JEE Main 2020 — Three Dimensional Geometry Question with Solution

We know that shortest distance between two skew lines exists along the line which is perpendicular to both the lines.

Now finding the equation of a plane $\left(\mathrm{P}\right)$, whose normal is perpendicular to the given first line and the line obtained by planes $x+y+z+1=0$, $2x-y+z+3=0$.

So, $\mathrm{P}$ is:

$x+y+z+1+\lambda (2x-y+z+3)=0$

It should be parallel to given line. $\Rightarrow 0(1+2\lambda )-1(1-\lambda )+1(1+\lambda )=0$
$\Rightarrow \lambda =0$
Thus, Plane, $\mathrm{P} \text{is} x+y+z+1=0$

So, the shortest distance between the lines will be same as distance between the plane $\mathrm{P}$ and given line $\frac{x-1}{0}=\frac{{\displaystyle y+1}}{{\displaystyle -1}}=\frac{{\displaystyle z}}{{\displaystyle 1}}$

Shortest distance $=$ Perpendicular distance of $(1,-1,0)$ from this plane $=\frac{|1-1+0+1|}{\sqrt{1^2+1^2+1^2}}=\frac{1}{\sqrt{3}}$