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40) which is meant to be expanded across the top row to reproduce the three components of C listed in Eqs. 38). 35) might be called a geometric definition of the vector product. Then Eqs. 38) would be an algebraic definition. To show the equivalence of Eq. 35) and the component definition, Eqs. 38), let us form A · C and B · C, using Eqs. 38). We have A · C = A · (A × B) = Ax (Ay Bz − Az By ) + Ay (Az Bx − Ax Bz ) + Az (Ax By − Ay Bx ) = 0. 41) Similarly, B · C = B · (A × B) = 0. 42) show that C is perpendicular to both A and B (cos θ = 0, θ = ±90◦ ) and therefore perpendicular to the plane they determine.
Using Eqs. 38) for the cross product and Eq. 24) for the dot product, we obtain A · B × C = Ax (By Cz − Bz Cy ) + Ay (Bz Cx − Bx Cz ) + Az (Bx Cy − By Cx ) =B·C×A=C·A×B = −A · C × B = −C · B × A = −B · A × C, and so on. 48) There is a high degree of symmetry in the component expansion. Every term contains the factors Ai , Bj , and Ck . If i, j , and k are in cyclic order (x, y, z), the sign is positive. If the order is anticyclic, the sign is negative. Further, the dot and the cross may be interchanged, A · B × C = A × B · C.
11 Specifically Eqs. 46) hold only for three-dimensional space. See D. Hestenes and G. Sobczyk, Clifford Algebra to Geometric Calculus (Dordrecht: Reidel, 1984) for a far-reaching generalization of the cross product. 2, then there is no problem identifying the cross product as a vector. The crossproduct operation maps the two triples A and B into a third triple, C, which by definition is a vector. We now have two ways of multiplying vectors; a third form appears in Chapter 2. But what about division by a vector?