The shortest distance between the lines $$\frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5}$$ and $$\frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3}$$ is

Given two lines are represented as: $ \frac{x-3}{2} = \frac{y+15}{-7} = \frac{z-9}{5} $ and $ \frac{x+1}{2} = \frac{y-1}{1} = \frac{z-9}{-3} $ The formula for the shortest distance between two lines is: $ d = \frac{|\left(\vec{a}_2 - \vec{a}_1\right) \cdot \left(\vec{b}_1 \times \vec{b}_2\right)|}{|\vec{b}_1 \times \vec{b}_2|} $ From the given lines: $ \vec{a}_1 = (3, -15, 9) $ $ \vec{a}_2 = (-1, 1, 9) $ $ \vec{b}_1 = (2, -7, 5) $ $ \vec{b}_2 = (2, 1, -3) $ Calculate the difference: $ \vec{a}_2

3D Geometry — Page 22 | JEE Mathematics

Q211

If the line

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

makes a right angle with the line

\dfrac{x+3}{3 \mu}=\dfrac{1-2 y}{6}=\dfrac{5-z}{7}

, then

4 \lambda+9 \mu

is equal to :

A 4

B 13

C 5

D 6

Correct Answer

Option D

Solution

\begin{aligned} & L_1: \frac{x-2}{(-3)}=\frac{y-\frac{2}{3}}{\left(\frac{4 \lambda+1}{3}\right)}=\frac{}{(-1)} \\ & L_2: \frac{x+3}{3 \mu}=\frac{y-\frac{1}{2}}{-3}=\frac{z-5}{-7} \\ & \because L_1 \perp L_2 \\ & \Rightarrow(-3)(3 \mu)+\left(\frac{4 \lambda+1}{3}\right)(-3)+(-1)(-7)=0 \\ & -9 \mu-4 \lambda-1+7=0 \\ & \Rightarrow 4 \lambda+9 \mu=6 \\ & \end{aligned}

Q212

Let

\mathrm{d}

be the distance of the point of intersection of the lines

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

and

\dfrac{x-7}{4}=\dfrac{y-9}{3}=\dfrac{z-4}{2}

from the point

(7,8,9)

. Then

\mathrm{d}^2+6

is equal to :

A 75

B 78

C 72

D 69

Correct Answer

Option A

Solution

\begin{aligned} & P_1:(3 k-6,2 k, k-1) \\ & P_2(4 \alpha+7,3 \alpha+9,2 \alpha+4) \\ & P_1 \equiv P_2 \\ & 3 k-6=4 \alpha+7 \Rightarrow 3 k-4 \alpha=13 \\ & 2 k=3 \alpha+9 \Rightarrow 2 k-3 \alpha=9 \\ & \therefore k=3, \alpha=-1 \\ & \therefore P_1:(3,6,2) \end{aligned}

Distance of

(3,6,2)

and

(7,8,9)

\begin{aligned} & =\sqrt{16+4+49}=\sqrt{69}=d \\ & d^2+6=69+6=75 \end{aligned}

Q213

Let

(\alpha, \beta, \gamma)

be the image of the point

(8,5,7)

in the line

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

. Then

\alpha+\beta+\gamma

is equal to :

A 16

B 20

C 18

D 14

Correct Answer

Option D

Solution

(x, y, z) \equiv(2 \lambda+1,3 \lambda+1,5 \lambda+2)

DR of

P Q

:

(2 \lambda-7,3 \lambda-6,5 \lambda-5)

DR of line :

(2,3,5)

\begin{array}{ll} \therefore & 2(2 \lambda-7)+3(3 \lambda-6)+5(5 \lambda-5)=0 \\ \Rightarrow & \lambda=\frac{3}{2} \\ \therefore \quad & Q\left(4, \frac{7}{2}, \frac{19}{2}\right) \\ \therefore \quad & (\alpha, \beta, y) \equiv(0,2,12) \quad\left(Q \text{ is mid point of } P \& P^{\prime}\right) \\ & \alpha+\beta+y \equiv 14 \end{array}

Q214

Let

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

be the image of the point

\mathrm{Q}(3,-3,1)

in the line

\dfrac{x-0}{1}=\dfrac{y-3}{1}=\dfrac{z-1}{-1}

and

\mathrm{R}

be the point

(2,5,-1)

. If the area of the triangle

\mathrm{PQR}

is

\lambda

and

\lambda^2=14 \mathrm{~K}

, then

\mathrm{K}

is equal to :

A 18

B 81

C 72

D 36

Correct Answer

Option B

Solution

\text{ Line } L \Rightarrow \frac{x-0}{1}=\frac{y-3}{1}=\frac{z-1}{-1}=\mu

Point

M(\mu, \mu+3,-\mu+1)

