Question

Let two planes $p_1:2x-y+z=2$, and $p_2:x+2y-z=3$ are given. The image of plane $P_1$ in the plane mirror $P_2$ is

• A. $x+7y-4z+5=0$
• B. $3x+4y-5z+9=0$
• C. $7x-y+2z-9=0$
• D. $7x+y+9z+9=0$

Answer 1

The correct option is C $7x-y+2z-9=0$

Let $P_3$ be the plane which is the image of $P_1$ in the plane mirror $P_2$.

The equation of $P_3$ can be written as $P_1+\lambda P_2=0$

$\Rightarrow 2x-y+z-2+\lambda (x+2y-z-3)=0$

By inspection, $(1,1,1)$ is a point on $P_1$.

Hence, the image of $(1,1,1)$ across $x+2y-z-3=0$ must lie on the image plane $P_3$.

Let the image of $(1,1,1)$ in the plane $P_2$ be $(x^{\prime },y^{\prime },z^{\prime })$

$\Rightarrow \frac{x^{\prime }-1}{1}=\frac{y^{\prime }-1}{2}=\frac{z^{\prime }-1}{-1}=\frac{-2(1+2-1-3)}{1^2+2^2+1^2}$

$\Rightarrow (x^{\prime },y^{\prime },z^{\prime })\equiv (\frac{4}{3},\frac{5}{3},\frac{2}{3})$

Substituting this in the equation of $P_3$, we get $-1+\lambda 3=0$

$\Rightarrow \lambda =\frac{1}{3}$

Hence, the equation of $P_3$ is

$2x-y+z-2+\frac{1}{3}(x+2y-z-3)=0$

$\Rightarrow 7x-y+2z-9=0$