\[\left\{ \mathbf{a}, \: \mathbf{b} , \: \mathbf{c} \right\}\]
for which we need to construct a reciprocal set of vectors \[\left\{ \mathbf{a'}, \: \mathbf{b'} , \: \mathbf{c'} \right\}\]
.Then
\[\mathbf{a} \cdot \mathbf{a'} = \mathbf{b} \cdot \mathbf{b'} = \mathbf{c} \cdot \mathbf{c'}=1\]
\[\mathbf{a} \cdot \mathbf{b'} = \mathbf{a} \cdot \mathbf{c'} = \mathbf{b} \cdot \mathbf{c'}=\mathbf{b} \cdot \mathbf{a'} = \mathbf{c} \cdot \mathbf{a'} = \mathbf{c'} \cdot \mathbf{b}=0\]
.Since
\[\mathbf{a'} \cdot \mathbf{b}= \mathbf{a} \cdot \mathbf{c'}=0\]
, we can write \[\mathbf{a'}= \alpha (\mathbf{b} \times \mathbf{c})\]
Take the dot product with
\[\mathbf{a}\]
to give\[\mathbf{a} \cdot \mathbf{a'}= \alpha (\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}))\]
\[\mathbf{a} \cdot \mathbf{a'}=1\]
so \[ \alpha (\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}))=1 \rightarrow \alpha = \frac{1}{\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})} \rightarrow \mathbf{a'} = \frac{\mathbf{b} \times \mathbf{c}}{\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})}\]
Similarly
\[\mathbf{b'} = \frac{\mathbf{c} \times \mathbf{a}}{\mathbf{b} \cdot (\mathbf{c} \times \mathbf{a})}, \: \mathbf{c'} = \frac{\mathbf{a} \times \mathbf{b}}{\mathbf{v} \cdot (\mathbf{a} \times \mathbf{b})}\]
If
\[\left\{ \mathbf{a}, \: \mathbf{b} , \: \mathbf{c} \right\}\]
are linearly dependent then \[\mathbf{a} \cdot (\mathbf{a} \times \mathbf{b}) =0\]
etc and no set of reciprocal vectors exists.