## Constructing a Set of Reciprocal Vectors in 3D

Suppose we have a set of vectors
$\left\{ \mathbf{a}, \: \mathbf{b} , \: \mathbf{c} \right\}$
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.