Let P be a plane passing through the points $(2,1,0),(4,1,1)$ and $(5,0,1)$ and $R$ be the point $(2,1,6)$. Then image of $R$ in the plane $P$ is

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Aicharya · Accepted Answer

Equation of the plane passing through $(2,1,0)$, $(4,1,1)$ and $(5,0,1)$ is $\begin{aligned} & \left|\begin{array}{lll} x-2 & y-1 & z-0 \ 4-2 & 1-1 & 1-0 \ 5-2 & 0-1 & 1-0 \end{array}ight|=0 \ & \Rightarrow x+y-2 z=3 \end{aligned}$ $\mathrm{R}^{\prime}(x, y, \mathrm{z})$ is image of $\mathrm{R}(2,1,6)$ w.r.t. to plane $x+y-2 z=3$ $\Rightarrow \frac{x-2}{1}=\frac{y-1}{1}=\frac{z-6}{-2}=\frac{-2[2+1-2(6)-3]}{1+1+4}$ $\Rightarrow \frac{x-2}{1}=\frac{y-1}{1}=\frac{z-6}{-2}=4$ $\Rightarrow x=6, y=5, \mathrm{z}=-2$ $	herefore \quad \mathrm{R}^{\prime}(x, y, \mathrm{z}) \equiv(6,5,-2)$

Let P be a plane passing through the points $(2,1,0),(4,1,1)$ and $(5,0,1)$ and $R$ be the point $(2,1,6)$.…

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