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

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
$-x+y+z-1=0$

$\frac{x_{1}-1}{-1}=\frac{y_{1}-2}{1}=\frac{z_{1}+1}{1}=\frac{1}{3}$
$x _{1}=\frac{2}{3}, y_{1}=\frac{7}{3}, z_{1}=\frac{-2}{3}$
$M \left(\frac{2}{3}, \frac{7}{3}, \frac{-2}{3}\right)$
$N \left( x _{2}, y _{2}, z _{2}\right)$
$\frac{ x _{2}-2}{-1}=\frac{ y _{2}+1}{1}=\frac{ z _{2}-3}{1}=\frac{1}{3}$
$x _{2}=\frac{5}{3}, y _{2}=\frac{-2}{3}, z _{2}=\frac{10}{3}$
$N =\left(\frac{5}{3}, \frac{-2}{3}, \frac{10}{3}\right)$
$d ^{2}=1^{2}+3^{2}+4^{2}=26$