Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2003 / Paper 1 / 1

Put $D=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})\ne 0$. The scalar triple product is unchanged by cyclic permutation, soThese are the diagonal pairings of a reciprocal basis.

The cross product $\mathbf{b}\times \mathbf{c}$ is orthogonal to both factors. Division by the nonzero scalar $D$ preserves that perpendicularity, giving $\boxed{\mathbf{A}\cdot \mathbf{b}=\mathbf{A}\cdot \mathbf{c}=0}$. Cyclic permutation gives the other off-diagonal pairings needed for the reciprocal basis.

Suppose $r\mathbf{A}+s\mathbf{B}+t\mathbf{C}=0$. Take its dot product successively with $\mathbf{a},\mathbf{b},\mathbf{c}$. The reciprocal pairings give $r=s=t=0$. Thus $\mathbf{A},\mathbf{B},\mathbf{C}$ are linearly independent and therefore noncoplanar in three dimensions.

For the numerical vectors, $\mathbf{b}\times \mathbf{c}=(-2,1,-2)$ and $D=(-1,1,0)\cdot (-2,1,-2)=3\ne 0$, proving noncoplanarity. The reciprocal basis isReciprocal dot products extract the coefficients of any vector in the original basis. Here $\mathbf{A}\cdot \mathbf{d}=-1/3$, $\mathbf{B}\cdot \mathbf{d}=2/3$, and $\mathbf{C}\cdot \mathbf{d}=5/3$, soSubstitution in Cartesian components recovers $(2,1,-1)$.