The shortest distance between the lines x-23=y+12=z-62 and x-63=1-y2=z+80 is equal to: Ans…

Question

The shortest distance between the lines $\dfrac{x-2}{3}=\dfrac{y+1}{2}=\dfrac{z-6}{2}$ and $\dfrac{x-6}{3}=\dfrac{1-y}{2}=\dfrac{z+8}{0}$ is equal to:

Answer: 14

Accepted Answer

Points $(2,-1,6)$, $(6,1,-8)$; directions $(3,2,2)$, $(3,-2,0)$.
$$(3,2,2)\times(3,-2,0)=(4,6,-12),\qquad |(4,6,-12)|=14.$$
Vector between points $(4,2,-14)$: dot with $(4,6,-12)=16+12+168=196$.
$$d=\frac{196}{14}=14.$$

Solutions · 1