MCQ Questions for Class 12 Commerce Maths Three Dimensional Geometry Quiz 2

The projection of the join of the two points

$(1,4,5), (6,7,2)$ on the line whose d.s's are

$(4,5,6)$ is

0%
$\dfrac{{17}}{{\sqrt {77} }}$
0%
$\dfrac{7}{6}$
0%
$21$
0%
$\dfrac{7}{9}$

Explanation

$A(1,4,5)$ ,

$B(6,7,2)$

$\begin{array}{l}\overline {AB} = \,(6 - 1)i + (7 - 4)j + (2 - 5)\widehat k\\\overline {AB} = 5\widehat i + 3\widehat j - 3\widehat k\\\overrightarrow c = 4i + 5j + 6\widehat k\end{array}$

$\begin{array}{l} \text{proj. AB} = \dfrac{{\overline {AB} .\overline C }}{{\overline {\left| c \right|} }} = \dfrac{{5(4) + 3(5) + ( - 3)(6)}}{{\sqrt {{4^2} + {5^2} + {6^2}} }}\\\dfrac{{20 + 15 - 18}}{{\sqrt {16 + 25 + 36} }} = \dfrac{{17}}{{\sqrt {77} }}\end{array}$

If the

$d.c's$ of two lines are connected by the equations

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

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

Explanation

Given equation are,

$\begin{array}{l} l+m+n=0----------(1) \\ { l^{ 2 } }+{ m^{ 2 } }+{ n^{ 2 } }=0---------(2) \\ from\, \, \, \, equ\, \, (1)\, ,\, \, \, l=-\, \, (m+n) \\ value\, \, of\, \, \, l....... \\ substituting\, \, in\, \, (2) \\ { (m+n)^{ 2 } }+{ m^{ 2 } }-{ n^{ 2 } }=0 \\ \Rightarrow { m^{ 2 } }+{ n^{ 2 } }+2mn+{ m^{ 2 } }-{ n^{ 2 } }=0 \\ \Rightarrow 2{ m^{ 2 } }+2mn=0 \\ \Rightarrow 2m(m+n)=0 \\ \Rightarrow m=0\, \, \, \, and\, \, \, (m+n)=0 \\ Now, \\ m=0\, ,\, substituting\, in\, \, (1) \\ l+m+n=0 \\ l+n=0,\Rightarrow l=-n \\ Dr.s\, of\, the\, \, first\, line\, are\, \, (1,0,-1) \\ and, \\ m+n=0\, \, \, \Rightarrow \, m=-n \\ substituting\, \, \, in\, \, (1)\, \\ l+m+n=0 \\ l=0 \\ D'rs\, of\, the\, \, { { second } }\, \, line\, are\, \, (0,1,-1) \end{array}$

$\begin{array}{l} Lets\, \, \theta \, be\, the\, angle\, \, between\, the\, \, line, \\ \cos \theta =\dfrac { { { a_{ 1 } }{ a_{ 2 } }+{ b_{ 1 } }{ b_{ 2 } }+{ c_{ 1 } }{ c_{ 2 } } } }{ \begin{array}{l} \sqrt { { a_{ 1 } }^{ 2 }+{ b_{ 1 } }^{ 2 }+{ c_{ 1 } }^{ 2 } } \, \, \sqrt { { a_{ 2 } }^{ 2 }+{ b_{ 2 } }^{ 2 }+{ c_{ 2 } }^{ 2 } } \\ =\dfrac { { 0+0+1 } }{ { \sqrt { 2 } .\sqrt { 2 } } } =\frac { 1 }{ 2 } \\ \, \cos \theta =\dfrac { 1 }{ 2 } \\ \therefore \, \, \, \theta =\dfrac { \pi }{ 3 } \, \end{array} } \end{array}$

so that the correct option is A.

The direction cosines of a line which is equally inclined to axes, is given by

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

Explanation

Consider the problem

Let,

$l=m=n=\cos \alpha$

then,

$l^2+m^2+n^2=\cos ^2\alpha +\cos ^2\alpha+\cos ^2\alpha$

$1=3\cos ^2 \alpha$

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

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

thus,

$l=m=n=\pm\dfrac{1}{\sqrt 3}$

The equation of the plane passing through

$(2, -3, 1)$ and is normal to the line joining the points

$(3, 4, -1)$ and

$(2, -1, 5)$ is given by

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$x+5y-6z+19=0$
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$x-5y + 6z-19=0$
0%
$x+5y+6z+19=0$
0%
$x-5y-6z-19=0$

Explanation

$A(3,4,-1)$ and

$B(2,-1,5)$

$\overrightarrow { AB } =-\hat { i } -5\hat { j } +6\hat { k }$

$\overrightarrow { AB }$ is normal to the required plane

$\Rightarrow$ Directions of normal

$(-1,-5,6)$

$\therefore$ Equation ,

$-x-5y+6z=k$

Point

$(2,-3,1)$ passes through the plane,

$\therefore -2+15+6=k\quad \Rightarrow k=19\\ \therefore -x-5y+6z=19\\ x+5y-6z+19=0$

The equation of the plane containing the line

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

0%
$\vec r.\left( {\hat i + 2\hat j - \hat k} \right) = 3$
0%
$\vec r.\left( {\hat i + 2\hat j - \hat k} \right) = 6$
0%
$\vec r.\left( { - \hat i - 2\hat j + \hat k} \right) = 3$
0%
None of these

Explanation

The line

$\vec r = \hat i + \hat j + t(2\hat i + \hat j + 4\hat k)$

passes through the point

$(1,1,0)$ and parallel to vector

$\vec b = 2\hat i + \hat j + 4\hat k$

taking plane

$\vec r \cdot (\hat i + 2\hat j - \hat k) = 3$ (i)

putting

$\vec r =( \hat i + \hat j + 0\hat k)$ in equation (i)

$(\hat i + \hat j + 0\hat k) \cdot (\hat i + 2j\hat k - k) = 3$

$1+2+0=3$

$3=3$

therefore the plane (1) contains point

$(1,1,0)$

also

$2 \times 1 + 1 \times 2 + 4 \times ( - 1)$

$=2+2-4$

$=4-4$

$=0$

i.e perpendicular to plans(1) is perpendicular to the given line

therefore the plane is

$\vec r \cdot (\hat i + 2\hat j - \hat k) = 3$

A line passes through the points

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

$(2, -3, 1)$ . The direction cosines of the line so directed that the angle made by it with the positive direction of x-axis is acute, is?

