

Therefore

$|\vec{OG}|=\sqrt{a^2+a^2+a^2}=a\sqrt{3}$ and

$|\vec{AD}|=\sqrt{(-a)^2+a^2+a^2}=a\sqrt{3}$

$\vec{OG} \cdot \vec{AD}=-a^2+a^2+a^2=a^2$

We know that angle between two vectors

$\vec{a},\vec{b}$ is given by

$\theta=cos^{-1}\dfrac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|}$

Thus angle between two diagonals

$\vec{OG}$ and

$\vec{AD}$ is

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

Equation of the plane through the mid-point of the line segment joining the points

$P(4, 5, -10), \,Q(-1, 2, 1)$ and perpendicular to

$PQ$ is

0%
$r. \left( \dfrac{}3{}2 \widehat i + \dfrac{7}{2} \widehat j - \dfrac{9}{2} \widehat k\right ) =45$
0%
$r. (- \widehat i + 2 \widehat j + \widehat k) = \dfrac{135}{2}$
0%
$r. (5 \widehat i + 3 \widehat j - 11 \widehat k) + \dfrac{135}{2} = 0$
0%
$r. (4 \widehat i + 5 \widehat j - 10 \widehat k) = 85$
0%
$r. (5\widehat i + 3 \widehat j - 11\widehat k) = \dfrac{135}{2}$

Explanation

Coordinate the mid-point of

$P(4, 5, -10)$ and

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

$\left( \dfrac{4 - 1}{2}, \dfrac{5 + 2}{2}, \dfrac{-10 + 1}{2}\right )$ . i.e.

$\left ( \dfrac{3}{2}. \dfrac{7}{2} , \dfrac{-9}{2} \right )$
Now, DR's of

$PQ$ is

$(-1 -4, 2 -5, 1 + 10)$
i.e.,

$(-5, -3, 11)$ or

$(5, 3, -11)$
Therefore, euation of plane passing through

$\left( \dfrac{3}{2}, \dfrac{7}{2}, \dfrac{-9}{2} \right )$ and having DR's

$(5, 3, -11)$ is

$5 \left( x - \dfrac{3}{2} \right ) + 3 \left( y - \dfrac{7}{2} \right ) - 11 \left( z - \dfrac{9}{2} \right ) = 0$

$\Rightarrow 5x + 3y - 11z = \dfrac{15}{2} + \dfrac{21}{2} + \dfrac{99}{2}$

$\Rightarrow 5x + 3y - 11z = \dfrac{135}{2}$
It is written by vector form

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

$(2, -1, 3)$ is the foot of the perpendicular drawn from the origin to the plane, then the equation of the plane is

0%
$2x + y - 3z + 6 = 0$
0%
$2x - y + 3z - 14 = 0$
0%
$2x - y + 3z - 13 = 0$
0%
None of these

Explanation

Let

$P(2,-1,3)$ be the foot of the perpendicular from the origin to the plane.

$\therefore \vec{OP}=\vec n=2\vec i-\vec j+3\vec k$ and

$p=|2\vec i-\vec j+3\vec k|=\sqrt{2^2+1^2+3^2}=\sqrt{4+1+9}=\sqrt{14}$

$\therefore \hat n=\dfrac{2\vec i-\vec j+3\vec k}{\sqrt{14}}$

The vector equation of the plane is

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

$\Rightarrow (x\vec i+y\vec j+z\vec k)\cdot \dfrac{2\vec i-\vec j+3\vec k}{\sqrt{14}}=\sqrt{14}$

$\Rightarrow (x\vec i+y\vec j+z\vec k)\cdot (2\vec i-\vec j+3\vec k)=14$

$\Rightarrow 2x-y+3z=14$

$\Rightarrow 2x-y+3z-14=0$

The equation of the line parallel to

$\cfrac { x-3 }{ 1 } =\cfrac { y+3 }{ 5 } =\cfrac { 2z-5 }{ 3 }$ and passing through the point

