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
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,
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