Generalizing the dot product to multivectors I am studying the book Linear and Geometric Algebra (Macdonald), and I've been stuck on a couple related, seemingly-elementary problems for a couple of days.
5.3.4. Suppose that $\mathbf{a} \bot \mathbf{b}$. Show that $\mathbf{a} \cdot (\mathbf{a} \land \mathbf{b}) = |\mathbf{a}|^2\mathbf{b}$.
5.3.5. Show that $\mathbf{e_1} \cdot (\mathbf{e_2} \land \mathbf{e_3}) = 0$
This early in the text, $\cdot$ has only been defined for vectors, not for multivectors. The only relevant identities seem to be:


*

*$\mathbf{a}\mathbf{b} = \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \land \mathbf{b}$ for pure vectors $\mathbf{a}$ and $\mathbf{b}$ (presented as the fundamental identity)

*The geometric product distributes over addition and is associative.

*Identities for the geometric product of vectors:
$$ \mathbf{a} \parallel \mathbf{b} \Rightarrow \mathbf{a} \mathbf{b} = \mathbf{b} \mathbf{a}$$
$$\mathbf{a} \bot \mathbf{b} \Rightarrow \mathbf{a} \mathbf{b} = - \mathbf{b} \mathbf{a}$$

*Scalar multiplication identities for $\land$ and the geometric product


I tried assuming that fundamental identity was not limited to 1-vectors, and applied to vector-bivector multiplication as well:
$$
\begin{align}
aB       & = a \cdot B + a \land B  && \text{unmotivated generalization} \\
Ba       & = B \cdot a + B \land a  && \text{unmotivated generalization} \\
aB + Ba  & = a \cdot B + B \cdot a + a \land B + B \land a && \text{add and rearrange} \\
aB + Ba  & = a \cdot B + a \cdot B + a \land B - a \land B && \text{assume} \cdot \text{commutes and} \land \text{anticommutes}\\
a \cdot B& = \frac 12 (aB + Ba) && \text{}\\
\end{align}
$$
But this doesn't work; as Guillermo Angeris shows in his answer, the correct form of $\cdot$ is actually $a \cdot B = \frac 12 (aB - Ba)$.
Both these problems seem to require a rule governing how $\cdot$ applies to vectors and bivectors (incorrect strawman: $\mathbf{a} \cdot (\mathbf{b}\mathbf{c}) = (\mathbf{a} \cdot \mathbf{b})\mathbf{c}$). However, this more-general $\cdot$ is not presented until the next chapter.
These two problems seem designed to teach or motivate something, but I don't see how they're soluble given the minimal set of definitions presented, at least not without the student trying to define their own generalization of the geometric product, $\cdot$, and/or $\land$ operators.
Is a solution derivable given what's already been presented?
 A: The rule you mentioned is actually $a\cdot(b\wedge c) = (a\cdot b) c - (a\cdot c) b$, which is (only kind of) derivable from the given rule.
Essentially you should just check (5.3.4) directly by finding both the antisymmetrized product and the symmetrized one, individually. We know that the product of a vector and a bivector can only have two results: one which is a trivector and one which is a vector. I'm not quite sure how Macdonald defines this, but essentially it boils down to showing the symmetric product vanishes
$$
a(ab) + (ab)a = (aa)b + a(ba) = |a|^2b - a(ab) = |a|^2b - |a|^2b = 0
$$
which we define as twice the wedge product between a vector and a bivector, and we have, for the antisymmetric part
$$
a(ab) - (ab)a = |a|^2b + |a|^2b = 2|a|^2b
$$
by the same argument as above. Hence we define, since it lowers the grade of the bivector $B$, the dot product as
$$
a\cdot B \equiv \frac{1}{2}(aB - Ba).
$$
I'm not familiar with the Macdonald book, so I may be missing something he does define in the chapter, but this is usually how the grade-lowering product is motivated.
A similar, direct argument works for (5.3.5).
A: My error: The problem cannot be solved at this point in the book. Add a Hint: In Chapter 6 we will learn that (vector inner bivector) is the vector part of 
(vector)(bivector). -- Alan Macdonald 
A: I don't have that book by MacDonald (although I recall liking the preprint of that I read eons ago), and I don't remember the lingo that he uses.
Define a k-vector as a quantity having a single grade.
Examples of a 2-vector:
$$\begin{aligned}   &\mathbf{e}_1 \mathbf{e}_2 + \mathbf{e}_3 \mathbf{e}_4 \\    &\mathbf{e}_1 \mathbf{e}_3\end{aligned}$$
The general dot product formula for two k-vectors $ a_r, b_s $, of grades r and s respectively, is typically defined as a grade selection of the following sort:
$$a_r \cdot b_s={\left\langle{{ a_r b_s }}\right\rangle}_{{\left\lvert {r - s} \right\rvert}}.$$
For 5.3.4 we have
$$\begin{aligned}   \mathbf{a} \cdot (\mathbf{a} \wedge \mathbf{b})   &=   {\left\langle{{ \mathbf{a} (\mathbf{a} \wedge \mathbf{b}) }}\right\rangle}_{1} \\    &=   {\left\langle{{ \mathbf{a} (\mathbf{a} \mathbf{b} - \mathbf{a} \cdot \mathbf{b}) }}\right\rangle}_{1} \\    &=   {\left\langle{{ \mathbf{a} \mathbf{a} \mathbf{b} }}\right\rangle}_{1}   -{\left\langle{{ \mathbf{a} }}\right\rangle}_{1} (\mathbf{a} \cdot \mathbf{b})  \\    &=   \mathbf{a}^2 {\left\langle{{ \mathbf{b} }}\right\rangle}_{1}   -{\left\langle{{ \mathbf{a} }}\right\rangle}_{1} (\mathbf{a} \cdot \mathbf{b})  \\    &=   \mathbf{a}^2 \mathbf{b}   -\mathbf{a} (\mathbf{a} \cdot \mathbf{b}) ,\end{aligned}$$
but $ \mathbf{a} \cdot \mathbf{b} = 0 $ for this problem, proving the desired result.
The dot product of a multivector is typically defined of the dot products of all the grade components of the multivector, as in
$$\begin{aligned}A &= \sum_k {\left\langle{{A}}\right\rangle}_{{k}} \\ B &= \sum_k {\left\langle{{B}}\right\rangle}_{{k}} \\ A \cdot B&=\sum_{r, s} {\left\langle{{A}}\right\rangle}_{{r}} \cdot {\left\langle{{B}}\right\rangle}_{{s}},\end{aligned}$$
so
$$\begin{aligned}   \mathbf{a} \cdot ( \mathbf{b} \mathbf{c} )   &=   \mathbf{a} \cdot ( \mathbf{b} \cdot \mathbf{c} + \mathbf{b} \wedge \mathbf{c} ) \\    &=   {\left\langle{{\mathbf{a} ( \mathbf{b} \cdot \mathbf{c})}}\right\rangle}_{{1-0}}   +   {\left\langle{{\mathbf{a} ( \mathbf{b} \wedge \mathbf{c})}}\right\rangle}_{{2-1}} \\    &=   (\mathbf{b} \cdot \mathbf{c}) \mathbf{a}   +   (\mathbf{a} \cdot \mathbf{b}) \mathbf{c}   -   (\mathbf{a} \cdot \mathbf{c}) \mathbf{b}.\end{aligned}$$
This is not generally equal to $ (\mathbf{a} \cdot \mathbf{b}) \mathbf{c} $.
