JEE Main 2022 (Online) 25th June Evening ShiftQuestion 1
Let $l_1$ be the line in $xy$-plane with $x$ and $y$ intercepts $\frac{1}{8}$ and $\frac{1}{4\sqrt{2}}$ respectively, and $l_2$ be the line in $zx$-plane with $x$ and $z$ intercepts $-\frac{1}{8}$ and $-\frac{1}{6\sqrt{3}}$ respectively. If $d$ is the shortest distance between the line $l_1$ and $l_2$, then $d^{-2}$ is equal to $\_\_\_\_\_\_$ .
3D Geometry
Answer$51$
Solution
$l_1$ passes through $(\tfrac18,0,0)$ and $(0,\tfrac1{4\sqrt2},0)$: direction $\parallel\vec d_1=(-1,\sqrt2,0)$. $l_2$ passes through $(-\tfrac18,0,0)$ and $(0,0,-\tfrac1{6\sqrt3})$: direction $\parallel\vec d_2=(3\sqrt3,0,-4)$. $\vec d_1\times\vec d_2=(-4\sqrt2,-4,-3\sqrt6)$, squared magnitude $32+16+54=102$. $\vec a_2-\vec a_1=(-\tfrac14,0,0)$, dot product $=\sqrt2$. $d=\frac{\sqrt2}{\sqrt{102}}=\frac1{\sqrt{51}}$, so $d^{-2}=51$.