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Label $\mathbf{a}$ opposite ${\mathbf{x}}_0$, $\mathbf{b}$ opposite ${\mathbf{x}}_1$, $\mathbf{c}$ opposite ${\mathbf{x}}_2$ and $\mathbf{d}$ opposite ${\mathbf{x}}_3$. Since the tetrahedron is nondegenerate, its scalar triple product $T={\mathbf{u}}_1\cdot ({\mathbf{u}}_2\times {\mathbf{u}}_3)$ is nonzero. Put $s=T/|T|$ to handle either orientation of a vector space. Then the required outward vectors areEach cross product has magnitude twice the area of its triangular face. The signs are fixed geometrically: $\mathbf{b}\cdot {\mathbf{u}}_1=-|T|<0$, so $\mathbf{b}$ points away from the opposite vertex; the same test with ${\mathbf{u}}_2$ and ${\mathbf{u}}_3$ handles $\mathbf{c}$ and $\mathbf{d}$. For the remaining face, the displacement toward its opposite vertex is $-{\mathbf{u}}_1$, and $\mathbf{a}\cdot (-{\mathbf{u}}_1)=-|T|<0$, again outward.

Expand the first cross product:ThereforeThis is closure of outward face area vectors, with every face area multiplied by the same factor two. Omitting $s$ would orient all four faces inward when $T<0$.