$$d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}$$
$\vec{b_1} \times \vec{b_2} = -4\hat{i} + 7\hat{j} - 13\hat{k}$
$$d = \frac{10}{\sqrt{59}}$$
Back to JEE Main 2022 Vectors & 3D Geometry
JEE Main 2022 — Mathematics Vectors & 3D Geometry
medium
mcq
2022
Official previous-year question
Verified 30 May 2026.
Question
The shortest distance between the lines $\frac{x-1}{2} = \frac{y+1}{3} = \frac{z}{1}$ and $\frac{x+1}{5} = \frac{y-2}{1} = \frac{z}{-1}$ is:
Options
- A
$\frac{10}{\sqrt{59}}$
- B
$\frac{10}{\sqrt{29}}$
- C
$\frac{5}{\sqrt{59}}$
- D
$\frac{20}{\sqrt{59}}$
Solution
Did you get this right?
Sign in to track your attempts and accuracy.
Your note
Sign in to keep a private note on this question. Nothing you write is ever public.
JEE Main Vectors & 3D Geometry in other years
More JEE Main Vectors & 3D Geometry Questions
The volume of the parallelepiped formed by vectors a=i+2j-k, b=2i-j+3k, c=3i+j+2k is:
2026
hard
mcq
The position vectors of points A and B are 2i+3j+k and 4i+j-2k. The position vector of the midpoint of AB is:
2025
easy
mcq
The image of the point (1,2,3) in the plane x+y+z=9 is:
2024
hard
mcq
The projection of vector a=2i-3j+6k on vector b=i+2j+2k is:
2024
medium
mcq
The scalar triple product [i j k] is:
2023
easy
mcq