Plücker coordinates for lines in 3d ed

Outer algebra ed

Description 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 plane \(L'\) in 4d, passing through the points\[ A' = \begin{pmatrix} A \\ 1 \end{pmatrix} = A + e_4, \quad B' = \begin{pmatrix} B \\ 1 \end{pmatrix}, \quad 0 \; . \]

Since \(L'\) passes through the origin, it is a (2d) linear subspace of \(\mathbb{R}^4\), i.e. it can be associated to the outer product\[ A' \wedge B' = (A + e_4) \wedge (B + e_4) = A \wedge B + (A - B) \wedge e_4 \; . \]This means, we can use the two vectors\[ u = A - B, \quad v = A \times B \]to describe the line \(L\) via\[ L' = u \wedge e_4 + \star_{3d} v \; . \]

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\[ \star_{3d} d + \star n \; . \]

General intersections ed

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

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

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

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

Intersections line/plane ed

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

\[ \star( \star(u \wedge e_4 + \star_{3d} v) \wedge \star(\star_{3d} d + \star n)) \]\[ = \star(( \star_{3d} u + v \wedge e_4) \wedge (d \, e_4 + n)) \]\[ = \star( d \, \star_{3d} u \wedge e_4 + \star_{3d} u \wedge n + v \wedge n \wedge e_4) \]\[ = d \, u + \langle u, n \rangle e_4 + v \times n \](maybe minus...)