Question

The shortest distance between the lines $\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-6}{2}$ and $\frac{x-6}{3}=\frac{1-y}{2}=\frac{z+8}{0}$ is equal to

Answer 1

Answer: 14

Shortest distance between the lines
$=\frac{\left|\begin{array}{ccc}4& 2& -14\\ 3& 2& 2\\ 3& -2& 0\end{array}\right|}{\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 2& 2\\ 3& -2& 0\end{array}\right|}$
$=\frac{16+12+168}{|-4\hat{i}+6\hat{j}-12\hat{k}|}=\frac{196}{14}=14$