Plücker coordinates for lines in 3d ed

Projective geometry ed

Arbitrary (1d) lines in 3d space can be easier handled, when seen as 2d-planes in 4d space. This is the magic of projective geometry. Let's say the line \(L\) goes through the points \(A, B \in \mathbb{R}^3\). Then we can find a plane \(L'\) in 4d, passing through the points\[ A' = \begin{pmatrix} A_x \\ A_y \\ A_z \\ 1 \end{pmatrix} = A + e_4, \quad B' = B + e_4, \quad 0 \; . \]

Why do we do this? - Since \(L'\) passes through the origin, it is now a (2d) linear subspace of \(\mathbb{R}^4\). Those can be easily expressed via outer products.

Outer algebra ed

Lines ed

The 2d linear subspace of \(L'\) can be magically associated with the outer product\[ A' \wedge B' = (A + e_4) \wedge (B + e_4) = A \wedge B + (A - B) \wedge e_4 \](well, any scaling \(\lambda \cdot A' \wedge B'\)).

This means, we can use the two 3d vectors\[ u = A - B, \quad v = A \times B \]as parameters to describe the line \(L\) via\[ L' = u \wedge e_4 + v^{\times} \; . \]

(\(v^{\times}\) here is the dual in 3d, i.e. \(e_x^{\times} = e_y \wedge e_z\))

Planes ed

A plane \(P\) in 3d through \(A,B,C\) turns to\[ A' \wedge B' \wedge C' = A \wedge B \wedge C + (A \wedge B + B \wedge C + C \wedge A) \wedge e_4 \]or with\[ d = \langle A \times B, C \rangle, \quad n = A \times B + B \times C + C \times A = (B-A) \times (C-A) \]we get\[ P' = \star(d \, e_4 + n) \; . \]

Formulas ed

General intersections ed

Let \(\psi\) be a \(k\)-vector, for a \(k\)-dimensional subspace.

The orthogonal complement is described by the dual\[ \psi^\perp = \star \psi \; . \]

The direct sum of two (linear independent) subspaces is described by the product\[ \psi \oplus \phi = \psi \wedge \phi \; . \]

Using those two rules, we can find the intersection of two subspaces as the twisted product\[ \psi \cap \phi = \star (\star \psi \wedge \star \phi) \; . \]

Intersections line/plane ed

Line \(L\) with plane \(P\):

\[ \star( \star(u \wedge e_4 + v^{\times}) \wedge \star \star(d \, e_4 + n)) \]\[ = \star((u^{\times} + v \wedge e_4) \wedge (d \, e_4 + n)) \]\[ = \star(d \, u^{\times} \wedge e_4 + u^{\times} \wedge n + v \wedge n \wedge e_4) \]\[ = d \, u + \langle u, n \rangle e_4 + v \times n \](maybe minus...)

This 4d point (well actually, this is a line in 4d) maps back into 3d as\[ L \cap P = \frac{d \, u + v \times n}{\langle u, n \rangle} \; . \]

Distance between lines ed

Given two lines \(L,M\), what is\[ L' \wedge M' \; ? \]...\[ (u \wedge e_4 + v^{\times}) \wedge (U \wedge e_4 + V^{\times}) \]\[ = (V^{\times} \wedge u + v^{\times} \wedge U) \wedge e_4 \]\[ = \star (\langle V, u \rangle + \langle v, U \rangle) \]

Too lazy to think why, but this is the signed distance between the lines!