Question

From a point $P (\lambda, \lambda, \lambda)$, perpendiculars $PQ$ and $PR$ are drawn respectively on the lines $y=x, z=1$ and $y=$ $-x, z=-1$. If $P$ is such that $\angle QPR$ is a right angle, then the possible value(s) of $\lambda$ is(are)

• A. $\sqrt{2}$
• B. $1$
• C. $-1$
• D. $-\sqrt{2}$

Answer 1

The Correct Option is C

Line 1: $\frac{x}{1}=\frac{y}{1}=\frac{z-1}{0}=r, Q(r, r, 1)$
Line 2: $\frac{x}{1}=\frac{y}{-1}=\frac{z+1}{0}= k , R(k,-k,-1)$
$\overrightarrow{P Q}=(\lambda-r) \hat{i}+(\lambda-r) \hat{j}+(\lambda-1) \hat{k}$
and $\lambda-r+\lambda-r=0$ as $\overrightarrow{P Q}$ is $\perp$ to $L_{1}$
$\Rightarrow 2 \lambda=2 r $
$\Rightarrow \lambda=r$
$\overline{ PR }=(\lambda- k ) \hat{ i }+(\lambda+ k ) \hat{ j }+(\lambda+1) \hat{ k }$
and $\lambda- k -\lambda- k =0$ as $\overrightarrow{ PR }$ is $\perp$ to $L _{2}$
$\Rightarrow k =0$
so $PQ \perp PR$
$(\lambda- r )(\lambda- k )+(\lambda- r )(\lambda+ k )+(\lambda-1)(\lambda+1)=0$
$\Rightarrow \lambda=1,-1$
For $\lambda=1$ as points $P$ and $Q$ coincide
$\Rightarrow \lambda=-1$