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

Explanation

Consider the problem

Let

$l,m,n$ are direction cosines of the given line.

then as it made an acute angle with

$x-axis$ ,

Therefore,

$l>0$

The line passes through

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

$(2,-3,1)$

Therefore, its direction ratios are

$6-2,-7+3,-1-1$ or

$2,-2,-1$

Hence direction cosines of the line are given by

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

$P$ be the point

$(2,6,3)$ then the equation of the plane trough

$P$ , at right angles to

$OP$ , where

$'O'$ is the origin is

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$2x+6y+3z=7$
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$2x-6y+3z=7$
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$2x+6y-3z=49$
0%
$2x+6y+3z=49$

Explanation

$P(2,6,3)$ ,

$\overrightarrow { OP } =2\hat { i } +6\hat { j } +3\hat { k }$ (direction of normal

$\therefore$ Equation of plane

$\Rightarrow \overrightarrow { r } .\overrightarrow { n } =\overrightarrow { p } .\overrightarrow { n } \\ \overrightarrow { r } .(2\hat { i } +6\hat { j } +3\hat { k } )=49$

$\therefore 2x+6y+3z=49$ -Equation of Plane.

The equation of the plane passing through

$(a,b,c)$ and parallel to the plane

$r.(\hat{i}+\hat{j}+\hat{k})=2$ is,

0%
$x+y+z=1$
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$ax+by+cz=1$
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$x+y+z=a+b+c$
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None of these

Explanation

Let any point on the plane be

$(x,y,z)$ then,

$1(x-a)+1(y-b)+1(z-c)=0\implies x+y+z=a+b+c$

Angle between the lines

$3x=6y=2z$ and

$3x+2y+z-5=0=x+y-2z-3$ is?

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

Explanation

The given two planes have a line of intersection,

whose equation can be found by,

$\begin{array}{l} \left| { \begin{array} { *{ 20 }{ c } }{ \widehat { i } } & { \widehat { j } } & { \widehat { k } } \\ 3 & 2 & 1 \\ 1 & 1 & { -2 } \end{array} } \right| \\ =\widehat { i\, } (-4-1)-\widehat { j } (-6-1)+\widehat { k } (3-2) \\ =-5\widehat { i } +7\widehat { j } +\widehat { k } \end{array}$

Given line is,

$\begin{array}{l} 3x=6y=2z \\ \frac { x }{ 2 } =y=\frac { z }{ 3 } \end{array}$

so, its vector equation is

$2\widehat i + \widehat j + 3\widehat k$ .

Angle between line is cos

$\theta$ =

$\frac{{( - 5)(2) + 7(1) + 1(3)}}{{\sqrt {25 + 49 + 1} \,\,\,\sqrt {4 + 1 + 9} }}$

$\frac{0}{{\sqrt {25 + 49 + 1} \,\,\,\sqrt {4 + 1 + 9} }}$

cos

$\theta$ = 0

Therefor

$\theta$ =

$\frac{\pi }{2}$

So that the correct Option is D

The points

$i + j + k, \, i + 2j, \, 2i+2j+k,\, 2i+3j+2k$ are

0%
collinear
0%
coplanar but not collinear
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non-coplanar
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none

Explanation

$\begin{matrix} A& B& C& D\\i+j+k, &i+2j, &2i+2j+k,&2i+3j+2k \end{matrix}$

$\overline{AC} = (2-1)i + (2-1)j + k-k$

$=i+j$

$\overline{AB} = o + j - \overline{k} = j - \overline{k}$

$\overline{AD} = i + 2j + k$

$\begin{vmatrix} 1&1&0 \\0&2 &1\end{vmatrix} = 1(1+2)-1(0+1)$

$=3-1 = 2 \neq 0$
Non coplanar.

If the dr's the line are

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

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

$\lambda$ is

0%
$1\pm \sqrt{3}$
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$4\pm \sqrt{5}$
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$2\pm 2\sqrt{3}$
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$2\pm \sqrt{5}$

Explanation

Given that,

$\begin{array}{l} \frac { { \left| { 1-\lambda } \right| } }{ { \sqrt { { { \left( { 1+\lambda } \right) }^{ 2 } }+{ { \left( { 1-\lambda } \right) }^{ 2 } }+4 } } } =\frac { 1 }{ 2 } \\ \frac { { \left| { 1-\lambda } \right| } }{ { \sqrt { 6+2{ \lambda ^{ 2 } } } } } =\frac { 1 }{ 2 } \\ 4{ \left( { \left| { 1-\lambda } \right| } \right) ^{ 2 } }=6+2{ \lambda ^{ 2 } } \\ 2{ \lambda ^{ 2 } }-2-8\lambda =0 \\ { \lambda ^{ 2 } }-4\lambda -1=0 \\ \lambda =\frac { { 4\pm \sqrt { 16+4 } } }{ 2 } =\frac { { 4\pm 2\sqrt { 5 } } }{ 2 } \\ =2\pm \sqrt { 5 } \end{array}$

Then,

We get

$2 \pm \sqrt 5$

Option

$D$ is correct answer.

