this approach extends to include points, lines, and a plane "at infinity", in the sense of projective geometry. Example. Let x = (2,3,7) and y = (2,1,0). Then (d:m) = (0:−2:−7:−7:14:−4). Alternatively, let the equations for points x of two distinct planes containing L be 0 = a + a ⋅ x 0 = b + b ⋅ x . Then their respective planes are perpendicular to vectors a and b, and the direction of L must be perpendicular to both. Hence we may set d = a × b, which is nonzero because a and b are neither zero nor parallel (the planes being distinct and intersecting). If point x satisfies both plane equations, then it also satisfies the linear combination {| |- | 0 || = a (b + b ⋅ x) − b (a + a ⋅ x) |- | || = (a b − b a) ⋅ x . |} That is, m = a b − b a is a vector perpendicular to displacements to points on L from the origin; it is, in fact, a moment consistent with the d previously defined from a and b. Proof 1: Need to show that m = a b − b a = r × d = r × (a × b).what is "r"? Without loss of generality, let a ⋅ a = b ⋅ b = 1. 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 . Since (by side-angle-side congruence), then . Since , let . By the inscribed angle theorem, , so . ; , therefore . Then DHF must be a right angle as well. Angles DCF and DHF are