# Reciprocal Vector Triple Product

I have got to show:

$[A,B,C] = 1/[a,b,c]$

Where $[n,m,k]$ denotes the scalar triple vector product and $A,B,C$ are reciprocal vectors to $a,b,c$ (non-coplanar, but not necessarily orthonormal). Does anybody know a simple way to do this without matrixes? I have tried manipulating it with vector algebra and (if I haven't done anything wrong) end up with

$(bxc) \cdot( a (cxa) \cdot b) = 1$

but from here I cannot proceed.

$A = \frac{(b\times c)}{[a ,b, c]}$
$B =\frac {(c\times a)}{[a, b, c]}$
$C = \frac{(a\times b)}{[a ,b ,c]}$

$(c \times a)\times(a \times b)$ = $((c \times a).b)a - [c, a, a]b$
$= ((c \times a).b)a - 0$
$= ((c \times a).b)a$

Now,
$[b \times c, c \times a, a \times b]$
$=(b\times c).((c\times a) \times (a\times b))$
$=(b \times c).((c \times a).b)a$
$=a.(b \times c).((c \times a).b)$
$=[a,b,c].[c,a,b]$
$=[a b c]^2$

$[A, B, C] = [ \frac{(b \times c)}{[a, b, c]} , \frac { (c \times a)}{[a ,b, c]} , \frac{ (a \times b)}{[a b c]}]$
$= \frac{1}{[a, b ,c]^3}*[b \times c, c \times a, a \times b]$
$= \frac{1}{[a, b ,c]^3}*[a b c]^2$
$= \frac{1}{[a, b, c]}$

You can also use the Levi-Civita tensor and expand using one identity involving the delta function as follows:

The identity used first:
$$\epsilon_{ijk} \epsilon_{ilm} = \delta_{jl} \delta_{km} - \delta_{jm} \delta_{kl}$$

The full method:

$$[A,B,C] = \frac{1}{[a, b ,c]^3} (\epsilon_{ijk} a_i b_j c_k)$$

$$= \frac{1}{[a, b ,c]^3} (\epsilon_{ijk} \epsilon_{ilm} b_l c_m \epsilon_{jno} c_n a_o \epsilon_{krs} a_r b_s)$$

$$= \frac{1}{[a, b ,c]^3} ((\delta_{jl} \delta_{km} - \delta_{jm} \delta_{kl}) b_l c_m \epsilon_{jno} c_n a_o \epsilon_{krs} a_r b_s)$$ using the identity above

$$= \frac{1}{[a, b ,c]^3} (b_j c_k \epsilon_{jno} c_n a_o \epsilon_{krs} a_r b_s - b_k c_j \epsilon_{jno} c_n a_o \epsilon_{krs} a_r b_s)$$

$$= \frac{1}{[a, b ,c]^3} ([b,c,a][c,a,b] - [c,c,a][b,a,b])$$ simplified using properties of the triple product

$$= \frac{1}{[a, b ,c]^3} ([a,b,c]^2)$$

$$= \frac{1}{[a, b ,c]}$$ as required