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For a three-dimensional basis with nonzero scalar triple product $\Delta$, its reciprocal basis satisfies ${\hat{e}}_2\times {\hat{e}}_3=e_1/\Delta$ and the cyclic analogues. Consequently its scalar triple product is $1/\Delta$, and taking the reciprocal basis again recovers the original basis. The coordinates of $V$ in the original basis are $V\cdot {\hat{e}}_i$; in the reciprocal basis they are $V\cdot e_i$.

For a rigidly rotating prescribed vacuum path, use spatial coordinates attached to the apparatus and laboratory time $t$. At a path point $\mathbf{r}$, the local apparatus velocity is $\mathbf{v}=\mathbf{\Omega }\times \mathbf{r}$. A light displacement has instantaneous laboratory displacement $d\mathbf{r}+\mathbf{v}dt$, up to the common rotation of the spatial axes. The null condition is thereforeWrite $d\ell =|d\mathbf{r}|$. To first order in $|\mathbf{v}|/c$, the positive root isReversing the path reverses the dot product term but leaves the leading path-length term unchanged. Subtract the two return times and use the scalar triple product identity:This is a line integral version of the Sagnac area law; the vector area identity also follows from Stokes theorem. For a mirror-defined polygon the actual ray paths move slightly as the mirrors rotate; the stationary optical path at zero rotation gives the same first-order result. The prescribed-path calculation is an idealization, not an exact finite-speed formula for every moving-mirror layout.

The dot product explains the geometry: only the projection of the oriented vector area along the angular velocity contributes. A planar loop whose normal vector is perpendicular to $\mathbf{\Omega }$ has zero leading delay. For repeated traversals with the same orientation, the signed vector areas add; oppositely oriented lobes can cancel even when the total unsigned area is large. Equal perimeter alone does not imply equal response. A constant translational velocity contributes nothing to this leading closed-path line integral, since ${\oint }_{C}d\mathbf{r}=0$.

For a nondegenerate convex minor spherical triangle, with angular sides $a,b,c$ opposite interior angles $A,B,C$, the spherical law of sines statesThe sines of sides refer to central angles, so a physical side length ${\ell }_a$ on a sphere of radius $R$ must enter as $\sin ({\ell }_a/R)$. All six angles lie in $(0,\pi )$, and the denominators are positive. The reciprocal form, $\sin a/\sin A=\sin b/\sin B=\sin c/\sin C$, is equivalent. This is Tatum's sine formula, equation 3.5.1, in Celestial Mechanics, section 3.5: spherical sine and cosine formulas.

A short proof explains the symmetry of the spherical law of sines. Let $u,v,w$ be the vertex unit vectors at $A,B,C$. At $u$, the outgoing unit tangent vectors towards $v,w$ are $t_B=(v-\cos cu)/\sin c$ and $t_C=(w-\cos bu)/\sin b$. Their angle is $A$, so the magnitude of their cross product is $\sin A$. Since both tangent vectors are orthogonal to $u$, the scalar triple product obeysDividing by $\sin a\sin b\sin c$ proves the spherical law of sines. Its beauty is that the same three-dimensional scalar triple product can be calculated at any vertex, producing all three ratios at once.

For the spherical triangle worked out under the spherical law of cosines, $A=90^{\circ }$, $b=c=60^{\circ }$ and $\cos a=1/4$. Hence $\sin a=\sqrt{15}/4$, and the spherical law of sines givesThe inverse sine gives $B\approx 63.434949^{\circ }$, but the supplementary angle $116.565051^{\circ }$ has the same sine. The spherical law of sines alone therefore leaves two candidates for this angle. The spherical law of cosines, using the known three sides, gives $\cos B=1/\sqrt{5}>0$ and selects $63.434949^{\circ }$; likewise $C=B$. When only two sides and an opposite angle are given, two distinct valid spherical triangles can remain. A cosine relation resolves an inverse sine branch only when enough additional data are known; it does not make every ambiguous data set unique.

Apply the vector triple product identity with first vector $b\times c$. Its contraction with $c$ vanishes, so $(b\times c)\times (c\times a)=[a\cdot (b\times c)]c$. Dotting with $a\times b$ gives the square of the scalar triple product. This is also the Gram determinant of $a,b,c$, hence is nonnegative even when the vectors are linearly dependent.

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.

For a continuously differentiable vector field $v$ on a bounded sufficiently regular region $V$, apply the divergence theorem to $a\times v$ for an arbitrary constant vector $a$. The vector calculus identity $\nabla \cdot (a\times v)=-a\cdot (\nabla \times v)$ and the cyclic scalar triple product identity show that the dot products of both sides of the displayed formula with $a$ agree. Since $a$ is arbitrary, the vector integrals agree. Here $n$ is the outward unit normal vector. Reversing the order on the boundary gives the equivalent positive-sign formula $\int n\times vdS=\int \nabla \times vdV$.