JEE Main 2024 — Three Dimensional Geometry Question with Solution

Given lines are $\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}$ and $\frac{x-5}{2}=\frac{y-1}{-3}=\frac{z-5}{6}$.

Let, $AB$ be a line parallel to $\frac{x-5}{2}=\frac{y-1}{-3}=\frac{z-5}{6}$ which intersects $\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}$ at point $B$, where $A\equiv \left(7,-2,11\right)$

So, equation of $AB$ is,

$\frac{x-7}{2}=\frac{y+2}{-3}=\frac{z-11}{6}=\lambda \left(\text{let}\right)$

$\Rightarrow x=2\lambda +7, y=-3\lambda -2, z=6\lambda +11$

$\Rightarrow B\equiv (2\lambda +7,-3\lambda -2,6\lambda +11)$

Also, the point $B$ lies on $\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}$.

$\Rightarrow \frac{2\lambda +7-6}{1}=\frac{-3\lambda -2-4}{0}=\frac{6\lambda +11-8}{3}$

$\Rightarrow -3\lambda -2-4=0$

$\Rightarrow -3\lambda =6$

$\Rightarrow \lambda =-2$

$\Rightarrow B\equiv (3,4,-1)$

$\Rightarrow \mathrm{AB}=\sqrt{(7-3{)}^2+(4+2{)}^2+(11+1{)}^2}$

$\Rightarrow \mathrm{AB}=\sqrt{16+36+144}$

$\Rightarrow \mathrm{AB}=\sqrt{196}$

$\Rightarrow \mathrm{AB}=14$