Question

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

Answer 1

Answer: 51

$8x+4\sqrt{2}y=1,z=0$
$\Rightarrow \frac{x-\frac{1}{8}}{1}=\frac{y-0}{-\sqrt{2}}=\frac{z-0}{0}=\lambda$
$-8x-6\sqrt{3}z=1,y=0$
$\Rightarrow \frac{x+\frac{1}{8}}{3\sqrt{3}}=\frac{y-0}{0}=\frac{z-0}{-4}$
$\left|\begin{array}{ccc}\frac{1}{4}& 0& 0\\ 1& -\sqrt{2}& 0\\ 3\sqrt{3}& 0& -4\end{array}\right|=\sqrt{2}$
$d=\frac{1}{\sqrt{51}}$
$\frac{1}{d^2}=51$