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

-1

Line $L_1$ is given by y = x; z = 1 can be expressed
$L_1:\frac{x}{1}=\frac{y}{1}=\frac{z-1}{0}=\alpha \left[say\right]\Rightarrow x=\alpha ,y=\alpha ,z=1$
Let the coordinates of Q on $L_1$ be $(\alpha ,\alpha ,1)$.
Line $L_2$ given by $y=-x,z=-1$ can be expressed as
$L_2:\frac{x}{1}=\frac{y}{-1}=\frac{z+1}{0}=\beta \left[say\right]$
$\Rightarrow x=\beta ,y=-\beta ,z=-1$
Let the coordinates of R on $L_2$ be $(\beta ,-\beta ,-1)$.
Direction ratios of PQ are $\lambda -\alpha ,\lambda -\alpha ,\lambda -1$.
Now, $PQ\perp L_1$



$\therefore 1(\lambda -\alpha )+1.(\lambda -\alpha )+0.(\lambda -1)=0\Rightarrow \lambda =\alpha$
Hence, $Q(\lambda ,\lambda ,1)$
Direction ratios of PR are $\lambda -\beta ,\lambda +\beta ,\lambda +1$.
Now, $PR\perp L_2$
$\therefore 1(\lambda -\beta )+(-1)(\lambda +\beta )+0(\lambda +1)=0\lambda -\beta -\lambda -\beta =0\Rightarrow \beta =0$
Hence, R(0, 0, -1)
Now, as $\angle QPR={90}^{\circ }$
[as $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$, If two lines with DR's $a_1,b_1,c_1;a_2,b_2,c_2$ are perpendicular]
$\therefore (\lambda -\lambda )(\lambda -0)+(\lambda -\lambda )(\lambda -0)+(\lambda -1)(\lambda +1)=0\Rightarrow (\lambda -1)(\lambda +1)=0\Rightarrow \lambda =1\text{or}\lambda =-1$
$\lambda =1$, rejected as P and Q are different points.
$\Rightarrow \lambda =-1$