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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

$\displaystyle \underline{x}\times \underline{y}$ $\displaystyle =$ $\displaystyle \left\vert \begin{array}{ccc}
\underline{e}_1 & \underline{e}_2 &...
...ine{e}_3\\ [2pt]
x_1 & x_2 & x_3\\ [2pt]
y_1 & y_2 & y_3
  $\displaystyle =$ $\displaystyle (x_2 y_3 - y_2 x_3)\underline{e}_1 + (x_3y_1 - y_3x_1) \underline{e}_2 + (x_1y_2- y_1x_2) \underline{e}_3$  
  $\displaystyle =$ $\displaystyle \left[\begin{array}{ccc}
0 & -x_3 & x_2\\ [2pt]
x_3 & 0 & -x_1\\ ...
\left[\begin{array}{c} x_1 \\ [2pt] x_2 \\ [2pt] x_3\end{array}\right]$  
    $\displaystyle \protect$ (B.15)

where $ \underline{e}_i$ denote the unit vectors in $ {\bf R}^3$. The cross-product is a vector in 3D that is orthogonal to the plane spanned by $ \underline{x}$ and $ \underline{y}$, 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 $ y\times x
= -\; x\times y$. 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

$\displaystyle \left\Vert\,\underline{x}\times \underline{y}\,\right\Vert \eqsp ...
...dot\left\Vert\,\underline{y}\,\right\Vert \cdot \vert\sin(\theta)\vert \protect$ (B.16)


$\displaystyle \sin(\theta)\eqsp \sqrt{1-\cos^2(\theta)}


$\displaystyle \cos(\theta)
\isdefs \frac{\left<\underline{x},\underline{y}\rig...

(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 $ \underline{x}\times\underline{y}= \mathbf{X}\underline{y}$ using the first matrix form in the third line of the cross-product definition in Eq.$ \,$(B.15) above. Then


where $ \mathbf{E}=[\underline{e}_1,\underline{e}_2,\underline{e}_3]$ denotes the identity matrix in $ {\bf R}^3$, $ {\cal P}_{\underline{x}}$ denotes the orthogonal-projection matrix onto $ \underline{x}$ [451], $ {\cal P}_{\underline{x}}$ denotes the projection matrix onto the orthogonal complement of $ \underline{x}$, $ \underline{y}_{\underline{x}^{\tiny\perp}}$ denotes the component of $ \underline{y}$ orthogonal to $ \underline{x}$, and we used the fact that orthogonal projection matrices $ {\cal P}$ are idempotent (i.e., $ {\cal P}^2 ={\cal P}$) and symmetric (when real, as we have here) when we replaced $ {\cal P}_{\underline{x}^{\tiny\perp}}$ by $ {\cal P}^T_{\underline{x}^{\tiny\perp}}{\cal P}_{\underline{x}^{\tiny\perp}}$ above. Finally, note that the length of $ \underline{y}_{\underline{x}^{\tiny\perp}}$ is $ \vert\vert\,\underline{y}\,\vert\vert \cdot\vert\sin(\theta)\vert$, where $ \theta$ is the angle between the 1D subspaces spanned by $ \underline{x}$ and $ \underline{y}$ in the plane including both vectors. Thus,

$\displaystyle (\underline{x}\times\underline{y})^2
\eqsp \vert\vert\,\underline...
...line{x}\,\vert\vert ^2 \vert\vert\,\underline{y}\,\vert\vert ^2\sin^2(\theta),

which establishes the desired result:

$\displaystyle \left\Vert\,\underline{x}\times\underline{y}\,\right\Vert = \vert...
...t\vert \cdot \vert\vert\,\underline{y}\,\vert\vert \cdot\vert\sin(\theta)\vert

Moreover, this proof gives an appealing geometric interpretation of the vector cross-product $ \underline{x}\times\underline{y}$ as having magnitude given by the product of $ \vert\vert\,\underline{x}\,\vert\vert $ times the norm of the difference between $ \underline{x}$ and the orthogonal projection of $ \underline{x}$ onto $ \underline{y}$ ( $ \vert\vert\,\underline{x}\times\underline{y}\,\vert\vert = \vert\vert\,\underl...
... \vert\vert\,\underline{x}-{\cal P}_{\underline{y}}(\underline{x})\,\vert\vert $) or vice versa ( $ \vert\vert\,\underline{x}\times\underline{y}\,\vert\vert = \vert\vert\,\underl...
... \vert\vert\,\underline{y}-{\cal P}_{\underline{x}}(\underline{y})\,\vert\vert $). 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 $ \underline{x}$ and $ \underline{y}$ according to the right-hand rule. This orthogonality can be checked by verifying that $ \mathbf{X}\underline{x}=\mathbf{Y}\underline{y}=\underline{0}$. The right-hand-rule parity can be checked by rotating the space so that $ \underline{x}'=[ \vert\vert\,\underline{x}\,\vert\vert ,0,0]^T$ and $ \underline{y}'=[ \vert\vert\,\underline{y}\,\vert\vert \cos(\theta), \vert\vert\,\underline{y}\,\vert\vert \sin(\theta),0]^T$ in which case $ \underline{x}'\times\underline{y}' =
\vert\vert\,\underline{x}\,\vert\vert [0,...
...}\,\vert\vert \vert\vert\,\underline{y}\,\vert\vert \sin(\theta)\underline{e}_3$. Thus, the cross product points ``up'' relative to the $ (\underline{x},\underline{y})$ plane for $ \theta
\in(0,\pi)$ and ``down'' for $ \theta\in(0,-\pi)$.

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 $ I = mR^2 = m\cdot
\vert\vert\,\underline{x}-{\cal P}_{\underline{\omega}}(\und...
...erline{\tilde{\omega}}^T\underline{x})\underline{\tilde{\omega}}\,\vert\vert ^2$, where $ \underline{\tilde{\omega}}\isdeftext \underline{\omega}/ \vert\vert\,\underline{\omega}\,\vert\vert $. In terms of the vector cross product, we can now express it as

$\displaystyle I \eqsp m\cdot(\underline{\tilde{\omega}}\times \underline{x})^2 ...
\eqsp mR^2

where $ R= \vert\vert\,\underline{x}\,\vert\vert \sin(\theta_{\underline{\tilde{\omega}}\underline{x}})$ is the distance from the rotation axis out to the point $ \underline{x}$ (which equals the length of the vector $ \underline{x}-{\cal P}_{\underline{\omega}}(\underline{x})$).

Tangential Velocity as a Cross Product

Referring again to Fig.B.4, we can write the tangential velocity vector $ \underline{v}$ as a vector cross product of the angular-velocity vector $ \underline{\omega}$B.4.11) and the position vector $ \underline{x}$:

$\displaystyle \underline{v}\eqsp \underline{\omega}\times \underline{x} \protect$ (B.17)

To see this, let's first check its direction and then its magnitude. By the right-hand rule, $ \underline{\omega}$ points up out of the page in Fig.B.4. Crossing that with $ \underline{x}$, again by the right-hand rule, produces a tangential velocity vector $ \underline{v}$ pointing as shown in the figure. So, the direction is correct. Now, the magnitude: Since $ \underline{\omega}$ and $ \underline{x}$ are mutually orthogonal, the angle between them is $ \pi /2$, so that, by Eq.$ \,$(B.16),

$\displaystyle \left\Vert\,\underline{\omega}\times \underline{x}\,\right\Vert \...
...,\right\Vert\cdot\left\Vert\,\underline{x}\,\right\Vert \eqsp \omega R \eqsp v

as desired.

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Angular Momentum
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Angular Velocity Vector