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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 points B, C, D, E lie on a circle (due to Thales's theorem). BD is the diameter of a that circle.

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

Angle BHF is a right angle due to the following argument. Let . Since (by side-angle-side congruence), then . Since , let . By the inscribed angle theorem, , so . , , therefore .

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.

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.