# 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.

• There is a vector $u$ orthogonal to both of them, and then $A=\pm(A\times u)\times u$ etc. – LutzL May 30 '14 at 20:57
• What exactly do you mean by A x u? How do you define the cross product between a bivector and a vector? – user137731 May 30 '14 at 21:00
• By identifying the exterior product of 3D vectors with the cross product. Or if you do not want that, take $u$ a vector in the intersection of the planes defined by $A$ and $B$ and $a$ and $b$ vectors inside the respective planes orthogonal to $u$. – LutzL May 30 '14 at 21:15

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.
If $A=a_1∧ a_2$ and $B=b_1∧ b_2$, then there is a vector $u$ in the intersection of the planes generated respectively by $a_1,a_2$ and $b_1,b_2$, a solution of the linear system $A∧u=0=B∧u$. In each plane there can now be found vectors $a$ and $b$ orthogonal to or at least not colinear with $u$. Then $A$ is a multiple of $a∧u$ and $B$ is a multiple of $b∧u$. By rescaling $a$ and $b$, identity can be achieved.