Question

If the distance of the point $P(1,-2,1)$ from the plane $x+2y-2z=\alpha$, where $\alpha >0$ is $5$, then the foot of the perpendicular from $p$ to the plane is

Answer 1

$P(1,-2,1),$ Plane$:x+2y-2z=\alpha$

distance of $P$ from plane is,

$\frac{1+2\times (-2)-2-\alpha }{\sqrt{1^2+2^2+2^2}}=\pm 5-5-\alpha =\pm 15\alpha =-18,8$

Foot of perpendicular $(1\pm \frac{5}{3},-2\pm \frac{2}{3},1\mp \frac{2}{3})$

$\therefore$ Foot of perpendicular are $(\frac{8}{3},\frac{-4}{3},\frac{1}{3})(\frac{-2}{3},\frac{-8}{3},\frac{5}{3})$