QJEEAdv14Mafff7: JEE Advanced 2014 Mathematics - Class 12 JEEAdv14Mafff7 Solution: From a point perpendicular and are drawn respectively on the…

Line $L_1$ given by $y=x ; z=1$ can be expressed

$L_1: \frac{X}{1}=\frac{Y}{1}=\frac{z-1}{0}$

$\begin{aligned}\frac{x}{1} & =\frac{y}{1}=\frac{z-1}{0}=\alpha (say)\\\\\Rightarrow x & =\alpha, y=\alpha, z=1\end{aligned}$

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

$L_2: \frac{x}{1}=\frac{y}{-1}=\frac{z+1}{0}$

$\frac{x}{1}=\frac{y}{-1}=\frac{z+1}{0}=\beta$ (say)

$\Rightarrow \quad x=\beta, y=-\beta, z=-1$

Let the coordinates of $R$ on $L_2$ be $(\beta,-\beta,-1)$ .

Direction ratios of $P Q$ are $\lambda-\alpha, \lambda-\alpha, \lambda-1$ .

Now, $P Q \perp L_1$

$\begin{aligned} & \therefore 1(\lambda-\alpha)+1 \cdot(\lambda-\alpha)+0 \cdot(\lambda-1)=0 \\\\ & \Rightarrow \lambda=\alpha \\\\ & \therefore Q(\lambda, \lambda, 1)\end{aligned}$

Direction ratio of $P R$ are

$\lambda-\beta, \lambda+\beta, \lambda+1 \text {. }$

Now, $P R \perp L_2$

$\begin{aligned} & \therefore 1(\lambda-\beta)+(-1)(\lambda+\beta)+0(\lambda+1)=0 \\\\ & \lambda-\beta-\lambda-\beta=0 \\\\ & \Rightarrow \beta=0 \\ & \end{aligned}$
$\therefore R(0,0,-1)$

Now, as $\angle Q P R=90^{\circ}$

$\text { (as } a_1 a_2+b_1 b_2+c_1 c_2=0 \text {, }$ if two lines with DR's $a_1, b_1, c_1 ; a_2, b_2, c_2$ are perpendicular)

$\begin{array}{cc}\therefore & (\lambda-\lambda)(\lambda-0)+(\lambda-\lambda)(\lambda-0) +(\lambda-1)(\lambda+1)=0 \\\\ &\Rightarrow (\lambda-1)(\lambda+1)=0 \\\\ &\Rightarrow \lambda=1 \text { or } \lambda=-1\end{array}$

$\lambda=1$ , rejected as $P$ and $Q$ are different points.

$\Rightarrow \lambda=-1$