Geometric Algebra Question Show that in 3D any pair of bivectors A and B have a common factor u such that A = au and B = bu. (a, b, u vectors -- au and bu are the geometric product) The only thing I can think of is to try putting everything in component form, but I keep getting really long equations with no way of simplifying the answer into either product of 2 vectors or product of pseudoscalar and vector form.

Let $A = im$ and $B = in$ for two vectors $m, n$. Then $C = m \wedge n$ is another bivector, and its dual $c = iC$ is a vector common to both $A$ and $B$:

$$A \wedge c = (im) \wedge [i(m \wedge n)] = i(m \cdot [i(m \wedge n)]) = -(m \wedge m \wedge n) = 0$$

and similarly for $B$.

Incidentally, $C = A \times B$, the commutator product of the bivectors. In vector algebra, this solution would be thought of as taking the cross product of the two normal vectors.