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?

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} $.
