Question

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

• A. 22
• B. $3$
• C. $-1$
• D. $-2$

Answer 1

The correct option is C $-1$

Line 1 : $\frac{x}{1}=\frac{y}{1}=\frac{z-1}{0}=r,Q(r,r,1)$

Line 2 : $\frac{x}{1}=\frac{y}{-1}=\frac{z+1}{0}=k,R(k,-k,-1)$

$\Rightarrow \overline{PQ}=(\lambda -r)1+(\lambda -r)j+(\lambda -1)k$

and $\lambda =r+\lambda -r=0$ as $\overline{PQ}$ is $1$ to $L_1$

$\Rightarrow 2\lambda =2r$

$\Rightarrow \lambda =r$

$\Rightarrow \overline{PR}=(\lambda -k)1+(\lambda +k)j+(\lambda +1)k$

and $\lambda -k-\lambda -k=0$ as $\overline{PR}$ is $1$ to $L_2$

$\Rightarrow x=0$

So, $PQ\perp PR$

$\Rightarrow (\lambda -r)(\lambda -k)+(\lambda -r)(\lambda +k)+(\lambda -1)(\lambda +1)=0$

$\Rightarrow \lambda =1,-1$

For $\lambda =1$ as points $P$ and $Q$ coincide.

$\Rightarrow \lambda =-1.$

Hence, the answer is $1,-1.$