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

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 $(2,1,-1)$.