$(1,3,5)$ in vector form, is:

0%
$\vec { r } =\left( \vec { i } +5\vec { j } +3\vec { k } \right) +t\left( \vec { i } +3\vec { j } +5\vec { k } \right)$
0%
$\vec { r } =\left( \vec { i } +3\vec { j } +5\vec { k } \right) +t\left( \vec { i } +5\vec { j } +3\vec { k } \right)$
0%
$\vec { r } =\left( \vec { i } +5\vec { j } +\cfrac { 3 }{ 2 } \vec { k } \right) +t\left( \vec { i } +3\vec { j } +5\vec { k } \right)$
0%
$\vec { r } =\left( \vec { i } +3\vec { j } +5\vec { k } \right) +t\left( \vec { i } +5\vec { j } +\cfrac { 3 }{ 2 } \vec { k } \right)$

Explanation

We have to find the equation of the line parallel to

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

$(1,3,5)$ vector form.

We know that the equation of the line passing through the point with position vector

$\vec{a}$ and parallel to the vector

$\vec{b}$ is

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

Consider

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

Rewriting we get

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

Thus vector representation is

$\vec{b}=\vec{i}+5\vec{j}+\dfrac{3}{2}\vec{k}$

Since line passes through

$(1,3,5)$ ,

$\vec{a}=\vec{i}+3\vec{j}+5\vec{k}$

Hence the required equation of the line is

$\vec{r}=(\vec{i}+3\vec{j}+5\vec{k})+t\left(\vec{i}+5\vec{j}+\dfrac{3}{2}\vec{k}\right)$

What is the angle between the lines

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

$\cfrac { x-1 }{ 1 } =\cfrac { 2y+3 }{ 3 } =\cfrac { z+5 }{ 2 } =?$

0%
$\cfrac { \pi }{ 2 }$
0%
$\cfrac { \pi }{ 3 }$
0%
$\cfrac { \pi }{ 6 }$
0%
None of the above

Explanation

The direction ratios of the line are:

$B=(a,b,c) = (1,-2,1)$

The normal to the plane has the coordinates:

$N=(-2, 1, 2)$

Let the angle between the line and the plane be

$'x'$

$\therefore, sinx= \dfrac{(B.N)}{( |B|\times|N| )}$

By substuting the values we get

$sin x =1$

$i.e.x=\dfrac {\pi}{2}$ .

A plane mirror is placed at the origin so that the direction ratios of its normal are

$(1,-1,1)$ . A ray of light, coming along the positive direction of the x-axis, strikes the mirror. The direction cosines of the reflected ray are

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

Explanation

Let the source of light be situated at A (a,0,0) where a is not equal to 0.

let OA be the incident ray and OB the reflected ray

ON is the normal to the mirror at O

D.R's of OA are

$0,0,1$

and its D.R's are

$1,0,0$

D.R's of ON are

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

therefore cos

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

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

let l,m,n be the D.R's of the reflected ray OA then,

$l$ =

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

$m$ =

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

$n$ =

$\dfrac {2}{ 3}$

$\dfrac {-1}{ 3}$ , m =

$\dfrac {-2}{ 3},$ n =

$\dfrac {2}{3}$

Hence the D.R's of the reflected ray are

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

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

$\dfrac {2}{3}$ .

The equation of the plane passing through (1, -2, 4), (3, -4, 5) and perpendicular to yz-plane is.

