Past exam of the mathematics course of the University of Cambridge / 2009 / ia / Paper 1 / 5C (Vectors and Matrices)

Use suffix notation, summing repeated indices from one to three. The contraction identity for the Levi-Civita symbol givesHere $a\cdot a=1$ because $a$ is a unit vector. For the second cross product identity, use ${ϵ}_{ijk}{ϵ}_{j\ell m}={\delta }_{im}{\delta }_{k\ell }-{\delta }_{i\ell }{\delta }_{km}$:The first term vanishes because $a_{\ell }{a}_n$ is symmetric while the Levi-Civita symbol is antisymmetric in those indices. The last equality uses cyclicity and antisymmetry of the scalar triple product.

On the unit sphere, a great circle has radius one, so the spherical distance is its shorter arc length, equal to its central angle in $[0,\pi ]$. The dot product of its endpoint unit vectors is the cosine of that angle. HenceFor the angular formulas, take the ordinary nondegenerate convex minor spherical triangle: no pair of vertices is antipodal and the three vertices do not lie on one great circle. These are needed for the displayed denominators and interior angles to be defined. Merely requiring three distinct points, as the PDF does, does not exclude degeneracy.

Put $p=\delta (A,B)$, $q=\delta (A,C)$ and $r=\delta (B,C)$. The outgoing unit tangent vectors at $A$ areTheir dot product is $\cos \alpha$. Equivalently, they are obtained from the normalized plane normals $a\times b$ and $a\times c$ by the same right-angle rotation in the tangent plane: $u=(a\times b)\times a/|a\times b|$, and similarly for $v$. Rotation preserves the dot product, givingSince $|a\times b|=\sin p$ and $|a\times c|=\sin q$, the first proved identity becomes the spherical law of cosines:For the spherical law of sines, take norms in the second proved identity. Writing $\tau =a\cdot (b\times c)$, and using $0<\alpha <\pi$, givesCyclic permutation gives the corresponding formulas at $B$ and $C$. Dividing each by the positive product $\sin p\sin q\sin r$ provesFinally, if all sides equal $s$, the spherical law of cosines givesbecause $0<s<\pi$ makes $1+\cos s>0$ and $\cos s<1$. Cosine is strictly decreasing on $(0,\pi )$, so every angle of a nondegenerate equilateral spherical triangle is greater than $\pi /3$.