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

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

Answer 1

The correct option is A

$(\frac{8}{3},\frac{4}{3},-\frac{7}{3})$

Distance of the point $P(1,-2,1)$ from the plane $x+2y-2z=\alpha$ is 5, then

$\left|\frac{1-4-2-\alpha }{3}\right|=5$

$\Rightarrow |-5-\alpha |=15\Rightarrow \alpha =10$

Let the foot of the perpendicular be $M,$ then

$\frac{x-1}{1}=\frac{y+2}{2}=\frac{z-1}{-2}=\lambda$

$\therefore M\equiv (\lambda +1,2\lambda -2,1-2\lambda )$ lies on the plane

$\Rightarrow \lambda +1+2(2\lambda -2)-2(1-2\lambda )-10=0$

$\Rightarrow \lambda +1+4\lambda -4-2+4\lambda -10=0$

$\Rightarrow 9\lambda =15$

$\Rightarrow \lambda =\frac{5}{3}\Rightarrow M(\frac{8}{3},\frac{4}{3},-\frac{7}{3})$