JEE PYQ: Vector 3D Question 38
Question 38 - 2020 (06 Sep Shift 1)
The shortest distance between the lines $\frac{x - 1}{0} = \frac{y + 1}{-1} = \frac{z}{1}$ and $x + y + z + 1 = 0$, $2x - y + z + 3 = 0$ is:
(1) $1$
(2) $\frac{1}{\sqrt{3}}$
(3) $\frac{1}{\sqrt{2}}$
(4) $\frac{1}{2}$