Vector Cross Product

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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

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

where

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

with

$\displaystyle \cos(\theta) \isdefs \frac{\left<\underline{x},\underline{y}\rig... ...c{x_1y_1+x_2y_2+x_3y_3}{\sqrt{x_1^2+x_2^2+x_3^2}\cdot\sqrt{y_1^2+y_2^2+y_3^2}}$

(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

$\begin{eqnarray*} (\underline{x}\times\underline{y})^2 &=& \underline{y}^T\mat... ...rline{x}^{\tiny\perp}}\underline{y}_{\underline{x}^{\tiny\perp}} \end{eqnarray*}$

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

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

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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