# The image of the point $P(1,3,4)$ in the plane $2x-y+z+3=0$ is ?

$\begin{array}{1 1} (a)\:(3,5,-2)\:\:\:\qquad\:\:(b)\:(-3,5,2)\:\:\:\qquad\:\:(c)\:(3,-5,2)\:\:\:\qquad\:\:(d)\:(3,5,2) \end{array}$

Let $Q$ be the image of $P(1,3,4)$.
$d.r.$ of normal to the plane is $\overrightarrow n=(2,-1,1)$
Since $PQ$ is $\perp$ to the plane, $d.r.$ of the line = $(2,-1,1)$
$\therefore$ The equation of the line $PQ$ is $\large\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}$$=\lambda Point Q can be expressed in terms of \lambda as (2\lambda+1,-\lambda+3,\lambda+4) Since Q is the image of P in the plane, the mid point of PQ lies on the plane. Mid point of PQ is A(\large\frac{2\lambda+2}{2},\frac{-\lambda+6}{2},\frac{\lambda+8}{2}) A lies on the plane. \therefore it satisfies the eqn. of the plane. \Rightarrow\:2(\large\frac{2\lambda+2}{2})$$-\large\frac{-\lambda+6}{2}$$+\large\frac{\lambda+8}{2}$$+3=0$
$\Rightarrow\:\lambda=-2$
Substituting the value of $\lambda$ we get the image $Q(-3,5,2)$
answered Jan 2, 2014