Direction vector of

Q M=(\mu-3, \mu+6,-\mu)

Direction vector of line

L=(1,1,-1)

Both direction vector are perpendicular so

\begin{aligned} & 1 \times(\mu-3)+1 \times(\mu+6)-1 \times(-\mu)=0 \\ & \mu=-1 \end{aligned}

Point

M(-1,2,2)

midpoint of

P Q

So point

P(\alpha, \beta, \gamma)=(-5,7,3)

Area of

\triangle P Q R=\frac{1}{2}|\overrightarrow{P Q} \times \overrightarrow{Q R}|

\begin{aligned} & \overrightarrow{P Q}=(8,-10,-2) \\ & \overrightarrow{Q R}=(-1,8,-2) \\ & \overrightarrow{P Q} \times \overrightarrow{Q R}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 8 & -10 & -2 \\ -1 & 8 & -2 \end{array}\right|=36 \hat{i}+18 \hat{j}+54 \hat{k} \end{aligned}

Area of

\triangle P Q R=\lambda

\lambda=\frac{1}{2} \sqrt{(36)^2+(18)^2+(54)^2}

\begin{aligned} & \lambda^2=\frac{1}{4}(4536) \\ & \lambda^2=1134 \\ & \lambda^2=14 K \text{ (given) } \\ & 14 K=1134 \\ & K=\frac{1134}{14}=81 \\ & K=81 \end{aligned}

Q215

If

A(3,1,-1), B\left(\dfrac{5}{3}, \dfrac{7}{3}, \dfrac{1}{3}\right), C(2,2,1)

and

D\left(\dfrac{10}{3}, \dfrac{2}{3}, \dfrac{-1}{3}\right)

are the vertices of a quadrilateral

A B C D

, then its area is

A

\dfrac{4 \sqrt{2}}{3}

B

\dfrac{2 \sqrt{2}}{3}

C

\dfrac{5 \sqrt{2}}{3}

D

2 \sqrt{2}

Correct Answer

Option A

Solution

A(3,1,-1), B\left(\frac{5}{3}, \frac{7}{3}, \frac{1}{3}\right), C(2,2,1), D\left(\frac{10}{3}, \frac{2}{3}, \frac{-1}{3}\right)

are vertices of a quadrilateral

\begin{aligned} & \overrightarrow{A C}=(2 \hat{i}+2 \hat{j}+\hat{k})-(3 \hat{i}+\hat{j}-\hat{k}) \\ & =-\hat{i}+\hat{j}+2 \hat{k} \\ & \overrightarrow{B D}=\left(\frac{10}{3} \hat{i}+\frac{2}{3} \hat{j}-\frac{1}{3} \hat{k}\right)-\left(\frac{5}{3} \hat{i}+\frac{7}{3} \hat{j}+\frac{1}{3} \hat{k}\right) \\ & \overrightarrow{B D}=\frac{5}{3} \hat{i}-\frac{5}{3} \hat{j}-\frac{2}{3} \hat{k} \\ & \text{ Area }=\frac{1}{2}|\overrightarrow{A C} \times \overrightarrow{B D}| \\ & =\frac{1}{2}\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 2 \\ \frac{5}{3} & \frac{-5}{3} & \frac{-2}{3} \end{array}\right| \\ & =\frac{1}{2} \sqrt{\left(\frac{8}{3}\right)^2+\left(\frac{8}{3}\right)^2}\left[\because \overrightarrow{A C} \times \overrightarrow{B D}=\frac{8}{3} \hat{i}+\frac{8}{3} \hat{j}\right] \\ & =\frac{1}{2} \times \frac{8}{3} \times \sqrt{2}=\frac{4 \sqrt{2}}{3} \\ \end{aligned}

Q216

The coordinates of the foot of the perpendicular from the point (1,

-

2, 1) on the plane containing the lines,

{{x + 1} \over 6} = {{y - 1} \over 7} = {{z - 3} \over 8}

and

{{x - 1} \over 3} = {{y - 2} \over 5} = {{z - 3} \over 7},

is :

A (2,

-

4, 2)

B (

-

1, 2,

-

1)

C (0, 0, 0)

D (1, 1, 1)

Correct Answer

Option C

Solution

\overrightarrow n

=

\overrightarrow {{n_1}} \times \overrightarrow {{n_2}}

=

\left| \begin{array}{lll}{\widehat i} & {\widehat j} & {\widehat k} \\ 6 & 7 & 8 \\ 3 & 5 & 7 \end{array} \right|

= (9, $-$ 18, 9) = (1, $-$ 2, 1) $\therefore$ Equation of plane is 1(x + 1) $-$ 2(y $-$ 1) + (z $-$ 3) = 0 $\Rightarrow$ x $-$ 2y + z = 0 foot to z