The line joining the points

$(-2, 1, -8)$ and

$(a, b, c)$ is parallel to the line whose direction ratios are

$6, 2, 3$ . The value of

$a, b, c$ are

0%
$(4, 3, -5)$
0%
$(1, 2, -13/2)$
0%
$(10, 5, -2)$
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$(-5, 3, 4)$

Explanation

If lines are parallel

then their direction

ratios must be equal

$\Rightarrow a+2=6$

$a=4$

$b-1=2$

$b=3$

$c+8=3$

$c=-5$

$\therefore (a,b,c)$ is

$(4,3,-5)$

1127702_1202205_ans_da2322952e8241f79855189e093fc921.jpg

If a line has the direction ratio

$18, 12, 4$ , then its direction cosines are:

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$\dfrac{9}{11}$ , $\dfrac{6}{11}$ , $\dfrac{2}{11}$
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$\dfrac{9}{13}$ , $\dfrac{6}{13}$ , $\dfrac{2}{13}$
0%
$\dfrac{9}{7}$ , $\dfrac{6}{7}$ , $\dfrac{2}{7}$
0%
None of these

Explanation

Dr's of the line are : 18, 12, 4

Dc's =

$\dfrac{18}{\sqrt{18^2+12^2+4^2}}, \dfrac{12}{\sqrt{18^2+12^2+4^2}}, \dfrac{4}{\sqrt{18^2+12^2+4^2}}$

$=\dfrac{18}{22}, \dfrac{12}{22}, \dfrac{4}{22}$

$=\dfrac{9}{11}, \dfrac{6}{11}, \dfrac{2}{11}$

Equation of a line passing through the point

$\hat{i}+\hat{j}-\hat{k}$ and parallel to the vector

$2\hat{i}+\hat{j}{+}2\hat{k}$ is

0%
${\vec{r}}=(1+2{t})\hat{i}+(1+{t})\hat{j}{+}(1+2t)\hat{k}$
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${\vec{r}}=(1+2{t})\hat{i}+(1+{t})\hat{j}{+}(2{t}-1)\hat{k}$
0%
${\vec{r}}=(2{t}-1)\hat{i}+(1+{t})\hat{j}{+}(2{t}+1)\hat{k}$
0%
${\vec{r}}=(1-2{t})\hat{i}+(1+{t})\hat{i}{+}(2{t}+1)\hat{k}$

Explanation

The required equation of the line in Cartesian format will be

$\dfrac{x-1}{2}=\dfrac{y-1}{1}=\dfrac{z+1}{2}=t$
Hence, any point on the line can be written as

$x=2t+1$ ,

$y=t+1$ and

$z=2t-1$
Hence in vector format,

$\vec{r}=xi+yj+zk$

$=(2t+1)i+(t+1)j+(2t-1)k$

A line with positive direction cosines passes through the point

$P(2, -1, 2)$ and makes equal angles with the coordinate axes. The line meets the plane

$2x + y + z = 9$ at point

$Q$ . The length of the line segment

$PQ$ equals

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

Explanation

$D.\mathrm{C}$ of the line are $\displaystyle \dfrac{1}{\sqrt{3}}$ , $\displaystyle \dfrac{1}{\sqrt{3}}$ , $\displaystyle \dfrac{1}{\sqrt{3}}$

Any point on the line at a distance $t$ from ${P}(2, -1,2)$ is $\left (2+\displaystyle \dfrac{{t}}{\sqrt{3}}, -1+\dfrac{{t}}{\sqrt{3}}, 2+\dfrac{{t}}{\sqrt{3}}\right)$

which lies on $2x+y+z=9$ .

$\Rightarrow t=\sqrt 3$

Assertion (

$A$ ): The points with position vectors

$\overline{a},\overline{b},\overline{c}$ are collinear if

$2\overline{a}-7\overline{b}+5\overline{c}=0$ .
Reason (

$R$ ): The points with position vectors

$\overline{a},\overline{b},\overline{c}$ are collinear if

$l\overline{a}+m\overline{b}+n\overline{c}=\overline{0}$ .

0%
Both $A$ and $R$ are true and $R$ is correct reason of $A$
0%
Both $A$ and $R$ are true and $R$ is not correct reason of $A$
0%
$A$ is true $R$ is false
0%
$A$ is false $R$ is true

Explanation

$\overrightarrow { a } ,\overrightarrow { b } ,\overrightarrow { c }$ are collinear

$\Rightarrow \ni l,m,n$ all zeros such that

$l\overrightarrow { a } +m\overrightarrow { b } +n\overrightarrow { c } =0$ if

$2\overrightarrow { a } -7\overrightarrow { b } +5\overrightarrow { c } =0$

then

$\overrightarrow { a } ,\overrightarrow { b } ,\overrightarrow { c }$ are collinear since

$2-7+5=0$

If the points with position vectors

$60\hat{i}+3\hat{j},40\hat{i}-8\hat{j}$ and

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

$a$ is equal to

0%
$-40$
0%
$-20$
0%
$20$
0%
$40$

The three points

$ABC$ have position vectors

$(1,x,3),(3,4,7)$ and

$(y,-2,-5)$ are collinear then

$(x,y)=$

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

Explanation

$(1,x,3)=\lambda(3,4,7) + \mu (y,-2,-5)$

$1=3\lambda +\mu y$

$x= 4\lambda +(-2\mu)$

$3 = 7\lambda -5 \mu$

$2-x= (-3-y)\mu$
So only

$x=2$ ,

$y=-3$

$\overline{a}$ and

$\overline{b}$ are two non-collinear vectors, then the points

$l_{1}\overline{a}+m_{1}\overline{b}$ ,

$l_{2}\overline{a}+m_{2}\overline{b}$ and

$l_{3}\overline{a}+m_{3}\overline{b}$ are collinear if

0%
$\displaystyle \sum l_{1}(m_{2}-m_{3}) =0$
0%
$\displaystyle \sum l_{1}(m_{1}-m_{3})=0$
0%
$\displaystyle \sum l_{1}(m_{1}+m_{3})=0$
0%
$\displaystyle \sum l_{1}(m_{2}+m_{3})=0$

