Find the coordinates of the foot of the perpendicular Q drawn from to the plane <img alt="2x-y+z

Find the coordinates of the foot of the perpendicular Q drawn from $P\left ( 3,2,1 \right )$ to the plane $2x-y+z+1= 0$ . Also, find the distance PQ and the image of the point P treating this plane as a mirror.

Answers (1)

The d.r's of normal to the plane are 2,-1,1
$\because$ PQ is $\perp^{r}$ to the plane so, its equation is:
$\frac{x-3}{2}= \frac{y-2}{-1}= \frac{z-1}{1}= \lambda$
The coordinates of any random point on the line PQ
$Q\left ( 2\lambda +3,-\lambda +2,\lambda +1 \right )$
$\because$ Q lines on the plane so, $2\left ( 2\lambda +3 \right )-\left ( \; \lambda+2 \right )+\left ( \lambda +1 \right )+1= 0$
$\Rightarrow 6\lambda +6= 0\; \; \therefore \lambda = -1$

$\therefore$ Foot of the perpendicular : $Q\left ( 1,3,0 \right )$
Distance $PQ\sqrt{\left ( 3-1 \right )^{2}+\left ( 2-3 \right )^{2}+\left ( 1-0 \right )^{2}}= \sqrt{6}\, units$
Let $M\left ( \alpha ,\beta ,\gamma \right )$ be the image of P in the plane
so, Q will be mid point of PM
ie $Q\left ( 1,3,0 \right )= Q\left ( \frac{\alpha +3}{2},\frac{\beta +2}{2},\frac{\gamma +1}{2} \right )$
On comparing the coordinates, we get : $\alpha = -1,\beta = 4,\gamma = -1$
$\therefore$ the image is $M\left ( -1,4,-1 \right ).$