How to obtain $(a \cdot b)^2 - (a \land b)^2 = a^2b^2$ and why $(a \land b)=|a||b||\sin(\phi)|$ if $(a \land b)^2=-|a|^2|b|^2\sin^2(\phi)$? (GA)

Question

I read in Wiki (https://en.wikipedia.org/wiki/Bivector) that antisymmetric part of geometric product can be represent as $(a \land b)$ and $(a \cdot b)^2 - (a \land b)^2 = a^2b^2$

I have 2 questions:

1. How to derive $(a \cdot b)^2 - (a \land b)^2 = a^2b^2$ ?

solved: $(ab)(ba) = (a \cdot b)^2 - (a \land b)^2$

2. Why $(a \land b)=|a||b||\sin(\phi)|$ if $(a \land b)^2=-|a|^2|b|^2\sin^2(\phi)$?

solved: $B=(a \land b)=|a|b||sin(\phi)|B*$ where $B*$ is unit bivector.

Thanks.

Answer 1

This is really two questions, first.

Square.

Given

$\mathbf{a} \cdot \mathbf{b} = \frac{1}{{2}} \left( { \mathbf{a} \mathbf{b} + \mathbf{b} \mathbf{a} } \right),$

$\mathbf{a} \wedge \mathbf{b} = \frac{1}{{2}} \left( { \mathbf{a} \mathbf{b} - \mathbf{b} \mathbf{a} } \right),$

we have

$\begin{aligned}\left( { \mathbf{a} \cdot \mathbf{b} } \right)^2 - \left( { \mathbf{a} \wedge \mathbf{b} } \right)^2 &=\frac{1}{{4}} \left( { \mathbf{a} \mathbf{b} + \mathbf{b} \mathbf{a} } \right)^2 - \frac{1}{{4}} \left( { \mathbf{a} \mathbf{b} - \mathbf{b} \mathbf{a} } \right)^2 \\ &=\frac{1}{{4}} \left( { \left( {\mathbf{a} \mathbf{b}} \right)^2 +\left( {\mathbf{b} \mathbf{a}} \right)^2+ \mathbf{a} \mathbf{b} \mathbf{b} \mathbf{a}+ \mathbf{b} \mathbf{a} \mathbf{a} \mathbf{b} } \right)-\frac{1}{{4}} \left( { \left( {\mathbf{a} \mathbf{b}} \right)^2+\left( {\mathbf{b} \mathbf{a}} \right)^2- \mathbf{a} \mathbf{b} \mathbf{b} \mathbf{a}- \mathbf{b} \mathbf{a} \mathbf{a} \mathbf{b} } \right) \\ &=\frac{1}{4} \left( { 2 \mathbf{a}^2 \mathbf{b}^2} \right)-\frac{1}{4} \left( { - 2 \mathbf{a}^2 \mathbf{b}^2} \right) \\ &= \mathbf{a}^2 \mathbf{b}^2.\end{aligned}$

Assuming a Euclidean vector space, you may introduce a bivector norm defined as

$\left\lVert {\mathbf{a} \wedge \mathbf{b}} \right\rVert = \sqrt{ -\left( { \mathbf{a} \wedge \mathbf{b}} \right)^2 },$

then we may write this identity is a slightly prettier form

$\mathbf{a}^2 \mathbf{b}^2 = \left( { \mathbf{a} \cdot \mathbf{b} } \right)^2 + \left\lVert {\mathbf{a} \wedge \mathbf{b}} \right\rVert{}^2.$

This is a generalization of the 3D identity

$\left\lVert \mathbf{a} \right\rVert{}^2 \left\lVert \mathbf{b} \right\rVert{}^2 = \left( { \mathbf{a} \cdot \mathbf{b} } \right)^2 + \left\lVert {\mathbf{a} \times \mathbf{b}} \right\rVert{}^2.$

Trig relations.

For the second question, let's express one vector in terms of length and direction, say

$\mathbf{a} = \left\lVert {\mathbf{a}} \right\rVert \mathbf{\hat{a}},$

and then express the second vector as a scaled rotation of that unit vector $ \mathbf{\hat{a}} $. That is

$\mathbf{b} = \left\lVert {\mathbf{b}} \right\rVert \mathbf{\hat{a}} e^{i\theta},$

where $ \theta $ is the angle from $ \mathbf{a} $ to $ \mathbf{b} $, and

$i = (\mathbf{a} \wedge \mathbf{b})/\left\lVert {\mathbf{a} \wedge \mathbf{b}} \right\rVert,$

is the unit pseudoscalar for the plane containing $ \mathbf{a}, \mathbf{b} $, and is oriented "from" $ \mathbf{a} $ "to" $ \mathbf{b} $ as sketched below. Also note that we did not need to use half angle sandwiched rotors, since we are rotating in the plane formed by the span of the two vectors.

You can now expand the product $ \mathbf{a} \mathbf{b} $ as

$\begin{aligned}\mathbf{a} \mathbf{b} &= \left\lVert {\mathbf{a}} \right\rVert \mathbf{\hat{a}} \left\lVert {\mathbf{b}} \right\rVert \mathbf{\hat{a}} e^{i\theta} \\ &= \left\lVert {\mathbf{a}} \right\rVert \left\lVert { \mathbf{b} } \right\rVert \mathbf{\hat{a}} \mathbf{\hat{a}} e^{i\theta} \\ &= \left\lVert {\mathbf{a}} \right\rVert \left\lVert { \mathbf{b} } \right\rVert e^{i\theta} \\ &= \left\lVert {\mathbf{a}} \right\rVert \left\lVert { \mathbf{b} } \right\rVert \left( { \cos\theta + i \sin\theta } \right).\end{aligned}$

Scalar and bivector grade selections show that

$\mathbf{a} \cdot \mathbf{b} = \left\lVert {\mathbf{a}} \right\rVert \left\lVert { \mathbf{b} } \right\rVert \cos\theta,$

and

$\mathbf{a} \wedge \mathbf{b} = \left\lVert {\mathbf{a}} \right\rVert \left\lVert { \mathbf{b} } \right\rVert i \sin\theta.$

In particular, we have

$\left( { \mathbf{a} \wedge \mathbf{b}} \right)^2 = \left\lVert {\mathbf{a}} \right\rVert^2 \left\lVert { \mathbf{b} } \right\rVert^2 (-1) \sin^2\theta,$

and

$\left\lVert { \mathbf{a} \wedge \mathbf{b}} \right\rVert^2 = \left\lVert {\mathbf{a}} \right\rVert^2 \left\lVert { \mathbf{b} } \right\rVert^2 \sin^2\theta,$

or

$\left\lVert { \mathbf{a} \wedge \mathbf{b}} \right\rVert = \left\lVert {\mathbf{a}} \right\rVert \left\lVert { \mathbf{b} } \right\rVert \left\lvert { \sin\theta } \right\rvert,$

which is what you meant to write.