Opposite edges of a tetrahedron

Two edges of a tetrahedron are opposite when they share no vertex. The six edges split into three opposite pairs.

Place one pair of opposite tetrahedron vertices on the line $b+\lambda a$ and the other pair on $d+\mu c$, with fixed nonzero direction vectors $a,c$. Let $\Delta =d-b$ and let $\delta \lambda ,\delta \mu$ be the parameter differences within the two pairs. The scalar triple product of the three edge vectors from one vertex gives the displayed volume. Terms parallel to $a$ or $c$ vanish in the triple product, leaving only $\Delta \cdot (a\times c)$. Sliding either pair along its own line while preserving its separation leaves this volume unchanged. When the two lines are skew lines, it can also be written $V=d_{\perp }{\ell }_1{\ell }_2|\sin \alpha |/6$, where $d_{\perp }$ is their shortest distance, ${\ell }_1,{\ell }_2$ the two edge lengths and $\alpha$ their direction angle. Intersecting or parallel lines give zero volume because all four vertices are coplanar.

Place one vertex of a tetrahedron at the origin and denote the other vertex vectors by $a,b,c$. Orthogonality of two pairs of opposite edges of a tetrahedron forces $a\cdot b=a\cdot c=b\cdot c$. It follows that the third pair is orthogonal too. The three sums of squared lengths of opposite edges all equal $|a{|}^2+|b{|}^2+|c{|}^2-2a\cdot b$. This gives a dot product test for the special geometry without solving for the face planes.