JEE Main 2024 (Online) 8th April Morning ShiftQuestion 1
If the shortest distance between the lines
$L_1:\vec{r}=(2+\lambda)\hat{i}+(1-3\lambda)\hat{j}+(3+4\lambda)\hat{k}$, $\lambda\in\mathbb{R}$
$L_2:\vec{r}=2(1+\mu)\hat{i}+3(1+\mu)\hat{j}+(5+\mu)\hat{k}$, $\mu\in\mathbb{R}$
is $\frac{m}{\sqrt{n}}$, where $\gcd(m,n)=1$, then the value of $m+n$ equals
3D Geometry
Solution
$L_1$: point $(2,1,3)$, $\vec d_1=(1,-3,4)$; $L_2$: point $(2,3,5)$, $\vec d_2=(2,3,1)$. Cross $=(-15,7,9)$, $|\cdot|^2=355$; joining $(0,2,2)$, dot $=32$. Distance $=\frac{32}{\sqrt{355}}$, so $m+n=387$.