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