Question

Find the shortest distance between the point $(–2,4,–5)$ and the line $\frac{x-2}{3}=\frac{y+2}{1}=\frac{z+1}{2}$.

Answer 1

Here, $P(–2,4,–5)$ is the given point.
$\frac{x-2}{3}=\frac{y+2}{1}=\frac{z+1}{2}=\lambda$
$\therefore$ Any point $Q$ on the line is $(3\lambda +2,\lambda -2,2\lambda -1)$.

$\Rightarrow \overrightarrow{PQ}=(3\lambda +4)\hat{i}+(\lambda -6)\hat{j}+(2\lambda +4)\hat{k}$

For the shortest distance, $\overrightarrow{PQ}\perp (3\hat{i}+\hat{j}+2\hat{k})$
$\Rightarrow (3\lambda +4)\times 3+(\lambda -6)\times 1+(2\lambda +4)\times 2=0$
$\Rightarrow 14\lambda +14=0$
$\Rightarrow \lambda =-1$

$\Rightarrow \overrightarrow{PQ}=\hat{i}-7\hat{j}+2\hat{k}$
$\Rightarrow \left|\overrightarrow{PQ}\right|=\sqrt{1^2+(-7{)}^2+2^2}$
$=\sqrt{54}=3\sqrt{6}$

$\therefore$ Shortest distance between the point $(–2,4,–5)$ and the line $\frac{x-2}{3}=\frac{y+2}{1}=\frac{z+1}{2}$ is $3\sqrt{6}$ units.