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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.

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2 Answers 2

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

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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

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