JEE Main 2022 · Mathematics · Three Dimensional Geometry · hard

The square of the shortest distance between the lines vec r=(hat i+2hat j+3hat k)+λ(hat i+hat j) and vec r=(2hat i+2hat j+5hat k)+μ(hat j+hat k) is

Numerical answer

The square of the shortest distance between the lines

\vec r=(\hat i+2\hat j+3\hat k)+\lambda(\hat i+\hat j)

and

\vec r=(2\hat i+2\hat j+5\hat k)+\mu(\hat j+\hat k)

Show answer and solution

Answer: 3

Direction vectors are

(1,1,0)

and

(0,1,1)

; their cross product is

\vec n=(1,-1,1)

with

|\vec n|^2=3

. The vector between points on the lines is

(1,0,2)

, and

(1,0,2)\cdot\vec n=1+2=3

. Shortest distance

=\frac{3}{\sqrt3}=\sqrt3

, so its square is

3

Original question written in the JEE Main 2022 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake