2 Definition of the vector product

We define the vector product of $\underset{̲}{a}$ and $\underset{̲}{b}$ , written $\underset{̲}{a}\times \underset{̲}{b}$ , as

$\underset{̲}{a}\times \underset{̲}{b}=\left|\underset{̲}{a}\right|\left|\underset{̲}{b}\right|sin\theta \underset{̲}{ê}$

By inspection of this formula note that this is a vector of magnitude $\left|\underset{̲}{a}\right|\left|\underset{̲}{b}\right|sin\theta$ in the direction of the vector $\underset{̲}{ê}$ , where $\stackrel{̂}{\underset{̲}{e}}$ is a unit vector perpendicular to the plane containing $\underset{̲}{a}$ and $\underset{̲}{b}$ in the sense defined by the right-handed screw rule. The quantity $\underset{̲}{a}\times \underset{̲}{b}$ is read as “ $\underset{̲}{a}$ cross $\underset{̲}{b}$ ”and is sometimes referred to as the cross product . The angle is chosen to lie between $0$ and $\pi$ . See Figure 40.

Figure 40 :

Formally we have

Note that $\left|\underset{̲}{a}\right|\left|\underset{̲}{b}\right|sin\theta$ gives the modulus of the vector product whereas $\stackrel{̂}{\underset{̲}{e}}$ gives its direction.

Now study Figure 41 which is used to illustrate the calculation of $\underset{̲}{b}\times \underset{̲}{a}$ . In particular note the direction of $\underset{̲}{b}\times \underset{̲}{a}$ arising through the application of the right-handed screw rule.

We see that $\underset{̲}{a}\times \underset{̲}{b}$ is not equal to $\underset{̲}{b}\times \underset{̲}{a}$ because their directions are opposite . In fact $\underset{̲}{a}\times \underset{̲}{b}=-\underset{̲}{b}\times \underset{̲}{a}$ .

Figure 41 :

Example 15

If $\underset{̲}{a}$ and $\underset{̲}{b}$ are parallel, show that $\underset{̲}{a}\times \underset{̲}{b}=\underset{̲}{0}$ .

Solution

If $\underset{̲}{a}$ and $\underset{̲}{b}$ are parallel then the angle $\theta$ between them is zero. Consequently $sin\theta =0$ from which it follows that $\underset{̲}{a}\times \underset{̲}{b}=\left|\underset{̲}{a}\right|\left|\underset{̲}{b}\right|sin\theta \stackrel{̂}{\underset{̲}{e}}=\underset{̲}{0}$ . Note that the result, $\underset{̲}{0}$ , is the zero vector .

Note in particular the important results which follow:

Example 16

Show that $\underset{̲}{i}\times \underset{̲}{j}=\underset{̲}{k}$ and find expressions for $\underset{̲}{j}\times \underset{̲}{k}$ and $\underset{̲}{k}\times \underset{̲}{i}$ .

Solution

Note that $\underset{̲}{i}$ and $\underset{̲}{j}$ are perpendicular so that the angle between them is $90^{\circ }$ . Also, the vector $\underset{̲}{k}$ is perpendicular to both $\underset{̲}{i}$ and $\underset{̲}{j}$ . Using Key Point 17, the modulus of $\underset{̲}{i}\times \underset{̲}{j}$ is $\left(1\right)\left(1\right)sin90^{\circ }=1.$ So $\underset{̲}{i}\times \underset{̲}{j}$ is a unit vector.The unit vector perpendicular to $\underset{̲}{i}$ and $\underset{̲}{j}$ in the sense defined by the right-handed screw rule is $\underset{̲}{k}$ as shown in Figure 42(a). Therefore $\underset{̲}{i}\times \underset{̲}{j}=\underset{̲}{k}$ as required.

Similarly you should verify that $\underset{̲}{j}\times \underset{̲}{k}=\underset{̲}{i}$ (Figure 42(b)) and $\underset{̲}{k}\times \underset{̲}{i}=\underset{̲}{j}$ (Figure 42(c)).