For the vectors in Fig. $\mathbf{3}\mathbf{-}\mathbf{32}$, with $\mathit{a}\mathbf{=}\mathbf{4}$, $\mathit{b}\mathbf{=}\mathbf{3}$, and $\mathit{c}\mathbf{=}\mathbf{5}$, what are (a) the magnitude and (b) the direction of $\overrightarrow{\mathbf{a}}\mathbf{\times }\overrightarrow{\mathbf{b}}$, (c) the magnitude and (d) the direction of$\overrightarrow{\mathbf{a}}\mathbf{\times }\overrightarrow{\mathbf{c}}$, and (e) the magnitude and (f) the direction of$\overrightarrow{\mathbf{b}}\mathbf{\times }\overrightarrow{\mathbf{c}}$? (The z-axis is not shown)

If the magnitude of the vectors and the angle between them is known, it is possible to find the magnitude of the cross product. The formula for the cross product is,

$\left|\overrightarrow{a}\times \overrightarrow{b}\right|=\left|a\right|\times \left|b\right|\sin \theta$ (i)

The direction of the cross product can be found using the right-hand rule.

The values of the vectors are given as, $\overrightarrow{a}=4,\overrightarrow{b}=3,\overrightarrow{c}=5$. The angle between $\overrightarrow{a}$and $\overrightarrow{b}$is $90°$.From the configuration of the figure,the sum of all the vectors is zero. Therefore,

$\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0$ (ii)

Calculations for$\left|\overrightarrow{a}\times \overrightarrow{b}\right|$

The angle between the$\overrightarrow{a}$ and $\overrightarrow{b}$is $90°$. Calculate the magnitude is of the cross product using the equation (i).

$$\begin{gathered} \left|\overrightarrow{a}\times \overrightarrow{b}\right|=\left(4\right)\times \left(3\right)\sin 90° \\ =12.00 \end{gathered}$$

Thus, the magnitude of cross product $\left|\overrightarrow{a}\times \overrightarrow{b}\right|$is 12.00 units .

Using the Right-HandRule, it can be seen that the cross product $\overrightarrow{a}\times \overrightarrow{b}$points in +z -direction.

First, calculate the $\overrightarrow{c}$from vector$\overrightarrow{a}$ and $\overrightarrow{b}$using,

$\overrightarrow{c}=\left(-\overrightarrow{a}-\overrightarrow{b}\right)$ (iii)

To calculate the cross product, substitute the value of vector $\overrightarrow{c}$from equation (iii)

$$\begin{gathered} \left|\overrightarrow{a}\times \overrightarrow{c}\right|=\left|\overrightarrow{a}\times \left(-\overrightarrow{a}-\overrightarrow{b}\right)\right| \\ =\left|-\left(\overrightarrow{a}\times \overrightarrow{b}\right)\right| \end{gathered}$$

(Since $\overrightarrow{a}\times \overrightarrow{a}=0$)

The magnitude of $\left(\overrightarrow{a}\times \overrightarrow{b}\right)$is 12.00 . Therefore,

$\left|\overrightarrow{a}\times \overrightarrow{c}\right|=12.00$

Therefore, the magnitude of$\left|\overrightarrow{a}\times \overrightarrow{c}\right|$ is 12.00 units .

Using the Right Hand Rule, it can be seen that $\overrightarrow{a}\times \overrightarrow{c}$points in -z-direction.

Use the value $\overrightarrow{c}$of from equation (iii) to calculate the magnitude of cross product $\overrightarrow{b}\times \overrightarrow{c}$.

$$\begin{gathered} \left|\overrightarrow{b}\times \overrightarrow{c}\right|=\left|\overrightarrow{b}\times \left(-\overrightarrow{a}-\overrightarrow{b}\right)\right| \\ =\left|-\left(\overrightarrow{b}\times \overrightarrow{a}\right)\right| \\ =\left|\overrightarrow{a}\times \overrightarrow{b}\right| \\ =12 \end{gathered}$$

Therefore, the magnitude of cross product of$\overrightarrow{b}\times \overrightarrow{c}$ is 12 units .

Using the Right Hand Rule, it can be seen that the cross product, $\overrightarrow{b}\times \overrightarrow{c}$points in +z-direction.