### Vector Cross Product

The *vector cross product* (or simply *vector product*, as
opposed to the *scalar product* (which is also called the
*dot product*, or *inner product*)) is commonly used in
*vector calculus*--a basic mathematical toolset used in
mechanics [270,258],
acoustics [349], electromagnetism [356], quantum
mechanics, and more. It can be defined symbolically in the form of
a *matrix determinant*:^{B.19}

where denote the

*unit vectors*in . The cross-product is a vector in 3D that is orthogonal to the plane spanned by and , and is oriented positively according to the

*right-hand rule*.

^{B.20}

The second and third lines of Eq.(B.15) make it clear that . This is one example of a host of identities that one learns in vector calculus and its applications.

#### 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 .

#### Mass Moment of Inertia as a Cross Product

In Eq.(B.14) above, the mass moment of inertia was expressed in terms of orthogonal projection as , where . In terms of the vector cross product, we can now express it as

#### Tangential Velocity as a Cross Product

Referring again to Fig.B.4, we can write the
*tangential velocity vector*
as a vector cross product of
the angular-velocity vector
(§B.4.11) and the position
vector
:

To see this, let's first check its direction and then its magnitude. By the right-hand rule, points up out of the page in Fig.B.4. Crossing that with , again by the right-hand rule, produces a tangential velocity vector pointing as shown in the figure. So, the direction is correct. Now, the magnitude: Since and are mutually orthogonal, the angle between them is , so that, by Eq.(B.16),

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Angular Momentum

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Angular Velocity Vector