Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 1 / 5B / ii / Solution

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 ${\lambda }_i$.