#### Cross-Product Magnitude

It is a straightforward exercise to show that the cross-product magnitude is equal to the product of the vector lengths times the sine of the angle between them:B.21

 (B.16)

where

with

(Recall that the vector cosine of the angle between two vectors is given by their inner product divided by the product of their norms [451].)

To derive Eq.(B.16), let's begin with the cross-product in matrix form as using the first matrix form in the third line of the cross-product definition in Eq.(B.15) above. Then

where denotes the identity matrix in , denotes the orthogonal-projection matrix onto [451], denotes the projection matrix onto the orthogonal complement of , denotes the component of orthogonal to , and we used the fact that orthogonal projection matrices are idempotent (i.e., ) and symmetric (when real, as we have here) when we replaced by above. Finally, note that the length of is , where is the angle between the 1D subspaces spanned by and in the plane including both vectors. Thus,

which establishes the desired result:

Moreover, this proof gives an appealing geometric interpretation of the vector cross-product as having magnitude given by the product of times the norm of the difference between and the orthogonal projection of onto ( ) or vice versa ( ). In this geometric picture it is clear that the cross-product magnitude is maximized when the vectors are orthogonal, and it is zero when the vectors are collinear. It is length times orthogonal length.''

The direction of the cross-product vector is then taken to be orthogonal to both and according to the right-hand rule. This orthogonality can be checked by verifying that . The right-hand-rule parity can be checked by rotating the space so that and in which case . Thus, the cross product points up'' relative to the plane for and down'' for .

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