User:AugPi/sandbox

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Proof： Need to show m = a b − b a = r×d = r×(a×b)

Let a•a = b•b = 1

Point B is the origin. Line L passes through point D and is orthogonal to the plane. The two planes pass through CD and DE and are both orthogonal to the plane. BD is the diameter of a circle.

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

Angles DHG = GED = 90º, so points D, H, E, G lie on a circle (see Thales's theorem), 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.