Past exam of the mathematical methods I course of the University of Oxford / 2021 resit / 9

Use $\mathbf{w}=-\mathbf{u}-\mathbf{v}$ and bilinearity and antisymmetry of the cross product:Consequently,If $\mathbf{u},\mathbf{v}$ are linearly independent, this linear span is the plane through the origin with normal vector $\mathbf{u}\times \mathbf{v}$. If they are dependent, all three nonzero vectors lie on one line through the origin, and their linear span has dimension one. For example, $\mathbf{u}=\mathbf{v}=(1,0,0)$ and $\mathbf{w}=(-2,0,0)$ obey the hypothesis but span only a line. Nonzero vectors alone do not force a two-dimensional linear span.

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 $T<0$.

Expand the dot product and cross product in Cartesian components:The two expressions are the same scalar triple product, or determinant with columns $\mathbf{u},\mathbf{v},\mathbf{w}$:The equality also holds when this determinant is zero.

Write $\Delta =⟨\mathbf{u},\mathbf{v},\mathbf{w}⟩=\mathbf{u}\cdot (\mathbf{v}\times \mathbf{w})$. The displayed definitions require $\Delta \ne 0$, equivalently linear independence of the original vectors. The vector triple product identity givesHere the first scalar triple product is $\Delta$ by cyclic permutation, and the second is zero because a cross product is orthogonal to both its factors. ThusThis is the reciprocal basis involution identity. When $\Delta =0$, the proposed reciprocal basis is undefined; for example, three vectors in the $xy$ plane cannot be inserted into the printed quotients.