On page 76 of Doran's Geometric Algebra for Physicists, the following identity is presented (Eq. (3.129)):
$$(x \cdot \Omega_B)^2 = \langle x \cdot \Omega_B x \Omega_B \rangle = - \Omega_B \cdot (x \wedge (x \cdot \Omega_B)) \quad,$$
where $x$ is a 1-vector and $\Omega_B$ is a 2-vector.
Going from the first expression to the second one was easy enough, but I can't for the life of me find out how to get the last one from the second one, in a series of reasonable, sensible steps. Expanding both expressions with respect to a basis shows that they are, in fact, equal, but that doesn't hint at how to conjure the last expression up from the second one in an intuitive way.