Question

The distance of the point $P(-2,3,1)$ from the line $QR$, through $Q(-3,6,2)$ which makes equal angles with the axes is

• A. $3$
• B. $8$
• C. $\sqrt{2}$
• D. $2\sqrt{2}$

Answer 1

The correct option is D $2\sqrt{2}$

QR makes equal angles with axis $\pm \frac{1}{\sqrt{3}}$

DR of $QR$ are $(\pm \frac{1}{\sqrt{3}},\pm \frac{1}{\sqrt{3}},\pm \frac{1}{\sqrt{3}})$

DR of $QR$ are $(\pm 1,\pm 1,\pm 1)$

$\Rightarrow QR:\frac{x+3}{1}=\frac{y-6}{1}=\frac{z-2}{1}=t$

Foot of perpendicular from $P$ be $P^{\prime }$ $(t-3,t+6,t+2)$

DR's of $P{P}^{\prime }$ are perpendicular to $QR$

DR's of $P{P}^{\prime }(t-3-1-2,t+6-3,t+2-1)$

$=(t-1,t+3,t+1)$

$\overrightarrow{P{P}^{\prime }}.\overrightarrow{QR}=0$

$\Rightarrow (t-1)+t+3+t+1=0$

$3t+3=0$

$t=-1$

$\Rightarrow P^{\prime }(-4,5,1)$

$\left|P{P}^{\prime }\right|=\sqrt{{(-2+4)}^2+{(3-5)}^2+{(1-1)}^2}=2\sqrt{2}$