= Tetrahedron volume under sliding opposite edges
{title2=$V=\tfrac16|\Delta\cdot(a\times c)|\,|\delta\lambda\,\delta\mu|$}
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.
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