Given $\textbf{a}\perp\textbf{b}$, $\textbf{a}\cdot(\textbf{a}\wedge\textbf{b})=\mid \textbf{a}\mid^{2} \textbf{b}$ (geometric algebra)

Question

I want to prove, given $\textbf{a}\perp\textbf{b}$, $$\textbf{a}\cdot(\textbf{a}\wedge\textbf{b})=\mid \textbf{a}\mid^{2} \textbf{b}$$ I realize this is just a matter of $$\textbf{a}\cdot(\textbf{a}\wedge\textbf{b})=\langle\textbf{aab}\rangle_{2-1}=\langle\mid\textbf{a}\mid^2\textbf{b}\rangle_1=\mid\textbf{a}\mid^2\textbf{b}$$ But I would like to know if there is anything wrong with the following line of reasoning: $$\mid\textbf{a}\mid^2\textbf{b}=(\textbf{aa})\textbf{b}\stackrel{1}{=}\textbf{aab}\stackrel{2}{=}\textbf{a}\cdot \textbf{ab}\stackrel{3}{=}\textbf{a}\cdot (\textbf{ab})\stackrel{4}{=}\textbf{a}\cdot (\textbf{a}\wedge\textbf{b})$$

1: Geometric Product is associative

2: In this case $\textbf{a}\cdot \textbf{a}$ is the geometric product aa

3: Geometric product takes precedent in the order of operations (Macdonald LAGA pg. 82)

4: In this case $\textbf{a}\wedge \textbf{b}$ is the geometric product ab

Answer 1

On the right side of (2) you need parentheses around $\mathbf{a} \cdot \mathbf{a}$. Then Step (3) becomes an assertion of a kind of associativity, which you do not justify.

Conventions are useful notationally, but they cannot prove anything. Other authors have different conventions. Would they be able to prove different things?

Answer 2

For precedence conventions, I think step 3 is problematic. In general, $$(A\cdot B)C \neq A \cdot (BC)$$ To see this, revert $\textbf{ab}$ to $\textbf{ba}$ in your step 3: $$0=(\textbf{a}\cdot \textbf{b})\textbf{a}\stackrel{3}{=}\textbf{a}\cdot (\textbf{ba}) = \textbf{a}\cdot (\textbf{b} \wedge \textbf{a}) \neq 0$$, which is a contradiction.