Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 2 / 11

The independent vectors $\mathbf{a},\mathbf{b}$ span the plane through the origin, and their cross product is a nonzero normal vector. Thus the required nonparametric equation is

The orientation assumption gives $\mathbf{c}\cdot (\mathbf{a}\times \mathbf{b})>0$. The same open half-space is therefore characterized byEquality would put the point in the separating plane, rather than strictly on the same side.

Take two edge vectors in the face, $\mathbf{b}-\mathbf{a}$ and $\mathbf{c}-\mathbf{a}$. Their cross product is the face normal vectorThe face plane is $(\mathbf{r}-\mathbf{a})\cdot \mathbf{n}=0$. Every term in $\mathbf{a}\cdot \mathbf{n}$ except $\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})$ vanishes, givingThe scalar triple product is nonzero, so this normal cannot vanish.

Set $D=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})>0$. Since these vectors form a basis, write $\mathbf{r}=\alpha \mathbf{a}+\beta \mathbf{b}+\gamma \mathbf{c}$. The reciprocal basis givesA point is in the closed tetrahedron precisely when it is a convex combination of the four vertices, with weights $1-\alpha -\beta -\gamma ,\alpha ,\beta ,\gamma$. Thus the necessary and sufficient conditions areThese four half-spaces are the four inward sides of its faces. This is tetrahedron membership via scalar triple products; making all inequalities strict describes the interior.

The opposite face lies in the plane spanned by $\mathbf{a},\mathbf{b}$ through the origin. Directly,The distance from a point to a plane is the absolute normal component, soHere $\mathbf{c}=(\mathbf{a}\times \mathbf{b})/2$, so the perpendicular foot is the origin, a vertex of the opposite triangular face. Thus the distance to the actual face equals the distance to its containing plane.