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
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 [454].)
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
[454],
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
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
.