If a point $\mathrm{P}(\alpha, \beta, \gamma)$ satisfying $$\left( {\matrix{ \alpha & \beta & \gamma \cr } } \right)\left( {\matrix{ 2 & {10} & 8 \cr 9 & 3 & 8 \cr 8 & 4 & 8 \cr } } \right) = \left( {

Point $\mathrm{P}(\alpha, \beta, \gamma)$ lies on the plane $2 x+4 y+3 z=5$, $$ \therefore $$ $2 \alpha+4 \beta+3 \gamma=5$ ........(1) Given, $$\left( {\matrix{ \alpha & \beta & \gamma \cr } } \right)\left( {\matrix{ 2 & {10} & 8 \cr 9 & 3 & 8 \cr 8 & 4 & 8 \cr } } \right) = \left( {\matrix{ 0 & 0 & 0 \cr } } \right)$$ $$ \therefore $$ $2 \alpha+9 \beta+8 \gamma=0$ .......(2) and $10 \alpha+3 \beta+4 \gamma=0$ ........(3) and $8 \alpha+8 \beta+8 \gamma=0$ ..........(4) Subtract (4) from (2) $-6

If a plane passes through the points $(-1, k, 0),(2, k,-1),(1,1,2)$ and is parallel to the line $\frac{x-1}{1}=\frac{2 y+1}{2}=\frac{z+1}{-1}$, then the value of $\frac{k^2+1}{(k-1)(k-2)}$ is :

Let $$P \equiv ( - 1,k,0),Q \equiv (2,k, - 1)$$ & $$R(1,1,2)$$ $$\overrightarrow P R = 2\widehat i + (1 - k)\widehat j + 2\widehat k$$ & $$\overrightarrow Q R = - \widehat i + (1 - k)\widehat j + 3\widehat k$$ $$\therefore$$ Normal to plane will be $$\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & {(1 - k)} & 2 \cr { - 1} & {(1 - k)} & 3 \cr } } \right| = \widehat i(1 - k) - \widehat j(8) + 3\widehat k(1 - k)$$ If line is parallel to this we have $$1(1 - k) + 1( - 8) + ( - 3)

3D Geometry — Page 15 | JEE Mathematics

Q141

The foot of perpendicular from the origin

\mathrm{O}

to a plane

\mathrm{P}

which meets the co-ordinate axes at the points

\mathrm{A}, \mathrm{B}, \mathrm{C}

is

(2, \mathrm{a}, 4), \mathrm{a} \in \mathrm{N}

. If the volume of the tetrahedron

\mathrm{OABC}

is 144 unit

^{3}

, then which of the following points is NOT on P ?

A

(3,0,4)

B

(0,6,3)

C

(0,4,4)

D

(2,2,4)

Correct Answer

Option A

Solution

Equation of Plane:

\begin{aligned} & (2 \hat{i}+a \hat{j}+4 \hat{\mathrm{k}}) \cdot[(\mathrm{x}-2) \hat{\mathrm{i}}+(\mathrm{y}-\mathrm{a}) \hat{\mathrm{j}}+(\mathrm{z}-4) \hat{\mathrm{k}}]=0 \\\\ & \Rightarrow 2 \mathrm{x}+\mathrm{ay}+4 \mathrm{z}=20+\mathrm{a}^{2} \\\\ & \Rightarrow \mathrm{A} \equiv\left(\frac{20+\mathrm{a}^{2}}{2}, 0,0\right) \\\\ & \mathrm{B} \equiv\left(0, \frac{20+\mathrm{a}^{2}}{\mathrm{a}}, 0\right) \\\\ & \mathrm{C} \equiv\left(0,0, \frac{20+\mathrm{a}^{2}}{4}\right) \\\\ & \Rightarrow \text{ Volume of tetrahedron } \\\\ & =\frac{1}{6}\left[ \begin{array}{lll}{\overrightarrow a } & {\overrightarrow b } & {\overrightarrow c } \end{array} \right] \\\\ & =\frac{1}{6} \overrightarrow{\mathrm{a}} \cdot(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}) \\\\ & \Rightarrow \frac{1}{6}\left(\frac{20+\mathrm{a}^{2}}{2}\right) \cdot\left(\frac{20+\mathrm{a}^{2}}{\mathrm{a}}\right) \cdot\left(\frac{20+\mathrm{a}^{2}}{4}\right)=144 \\\\ & \Rightarrow\left(20+\mathrm{a}^{2}\right)^{3}=144 \times 48 \times \mathrm{a} \\\\ & \Rightarrow \mathrm{a}=2 \\\\ & \Rightarrow \text{ Equation of plane is } 2 \mathrm{x}+2 \mathrm{y}+4 \mathrm{z}=24 \\\\ & \text{ Or } \mathrm{x}+\mathrm{y}+2 \mathrm{z}=12 \\\\ & \Rightarrow(3,0,4) \quad \text{ Not } \text{ lies on } \text{ the } \\\\ & \mathrm{x}+\mathrm{y}+2 \mathrm{z}=12 \end{aligned}

