Using vector basis to find the shortest distance between two segments in a parallelepiped

Question

We are given a parallelepiped with edges $a,b,c$. Find the shortest distance between the cross-diagonals of two adjacent walls.

I have made a diagram but do not know how to use the vectors to solve the problem.

Answer 1

Let $A(0,0,0)$, $B(a,0.0),$ $N(a,b,c),$ $P(0,b,c)$ and $M(a,0,c)$.

Thus, since $AP||BN,$ our distance it's the distance between $B$ and the plane $(APM).$

Easy to see that the normal vector of this plane is perpendicular to

$\vec{AP}(0,b,c)$ and to $\vec{AM}(a,0,c)$ and it's $$\vec{n}(bc,ac,-ab)$$ and an equation of the plane it's $$bcx+acy-abz=0$$ and the needed distance it's: $$\frac{|bc\cdot a+ac\cdot0-ab\cdot0|}{\sqrt{(bc)^2+(ac)^2+(-ab)^2}}=\frac{abc}{\sqrt{a^2b^2+a^2c^2+b^2c^2}}.$$