The vector product

We saw earlier, in Sect. 3.10, that it is possible to combine two vectors multiplicatively, by means of a scalar product, to form a scalar. Recall that the scalar product ${\bf a}\!\cdot\!{\bf b}$ of two vectors ${\bf a}=(a_x,a_y,a_z)$ and ${\bf b}=(b_x,b_y,b_z)$ is defined

$${\bf a} \!\cdot\!{\bf b} = a_x b_x+ a_y b_y + a_z b_z = \vert{\bf a}\vert \vert{\bf b}\vert \cos\theta,$$ (319)

where $\theta$ is the angle subtended between the directions of ${\bf a}$ and ${\bf b}$.

Is it also possible to combine two vector multiplicatively to form a third (non-coplanar) vector? It turns out that this goal can be achieved via the use of the so-called vector product. By definition, the vector product, ${\bf a}\times{\bf b}$, of two vectors ${\bf a}=(a_x,a_y,a_z)$ and ${\bf b}=(b_x,b_y,b_z)$ is of magnitude

$$\vert{\bf a}\times{\bf b} \vert =\vert{\bf a}\vert \vert{\bf b}\vert \sin\theta.$$ (320)

The direction of ${\bf a}\times{\bf b}$ is mutually perpendicular to ${\bf a}$ and ${\bf b}$, in the sense given by the right-hand grip rule when vector ${\bf a}$ is rotated onto vector ${\bf b}$ (the direction of rotation being such that the angle of rotation is less than $180^\circ$). See Fig. 70. In coordinate form,

$${\bf a}\times{\bf b} = (a_y b_z-a_z b_y, a_z b_x - a_x b_z, a_x b_y - a_y b_x).$$ (321)

Figure 70: The vector product.

There are a number of fairly obvious consequences of the above definition. Firstly, if vector ${\bf b}$ is parallel to vector ${\bf a}$, so that we can write ${\bf b} = \lambda {\bf a}$, then the vector product ${\bf a}\times{\bf b}$ has zero magnitude. The easiest way of seeing this is to note that if ${\bf a}$ and ${\bf b}$ are parallel then the angle $\theta$ subtended between them is zero, hence the magnitude of the vector product, $\vert{\bf a}\vert \vert{\bf b}\vert \sin\theta$, must also be zero (since $\sin 0^\circ=0$). Secondly, the order of multiplication matters. Thus, ${\bf b}\times {\bf a}$ is not equivalent to ${\bf a}\times{\bf b}$. In fact, as can be seen from Eq. (321),

$${\bf b}\times {\bf a} = - {\bf a}\times {\bf b}.$$ (322)

In other words, ${\bf b}\times {\bf a}$ has the same magnitude as ${\bf a}\times{\bf b}$, but points in diagrammatically the opposite direction.

Now that we have defined the vector product of two vectors, let us find a use for this concept. Figure 71 shows a rigid body rotating with angular velocity $\mbox{\boldmath$\omega$}$. For the sake of simplicity, the axis of rotation, which runs parallel to $\mbox{\boldmath$\omega$}$, is assumed to pass through the origin $O$ of our coordinate system. Point $P$, whose position vector is ${\bf r}$, represents a general point inside the body. What is the velocity of rotation ${\bf v}$ at point $P$? Well, the magnitude of this velocity is simply

$$v=\sigma \omega = \omega r \sin\theta,$$ (323)

where $\sigma$ is the perpendicular distance of point $P$ from the axis of rotation, and $\theta$ is the angle subtended between the directions of $\mbox{\boldmath$\omega$}$ and ${\bf r}$. The direction of the velocity is into the page. Another way of saying this, is that the direction of the velocity is mutually perpendicular to the directions of $\mbox{\boldmath$\omega$}$ and ${\bf r}$, in the sense indicated by the right-hand grip rule when $\mbox{\boldmath$\omega$}$ is rotated onto ${\bf r}$ (through an angle less than $180^\circ$). It follows that we can write

$${\bf v} = \mbox{\boldmath$\omega$}\times {\bf r}.$$ (324)

Note, incidentally, that the direction of the angular velocity vector $\mbox{\boldmath$\omega$}$ indicates the orientation of the axis of rotation--however, nothing actually moves in this direction; in fact, all of the motion is perpendicular to the direction of $\mbox{\boldmath$\omega$}$.

Figure 71: Rigid rotation.