Let a and b be two vectors in \mathbb{R}^3, and let (a\times b) be their cross product. I want to simplify (a\times b)\cdot(a\times b) using geometric algebra. I know that
(a\times b)\cdot (a\times b)=a^2 b^2-(a\cdot b)^2,
where the square of a vector a^2 is given by a\cdot b.
To evaluate this using geometric algebra, we write
a\times b=-I(a\wedge b),
where I is the pseudoscalar I^2=-1. Then:
(a\times b)\cdot(a\times b)=[-I(a\wedge b)]\cdot[-I(a\wedge b)].
We then simplify this using geometric algebra:
linearity
=[I(a\wedge b)]\cdot[I(a\wedge b)],
(Ia)\cdot b=I(a\wedge b)
=I[(a\wedge b)\wedge(I(a\wedge b))]
a\wedge(Ib)=I(a\cdot b)
=I^2(a\wedge b)\cdot(a\wedge b)
=I^2(a^2b^2-(a\cdot b)^2)
=-(a^2b^2-(a\cdot b)^2)
This however is the negative of the expected result. Can anyone see where I have gone wrong?