Question

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

• A. $\left(\frac{8}{3}, \frac{4}{3},-\frac{7}{3}\right)$
• B. $\left(\frac{4}{3},-\frac{4}{3}, \frac{1}{3}\right)$
• C. $\left(\frac{1}{3}, \frac{2}{3}, \frac{10}{3}\right)$
• D. $\left(\frac{2}{3},-\frac{1}{3}, \frac{5}{2}\right)$

Answer 1

Answer: (A)

Distance of point $(1,-2,1)$ from plane $x +2 y -2 z =\alpha$ is $5 $
$\Rightarrow \alpha=10$
Equation of PQ $\frac{x-1}{1}=\frac{y+2}{2}=\frac{z-1}{-2}=t$
$Q \equiv( t +1,2 t -2,-2 t +1)$ and $PQ =5 $
$\Rightarrow t =\frac{5+\alpha}{9}=\frac{5}{3} $
$\Rightarrow Q \equiv\left(\frac{8}{3}, \frac{4}{3}, \frac{-7}{3}\right)$