JEE 2025 — Mathematics PYQ

$L_1:\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$

$L_2:\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$

$\overrightarrow{PQ}=(3\mu -2\lambda +1)\hat{i}+(4\mu -3\lambda +2)\hat{j}+(5\mu -4\lambda +2)\hat{k}$

$\therefore \overrightarrow{PQ}\perp {\vec{b}}_1\text{ and }\overrightarrow{PQ}\perp {\vec{b}}_2$

${\vec{b}}_1\times {\vec{b}}_2=<br>\left|\begin{array}{ccc}<br>\hat{i}& amp;\hat{j}& amp;\hat{k}\\ <br>2& amp;3& amp;4\\ <br>3& amp;4& amp;5<br>\end{array}\right|<br>=-\hat{i}+2\hat{j}-\hat{k}$

$\therefore \overrightarrow{PQ}\parallel {\vec{b}}_1\times {\vec{b}}_2$

$\frac{3\mu -2\lambda +1}{-1}<br>=\frac{4\mu -3\lambda +2}{2}<br>=\frac{5\mu -4\lambda +2}{-1}$

$6\mu -4\lambda +2=-4\mu +3\lambda -2$

$-4\mu +3\lambda -2=10\mu -8\lambda +4$

$10\mu -7\lambda +4=0\text{and}-14\mu +11\lambda -6=0$

$70\mu -49\lambda +28=0$

$-70\mu +55\lambda -30=0$

$6\lambda -2=0$

$\Rightarrow \lambda =\frac{1}{3}$

$\mu =\left(\frac{7}{3}-4\right)\frac{1}{10}<br>=\frac{-5}{3\times 10}<br>=-\frac{1}{6}$

$\therefore P=\left(2\left(\frac{1}{3}\right)+1,3\left(\frac{1}{3}\right)+2,4\left(\frac{1}{3}\right)+3\right)$

$=\left(\frac{5}{3},3,\frac{13}{3}\right)$

$\text{Equation of }PQ:$

$\frac{x-\frac{5}{3}}{-1}<br>=\frac{y-3}{2}<br>=\frac{z-\frac{13}{3}}{-1}$

$\text{For }\left(\frac{14}{3},-3,\frac{22}{3}\right)\text{ satisfies the equation.}$