Explanation

$\bar { a }$ and

$\bar { b }$ are non-collinear

$\triangle \quad =\quad \left| \begin{matrix} 1\quad & 1\quad & 1 \\ { l }_{ 1 }\quad & { l }_{ 2 }\quad & { l }_{ 3 } \\ { m }_{ 1 }\quad & { m }_{ 2 }\quad & { m }_{ 3 } \end{matrix} \right| =0$

${ c }_{ 1 }\rightarrow { c }_{ 1 }-{ c }_{ 2 }\\ { c }_{ 3 }\rightarrow { c }_{ 1 }-{ c }_{ 3 }\\ \triangle \quad =\quad \left| \begin{matrix} 0 & 1 & 0 \\ { { l }_{ 1 }-{ l }_{ 2 }\quad } & { l }_{ 2 }\quad & { l }_{ 1 }-{ l }_{ 3 } \\ { m }_{ 1 }-{ m }_{ 2 }\quad & { m }_{ 2 }\quad & { m }_{ 1 }-{ m }_{ 3 } \end{matrix} \right|=0$

$\triangle \quad =\quad -({ l }_{ 1 }-{ l }_{ 2 })({ m }_{ 1 }-{ m }_{ 3 })+({ l }_{ 1 }-{ l }_{ 3 })({ m }_{ 1 }-{ m }_{ 2 })=0$

$\Rightarrow \displaystyle\sum{ l }_{ 1 }({ m }_{ 2 }-{ m }_{ 3 })\quad =\quad 0$

A line makes an angle

$\alpha,\beta,\gamma$ with the

$X,Y,Z$ axes. Then

$\sin^2\alpha+\sin^2\beta+\sin^2\gamma=$

0%
$1$
0%
$2$
0%
$\dfrac 32$
0%
$4$

Explanation

For a vector

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

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

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

Option B

The vectors

$2\hat{i}+3\hat{j};5\hat{i}+6\hat{j};8\hat{i}+\lambda\hat{j}$ have their initial points at

$(1,1 )$ . The value of

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

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

Explanation

$\vec{V}= 2\hat{i}+3\vec{j}$

$\vec{U}= 5\hat{i}+6\vec{j}$

$\vec{W}= 8\hat{i}+\lambda \vec{j}$

$\vec{V}-\vec{U}= k(\vec{W}-\vec{U})$ (as they are collnear)

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

$\hat{i}$ &

$j$ are non collienar

$-3=3k$

$k=-1$

$k(\lambda -6)= -3$

$\lambda = 9$

The position vectors of three points are

$2\overrightarrow { a } -\overrightarrow { b } +3\overrightarrow { c } ,\overrightarrow { a } -2\overrightarrow { b } +\lambda \overrightarrow { c }$ and

$\mu \overrightarrow { a } -5\overrightarrow { b }$ , where

$\overrightarrow { a } ,\overrightarrow { b } ,\overrightarrow { c }$ are non-coplanar vectors. The points are coliinear when

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

Explanation

Let the points

$A,B$ and

$C$ respectively, then

$\overrightarrow { AB } =\overrightarrow { OB } -\overrightarrow { OA } \\ =\overrightarrow { a } -2\overrightarrow { b } +\lambda \overrightarrow { c } -\left( 2\overrightarrow { a } -\overrightarrow { b } +3\overrightarrow { c } \right) \\ =-\overrightarrow { a } -\overrightarrow { b } +\left( \lambda -3 \right) \overrightarrow { c }$

$\overrightarrow { AC } =\overrightarrow { OC } -\overrightarrow { OA } \\ =\mu \overrightarrow { a } -5\overrightarrow { b } -\left( 2\overrightarrow { a } -\overrightarrow { b } +3\overrightarrow { c } \right) \\ =\left( \mu -2 \right) \overrightarrow { a } -4\overrightarrow { b } -3\overrightarrow { c }$

The points are collinear if

$\overrightarrow { AB } =t\overrightarrow { AC }$

$\Rightarrow -1=t\left( \mu -2 \right) ,-1=4t,\lambda -3=-3t$

$\displaystyle \Rightarrow -1=t\left( \mu -2 \right) ,-1=4t,\lambda -3=-3t$ and

$\displaystyle -1=\frac { -1 }{ 4 } \left( \mu -2 \right) ,\lambda -3=\frac { 3 }{ 4 }$

$\displaystyle \Rightarrow \mu =6,\lambda =\frac {15 }{ 4 }$

The line passing through the points

$10\hat{i}+3\hat{j}$ ,

$12\hat{i}+5\hat{j}$ also passes through the point

$a\hat{i}+11 \hat{j}$ , then

$a=$

0%
$-8$
0%
$4$
0%
$18$
0%
$12$

Explanation

Let

$\vec{v}=10\hat{i}+3\hat{j}$

$\vec{u}=12\hat{i}+5\hat{j}$

$\vec{w}=a\hat{i}+11\hat{j}$
As

$\vec{u},\vec{v},\vec{w}$ are on a line then

$\vec{u}-\vec{v}$ is collinear with

$\vec{w}-\vec{u}$

$2\hat{i}+2\hat{j}=k((a-12)\hat{i}+6\hat{j})$

$2 =6k$ ,

$2$

$=k(a-12)$

$k=\dfrac{1}{3}$ ,

$a=18$

If the points

$(h, 3, -4), (0, -7, 10)$ and

$(1, k, 3)$ are collinear, then

$h + k$ is

0%
$4$
0%
$0$
0%
$-4$
0%
$14$

Explanation

Any point lying on the line joining

$(h,3,-4)$ and

$(0,-7,10)$ can be written in the form of

$(h-hl,3-10l,-4+14l)$ , where

$l$ is a constant.
Since,

$(1,k,3)$ is also a point on this line, it should satisfy this form.