{{x - 1} \over 1}

=

{{y + 2} \over { - 2}}

=

{{z - 1} \over 1}

=

- {{\left[ {1 + 4 + 1} \right]} \over 6}

x = 0, y = 0, z = 0

Q217

The shortest distance between the lines

\dfrac{x-3}{2}=\dfrac{y+15}{-7}=\dfrac{z-9}{5}

and

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

is

A

8 \sqrt{3}

B

6 \sqrt{3}

C

5 \sqrt{3}

D

4 \sqrt{3}

Correct Answer

Option D

Solution

Given two lines are represented as: $\dfrac{x-3}{2} = \dfrac{y+15}{-7} = \dfrac{z-9}{5}$ and $\dfrac{x+1}{2} = \dfrac{y-1}{1} = \dfrac{z-9}{-3}$ The formula for the shortest distance between two lines is: $d = \dfrac{|\left(\vec{a}_2 - \vec{a}_1\right) \cdot \left(\vec{b}_1 \times \vec{b}_2\right)|}{|\vec{b}_1 \times \vec{b}_2|}$ From the given lines: $\vec{a}_1 = (3, -15, 9)$ $\vec{a}_2 = (-1, 1, 9)$ $\vec{b}_1 = (2, -7, 5)$ $\vec{b}_2 = (2, 1, -3)$ Calculate the difference: $\vec{a}_2 - \vec{a}_1 = (-1-3, 1+15, 9-9) = (-4, 16, 0)$ Next, compute the cross product: $\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -7 & 5 \\ 2 & 1 & -3 \end{vmatrix} = (16\hat{i} + 16\hat{j} + 16\hat{k})$ The magnitude of the cross product is: $|\vec{b}_1 \times \vec{b}_2| = |(16\hat{i} + 16\hat{j} + 16\hat{k})| = 16\sqrt{3}$ Calculate the dot product: $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (-4\hat{i} + 16\hat{j} + 0\hat{k}) \cdot (16\hat{i} + 16\hat{j} + 16\hat{k}) = -64 + 256 + 0 = 192$ Finally, find the shortest distance: $d = \dfrac{|192|}{16\sqrt{3}} = \dfrac{192}{16\sqrt{3}} = 4\sqrt{3}$ Thus, the shortest distance is: $\boxed{4 \sqrt{3}}$

Q218

Let

\mathrm{L}_1: \dfrac{x-1}{1}=\dfrac{y-2}{-1}=\dfrac{z-1}{2}

and

\mathrm{L}_2: \dfrac{x+1}{-1}=\dfrac{y-2}{2}=\dfrac{z}{1}

be two lines. Let

L_3

be a line passing through the point

(\alpha, \beta, \gamma)

and be perpendicular to both

L_1

and

L_2

. If

L_3

intersects

\mathrm{L}_1

, then

|5 \alpha-11 \beta-8 \gamma|

equals :

A 25

B 20

C 16

D 18

Correct Answer

Option A

Solution

Step 1. Identify the Given Lines The line

\mathrm{L}_1:\; \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2}

passes through the point

P_1=(1,2,1)

with direction vector

\mathbf{u}=(1,-1,2).

Similarly, the line

\mathrm{L}_2:\; \frac{x+1}{-1}=\frac{y-2}{2}=\frac{z}{1}

passes through the point

P_2=(-1,2,0)

with direction vector

\mathbf{v}=(-1,2,1).

Step 2.

Determine the Direction of $L_3$ Since $L_3$ is perpendicular to both $L_1$ and $L_2$ , its direction vector must be parallel to the cross product of $\mathbf{u}$ and $\mathbf{v}$ .

Compute:

\mathbf{d} = \mathbf{u}\times\mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 2 \\ -1 & 2 & 1 \end{vmatrix}.

Using the determinant, we obtain:

\begin{aligned} d_x &= (-1)(1) - (2)(2) = -1-4=-5,\\[1mm] d_y &= (2)(-1) - (1)(1) = -2-1=-3,\\[1mm] d_z &= (1)(2) - (-1)(-1) = 2-1=1. \end{aligned}

Thus, a valid direction vector is

\mathbf{d}=(-5,-3,1).

Step 3.

Express $L_3$ in Terms of Its Intersection with $L_1$ Since $L_3$ intersects $L_1$ , let the intersection point (on $L_1$ ) be expressed using a parameter $t$ as:

A(t) = (1+t,\, 2-t,\, 1+2t).

Since $L_3$ passes through $A(t)$ and also through the given point

Q=(\alpha,\beta,\gamma),

its parametric form can be written as:

(\alpha,\beta,\gamma)=A(t) + s\,\mathbf{d} = (1+t,\,2-t,\,1+2t) + s(-5,-3,1)

for some parameters $s$ and $t$ .

Step 4.

