QJEEAdv13Mac194: JEE Advanced 2013 Mathematics - Class 12 JEEAdv13Mac194 Solution: A line passing through the origin is perpendicular to the lines …

Given two lines.

$\begin{gathered}l_1:(3+t) \hat{i}+(-1+2 t) \hat{j}+(4+2 t) \hat{k} \text { and } \\l_2:(3+2 s) \hat{i}+(3+2 s) \hat{j}+(2+s) \hat{k} \\\Rightarrow l_1: \frac{x-3}{1}=\frac{y+1}{2}=\frac{z-4}{2}=t \text { and } \\\quad l_2: \frac{x-3}{2}=\frac{y-3}{2}=\frac{z-2}{1}=s\end{gathered}$

Given, a line $l$ is perpendicular to $l_1$ and $l_2$

$\therefore \quad l$ is parallel to $l_1 \times l_2$

$\begin{aligned}& \Rightarrow \vec{l}_1 \times \vec{l}_2=\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\1 & 2 & 2 \\2 & 2 & 1\end{array}\right| \\& \Rightarrow \vec{l}_1 \times \vec{l}_2=-2 \hat{i}+3 \hat{j}-2 \hat{k}\end{aligned}$

The equation of line $l$ passing through origin and parallel to $-2 \hat{i}+3 \hat{j}-2 \hat{k}$ is

$l: \frac{x}{-2}=\frac{y}{3}=\frac{z}{-2}=\lambda$

Let P is the point of intersection of $l$ and $l_1$

$\begin{aligned}& \Rightarrow \mathrm{P}=(-2 \lambda, 3 \lambda,-2 \lambda)=(3+t,-1+2 t, 4+2 t) \\& \Rightarrow \lambda=\frac{3+t}{-2}=\frac{-1+2 t}{3}=\frac{4+2 t}{-2} \\& \Rightarrow t=-1, \lambda=-1\end{aligned}$

$\text { So, } \mathrm{P}=(2,-3,2)$

Let Q is a point on the line $l_2$

$Q=(3+2 s, 3+2 s, 2+s)$

Apply $|\mathrm{PQ}|=\sqrt{17}$

$\begin{aligned}& \Rightarrow \sqrt{(3+2 s-2)^2+(3+2 s+3)^2+(2+s-2)^2} \\& =\sqrt{17} \\& \Rightarrow \sqrt{9 s^2+28 s+37}=\sqrt{17} \\& \Rightarrow \quad 9 s^2+28 s+37=17 \\& \Rightarrow \quad 9 s^2+28 s+20=0 \\& \Rightarrow \mathrm{s}=\frac{-28 \pm \sqrt{28^2-4 \times 9 \times 20}}{2.9} \\& \Rightarrow \mathrm{s}=\frac{-28 \pm 8}{18} \\& \Rightarrow \mathrm{s}=\frac{-10}{9},-2 \\\end{aligned}$

If $s=-2$ , then $Q=(-1,-1,0)$

$\text { If } s=-\frac{10}{9} \text {, then } Q=\left(\frac{7}{9}, \frac{7}{9}, \frac{8}{9}\right)$

Hints :

(i) If $l$ is perpendicular to $\vec{l}_1$ and $\vec{l}_2$ , then $l$ is parallel to $\vec{l}_1 \times \vec{l}_2$ .

(ii) A general point on the line

$\begin{gathered}\quad \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=\lambda \text { is } \\\left(x_1+\lambda a, y_1+\lambda b, z_1+\lambda c\right) .\end{gathered}$