Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 1 / 5B (Vectors and Matrices)

Taking the parallelogram spanned by $e_2,e_3$ as base gives base area $|e_2\times e_3|$ and height $|e_1\cdot (e_2\times e_3)|/|e_2\times e_3|$. A tetrahedron has one third of the corresponding pyramid volume and the triangle has half the parallelogram base, henceThe reciprocal vectors arewhich directly satisfy $f_i\cdot e_j={\delta }_{ij}$.

The final face contains $v_i=v+e_i$, so $c\cdot (v+e_i)+d=0$. Therefore $e_i\cdot c=-(c\cdot v+d)$.

The $i$th face through $v$ contains the two edge directions $e_j$ with $j\ne i$, so $a_i\cdot e_j=0$ for those $j$. The one-dimensional space with these two orthogonality conditions is spanned by $f_i$, hence $a_i={\lambda }_i{f}_i$ for some nonzero real ${\lambda }_i$.

Dotting $c={\sum }_j{\gamma }_j{a}_j$ with $e_i$ gives $e_i\cdot c={\gamma }_i{\lambda }_i$, proving ${\gamma }_i=(e_i\cdot c)/{\lambda }_i$. AlsoCombining these with $V=|\Delta |/6$ gives the displayed face formula, with the orientation chosen so its signed right-hand side is positive.