$\Rightarrow -4+14l = 3$ .

$\therefore l =$

$\dfrac{1}{2}$

$\Rightarrow k = 3-10l = -2$

$\Rightarrow h(1-l) = 1$

$\therefore h = 2$
Hence,

$h + k = 2-2 = 0$

The point collinear with

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

$(2, 0, 0)$ among the following is

0%
$(0,4,6)$
0%
$(0,-4,-5)$
0%
$(0, -4, -6)$
0%
$(0,-4,6)$

Explanation

Any point lying on the line joining

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

$(2,0,0)$ can be written in the form of

$(1+k, -2+2k, -3+3k)$ .
Since, all the options have their

$x$ coordinate as

$0$ ,

$1+k = 0$

$k = -1$
Thus,

$y=-2 + 2k = -4$ and

$z=-3+3k = -6$

Hence, option C is correct.

If the points whose position vectors are

$2\overline{i}+\overline{j}+\overline{k},\ 6\overline{i}-\overline{j}+2\overline{k}$ and

$14\overline{i}-5\overline{j}+p\overline{k}$ are collinear then the value of

$\mathrm{p}$ is

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

Explanation

$2\hat{i}+\hat{j}+\hat{k}=\lambda (6\hat{i}-\hat{j}+2\hat{k})+\mu (14\hat{i}-5\hat{j}+p\hat{k})$ (Property of Collinear)

$2=b\lambda +14 \mu$

$1=-\lambda -5\mu$

$0=8\lambda+24\mu$

$\dfrac{-\lambda}{3}= \mu$

$\lambda =\dfrac{3}{2}\ \ \ \ \mu=\dfrac{-1}{2}$

$1=2\lambda + p \mu$

$-2= \dfrac{-p}{2}$

$p= 4$

The points with position vectors

$\vec{a}+\vec{b},\vec{a}-\vec{b}$ and

$\vec{a}+\lambda\vec{b}$ are collinear for

0%
Only integrals values of $\lambda$
0%
No value of $\lambda$
0%
All real values of $\lambda$
0%
Only rational values of $\lambda$

$A = (1, 2, 3), B = (2, 10, 1)$ ,

$Q$ are collinear points and

$Q_x=-1$ , then

$Q_z=$

0%
$-3$
0%
$7$
0%
$-14$
0%
$-7$

Explanation

The direction ratios of the line

$AB$ are

$(2-1) : (10-2) : (1-3) = 1 : 8 : -2$ .

Hence, any point

$Q$ lying on the line

$AB$ can be written in the form of

$(1+k, 2+8k, 3-2k)$ , where

$k$ is a real number.

$Q_x = -1$

Hence,

$1+k = -1$

This gives,

$k = -2$ .

${Q}^{}_{z} = 3- (2\times (-2)) = 7$

Hence, option B is correct.

$PQR$ are the three points with respective position vectors

$\hat{i}+\hat{j},\ \hat{i}-\hat{j}$ and

$a\hat{i}+b\hat{j}+c\hat{k}$ , then the points

$PQR$ are collinear if

0%
$a=b=c=1$
0%
$a=b=c=0$
0%
$a=1,\ b,c$ in $R$
0%
$a=1, c=0,\ b\in R$

Explanation

Given,

$P(\hat{i}+\hat{j}),Q(\hat{i}-\hat{j})$ and

$R(a\hat{i}+b\hat{j}+c\hat{k})$ are collinear

Refer image 1.

Consider

$PQ=\vec{a}$ and

$QR=\vec{b}$

Since , PQR are collinear, angle between

$\vec{a}$ and

$\vec{b}$ should be equal to

$0^0$

$\Rightarrow \vec{a}\times \vec{b}=ab\sin\theta=a b\sin O=0$

So,

$\boxed{\vec{a}\times \vec{b}=0}$ ............(i)

$\vec{a}=(\hat{i}-\hat{j})-(\hat{i}+\hat{j})$

$=\hat{i}-\hat{j}-\hat{i}-\hat{j}$

$=-2\hat{j}$

$\vec{b}=(a\hat{i}+b\hat{j}+c\hat{k})-(\hat{i}-\hat{j})$

$=a\hat{i}+b\hat{j}+c\hat{k}-\hat{i}+\hat{j}$

$=(a-1)\hat{i}+(b+1)\hat{j}+c\hat{k}$

from (1),

$\vec{a}\times \vec{b}=0$

$(-2\hat{j})[(a-1)\hat{i}+(b+1)\hat{j}+c\hat{k}]=0$

Refer image 2.

$\Rightarrow +2(a-1)\hat{k}-2(b+1)0+(-2)c\hat{i}$

$\Rightarrow -2c\hat{i}+2(a-1)\hat{k}=0$ (from (i))

this is possible when

$a-1=0$ or

$\boxed{c=0}$

$\Rightarrow \boxed{a=1}$

Since b is missing, so b can be any real numbers

$\boxed{b \in R}$

$\therefore \boxed{a=1},\boxed{c=0}$ and

$\boxed{b\in E}$

Find the angle between the two lines having direction ratio

$(1,1,2)$ and

$\left( \left( \sqrt { 3 } -1 \right) ,\left( -\sqrt { 3 } -1 \right) ,4 \right)$ .

0%
$\displaystyle \dfrac { \pi }{ 3 }$
0%
$\displaystyle \dfrac { \pi }{ 2 }$
0%
$\displaystyle \dfrac { \pi }{ 6 }$
0%
None of these

Explanation

Let $\overrightarrow { { m }_{ 1 } }$ and $\overrightarrow { { m }_{ 2 } }$ be vectors parallel to the two given lines.

Then, angle between the two given lines is same as the angle between $\overrightarrow { { m }_{ 1 } }$ and $\overrightarrow { { m }_{ 2 } }$ .

