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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 $f_i\cdot e_j={\delta }_{ij}$.