#### 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}

where

*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,

*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 .

**Next Section:**

Mass Moment of Inertia as a Cross Product

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Striking One of the Masses