User:AugPi/sandbox

proof：need to show m = a b − b a = r×d = r× a×b, let a•a = b•b = 1

plane orthogonal to the line and include the origin

point B is the origin. the line is pass point D and orthogonal to the plane, the 2 plane pass CD and DE both orthogonal to the plane, BD is diameter

a = BE/ ||BE||, b = BC/ ||BC||，r = BD, −a = ||BF|| = ||BE||，−b = ||BG|| = ||BC||, m= ab−ba = FG, ||d || = || a×b || = sin(FBG)

angle BCE = BDE = BGF, so points D,G,E,H on a circle, and angle GHG is right angle, FG orthogonal to BD, so 4 points C, D, H, F on a circle, and

||BF|| ||BC|| = ||BH|| ||BD||, that is ab sin(FBG) = ||BH|| ||r|| sin(FBG), 2·area of triangle BFG = ab·sin(FBG) = ||BH|| ||FG|| = ||BH|| ||r|| sin(FBG), ||m|| = ||FG|| = ||r|| sin(FBG) = ||r|| ||d ||, check direction and m = r×d.

when ||r|| = 0, the line is the one pass origin with direction d; if ||r|| > 0, the line is with direction d, the plane including the origin and the line has normal vector m, the line is tangent to a circle on the plane centered origin and with radius ||r|| at point r.

Example. Let a₀ = 2, a = (−1,0,0) and b₀ = −7, b = (0,7,−2). Then (d:m) = (0:−2:−7:−7:14:−4).

Although the usual algebraic definition tends to obscure the relationship, (d:m) are the Plücker coordinates of L.