JEE Main 2022 (Online) 30th July Evening ShiftQuestion 1
Let $L_1$ be a line in the $yz$-plane with $y$ and $z$ intercepts $\frac14$ and $\frac1C$ ($C>0$) respectively. Let $L_2$ be the line in the $xz$-plane with $x$ and $z$ intercepts $\frac13$ and $-\frac1C$ respectively. If the shortest distance between $L_1$ and $L_2$ is $\frac15$, then $C^2$ is equal to ________.
3D Geometry
Answer$75$
Solution
$L_1$: $x=0,\ 4y+Cz=1$; point $(0,\tfrac14,0)$, direction $(0,C,-4)$. $L_2$: $y=0,\ 3x-Cz=1$; point $(\tfrac13,0,0)$, direction $(C,0,3)$. Cross product: $(3C,\,-4C,\,-C^2)$, magnitude $C\sqrt{25+C^2}$. $(\tfrac13,-\tfrac14,0)\cdot(3C,-4C,-C^2)=C+C=2C$. Distance $=\dfrac{2C}{C\sqrt{25+C^2}}=\dfrac{2}{\sqrt{25+C^2}}=\dfrac15$. $25+C^2=100\Rightarrow C^2=75$.