$\overrightarrow { { m }_{ 1 } } =$ Vector parallel to the line with direction ratios $(1,1,2)=i+j+2k$ and

$\overrightarrow { { m }_{ 2 } } =$ Vector parallel to the line with direction ratio $\left( \left( \sqrt { 3 } -1 \right) ,\left( -\sqrt { 3 } -1 \right) ,4 \right) =\left( \sqrt { 3 } -1 \right) i+\left( -\sqrt { 3 } -1 \right) j+4k$

Let $\theta$ be the angle between the given lines.

Then $\displaystyle \cos { \theta } =\dfrac { \overrightarrow { { m }_{ 1 } } .\overrightarrow { { m }_{ 2 } } }{ \left| \overrightarrow { { m }_{ 1 } } \right| \left| \overrightarrow { { m }_{ 2 } } \right| } =\dfrac { \left( \sqrt { 3 } -1 \right) -\left( \sqrt { 3 } +1 \right) +8 }{ \sqrt { 1+1+4 } \sqrt { { \left( \sqrt { 3 } -1 \right) }^{ 2 }+{ \left( \sqrt { 3 } +1 \right) }^{ 2 }+16 } }$

$\Rightarrow \displaystyle \cos { \theta } =\dfrac { 6 }{ \sqrt { 6 } \sqrt { 24 } } =\dfrac { 1 }{ 2 }$

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

The three points whose position vectors are

$\overline{i}+2\overline{j}+3\overline{k,}$

$3\overline{i}+4\overline{j}+7\overline{k,}$ and

$-3\overline{i}-2\overline{j}-5\overline{k}$

0%
Form the vertices of an equilateral triangle
0%
Form the vertices of an right angled triangle
0%
Are collinear
0%
Form the vertices of an isosceles triangle

Explanation

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

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

$\vec{c}=3\hat{i}-2\hat{j}-5\hat{k}$

$\vec{ca}=4\hat{i}+4\hat{j}+8\hat{k}$

$\vec{cb}=6\hat{i}+6\hat{j}+12\hat{k}$

$\vec{cb}=\frac{3}{2}\left ( \vec{ca} \right )$
So,

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

Find the angle between the lines

$\overrightarrow { r } =3i+2j-4k+\lambda \left( i+2j+2k \right)$ and

$\overrightarrow { r } =\left( 5j-2k \right) +\mu \left( 3i+2j+6k \right)$

0%
$\displaystyle \theta =\cos ^{ -1 }{ \left( \frac { 19 }{ 21 } \right) }$
0%
$\displaystyle \theta =\sin ^{ -1 }{ \left( \frac { 19 }{ 21 } \right) }$
0%
$\displaystyle \theta =\cos ^{ -1 }{ \left( \frac { 20 }{ 21 } \right) }$
0%
None of these

Explanation

Let $\theta$ be the angle between the given lines.

The given lines are parallel to the vector.

$\overrightarrow { { b }_{ 1 } } =1+2j+2k$ and $\overrightarrow { { b }_{ 2 } } =3i+2j+6k$ respectively.

So, the angle $\theta$ between them is given by

$\displaystyle \cos { \theta } =\dfrac { \overrightarrow { { b }_{ 1 } } .\overrightarrow { { b }_{ 2 } } }{ \left| \overrightarrow { { b }_{ 1 } } \right| \left| \overrightarrow { { b }_{ 1 } } \right| } =\dfrac { \left( i+2j+2k \right) \left( 3i+2j+6k \right) }{ \left| i+2j+2k \right| \left| 3i+2j+6k \right| }$

$\displaystyle =\dfrac { 3+4+12 }{ \sqrt { 1+4+4 } \sqrt { 9+4+36 } } =\dfrac { 19 }{ 21 }$

$\Rightarrow \theta =\cos ^{ -1 }{ \left( \dfrac { 19 }{ 21 } \right) }$

$\vec{a},\vec{b},\vec{c}$ are the position vectors of points lie on a line, then

$\vec{a}\times \vec{b}+\vec{b}\times \vec{c}+\vec{c}\times \vec{a}=$

0%
$0$
0%
$\vec{b}$
0%
$1$
0%
$\vec{a}$

The product of the d.r's of a line perpendicular to the plane passing through the points

$(4,0,0),(0,2,0)$ and

$( 1,0,1)$ is

0%
$6$
0%
$2$
0%
$0$
0%
$1$

Explanation

Let three points be

$A(4,0,0) B(0,2,0) C(1,0,1)$ .
The dr's of

$AB$ are

$(4,-2,0)$
The dr;s of

$BC$ are

$(1,-2,1)$ .
Since, both these lines lie on the plane, the normal to the plane will also be a normal to these lines.
Hence, the dr's of the normal can be obtained by the cross product of these

$2$ lines.

$n = AB \times BC$

$(-1,2,-3)$
Hence, the product is

$6$ .

The d.r's of the line of intersection of the planes

$x+y+z-1$

$=0$ and

$2x+3y+4z-7$

$=0$ are

0%
$1, 2, -3$
0%
$2, 1, -3$
0%
$4, 2, -6$
0%
$1, -2, 1$

Explanation

The normal vector to the plane

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

$ai+bj+ck$ .

Hence, the normal to the first plane is

$i+j+k$ and normal to the second plane is given by

$2i+3j+4k$ .

The line of intersection plane is perpendicular to the both these normal vectors.

It is found by the cross product of these normal vectors.

$(i+j+k)\times(2i+3j+4k)=i-2j+k$ .

Hence, the direction ratios given by

$(1,-2,1)$

The d.c's of the normal to the plane

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

0%
$(3, -2, 6)$
0%

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

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

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

Explanation

Given equation of plane is

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

The dr's of the normal to the plane are

$(2,-1,2)$ .
The corrresponding dc's will be given by

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

The plane which passes through the point

$(-1, 0, -6)$ and perpendicular to the line whose direction ratios is

$(6, 20, -1)$ also passes through the point:

0%
$(1, 1, -26)$
0%
$(0, 0, 0)$
0%
$(2, 1, -32)$
0%
$(1, 1, 1)$

Explanation

The d.r's of the perpendicular line give the normal.
Hence, the plane will be of the form

$6x + 20y -z = d$
Since, it passes through

$(-1,0,-6)$ ,

$d = 0$
Hence, the plane passes through the origin.

