Question

The shortest distance between the lines
$\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}$ and
$\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}$ is:

• A. $2\sqrt{30}$
• B. $\frac{7}{2}\sqrt{30}$
• C. $3$
• D. $3\sqrt{30}$

Answer 1

The correct option is D $3\sqrt{30}$

$\overrightarrow{AB}=-3\hat{i}-7\hat{j}+6\hat{k}-(3\hat{i}+8\hat{j}+3\hat{k})=-6\hat{i}-15\hat{j}+3\hat{k}$
$\overrightarrow{p}=3\hat{i}-\hat{j}+\hat{k}$
$\overrightarrow{q}=-3\hat{i}+2\hat{j}+4\hat{k}$
$\overrightarrow{p}\times \overrightarrow{q}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -1& 1\\ -3& 2& 4\end{array}\right|=-6\hat{i}-15\hat{j}+3\hat{k}$

Shortest distance $=\frac{|\overrightarrow{AB}\cdot (\overrightarrow{p}\times \overrightarrow{q})|}{|\overrightarrow{p}\times \overrightarrow{q}|}=\frac{|36+225+9|}{\sqrt{36+225+9}}=3\sqrt{30}$.