JEE Advanced 2024 Paper 2 · Mathematics · Three Dimensional Geometry · medium

(x-1)/(1)=(y-2)/(2)=(z-3)/(2) and L2: (x-2)/(2)=(y-1)/(-1)=z/2.

One or more options correct

Consider the lines

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

and

L_2:\ \dfrac{x-2}{2}=\dfrac{y-1}{-1}=\dfrac{z}{2}

. Which of the following statements is/are correct?

A. $L_1$ and $L_2$ are skew lines.
B. The shortest distance between $L_1$ and $L_2$ is $\dfrac{19}{\sqrt{65}}$ .
C. The plane containing $L_1$ and parallel to $L_2$ is $6x+2y-5z+5=0$ .
D. The acute angle $\theta$ between $L_1$ and $L_2$ satisfies $\cos\theta=\dfrac{4}{7}$ .

Show answer and solution

Answer: A, B, C

Direction vectors

\vec d_1=(1,2,2)

\vec d_2=(2,-1,2)

;

\vec d_1\times\vec d_2=(6,2,-5)

with magnitude

\sqrt{65}

. The vector joining points

(1,2,3)

and

(2,1,0)

(1,-1,-3)

and its dot product with the cross product is

6-2+15=19\neq0

, so the lines are skew (A true) and the shortest distance is

19/\sqrt{65}

(B true). The plane through

(1,2,3)

with normal

(6,2,-5)

6x+2y-5z=6+4-15=-5

, i.e.

6x+2y-5z+5=0

(C true). Finally

\cos\theta=\dfrac{|\vec d_1\cdot\vec d_2|}{|\vec d_1||\vec d_2|}=\dfrac{4}{3\cdot3}=\dfrac49

, not

\dfrac47

(D false).

Original question written in the JEE Advanced 2024 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake