Question

Find the shortest distance between the skew lines :

$l_1:\frac{x-1}{2}=\frac{y+1}{1}=\frac{z-2}{4}$

$l_2:\frac{x+2}{4}=\frac{y-0}{-3}=\frac{z+1}{1}$

Answer 1

Line $l_1$ passes through the point $P(1,-1,2)$.

The equation of the plane containing line $l_2$ is

$a(x+2)+b(y-0)+c(z+1)=0$ .........(1)

where, $4a-3b+c=0$

If it is parallel to line $l_1$, then

$2a+b+4c=0$

Therefore, $\frac{a}{-13}=\frac{b}{-14}=\frac{c}{10}$

Substituting values of a, b. c in equation 1 , we get,

$13x+14y-10z+16=0$

This is the equation of the plane containing line $l_2$ and parallel to line $l_1$.

Shortest distance between lines $l_1$ and $l_2$ = $|\frac{13-14-20+16}{\sqrt{{13}^2+{14}^2+(-10{)}^2}}|=\frac{5}{\sqrt{465}}$