Question

Find the length of the perpendicular from $(2,-3,1)$ to the line $\frac{x+1}{2}=\frac{y-3}{3}=\frac{z+2}{-1}$.

Answer 1

Given:

Point: $P(2,-3,1)$

Line: $\frac{x+1}{2}=\frac{y-3}{3}=\frac{z+2}{-1}=\lambda$ (say)

Any point on this line will be

$2\lambda -1,3\lambda +3,-\lambda -2$

Let $Q\left(2\lambda -1,3\lambda +3,-\lambda -2\right)$ be the foot of perpendicular

Directional ratio of perpendicular $\overrightarrow{PQ}$ is $\left(2\lambda -1-2,3\lambda +3-\left(-3\right),-\lambda -2-1\right)$

$\Rightarrow \left(2\lambda -3,3\lambda +6,-\lambda -3\right)$

Since $\overrightarrow{PQ}$ is perpendicular to the given line

$\begin{array}{c}\left(2,-3,1\right).\left(\overrightarrow{PQ}\right)=0\\ \left(2\lambda -3\right)2+\left(3\lambda +6\right)3-1\left(-\lambda -3\right)=0\\ \lambda =\frac{-15}{14}\end{array}$

Point $Q\left(\left(\frac{-30}{14}-1\right),\left(3\left(\frac{-15}{14}\right)+3\right),\left(-\left(-\frac{15}{14}\right)-2\right)\right)$

$\begin{array}{c}Q\left(\frac{-44}{14},\frac{-3}{14},\frac{-13}{14}\right)\\ \Rightarrow \sqrt{{\left(\frac{-44-28}{14}\right)}^2+{\left(\frac{-3+42}{14}\right)}^2+{\left(\frac{-13-14}{14}\right)}^2}\\ \Rightarrow \frac{86}{14}\approx 6\end{array}$