0%
2y + z = 0
0%
y + 2y + 6 = 0
0%
y + 2y - 6 = 0
0%
3y + 2z - 2 = 0

$L_{1}$ and

$L_{2}$ are two lines whose vector equations are

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

$L_{2} = \vec {r} = \mu (a\hat {i} + b\hat {j} + c\hat {k})$ , 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%
$\dfrac {\pi}{6}$
0%
$\dfrac {\pi}{4}$
0%
$\dfrac {\pi}{3}$
0%
None of these

If points

$P\left( 4,5,x \right) ,Q\left( 3,y,4 \right)$ and

$R\left( 5,8,0 \right)$ are colinear, then the value of

$x+y$ is

0%
$-4$
0%
$3$
0%
$5$
0%
$4$

Explanation

$\vec{PR}=(5-4)\hat{i}+(8-5)\hat{j}+(0-x)\hat{k}$

$\implies \vec{PR}=\hat{i}+3\hat{j}-x\hat{k}$

Again,

$\vec{QR}=(5-3)\hat{i}+(8-y)\hat{j}+(0-4)\hat{k}$

$\implies \vec{QR}=2\hat{i}+(8-y)\hat{j}-4\hat{k}$

Since, P,Q and R are co-linear points, hence

$\dfrac{1}{2}=\dfrac{3}{8-y}=\dfrac{-x}{-4}$

On Solving, we get:-

$x=2,y=2$

$\implies x+y=4$

Hence, answer is option-(D).

$\alpha,\beta,\gamma\in[0,2\pi]$ , then the sum of all possible values of

$\alpha, \beta,\gamma$ if

$\sin \alpha=-\dfrac{1}{\sqrt{2}}$ ,

$\cos \beta=-\dfrac{1}{2}$ ,

$\tan \gamma=-\sqrt{3}$ , is

0%
$\dfrac{22\pi}{3}$
0%
$\dfrac{21\pi}{3}$
0%
$\dfrac{20\pi}{3}$
0%
$8\pi$

Explanation

$\sin\alpha = \dfrac{-1}{\sqrt2}$

(-ve) value lies on III, IV quadrant for sin

$\therefore \alpha = \pi + \dfrac{\pi}{4}, 2\pi - \dfrac{\pi}{4}$

$= \dfrac{5\pi}{4}, \dfrac{7\pi}{4}$

$\cos\beta = \dfrac{-1}{2}$

(-ve) value lies on II, III quadrant for cosine

$\beta = \pi - \dfrac{\pi}{3}, \pi + \dfrac{\pi}{3}$

$= \dfrac{2\pi}{3}, \dfrac{4\pi}{3}$

$\tan\gamma = -\sqrt3$

(-ve) value lies on II, IV quadrant for tan :

$\gamma = -\sqrt3$

$\gamma = \pi - \dfrac{\pi}{3}, 2\pi - \dfrac{\pi}{3}$

The sum of all values

$(\alpha + \beta + \gamma)= \dfrac{2\pi}{3}, \dfrac{5\pi}{3}$

$= \dfrac{5\pi}{4} + \dfrac{7\pi}{4} + \dfrac{2\pi}{3} + \dfrac{4\pi}{3} + \dfrac{2\pi}{3} + \dfrac{7\pi}{3}$

$= 3\pi + 2\pi + \dfrac{7\pi}{3}$

$= 5\pi + \dfrac{7\pi}{3}$

$= \dfrac{22\pi}{3}$

$If\,\ell ,\,m,\,n\,\& \,\ell ',\,m',\,n'\,$ be the cosine of two lines which include then

0%
$\cos \,\theta = \ell \ell ' + mm' + nn'$
0%
$\sin \,\theta = \ell \ell ' + mm' + nn'$
0%
$\cos \,\theta \, = mn' + m'n + n\ell ' + n'\ell + \ell m' + \ell 'm$
0%
$\sin \,\theta \, = mn' + m'n + n\ell ' + n'\ell + \ell m' + \ell 'm$

$\Delta ABC, |\bar{CB}| = a, |\bar{CA}| = b, |\bar{AB}| = c$ .

$CD$ is median through the vertex

$C$ . Then

$\bar{CA}.\bar{CD}$ equals.

