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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

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


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


$\displaystyle \cos(\theta)
\isdefs \frac{\left<\underline{x},\underline{y}\right>}{\left\Vert\,\underline{x}\,\right\Vert\cdot\left\Vert\,\underline{y}\,\right\Vert}.

(Recall that the vector cosine of the angle between two vectors is given by their inner product divided by the product of their norms [454].)

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

\eqsp \underline{y}^T\left( \vert\vert\,\underline{x}\,\vert\vert ^2\cdot\mathbf{E}- \underline{x}\underline{x}^T\right)\underline{y}
\eqsp \underline{y}^T \vert\vert\,\underline{x}\,\vert\vert ^2\left(\mathbf{E}- {\cal P}_{\underline{x}}\right)\underline{y}\\ [5pt]
&\isdef & \vert\vert\,\underline{x}\,\vert\vert ^2\underline{y}^T{\cal P}_{\underline{x}^{\tiny\perp}}\underline{y},
\eqsp \vert\vert\,\underline{x}\,\vert\vert ^2\underline{y}^T{\cal P}^T_{\underline{x}^{\tiny\perp}}{\cal P}_{\underline{x}^{\tiny\perp}}\underline{y},
\isdefs \vert\vert\,\underline{x}\,\vert\vert ^2\underline{y}^T_{\underline{x}^{\tiny\perp}}\underline{y}_{\underline{x}^{\tiny\perp}}

where $ \mathbf{E}=[\underline{e}_1,\underline{e}_2,\underline{e}_3]$ denotes the identity matrix in $ \mathbb{R}^3$ , $ {\cal P}_{\underline{x}}$ denotes the orthogonal-projection matrix onto $ \underline {x}$ [454], $ {\cal P}_{\underline{x}^{\tiny\perp}}$ 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{x}\,\vert\vert ^2\underline{y}^T_{\underline{x}^{\tiny\perp}}\underline{y}_{\underline{x}^{\tiny\perp}}
\eqsp \vert\vert\,\underline{x}\,\vert\vert ^2 \vert\vert\,\underline{y}_{\underline{x}^{\tiny\perp}}\,\vert\vert ^2
\eqsp \vert\vert\,\underline{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\vert\,\underline{x}\,\vert\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\,\underline{x}\,\vert\vert \cdot \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\,\underline{y}\,\vert\vert \cdot \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 [0,0, \vert\vert\,\underline{y}\,\vert\vert \sin(\theta)]^T =
\vert\vert\,\underline{x}\,\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)$ .

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``Physical Audio Signal Processing'', by Julius O. Smith III, W3K Publishing, 2010, ISBN 978-0-9745607-2-4
Copyright © 2023-08-20 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University