Volume of tetrahedron with regard to opposite sides

Question

Prove that the volume of the tetrahedron $ABCD$ is $\frac{1}{6}AB\cdot CD\cdot EF \sin x$ where $EF$ is the shortest distance between $AB$ and $CD,$ and $x$ is the angle between these two lines. I have attempted the proof using vector methods, but have not got very far. Any help would be appreciated.

Answer 1

Let $AD'BC'A'DB'C$ be a parallelepiped.

Thus, $$V_{ABCD}=\frac{1}{3}V_{AD'BC'A'DB'C}=\frac{1}{3}S_{AD'BC'}\cdot\rho((A'DB'C),(AD'BC'))=$$ $$=\frac{1}{3}\cdot\frac{1}{2}AB\cdot D'C'\cdot\sin{x}\cdot\rho(AB,DC)=\frac{1}{6}AB\cdot CD \cdot EF\sin{x}.$$