0%
$\dfrac{1}{4}(3a^2 + b^2 - c^2)$
0%
$\dfrac{1}{4}(a^2 + 3b^2 - c^2)$
0%
$\dfrac{1}{4}(a^2 + b^2 - 3c^2)$
0%
$\dfrac{1}{4}(-3a^2 + b^2 - c^2)$

Explanation

$\overrightarrow{CD}=\overrightarrow{CA}+\overrightarrow{AD}$

$\overrightarrow{AD}=\dfrac{1}{2}\overrightarrow{AB}$

$\overrightarrow{CD}=\overrightarrow{CA}+\dfrac{1}{2}\overrightarrow{AB}$

$\overrightarrow{CD}.\overrightarrow{CA}=\left(\overrightarrow{CA}+\dfrac{1}{2}\overrightarrow{AB}\right)\overrightarrow{CA}$

$|\overrightarrow{CA}|^2+\dfrac{1}{2}\overrightarrow{AB}.\overrightarrow{CA}$

$b^2+\dfrac{1}{2}\overrightarrow{AB}.\overrightarrow{CA}$

$(|\overrightarrow{CA}|=b)$

$=\overrightarrow{b}+\dfrac{1}{2}|\overrightarrow{AB}||\overrightarrow{CA}|\cos (\pi-A)$

$b^2-\dfrac{1}{2}bc\left(\dfrac{b^2+c^2-a^2}{2ac}\right)$

$b^2-\dfrac{b^2}{4}-\dfrac{c^2}{4}+\dfrac{a^2}{4}$

$\dfrac{3b^2}{4}-\dfrac{c^2}{4}+\dfrac{a^2}{4}$

$\dfrac{1}{4}(a^2+3b^2-c^2)$

2114514_1061101_ans_638f198bedf1447e82112bbeb19c78d0.png

The equation of plane passing through a point

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

$a= (3, 0, - 1)$ and

$b=(- 3, 2, 2)$ is:

0%
$2x - 3y + 6z - 25 = 0$
0%
$2x - 3y + 6z + 25 = 0$
0%
$3x - 2y + 6z - 25 = 0$
0%
$3x - 2y + 6z + 25 = 0$

Explanation

Let the DR of the normal of required plane be

$<a, b, c>$

Since plane is parallel to

$(3, 0, -1)$

$\therefore$ normal must be perpendicular

$\therefore 3a + 0b - c = 0$ ...

$(1)$

Also, it is parallel to

$(-3, 2, 2)$

$\therefore -3a + 2b + 2c = 0$ ...

$(2)$

From

$(1)$ and

$(2)$

$\dfrac{a}{2} = \dfrac{-b}{6 - 3} = \dfrac{c}{6}$

$a = 2, b = -3, c = 6$

It passes through

$(2, -1, 3)$

$2(x - 2) - 3(y + 1) + 6(z - 3) = 0$

$\Rightarrow 2x - 4 - 3y - 3 + 6z - 18 = 0$

$\Rightarrow 2x - 3y + 6z - 25 = 0$

If a = 4i + 3j and b be two vectors perpendicular to each other on the xy- plane. Then, a vector in the same plane having projections 1 and 2 along a and b respectively, is

0%
i + 2j
0%
2i - j
0%
2i + j
0%
None of these

Explanation

Let

$\vec c$ be the required vector

Given that

$\vec a =4\hat i+3\hat j$

Since, you have not given

$\vec b$ , it is assumed to be

$=-3\hat i+4\hat j$

Projection of

$\vec c \theta n \vec a=\dfrac {\vec c.\vec a}{|\vec a|}=1$

$\boxed {\vec c.\vec a=5}$

Probability of

$\vec c$ on

$\vec b=\dfrac {\vec c.\vec b}{(\vec b)}=2, \vec c.\vec b=10$

Let

$\vec c=n\hat i+y\hat j$

$\vec c.\vec b=100, -3x+y=10$

$x=\dfrac {-2}{5}, y=\dfrac {11}{5}$

$\boxed {\vec =\dfrac {-2}{5}\hat i+\dfrac {11}{5}\hat j}$

If a line makes angle

$90^{o},135^{o},45^{o}$ with the

$X-$ ,

$Y-$ and

$Z-$ axes respectively, then its direction cosines are

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

In the isosceles

$\triangle$ ABC,

$|AB| = |BC|=8$ and point E divides AB internally in the ratio 1 : 3 then the cosine of angel between CE and CA is (where, |CA| = 12) ?

