= Solution
Let $\Delta=[\mathbf a,\mathbf b,\mathbf c]\ne0$. These three <vectors> are linearly independent, so write $\mathbf d=\alpha\mathbf a+\beta\mathbf b+\gamma\mathbf c$. Taking the <dot product> with $\mathbf b\times\mathbf c$ gives $\alpha=[\mathbf b,\mathbf c,\mathbf d]/\Delta$. Likewise, the <dot products> with $\mathbf c\times\mathbf a$ and $\mathbf a\times\mathbf b$ give
$$
\beta=\frac{[\mathbf c,\mathbf a,\mathbf d]}\Delta=-\frac{[\mathbf c,\mathbf d,\mathbf a]}\Delta,
\qquad\gamma=\frac{[\mathbf a,\mathbf b,\mathbf d]}\Delta.
$$
Here the cyclic invariance and sign change under an interchange of the <scalar triple product> fix every sign. Substituting the three coefficients proves
$$
\boxed{\mathbf d=\frac{[\mathbf b,\mathbf c,\mathbf d]}\Delta\mathbf a
-\frac{[\mathbf c,\mathbf d,\mathbf a]}\Delta\mathbf b
+\frac{[\mathbf a,\mathbf b,\mathbf d]}\Delta\mathbf c.}
$$
This is precisely coordinate extraction using the <reciprocal basis> $(\mathbf b\times\mathbf c,\mathbf c\times\mathbf a,\mathbf a\times\mathbf b)/\Delta$.
Back to article page
