3D Geometry — Page 12 | JEE Mathematics

Q111

Let the foot of the perpendicular from the point (1, 2, 4) on the line

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

be P. Then the distance of P from the plane

3x + 4y + 12z + 23 = 0

is :

A 5

B

{{50} \over {13}}

C 4

D

{{63} \over {13}}

Correct Answer

Option A

Solution

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

Let P = (4t $-$ 2, 2t + 1, 3t $-$ 1) $\because$ P is the foot of perpendicular of (1, 2, 4) $\therefore$

4(4t - 3) + 2(2t - 1) + 3(3t - 5) = 0

\Rightarrow 29t = 29 \Rightarrow t = 1

$\therefore$ P = (2, 3, 2) Now, distance of P from the plane

3x + 4y + 12z + 23 = 0

, is

\left| {{{6 + 12 + 24 + 23} \over {\sqrt {9 + 16 + 144} }}} \right| = {{65} \over {13}} = 5

Q112

The shortest distance between the lines

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

and

{{x + 3} \over 2} = {{y - 6} \over 1} = {{z - 5} \over 3}

, is :

A

{{18} \over {\sqrt 5 }}

B

{{22} \over {3\sqrt 5 }}

C

{{46} \over {3\sqrt 5 }}

D

6\sqrt 3

Correct Answer

Option A

Solution

{L_1}:{{x - 3} \over 2} = {{y - 2} \over 3} = {{z - 1} \over { - 1}}

{L_2}:{{x + 3} \over 2} = {{y - 6} \over 1} = {{z - 5} \over 3}

Now,

\overrightarrow p \times \overrightarrow q = \left| \begin{array}{lll}{\widehat i} & {\widehat j} & {\widehat k} \\ 2 & 3 & { - 1} \\ 2 & 1 & 3 \end{array} \right| = 10\widehat i - 8\widehat j - 4\widehat k

and

{\overrightarrow a _2} - {\overrightarrow a _1} = 6\widehat i - 4\widehat j - 4\widehat k

$\therefore$

S.D. = \left| {{{60 + 32 + 16} \over {\sqrt {100 + 64 + 16} }}} \right| = {{108} \over {\sqrt {180} }} = {{18} \over {\sqrt 5 }}

Q113

If two straight lines whose direction cosines are given by the relations

l + m - n = 0

,

3{l^2} + {m^2} + cnl = 0

are parallel, then the positive value of c is :

A 6

B 4

C 3

D 2

Correct Answer

Option A

Solution

Given that the direction cosines satisfy $l + m - n = 0$ , we find that $n = l + m$ .

The other equation is $3l^2 + m^2 + cnl = 0$ , and substituting $n = l + m$ gives $3l^2 + m^2 + cl(l + m) = 0$ .

This simplifies to $(3 + c)l^2 + clm + m^2 = 0$ .

As the lines are parallel, they share the same direction ratios, so we can express $l$ in terms of $m$ , say $l = km$ .

Substituting this into our equation gives $(3 + c)(km)^2 + ckm^2 + m^2 = 0$ .

This simplifies to $m^2[k^2(3 + c) + kc + 1] = 0$ .

Since $m \neq 0$ , we must have $k^2(3 + c) + kc + 1 = 0$ .

Here, we consider the ratio $k = \dfrac{l}{m}$ to be constant, since the lines are parallel.

The equation then becomes a quadratic equation in $k$ .

As the lines are parallel, the discriminant of the quadratic equation must be equal to zero for the equation to have equal roots.

Hence, the discriminant $D = (c^2 - 4(3 + c)) = 0$ .

Solving this quadratic equation gives $c^2 - 4c - 12 = 0$ .

Factoring this equation gives $(c - 6)(c + 2) = 0$ .

Solving for $c$ gives $c = 6, -2$ .

However, we are looking for the positive value of $c$ , so $c = 6$ .

Therefore, the correct answer is 6

Q114

If the two lines

{l_1}:{{x - 2} \over 3} = {{y + 1} \over {-2}},\,z = 2

and

{l_2}:{{x - 1} \over 1} = {{2y + 3} \over \alpha } = {{z + 5} \over 2}

are perpendicular, then an angle between the lines l2 and

{l_3}:{{1 - x} \over 3} = {{2y - 1} \over { - 4}} = {z \over 4}

is :

A

{\cos ^{ - 1}}\left( {{{29} \over 4}} \right)

B

{\sec ^{ - 1}}\left( {{{29} \over 4}} \right)

C

{\cos ^{ - 1}}\left( {{2 \over {29}}} \right)

D

{\cos ^{ - 1}}\left( {{2 \over {\sqrt {29} }}} \right)