Express $\alpha,\beta,\gamma$ in Terms of $t$ and $s$ Comparing coordinates, we have:

\begin{cases} \alpha = 1 + t - 5s,\\[1mm] \beta = 2 - t - 3s,\\[1mm] \gamma = 1 + 2t + s. \end{cases}

Step 5. Compute $5\alpha - 11\beta - 8\gamma$ Substitute the expressions for $\alpha$ , $\beta$ , and $\gamma$ :

\begin{aligned} 5\alpha - 11\beta - 8\gamma &= 5(1+t-5s) - 11(2-t-3s) - 8(1+2t+s)\\[1mm] &= (5 + 5t - 25s) - (22 - 11t - 33s) - (8 + 16t + 8s)\\[1mm] &= \left(5 - 22 - 8\right) + \left(5t + 11t - 16t\right) + \left(-25s + 33s - 8s\right)\\[1mm] &= -25 + 0t + 0s\\[1mm] &= -25. \end{aligned}

Taking the absolute value yields:

\left|5\alpha-11\beta-8\gamma\right| = 25.

Conclusion The value of

\left|5\,\alpha - 11\,\beta - 8\,\gamma\right|

is

\boxed{25}.

Q219

Let a straight line

L

pass through the point

P(2, -1, 3)

and be perpendicular to the lines

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

and

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

. If the line

L

intersects the

yz

-plane at the point

Q

, then the distance between the points

P

and

Q

is:

A

\sqrt{10}

B

2

C

2\sqrt{3}

D

3

Correct Answer

Option D

Solution

To find the distance between the points $P$ and $Q$ , let's consider the following steps: Determine the direction vector of line $L$ : For the line $L$ to be perpendicular to both given lines, we find the direction vector of $L$ using the cross product of the direction vectors of the given lines.

The first line has direction vector $\langle 2, 1, -2 \rangle$ and the second line has direction vector $\langle 1, 3, 4 \rangle$ .

The cross product is: $\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 3 & 4 \end{vmatrix} = 10\hat{i} - 10\hat{j} + 5\hat{k} = 5(2\hat{i} - 2\hat{j} + \hat{k})$ Thus, the direction vector of line $L$ is $\langle 2, -2, 1 \rangle$ .

Equation of line $L$ : Since $L$ passes through point $P(2, -1, 3)$ with direction vector $\langle 2, -2, 1 \rangle$ , its parametric form is: $\dfrac{x-2}{2} = \dfrac{y+1}{-2} = \dfrac{z-3}{1} = \lambda$ Find the intersection with the $yz$ -plane: On the $yz$ -plane, $x = 0$ .

Substitute into the line equation to find $\lambda$ : $2\lambda + 2 = 0 \implies \lambda = -1$ Substitute $\lambda = -1$ back into the parametric equations to find $Q$ : $x = 2(-1) + 2 = 0, \quad y = -2(-1) - 1 = 1, \quad z = (-1) + 3 = 2$ Therefore, the intersection point $Q$ is $(0, 1, 2)$ .

Calculate the distance between $P$ and $Q$ : $\text{d}(P, Q) = \sqrt{(2-0)^2 + (-1-1)^2 + (3-2)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$ Thus, the distance between points $P$ and $Q$ is 3.

Q220

Let P be the foot of the perpendicular from the point

\mathrm{Q}(10,-3,-1)

on the line

\dfrac{x-3}{7}=\dfrac{y-2}{-1}=\dfrac{z+1}{-2}

. Then the area of the right angled triangle

P Q R

, where

R

is the point

(3,-2,1)

, is

A

\sqrt{30}

B

9 \sqrt{15}

C

3 \sqrt{30}

D

8 \sqrt{15}

Correct Answer

Option C

Solution

\begin{aligned} &\begin{aligned} & \frac{x-3}{7}=\frac{y-2}{-1}=\frac{z+1}{-2}=\lambda \\ & \Rightarrow 7 \lambda+3,-\lambda+2,-2 \lambda-1 \\ & \text{ dr's of QP } \Rightarrow \\ & 7 \lambda-7,-\lambda+5,-2 \lambda \end{aligned}\\ &\text{ Now }\\ &\begin{aligned} & (7 \lambda-7) \cdot 7-(-\lambda+5)+(2 \lambda) \cdot 2=0 \\ & 54 \lambda-54=0 \Rightarrow \lambda=1 \\ & \therefore \mathrm{P}=(10,1,-3) \\ & \overrightarrow{\mathrm{PQ}}=-4 \hat{\mathrm{j}}+2 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{PR}}=-7 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \end{aligned} \end{aligned}

Area

=\left| {{1 \over 2}\left| \begin{array}{lll}i & j & k \\ 0 & { - 4} & 2 \\ { - 7} & { - 3} & 4 \end{array} \right|} \right| = 3\sqrt {30}