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 form 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. $(\frac{2}{3},\frac{1}{3},\frac{5}{2})$

Answer 1

The correct option is A

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

Distance of point P from plane = 5
$\therefore 5=\left|\frac{1-4-2-\alpha }{3}\right|$

$\Rightarrow \alpha =10$
Foot of perpendicular
$\frac{x-1}{1}=\frac{y+2}{2}=\frac{z-1}{-2}=\lambda$

So, $x=\lambda +1,y=2\lambda -2,z=-2\lambda +1)$ and these must satisfy the equation of plane,$x+2y-2z=10$

$\Rightarrow x=\frac{8}{3},y=\frac{4}{3},z=-\frac{7}{3}$
Thus, the foot of the perpendicular is $A(\frac{8}{3},\frac{4}{3},-\frac{7}{3})$