Correct Answer

Option B

Solution

$\because$ L1 and L2 are perpendicular, so

3 \times 1 + ( - 2)\left( {{\alpha \over 2}} \right) + 0 \times 2 = 0

\Rightarrow \alpha = 3

Now angle between l2 and l3,

\cos \theta = {{1( - 3) + {\alpha \over 2}( - 2) + 2(4)} \over {\sqrt {1 + {{{\alpha ^2}} \over 4} + } 4\sqrt {9 + 4 + 16} }}

\Rightarrow \cos \theta = {2 \over {{{29} \over 2}}} \Rightarrow \theta = {\cos ^{ - 1}}\left( {{4 \over {29}}} \right) = {\sec ^{ - 1}}\left( {{{29} \over 4}} \right)

Q115

Let the plane 2x + 3y + z + 20 = 0 be rotated through a right angle about its line of intersection with the plane x

-

3y + 5z = 8. If the mirror image of the point

\left( {2, - {1 \over 2},2} \right)

in the rotated plane is B(a, b, c), then :

A

{a \over 8} = {b \over 5} = {c \over { - 4}}

B

{a \over 4} = {b \over 5} = {c \over { - 2}}

C

{a \over 8} = {b \over { - 5}} = {c \over 4}

D

{a \over 4} = {b \over 5} = {c \over 2}

Correct Answer

Option A

Solution

Consider the equation of plane,

P:(2x + 3y + z + 20) + \lambda (x - 3y + 5z - 8) = 0

P:(2 + \lambda )x + 3(3 - 3\lambda )y + 1(1 + 5\lambda )z + (20 - 8\lambda ) = 0

$\because$ Plane P is perpendicular to

2x + 3y + z + 20 = 0

So,

4 + 2\lambda + 9 - 9\lambda + 1 + 5\lambda = 0

\Rightarrow \lambda = 7

P:9x - 18y + 36z - 36 = 0

or

P:x - 2y + 4z = 4

If image of

\left( {2, - {1 \over 2},2} \right)

in plane P is (a, b, c) then

{{a - 2} \over 1} = {{b + {1 \over 2}} \over { - 2}} = {{c - 2} \over 4}

and

\left( {{{a + 2} \over 2}} \right) - 2\left( {{{b - {1 \over 2}} \over 2}} \right) + 4\left( {{{c + 2} \over 2}} \right) = 4

Clearly

a = {4 \over 3}

,

b = {5 \over 6}

and

c = - {2 \over 3}

So,

a:b:c = 8:5: - 4

Q116

If the plane

2x + y - 5z = 0

is rotated about its line of intersection with the plane

3x - y + 4z - 7 = 0

by an angle of

{\pi \over 2}

, then the plane after the rotation passes through the point :

A (2,

-

2, 0)

B (

-

2, 2, 0)

C (1, 0, 2)

D (

-

1, 0,

-

2)

Correct Answer

Option C

Solution

{P_1}:2x + y - 52 = 0

,

{P_2}:3x - y + 4z - 7 = 0

Family of planes P1 and P2

P:{P_1} + \lambda {P_2}

$\therefore$

P:(2 + 3\lambda )x + (1 - \lambda )y + ( - 5 + 4\lambda )z - 7\lambda = 0

$\because$

P \bot {P_1}

$\therefore$

4 + 6\lambda + 1 - \lambda + 25 - 20\lambda = 0

\lambda = 2

$\therefore$

P:8x - y + 32 - 14 = 0

It passes through the point (1, 0, 2)

Q117

Let

\overrightarrow a = \widehat i + \widehat j + 2\widehat k

,

\overrightarrow b = 2\widehat i - 3\widehat j + \widehat k

and

\overrightarrow c = \widehat i - \widehat j + \widehat k

be three given vectors. Let

\overrightarrow v

be a vector in the plane of

\overrightarrow a

and

\overrightarrow b

whose projection on

\overrightarrow c

is

{2 \over {\sqrt 3 }}

. If

\overrightarrow v \,.\,\widehat j = 7

, then

\overrightarrow v \,.\,\left( {\widehat i + \widehat k} \right)

is equal to :

A 6

B 7

C 8

D 9

Correct Answer

Option D

Solution

Let

\overrightarrow v = {\lambda _1}\overrightarrow a + {\lambda _2}\overrightarrow b

, where

{\lambda _1},\,{\lambda _2} \in R

.

