Question

Let $d$ be the distance between the foot of perpendiculars of the points $P(1,2-1)$ and $Q(2,-1,3)$ on the plane $-x+y+z=1$. Then $d^2$ is equal to _______.

Answer 1

Answer: 26

Points $P(1,2,-1)$ and $Q(2,-1,3)$ lie on same side of the plane.
Perpendicular distance of point $P$ from plane is
$\left|\frac{-1+2-1-1}{\sqrt{1^2+1^2+1^2}}\right|=\frac{1}{\sqrt{3}}$
Perpendicular distance of point $Q$ from plane is
$=\left|\frac{-2-1+3-1}{\sqrt{1^2+1^2+1^2}}\right|=\frac{1}{\sqrt{3}}$
$\Rightarrow \overrightarrow{PQ}$ is parallel to given plane. So, distance between $P$ and $Q=$ distance between their foot of perpendiculars.
$\Rightarrow |\overrightarrow{PQ}|=\sqrt{(1-2{)}^2+(2+1{)}^2+(-1-3{)}^2}$
$=\sqrt{26}$
$|\overrightarrow{PQ}{|}^2=26=d^2$
Alternate
$-x+y+z-1=0$

$\frac{x_1-1}{-1}=\frac{y_1-2}{1}=\frac{z_1+1}{1}=\frac{1}{3}$
$x_1=\frac{2}{3},y_1=\frac{7}{3},z_1=\frac{-2}{3}$
$M\left(\frac{2}{3},\frac{7}{3},\frac{-2}{3}\right)$
$N\left(x_2,y_2,z_2\right)$
$\frac{x_2-2}{-1}=\frac{y_2+1}{1}=\frac{z_2-3}{1}=\frac{1}{3}$
$x_2=\frac{5}{3},y_2=\frac{-2}{3},z_2=\frac{10}{3}$
$N=\left(\frac{5}{3},\frac{-2}{3},\frac{10}{3}\right)$
$d^2=1^2+3^2+4^2=26$