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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} $

1 Answer

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.
Substituting the value of $\lambda$ we get the image $Q(-3,5,2)$
answered Jan 2, 2014 by rvidyagovindarajan_1

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