= ({\lambda _1} + 2{\lambda _2})\widehat i + ({\lambda _1} - 3{\lambda _2})\widehat j + (2{\lambda _1} + {\lambda _2})\widehat k

$\because$ Projection of

\overrightarrow v

on

\overrightarrow c

is

{2 \over {\sqrt 3 }}

$\therefore$

{{{\lambda _1} + 2{\lambda _2} - {\lambda _1} + 3{\lambda _2} + 2{\lambda _1} + {\lambda _2}} \over {\sqrt 3 }} = {2 \over {\sqrt 3 }}

$\therefore$

{\lambda _1} + 3{\lambda _2} = 1

..... (i) and

\overrightarrow v \,.\,\widehat j = 7 \Rightarrow {\lambda _1} - 3{\lambda _2} = 7

... (ii) from equation (i) and (ii)

{\lambda _1} = 4

,

{\lambda _2} = - 1

$\therefore$

\overrightarrow v = 2\widehat i + 7\widehat j + 7\widehat k

$\therefore$

\overrightarrow v \,.\,(\widehat i + \widehat k) = 2 + 7

= 9

Q118

Let p be the plane passing through the intersection of the planes

\overrightarrow r \,.\,\left( {\widehat i + 3\widehat j - \widehat k} \right) = 5

and

\overrightarrow r \,.\,\left( {2\widehat i - \widehat j + \widehat k} \right) = 3

, and the point (2, 1,

-

2). Let the position vectors of the points X and Y be

\widehat i - 2\widehat j + 4\widehat k

and

5\widehat i - \widehat j + 2\widehat k

respectively. Then the points :

A X and X + Y are on the same side of P

B Y and Y

-

X are on the opposite sides of P

C X and Y are on the opposite sides of P

D X + Y and X

-

Y are on the same side of P

Correct Answer

Option C

Solution

Let the equation of required plane

\pi :(x + 3y - z - 5) + \lambda (2x - y + z - 3) = 0

$\because$ (2, 1, $-$ 2) lies on it so,

2 + \lambda ( - 2) = 0

\Rightarrow \lambda = 1

Hence,

\pi :3x + 2y - 8 = 0

$\because$

{\pi _x} = - 9

,

{\pi _y} = 5

,

{\pi _{x + y}} = 4

{\pi _{x - y}} = - 22

and

{\pi _{y - x}} = 6

Clearly X and Y are on opposite sides of plane $\pi$

Q119

Let Q be the mirror image of the point P(1, 0, 1) with respect to the plane S : x + y + z = 5. If a line L passing through (1,

-

1,

-

1), parallel to the line PQ meets the plane S at R, then QR2 is equal to :

A 2

B 5

C 7

D 11

Correct Answer

Option B

Solution

As L is parallel to PQ d.r.s of S is $\therefore$

L \equiv {{x - 1} \over 1} = {{y + 1} \over 1} = {{z + 1} \over 1}

Point of intersection of L and S be $\lambda$

\Rightarrow (\lambda + 1) + (\lambda - 1) + (\lambda - 1) = S

\Rightarrow \lambda = 2

$\therefore$

R \equiv (3,1,1)

Let

Q(\alpha ,\beta ,\gamma )

\Rightarrow {{\alpha - 1} \over 1} = {\beta \over 1} = {{\gamma - 1} \over 1} = {{ - 2( - 3)} \over 3}

\Rightarrow \alpha = 3,\,\beta = 2,\,\gamma = 3

\Rightarrow Q \equiv (3,2,3)

{(QR)^2} = {0^2} + {(1)^2} + {(2)^2} = 5

Q120

The shortest distance between the lines

\dfrac{x-3}{4}=\dfrac{y+7}{-11}=\dfrac{z-1}{5}

and

\dfrac{x-5}{3}=\dfrac{y-9}{-6}=\dfrac{z+2}{1}

is:

A

\dfrac{185}{\sqrt{563}}

B

\dfrac{187}{\sqrt{563}}

C

\dfrac{178}{\sqrt{563}}

D

\dfrac{179}{\sqrt{563}}

Correct Answer

Option B

Solution

Given lines are

\frac{x-3}{4}=\frac{y-(-7)}{-11}=\frac{z-1}{5}

and

\frac{x-5}{3}=\frac{y-9}{-6}=\frac{z-(-2)}{1}

Shortest distance between two lines,

\begin{aligned} & d=\frac{\left|\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)\right|}{\left|\left(\vec{b}_1 \times \vec{b}_2\right)\right|} \\ & \vec{a}_2-\vec{a}_1=2 \hat{i}+16 \hat{j}-3 \hat{k} \text{ and } \\ & \vec{b}_1 \times \vec{b}_2=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 4 & -11 & 5 \\ 3 & -6 & 1 \end{array}\right|=19 \hat{i}+11 \hat{j}+9 \hat{k} \\ & \therefore \quad d=\frac{187}{\sqrt{563}} \end{aligned}