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Physical Audio Signal Processing
    Elementary Physics, Mechanics, and Acoustics
       Rigid-Body Dynamics
          Vector Cross Product
             Cross-Product Magnitude


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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)$.

Previous: Vector Cross Product
Next: Mass Moment of Inertia as a Cross Product

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About the Author: Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See http://ccrma.stanford.edu/~jos/ for details.


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