Q142

Let

P

be the plane, passing through the point

(1,-1,-5)

and perpendicular to the line joining the points

(4,1,-3)

and

(2,4,3)

. Then the distance of

P

from the point

(3,-2,2)

is :

A 5

B 4

C 6

D 7

Correct Answer

Option A

Solution

Equation of Plane :

2(x-1)-3(y+1)-6(z+5)=0

or $2 x-3 y-6 z=35$ $\Rightarrow$ Required distance $=$

\frac{|2(3)-3(-2)-6(2)-35|}{\sqrt{4+9+36}}

= 5

Q143

If a point

\mathrm{P}(\alpha, \beta, \gamma)

satisfying

\left( \begin{array}{lll}\alpha & \beta & \gamma \end{array} \right)\left( \begin{array}{lll}2 & {10} & 8 \\ 9 & 3 & 8 \\ 8 & 4 & 8 \end{array} \right) = \left( \begin{array}{lll}0 & 0 & 0 \end{array} \right)

lies on the plane

2 x+4 y+3 z=5

, then

6 \alpha+9 \beta+7 \gamma

is equal to :

A

\dfrac{11}{5}

B 11

C

-1

D

\dfrac{5}{4}

Correct Answer

Option B

Solution

Point $\mathrm{P}(\alpha, \beta, \gamma)$ lies on the plane $2 x+4 y+3 z=5$ , $\therefore$ $2 \alpha+4 \beta+3 \gamma=5$ ........(

1) Given,

\left( \begin{array}{lll}\alpha & \beta & \gamma \end{array} \right)\left( \begin{array}{lll}2 & {10} & 8 \\ 9 & 3 & 8 \\ 8 & 4 & 8 \end{array} \right) = \left( \begin{array}{lll}0 & 0 & 0 \end{array} \right)

$\therefore$ $2 \alpha+9 \beta+8 \gamma=0$ .......(2) and $10 \alpha+3 \beta+4 \gamma=0$ ........(3) and $8 \alpha+8 \beta+8 \gamma=0$ ..........(

4) Subtract (4) from (2) $-6 \alpha+\beta=0$ $\beta=6 \alpha$ From equation (4) $8 \alpha+48 \alpha+8 \gamma=0$ $\gamma=-7 \alpha$ From equation (1) $2 \alpha+24 \alpha-21 \alpha=5$ $5 \alpha=5$ $\alpha=1$ $\beta=+6, \quad \gamma=-7$ $\therefore 6 \alpha+9 \beta+7 \gamma$ $=6+54-49$ $=11$

Q144

The shortest distance between the lines

{{x - 5} \over 1} = {{y - 2} \over 2} = {{z - 4} \over { - 3}}

and

{{x + 3} \over 1} = {{y + 5} \over 4} = {{z - 1} \over { - 5}}

is :

A

7\sqrt 3

B

5\sqrt 3

C

4\sqrt 3

D

6\sqrt 3

Correct Answer

Option D

Solution

\begin{aligned} & \mathrm{L}_1: \frac{x-5}{1}=\frac{y-2}{2}=\frac{z-4}{-3} \\\\ & \overrightarrow{a_1}=5 \hat{i}+2 \hat{j}+4 \hat{k} \\\\ & \overrightarrow{r_1}=\hat{i}+2 \hat{j}-3 \hat{k} \\\\ & \mathrm{~L}_2: \frac{x+3}{1}=\frac{y+5}{4}=\frac{z-1}{-5} \\\\ & \overrightarrow{a_2}=-3 \hat{i}-5 \hat{j}+\hat{k} \\\\ & \overrightarrow{r_2}=\hat{i}+4 \hat{j}-5 \hat{k} \\\\ & \overrightarrow{r_1} \times \overrightarrow{r_2}=\left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 1 & 4 & -5 \end{array}\right| \\\\ & =2 \hat{i}+2 \hat{j}+2 \hat{k} \end{aligned}

\begin{aligned} & \text{ Shortest distance (d) }=\left|\frac{\left(\overrightarrow{r_1} \times \overrightarrow{r_2}\right) \cdot\left(\overrightarrow{a_1}-\overrightarrow{a_2}\right)}{\left|\overrightarrow{r_1} \times \overrightarrow{r_2}\right|}\right| \\\\ & =\frac{36}{2 \sqrt{3}}=6 \sqrt{3} \end{aligned}