0%
$-\dfrac { \sqrt [ 3 ]{ 7 } }{ 8 }$
0%
$\dfrac { \sqrt [ 3 ]{ 8 } }{ 17 }$
0%
$\dfrac { \sqrt [ 3 ]{ 7 } }{ 8 }$
0%
$-\dfrac { \sqrt [ 3 ]{ 8 } }{ 17 }$

The direction cosines of the line

$x=44z+3; y=2-2\sqrt{19}z$ is...

0%
$\dfrac{44}{\sqrt{2013}};\dfrac{-2\sqrt{19}}{\sqrt{2013}};\dfrac{1}{\sqrt{2013}}$
0%
$\dfrac{-44}{\sqrt{2013}};\dfrac{2\sqrt{19}}{\sqrt{2013}};\dfrac{1}{\sqrt{2013}}$
0%
$\dfrac{44}{\sqrt{2013}};\dfrac{2\sqrt{19}}{\sqrt{2013}};\dfrac{1}{\sqrt{2013}}$
0%
$\dfrac{-44}{\sqrt{2013}};\dfrac{-2\sqrt{19}}{\sqrt{2013}};\dfrac{-1}{\sqrt{2013}}$

The distance of the point

$3\hat {i}+5\hat {k}$ from the line parallel to

$6\hat {i}+\hat {j} 2\hat {k}$ and passing through the point

$8\hat {i}+3\hat {j}+\hat {k}$ is

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

Three points whose position vectors are

$x\bar{i}+y\bar{j}+z\bar{k}$ ,

$\bar{i}+2\bar{j}$ and

$-\bar{i}-\bar{j}$ are collinear, then relation between

$x, y, z$ is?

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

A line makes equal angles with the coordinate axis. The direction cosines of this line are

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

Explanation

Let the direction cosines of the line make an angle α with each of the coordinate axes.

$∴l=cosα,m=cosα,n=cosα$

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

${ cos }^{ 2 }α+{ cos }^{ 2 }α+{ cos }^{ 2 }α=1$

$3{ cos }^{ 2 }α=1$

${ cos }^{ 2 }α=\frac { 1 }{ 3 }$

$cosα=±\frac { 1 }{ \sqrt { 3 } }$

Thus,the direction cosines of the line,which is equally inclined to the coordinate axes,are

$±\frac { 1 }{ \sqrt { 3 } } ,±\frac { 1 }{ \sqrt { 3 } } ,and±\frac { 1 }{ \sqrt { 3 } } .$

If vector

$\overrightarrow{a}+\overrightarrow{b}$ bisects the between

$\overrightarrow{a}$ and

$\overrightarrow{b}$ , then

$\overrightarrow|{a}|=\overrightarrow|{b}|$ .

0%
True
0%
False

The direction Ratio's of normal of the plane through

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

$\pi/4$ with plane

$x+y =3$ are

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

A line passes through the point

$(6, -7, -1)$ and

$(2, -3, 1)$ . Then the sum of the direction cosines of the line, if the line makes acute angle with positive direction of x-axis, is?

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

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

$3,-2,4$ and

$1,3,-2$

0%
$2,-5,6$
0%
$4,-10,12$
0%
$-8,10,11$
0%
$-8,10,-11$

CBSE Questions for Class 12 Commerce Maths Three Dimensional Geometry Quiz 10 - 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 10 - MCQExams.com

0:0:2

Practice Class 12 Commerce Maths Quiz Questions and Answers

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