User:AugPi/sandbox

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

Let a•a = b•b = 1.

Plane orthogonal to line L and including the origin.

Point B is the origin. Line L passes through point D and is orthogonal to the plane of the picture. The two planes pass through CD and DE and are both orthogonal to the plane of the picture. Points C and E are the closest points on those planes to the origin B, therefore angles BCD and BED are right angles and so the points B, C, D, E lie on a circle (due to a corollary of Thales's theorem). BD is the diameter of that circle.

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

Angle BHF is a right angle due to the following argument. Let $\epsilon :=\angle BEC$. Since $\Delta BEC\cong \Delta BFG$ (by side-angle-side congruence), then $\angle BFG=\epsilon$. Since $\angle BEC+\angle CED=90^{\circ }$, let $\epsilon ':=90^{\circ }-\epsilon =\angle CED$. By the inscribed angle theorem, $\angle DEC=\angle DBC$, so $\angle DBC=\epsilon '$. $\angle HBF+\angle BFH+\angle FHB=180^{\circ }$; $\epsilon '+\epsilon +\angle FHB=180^{\circ }$, $\epsilon +\epsilon '=90^{\circ }$ therefore $\angle FHB=90^{\circ }$. Then DHF must be a right angle as well.

Angles DCF and DHF are right angles, so the four points C, D, H, F lie on a circle, and (by the intersecting secants theorem)

||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.     ∎

Proof 2:

Let a · a = b · b = 1. According to the vector triple product formula,

r × (a × b) = (r · b) a − (r · a) b

Then

r × (a × b)	=	a ||r|| ||b|| cos(∠DBC) − b ||r|| ||a|| cos(∠DBE)
	=	a ||r|| cos(∠DBC) − b ||r|| cos(∠DBE)
	=	a ||BC|| − b ||BE||
	=	−b a − (−a) b
	=	a b − b a     ∎

When ||r|| = 0, the line L passes the origin with direction d. If ||r|| > 0, the line has direction d; the plane that includes the origin and the line L has normal vector m; the line is tangent to a circle on that plane (normal to m and perpendicular to the plane of the picture) centered at the origin and with radius ||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.