If the distance of the point $P(1,-2,1)$ from the plane $x+2 y-2 z=\alpha$, where $\alpha0$, is 5 , then the foot of the perpendicular from $P$ to the plane is

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

Aicharya · Accepted Answer

$ 	ext { Distance of point } P 	ext { from plane }=5 $ $ \begin{gathered} 	herefore=\left|\frac{1-4-2-\alpha}{3}ight| \ \alpha=10 \end{gathered} $ Foot of perpendicular $ \begin{aligned} \quad \frac{x-1}{1} & =\frac{y+2}{2}=\frac{z-1}{-2}=\frac{5}{3} \ \Rightarrow \quad \quad \quad x & =\frac{8}{3}, y=\frac{4}{3}, z=-\frac{7}{3} \end{aligned} $ Thus, the foot of the perpendicular is $ A\left(\frac{8}{3}, \frac{4}{3},-\frac{7}{3}ight) $

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…

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