Question

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:

• A. $1$
• B. $\frac{1}{\sqrt{2}}$
• C. $\frac{1}{\sqrt{3}}$
• D. $\frac{1}{2}$

Answer 1

The correct option is C $\frac{1}{\sqrt{3}}$

Plane through line of intersection is
$x+y+z+1+\lambda (2x-y+z+3)=0$

It should be parallel to given line
$0(1+2\lambda )-1(1-\lambda )+1(1+\lambda )=0$
$\Rightarrow \lambda =0$
Plane $:x+y+z+1=0$
Shortest distance of $(1,-1,0)$ from this plane
$=\frac{\left|\begin{array}{c}1-1+0+1\end{array}\right|}{\sqrt{1^2+1^2+1^2}}=\frac{1}{\sqrt{3}}$