In the context of Geometric Algebra, in $ \mathbb{R}^3 $: Let $A$ and $B$ be bivectors (representing planes). Show that $ (\langle AB \rangle_2)^* $ is a vector in the line of intersection of the two planes, where the $ ^* $ represents the dual.
My approach: $ A^*, B^* $ are vectors orthogonal to $A$, $B$; $ (A^* \wedge B^*) $ represents the plane orthogonal to both $A$ and $B$, and $ (A^* \wedge B^*)^* $ is a vector orthogonal to this plane, thus lying in the line of intersection of $A$ and $B$. Using the duality of the inner and outer products: $$ (A^* \wedge B^*)^* = A^*\cdot B^{**} = A^*.(-B) = - A^*\cdot B$$
But not quite sure how to proceed from here. Is there a general commutativity property of the dot product of a vector and a bivector? Also, what is the geometric interpretation of this result?
In general, vector-bivector contractions anti-commute. You can use this to break the contraction down into geometric products. You have, within a minus sign,
$$(IA) \cdot B = \frac{1}{2} (IAB - BIA)$$
where $I$ is the 3d pseudoscalar. You should use at this point that $I$ commutes with all blades in 3d, which allows you to factor it out, and you should then be able to recognize the rest of the expression is in fact $\langle AB \rangle_2 = A \times B$, the commutator product (MacDonald has the commutator product defined, I hope, right?).
Edit: remember, the dot product of a vector $c$ and a bivector $D$ gives the vector in $D$ that is orthogonal to $c$.
Edit edit: as always, if you don't know whether a given product commutes, you can use grade projection and associativity instead.
$$\langle IAB \rangle_1 = \langle (IA) B \rangle_1 = (IA) \cdot B$$
but using different associative grouping,
$$\langle IAB \rangle_1 = \langle I(AB) \rangle_1 = I \langle AB \rangle_2$$
$AB$ has only grade-0 and grade-2 terms, so this is exhaustive.