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
![$\displaystyle \sin(\theta)\eqsp \sqrt{1-\cos^2(\theta)}
$](http://www.dsprelated.com/josimages_new/pasp/img2819.png)
![$\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}}
$](http://www.dsprelated.com/josimages_new/pasp/img2820.png)
To derive Eq.(B.16), let's begin with the cross-product in matrix
form as
using the first matrix form in the
third line of the cross-product definition in Eq.
(B.15) above. Then
![\begin{eqnarray*}
(\underline{x}\times\underline{y})^2
&=&
\underline{y}^T\mat...
...rline{x}^{\tiny\perp}}\underline{y}_{\underline{x}^{\tiny\perp}}
\end{eqnarray*}](http://www.dsprelated.com/josimages_new/pasp/img2822.png)
where
denotes the identity matrix in
,
denotes the orthogonal-projection matrix onto
[451],
denotes the projection matrix onto
the orthogonal complement of
,
denotes the component of
orthogonal to
, and we used the fact that orthogonal projection matrices
are idempotent (i.e.,
) and
symmetric (when real, as we have here) when we replaced
by
above. Finally,
note that the length of
is
, where
is the angle
between the 1D subspaces spanned by
and
in the plane
including both vectors. Thus,
![$\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),
$](http://www.dsprelated.com/josimages_new/pasp/img2832.png)
![$\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
$](http://www.dsprelated.com/josimages_new/pasp/img2833.png)
![$ \underline{x}\times\underline{y}$](http://www.dsprelated.com/josimages_new/pasp/img2812.png)
![$ \vert\vert\,\underline{x}\,\vert\vert $](http://www.dsprelated.com/josimages_new/pasp/img2834.png)
![$ \underline{x}$](http://www.dsprelated.com/josimages_new/pasp/img260.png)
![$ \underline{x}$](http://www.dsprelated.com/josimages_new/pasp/img260.png)
![$ \underline{y}$](http://www.dsprelated.com/josimages_new/pasp/img780.png)
![$ \vert\vert\,\underline{x}\times\underline{y}\,\vert\vert = \vert\vert\,\underl...
... \vert\vert\,\underline{x}-{\cal P}_{\underline{y}}(\underline{x})\,\vert\vert $](http://www.dsprelated.com/josimages_new/pasp/img2835.png)
![$ \vert\vert\,\underline{x}\times\underline{y}\,\vert\vert = \vert\vert\,\underl...
... \vert\vert\,\underline{y}-{\cal P}_{\underline{x}}(\underline{y})\,\vert\vert $](http://www.dsprelated.com/josimages_new/pasp/img2836.png)
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
.
Next Section:
Mass Moment of Inertia as a Cross Product
Previous Section:
Striking One of the Masses