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Mathematics Question on Three Dimensional Geometry

If the distance of the point P(1,−2,1)P (1, - 2,1) from the plane x+2y−2z=ax + 2y - 2z = a, where a>0a > 0, is 55, then the foot of the perpendicular form P P to the plane is

A

(83,43,−73)\bigg(\frac{8}{3},\frac{4}{3},-\frac{7}{3}\bigg)

B

(43,43,13)\bigg(\frac{4}{3},\frac{4}{3},\frac{1}{3}\bigg)

C

(13,23,103)\bigg(\frac{1}{3},\frac{2}{3},\frac{10}{3}\bigg)

D

(23,13,52)\bigg(\frac{2}{3},\frac{1}{3},\frac{5}{2}\bigg)

Answer

(83,43,−73)\bigg(\frac{8}{3},\frac{4}{3},-\frac{7}{3}\bigg)

Explanation

Solution

Distance of point P from plane = 5
\therefore \hspace15mm 5= \bigg|\frac{1-4-2-\alpha}{3}\bigg|
\Rightarrow \hspace15mm \alpha =0
Foot of perpendicular
\hspace30mm \frac{x-1}{1} = \frac{y+2}{2} = \frac{z-1}{-2}= \frac{5}{3}
\Rightarrow \hspace30mm x=\frac{8}{3}, y = \frac{4}{3},z = - \frac{7}{3}
Thus, the foot of the perpendicular is A(83,43,−73). A \bigg(\frac{8}{3}, \frac{4}{3},- \frac{7}{3}\bigg).