Q145

Let the image of the point

P(2,-1,3)

in the plane

x+2 y-z=0

be

Q

. Then the distance of the plane

3 x+2 y+z+29=0

from the point

Q

is :

A

2\sqrt{14}

B

\dfrac{22\sqrt2}{7}

C

\dfrac{24\sqrt2}{7}

D

3\sqrt{14}

Correct Answer

Option D

Solution

Equation of line PM $\dfrac{x-2}{1}=\dfrac{y+1}{2}=\dfrac{z-3}{-1}=\lambda$ Any point on line $=(\lambda+2,2 \lambda-1,-\lambda+3)$ For point ' $m$ ' $(\lambda+2)+2(2 \lambda-1)-(3-\lambda)=0$ $\Rightarrow$

\lambda=\frac{1}{2}

Point $\mathrm{m}\left(\dfrac{1}{2}+2,2 \times \dfrac{1}{2}-1, \dfrac{-1}{2}+3\right)$

=\left(\frac{5}{2}, 0, \frac{5}{2}\right)

For Image $\mathrm{Q}(\alpha, \beta, \gamma)$

\begin{aligned} & \frac{\alpha+2}{2}=\frac{5}{2}, \frac{\beta-1}{2}=0, \\\\ & \frac{\gamma+3}{2}=\frac{5}{2} \end{aligned}

\begin{aligned} & Q:(3,1,2) \\\\ & d=\left|\frac{3(3)+2(1)+2+29}{\sqrt{3^2+2^2+1^2}}\right| \\\\ & \Rightarrow d=\frac{42}{\sqrt{14}}=3 \sqrt{14} \end{aligned}

Q146

Let the shortest distance between the lines

L: \dfrac{x-5}{-2}=\dfrac{y-\lambda}{0}=\dfrac{z+\lambda}{1}, \lambda \geq 0

and

L_{1}: x+1=y-1=4-z

be

2 \sqrt{6}

. If

(\alpha, \beta, \gamma)

lies on

L

, then which of the following is NOT possible?

A

\alpha+2 \gamma=24

B

2 \alpha+\gamma=7

C

\alpha-2 \gamma=19

D

2 \alpha-\gamma=9

Correct Answer

Option A

Solution

$\dfrac{x-5}{-2}=\dfrac{y-\lambda}{0}=\dfrac{z+\lambda}{1}, \lambda \geq 0$

\begin{aligned} & \frac{x+1}{1}=\frac{y-1}{1}=\frac{z-4}{-1} \\\\ & \vec{a}_{1}=5 \hat{i}+\lambda \hat{j}-\lambda \hat{k}_{,} \vec{a}_{2}=-\hat{i}+\hat{j}+4 \hat{k} \\\\ & \vec{a}_{1}-\vec{a}_{2}=6 \hat{i}+(\lambda-1) \hat{j}-(\lambda+4) \hat{k} \\\\ & \vec{b}_{1}=-2 \hat{i}+\hat{k}, \vec{b}=\hat{i}+\hat{j}-\hat{k} \\\\ & \vec{b}_{1} \times \vec{b}_{2}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ -2 & 0 & 1 \\ 1 & 1 & -1 \end{array}\right| \\ & =-\hat{i}-\hat{j}-2 \hat{k} \end{aligned}

$\left(\vec{a}_{1}-\vec{a}_{2}\right) \cdot \vec{b}_{1} \times \vec{b}_{2}=-6+1-\lambda+2 \lambda+8=\lambda+3$ and $\left|\vec{b}_{1} \times \vec{b}_{2}\right|=\sqrt{6}$ $\because \dfrac{|\lambda+3|}{\sqrt{6}}=2 \sqrt{6}$ $\therefore \lambda=9, \because \lambda \geq 0$ $\therefore \quad L: \dfrac{x-5}{-2}=\dfrac{y-9}{0}=\dfrac{z+9}{1}=k$ $\therefore \quad \alpha=-2 k+5, \beta=9, \gamma=k-9$ Here $k$ is real then $\alpha+2 \gamma=-13 \neq 24$ .

But all other are in terms of $k$ hence possible.

