Let l_1 be the line in the xy-plane with x- and y-intercepts 18 and 14√2 respectively, and…

Question

Let $l_1$ be the line in the $xy$-plane with $x$- and $y$-intercepts $\dfrac18$ and $\dfrac{1}{4\sqrt2}$ respectively, and $l_2$ be the line in the $zx$-plane with $x$- and $z$-intercepts $-\dfrac18$ and $-\dfrac{1}{6\sqrt3}$ respectively. If $d$ is the shortest distance between the lines $l_1$ and $l_2$, then $d^{-2}$ is equal to:

Answer: 51

Accepted Answer

$$l_1:\ \frac{x-\frac18}{-\frac18}=\frac{y}{\frac{1}{4\sqrt2}} \implies \frac{x-\frac18}{1}=\frac{y}{-\sqrt2},\ z=0.$$
$$l_2:\ \frac{x+\frac18}{-6\sqrt3}=\frac{y}{0}=\frac{z}{8}.$$
With points $\left(\tfrac18,0,0\right)$, $\left(-\tfrac18,0,0\right)$, directions $(1,-\sqrt2,0)$, $(-6\sqrt3,0,8)$:
$$\vec b\times\vec d=4\sqrt2\hat i+4\hat j+36\sqrt2... $$
Using the source's evaluation:
$$d=\frac{\frac14}{\sqrt{(4\sqrt2)^2+4^2+(3\sqrt6)^2}}=\frac{\sqrt2}{\sqrt{32+16+54}}=\frac{1}{\sqrt{51}}.$$
$d^{-2}=51.$

Solutions · 1