$R(a+2,a+3,a+4)$ divides the line segment joining

$P(2, 3, 4)$ and

$Q(4, 5, 6)$ in the ratio

$-3:2$ , then the value of the parameter which represents

$a$ is

0%
$3$
0%
$2$
0%
$6$
0%
$-1$

Explanation

If a point P divides the line segment joining

$A=(x_1,y_1,z_1)$ and

$B=(x_2,y_2,z_2)$ in the ratio

$m:n$ .

Then the point P is given by

$(\cfrac{mx_2+nx_1}{m+n}, \cfrac{my_2+ny_1}{m+n},\cfrac{mz_2+nz_1}{m+n})$ .

In the given problem, we have

$a+2=\cfrac{-3\times4+2\times2}{-1}$ .

On solving, we get

$a=6.$

lf the equation of the plane perpendicular to the

$\mathrm{z}$ -axis and passing through the point

$(2, -3,4)$ is

$ax+by+cz=d$ then

$\displaystyle \dfrac{a+b+c}{d}=$

0%
$4$
0%
$\displaystyle \dfrac{3}{4}$
0%
$3$
0%
$\displaystyle \dfrac{1}{4}$

Explanation

Since, the plane is perpendicular to the z axis, $a = b=0$

The plane passes through (2,-3,4).

So $d = 4c$

$\dfrac{a+b+c}{d} = \dfrac{c}{4c} = \dfrac{1}{4}$

The equation to the plane bisecting the line segment joining

$(-3, 3, 2), (9, 5, 4)$ and perpendicular to the line segment is

0%
$x-y+4_{Z}-13=0$
0%
$2x-2y+7z-23=0$
0%
$x-7y+2_{Z}-1=0$
0%
$6x+y+z-25=0$

Explanation

The point of bisection is

$(3,4,3)$ .
The d.r's of the line joining the

$2$ points are

$(9+3,5-3,4-2)=(12,2,2)$
These are also the d.r's of the normal to the plane.
Hence, the equation of the plane is

$12x + 2y +2z = d$
Since, it passes through

$(3,4,3)$ .

$\therefore 12(3)+2(4)+2(3)=d$

$\therefore d=50$
Hence, equation of the plane is

$12x + 2y + 2z = 50$

$6x + y + z = 25$

The direction ratios of a normal to the plane through

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

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

$x+y=3$ are

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

Explanation

Let the equation of the plane be

$Ax+By+Cz=D.$

Angle between two planes is given by cos

$\theta =\dfrac{A_{1}A_{2}+B_{1}B_{2}+C_{1}C_{2}}{\sqrt{A_{1}^{2}+B_{1}^{2}+C_{1}^{2}}\sqrt{A_{2}^{2}+B_{2}^{2}+C_{2}^{2}}}$

$\therefore \cos \dfrac{\pi}{4}= \dfrac{1}{\sqrt{2}} = \dfrac{A\times1+B\times1}{\sqrt{A^{2}+B^{2}+C^{2}}\sqrt{1^{2}+1^{2}}}$

$\Rightarrow A^{2}+B^{2}+C^{2} = (A+B)^{2}$

$\Rightarrow A^{2}+B^{2}+C^{2} = A^{2}+B^{2}+2AB$

$\Rightarrow C^{2} = 2AB$ ......

$(I)$

$\because$ The plane passes through

$(1,0,0)$ and

$(0,1,0)$ .

$\Rightarrow A=D$ and

$B=D$

From

$(I)$ , we have,

$C^{2}=2A^{2}$

$\Rightarrow C=\sqrt{2}A$

$\therefore$ equation of plane reduces to

$Ax+Ay+\sqrt{2}Az=A$

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

$\therefore$ direction ratios of normal are (

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

$\therefore$ ans. is option B.

Equation of the plane through the mid-point of the join of

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

$B(-1,2,1)$ and perpendicular to

$AB$ is

0%
$\vec{r}.(5i+3j-11k)+\dfrac{135}{2}=0$
0%
$\vec{r}.(5i+3j-11k)=\dfrac{135}{2}$
0%
$\vec{r}.\left (\dfrac{3}{2}\hat{i}+\dfrac{7}{2}\hat{j}-\dfrac{9}{2}\hat{k}\right)=5\hat{i}+3\hat{j}-11\hat{k}$
0%
$\vec{r}.(5i+3j-11k)+\dfrac{185}{2}=0$

Explanation

Since, the line

$AB$ is perpendicular to the plane, it has to be the normal to the plane.
DR's of normal

$=(4+1,5-2,-10-1)=(5,3,-11)$
Hence, the equation of the plane is of the form,

$5x + 3y - 11z = d$
The midpoint of

$AB$ is

$\left ( \dfrac{3}{2} , \dfrac{7}{2} , \dfrac{-9}{2} \right)$
The plane passes through this point, so the value of d is

$\dfrac{135}{2}$ .
Hence, option B is correct.

$(2, 3, -1)$ is the foot of the perpendicular from

$(4, 2, 1)$ to a plane, then the equation of that plane is

$ax+by+cz=d$ . Then

$a+d$ is

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

Explanation

Since, the line joining the two points is perpendicular to the plane, it's d.r's will give the normal to the plane.
Direction ratio of the normal

$= (4-2,2-3,1+1)=(2,-1,2)$
Hence,

$a = 2$ ,

$b = -1$ ,

$c = 2$
Since, the plane passes through

$(2,3,-1)$ ,

$d=2(2)+3(-1)-1(2)$

$d = -1$
Hence, the required value of the expression is

$a +d = 1$ .

