Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2004 / Paper 2 / 1

Choose a common origin at vertex $C$, set $\mathbf{a}=\overrightarrow{CB}$ and $\mathbf{b}=\overrightarrow{CA}$, and orient the remaining edge as $\mathbf{c}=\overrightarrow{AB}$. ThenThis convention makes the interior angle between $\mathbf{a}$ and $\mathbf{b}$ equal to $C$, and their lengths equal to the opposite side lengths $a$ and $b$.

Take the dot product of the difference with itself. Since $\mathbf{a}\cdot \mathbf{b}=ab\cos C$, the law of cosines follows:The magnitude of the cross product of two edge vectors is twice the triangle's area $K$. Applying this at each vertex gives $2K=ab\sin C=bc\sin A=ca\sin B$. Divide by the nonzero product $abc$ to obtain the law of sines:

Let $\Omega$ be the interior of the tetrahedron, and take any constant vector field $\mathbf{q}$. Its divergence is zero, so the divergence theorem givesChoosing $\mathbf{q}$ to be each Cartesian basis vector proves ${\sum }_i{\mathbf{V}}_i=0$. This is the general fact that vector areas of a closed polyhedron sum to zero, specialized to four faces. The outward orientations are essential: the scalar face areas are all positive and do not sum to zero.