In 3D-projective geometry the homogeneous coordinates of the line connecting two points with homogeneous coordinates $(x_0 : x_1 : x_2 : x_3)$ and $(y_0 : y_1 : y_2 : y_3)$ can be calculated as a kind of exterior product. With $\Delta_{ij} = x_iy_j-x_jx_i$ one gets the six Plücker coordinates of the line: $(\Delta_{01} : \Delta_{02} : \Delta_{03} : \Delta_{23} : \Delta_{31} : \Delta_{12})$.
Suppose now a plane has homogeneous coordinates $(p_0 : p_1 : p_2 : p_3)$.
Does the same exterior product $\Delta_{ij} = x_ip_j-x_jp_i$ in this case have any geometrical or other meaning at all? It would be as if in geometric algebra one calculates the bivector of a vector and another vector of the dual space.
I only see that the plane is treated as if it is a point with the same coordinates, but surely there is no interpretation for the mapping $$\text{Plane}(p_0 : p_1 : p_2 : p_3) \mapsto \text{Point}(p_0 : p_1 : p_2 : p_3)$$ is there?
I ask this because I noticed that in the definition of relativistic angular momentum in physics one defines a four-vector effectively in such a way.
One can construct a correspondence between projective geometry and geometric algebra such that 1-blades correspond to points, 2-blades to lines, 3-blades to planes, and 4-blades to volumes. Since planes correspond to 3-blades, and you can't form an antisymmetric wedge product on two 3-blades, there's no meaningful result to find.
Furthermore, dualizing those 3-blades to 1-blades is not meaningful; you cannot use full duals in projective geometry, as that involves dualizing over the projective dimension--which is inherently non-metrical and thus does not have a concept of a natural correspondence between vectors and dual vectors. More succinctly, there is no unique point corresponding to a given affine plane. How could there be?
Angular momentum in physics can be defined as a grade-2 multivector--a plane or linear combination of planes--but that is where the correspondence ends.