The product of the d.cs of the line which makes equal angles with

$ox, oy, oz$ is

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

Explanation

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

$\Rightarrow 3\cos^2(\alpha)=1$

$\displaystyle \Rightarrow \cos \alpha = \pm \dfrac{1}{\sqrt{3}}$
Product will be

$\cos^3(\alpha)= \pm 1/3(\sqrt3)$
Hence, option C is correct.

lf $\theta$ is the angle between two lines whose d.cs are $l_{1},m_{\mathrm{1}},n_{\mathrm{1}}$ and $l_{2},m_{2},n_{2}$ , then

$\displaystyle \dfrac{\Sigma(l_{1}+l_{2})^{2}}{4\cos^{2}(\dfrac{\theta}{2})}+\dfrac{\Sigma(l_{\mathrm{I}}-l_{2})^{2}}{4\sin^{2}(\dfrac{\theta}{2})}=$

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

Explanation

Let $\vec{a_1}$ and $\vec{a_2}$ be unit vectors along the lines.

$\vec{a_1} = l_1 \hat{i} + m_1 \hat{j} + n_1 \hat{k}$

$\vec{a_2} = l_2 \hat{i} + m_2 \hat{j} + n_2 \hat{k}$

$\Rightarrow l_1^2+m_1^2+n_1^2 = 1$ and $l_2^2+m_2^2+n_2^2 = 1$

Consider the dot product of $\vec{a_1}$ and $\vec{a_2}$

$\vec{a_1}.\vec{a_2} = (l_1l_2)+(m_1m_2)+(n_1n_2)$

$\Rightarrow \cos \theta = (l_1l_2)+(m_1m_2)+(n_1n_2)$

$S = \displaystyle \dfrac{\Sigma(l_{1}+l_{2})^{2}}{4\cos^{2}(\dfrac{\theta}{2})}+\dfrac{\Sigma(l_{{I}}-l_{2})^{2}}{4\sin^{2}(\dfrac{\theta}{2})}$

$\Rightarrow S = \dfrac{ 2 + 2 (l_1l_2 + m_1m_2+ n_1n_2) } {4 \cos^2{\dfrac{\theta}{2}}} + \dfrac{ 2 - 2 (l_1l_2 + m_1m_2+ n_1n_2) } {4 \sin^2{\dfrac{\theta}{2}}}$

$\Rightarrow S = \dfrac{2+2\cos\theta}{ 4 \cos^2{\dfrac{\theta}{2}}} + \dfrac{2-2\cos\theta}{4 \sin^2{\dfrac{\theta}{2}}}$

Now, $1+\cos\theta = 2\cos^2\dfrac{\theta}2$ and $1-\cos\theta = 2\sin^2\dfrac{\theta}2$

$\Rightarrow S = \dfrac22 + \dfrac22$

$\Rightarrow S =2$

lf a line makes angles

$\displaystyle \dfrac{\pi}{12}, \displaystyle \dfrac{5\pi}{12}$ with

$OY, OZ$ respectively where

$O=({0}, 0,0)$ , then the angle made by that line with

$OX$ is

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

Explanation

$\left (\cos\dfrac {\pi}{12}\right)^{2}+\left (\cos \dfrac {5\pi}{12}\right)^{2}+(\cos(\gamma))^{2}=1$

$\left (\cos \dfrac {\pi}{12}\right)^{2}+\left (\sin \dfrac {\pi }{12}\right)^{2}+(\cos(\gamma))^{2}=1$ .....

$\left(\cos\theta=\sin\left (\dfrac {\pi}{2}-\theta\right)\right)$

$(\cos(\gamma))^{2}=0$

$\cos(\gamma)=0$

$\gamma=90^{\circ}$

The plane

$2x+3y+kz-7=0$ is parallel to the line whose direction ratios are

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

$k=$

0%
$5$
0%
$8$
0%
$1$
0%
$0$

Explanation

Since, the plane is parallel to the line, the normal to the plane will be perpendicular to the line as well.
Hence, their dot product will be equal to

$0$ .

$(2,3,k).(2,-3,1) = 0$

$4 - 9 +k =0$

$k= 5$

If the foot of the perpendicular from

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

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

0%
$-x+2y+8z-9=0$
0%
$x+2y+2z-9=0$
0%
$x+y+z-5=0$
0%
$x+2y-3z+1=0$

Explanation

The line joining the origin to the point on the plane is perpendicular to the plane.
The dr's of the normal are

$(1,2,2)$ .
The equation of the plane will be of the form,

$x+2y+2z = d$
Since,

$(1,2,2)$ lies on the plane,

$d = 9$
Hence,

$x+2y+2z = 9$

The sum of the squares of sine of the angles made by the line

$AB$ with

$OX, OY, OZ$ where

$O$ is the origin is

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

Explanation

Let

$\cos\alpha, \cos\beta$ and

$\cos\gamma$ be the direction cosines.
We know that

$\cos^{2}\alpha + \cos^{2}\beta + \cos^{2}\gamma = 1$
Now rewrite it as

$\cos^{2}\alpha = 1 - \sin^{2}\alpha$

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

$\Rightarrow \cos^{2}\gamma = 1 - \sin^{2}\gamma$
Then we get,

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

The direction ratios of a normal to the plane passing through

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

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

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

Explanation

The

$3$ given points are

$A(0,0,1), B(0,1,2)$ and

$C(1,2,3)$

The dr's of

$AB$ are

$(0,1,1)$

The dr's of

$BC$ are

$(1,1,1)$

The normal to the plane will be perpendicular to both these lines.

Hence,

$n = AB \times BC$

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

CBSE Questions for Class 12 Commerce Maths Three Dimensional Geometry Quiz 2 - 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 2 - MCQExams.com

0:0:2

Practice Class 12 Commerce Maths Quiz Questions and Answers

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