Question

Let l 1 be the line in xy -plane with x and y intercepts ⅛ and $\frac{1}{4√2}$ respectively and l 2 be the line in zx -plane with x and z intercepts -⅛ and $−\frac{1}{6√3}$ respectively. If d is the shortest distance between the line l 1 and l 2, then d –2 is equal to ______.

Answer 1

The correct answer is: 51.

$\frac{x-\frac{1}{8}}{\frac{1}{8}}=\frac{y}{-\frac{1}{4\sqrt2}}=\frac{z}{0}....L_1$

or

$\frac{x-\frac{1}{8}}{1}=\frac{y}{-\sqrt2}=\frac{z}{0}....(i)$

$=\frac{\sqrt2}{\sqrt{32+16+54}}=\frac{!}{\sqrt51}$

d –2 = 51