# Find the coordinates of the foot Q of the perpendicular drawn from the point P(1,3,4) to the plane $2x-y+z+3= 0\cdot$ Find the distance PQ and yhe image of P treating the plane as a mirror.

Consider the figure.

Equation of PQ : $\frac{x-1}{2}= \frac{y-3}{-1}= \frac{z-4}{1}= \lambda$
coordinates of the random point on the lines is $Q\left ( 2\lambda +1,3-\lambda ,\lambda +4 \right )$. As Q lies on the plane $2x-y+z+3= 0$
$\lambda = -1$
The foot of perpendicular is $Q\left ( -1,4,3 \right )$
$\therefore PQ= \sqrt{\left ( 1+1 \right )^{2}+\left ( 3-4 \right )^{2}+\left ( 4-3 \right )^{2}}= \sqrt{6}\: units$
let $M\left ( h,p,s \right )$  be  the image of P in the plane.
clearly Q is the midpoint of PM so
$Q\left ( \frac{h+1}{2},\frac{p+3}{2},\frac{s+4}{2} \right )= Q\left ( -1,4,3 \right )$
On comparing, the coordinates we get $= h= -3,p= 5\; and\; s= 2$
$\therefore The \; image \; is\; M\left ( -3,5,2 \right )$

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