Question

Let $Q$ be the mirror image of the point $P(1,0,1)$ with respect to the plane $S:x+y+z=5$. If a line $L$ passing through $(1,-1,-1)$, parallel to the line $PQ$ meets the plane $S$ at $R$, then $Q{R}^2$ is equal to:

• A. 2
• B. 5
• C. 7
• D. 11

Answer 1

Answer: (B)

Let parallel vector of $L=\vec{b}$
mirror image of $Q$ on given plane $x+y+z=5$
$\frac{a-1}{1}=\frac{b-0}{1}=\frac{c-1}{1}=\frac{-2(2-5)}{3}$
$a=3,b=2,c=3$
$Q\equiv (3,2,3)$
$\because \vec{b}||\overrightarrow{PQ}$
so,$\vec{b}=(1,1,1)$
Equation of line
$L:\frac{x-1}{1}=\frac{y+1}{1}=\frac{z+1}{1}$
Let point $R,(\lambda +1,\lambda -1,\lambda -1)$
lying on plane $x+y+z=5$,
so, $3\lambda -1=5$
$\Rightarrow \lambda =2$
Point $R$ is $(3,1,1)$
$Q{R}^2=5$