Q147

A vector

\vec{v}

in the first octant is inclined to the

x

-axis at

60^{\circ}

, to the

y

-axis at 45 and to the

z

-axis at an acute angle. If a plane passing through the points

(\sqrt{2},-1,1)

and

(a, b, c)

, is normal to

\vec{v}

, then :

A

a+b+\sqrt{2} c=1

B

\sqrt{2} a+b+c=1

C

\sqrt{2} a-b+c=1

D

a+\sqrt{2} b+c=1

Correct Answer

Option D

Solution

l = {1 \over 2},m = {1 \over {\sqrt 2 }},n = \cos \theta

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

\Rightarrow {1 \over 4} + {1 \over 2} + {n^2} = 1 \Rightarrow {n^2} = {1 \over 4} \Rightarrow n = \, + \,{1 \over 2}

$\theta$ is acute $\therefore$

n = {1 \over 2}

$\therefore$

\overrightarrow v = k\left( {{1 \over 2}\widehat i + {1 \over {\sqrt 2 }}\widehat j + {1 \over 2}\widehat k} \right),k \in R

\overrightarrow v .\,(\overrightarrow a - \overrightarrow b ) = 0

(\sqrt 2 - a){1 \over 2} + ( - 1 - b){1 \over {\sqrt 2 }} + (1 - c){1 \over 2} = 0

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

\Rightarrow a + \sqrt 2 b + c = 1

Q148

If a plane passes through the points

(-1, k, 0),(2, k,-1),(1,1,2)

and is parallel to the line

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

, then the value of

\dfrac{k^2+1}{(k-1)(k-2)}

is :

A

\dfrac{17}{5}

B

\dfrac{6}{13}

C

\dfrac{13}{6}

D

\dfrac{5}{17}

Correct Answer

Option C

Solution

Let

P \equiv ( - 1,k,0),Q \equiv (2,k, - 1)

&

R(1,1,2)

\overrightarrow P R = 2\widehat i + (1 - k)\widehat j + 2\widehat k

&

\overrightarrow Q R = - \widehat i + (1 - k)\widehat j + 3\widehat k

$\therefore$ Normal to plane will be

\left| \begin{array}{lll}{\widehat i} & {\widehat j} & {\widehat k} \\ 2 & {(1 - k)} & 2 \\ { - 1} & {(1 - k)} & 3 \end{array} \right| = \widehat i(1 - k) - \widehat j(8) + 3\widehat k(1 - k)

If line is parallel to this we have

1(1 - k) + 1( - 8) + ( - 3)(1 - k) = 0

\Rightarrow 2(1 - k) = - 8

\Rightarrow 1 - k = - 4 \Rightarrow k = 5

$\therefore$

{{{k^2} + 1} \over {(k - 1)(k - 2)}} = {{26} \over {4.3}} = {{13} \over 6}

Q149

The line

l_1

passes through the point (2, 6, 2) and is perpendicular to the plane

2x+y-2z=10

. Then the shortest distance between the line

l_1

and the line

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

is :

A 9

B 7

C

\dfrac{19}{3}

D

\dfrac{13}{3}

Correct Answer

Option A

Solution

Equation of

{l_1} = {{x - 2} \over 2} = {{y - 6} \over 1} = {{z - 2} \over { - 2}}

Shortest distance with

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

is S.d

= \left| {{\begin{array}{lll}3 & {10} & 2 \\ 2 & 1 & { - 2} \\ 2 & { - 3} & 2 \end{array} \over {\left| { - 4\widehat i - 8\widehat j - 8\widehat k} \right|}}} \right| = \left| {{{( - 12) - 10(8) + 2( - 8)} \over {12}}} \right|

= 9 units

Q150

The plane

2x-y+z=4

intersects the line segment joining the points A (

a,-2,4)

and B (

2,b,-3)

at the point C in the ratio 2 : 1 and the distance of the point C from the origin is

\sqrt5

. If $$ab

A

\dfrac{17}{3}

B

\dfrac{97}{3}

C

\dfrac{16}{3}

D

\dfrac{73}{3}

Correct Answer

Option A

Solution

C:\left( {{{a + 4} \over 3},{{ - 2 + 2b} \over 3},{{4 - 6} \over 3}} \right)

= \left( {{{a + 4} \over 3},{{ - 2 + 2b} \over 3},{{ - 2} \over 3}} \right)

C lies on plane

2x - y + z = 4

2\left( {{{a + 4} \over 3}} \right) - \left( {{{2b - 2} \over 3}} \right) - {2 \over 3} = 4

\Rightarrow 2a + 8 - 2b + 2 - 2 = 12

\Rightarrow a - b = 2

..... (i) Now,

OP = \sqrt 5

{\left( {{{a + 4} \over 3}} \right)^2} + {\left( {{{2b - 2} \over 3}} \right)^2} + {4 \over 9} = 5

and using (i)

a = {{11} \over 5},1

\Rightarrow b = {1 \over 5}, - 1

as also

\Rightarrow a = 1,b = - 1

P(2, - 1, - 3),c\left( {{5 \over 3},{{ - 4} \over 3},{{ - 2} \over 3}} \right)

C{P^2} = {1 \over 9} + {1 \over 9} + {{49} \over 9} = {{17} \over 3}