91Ó°ÊÓ

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,

$\left|\stackrel{→}{a}×\stackrel{→}{b}\right|=\left|a\right|×\left|b\right|\sin \mathrm{θ}$ (i)

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

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

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

Calculations for$\left|\stackrel{→}{a}×\stackrel{→}{b}\right|$

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

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

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

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

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

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

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

$$\begin{gathered} \left|\stackrel{→}{a}×\stackrel{→}{c}\right|=\left|\stackrel{→}{a}×\left(-\stackrel{→}{a}-\stackrel{→}{b}\right)\right| \\ =\left|-\left(\stackrel{→}{a}×\stackrel{→}{b}\right)\right| \end{gathered}$$

(Since $\stackrel{→}{a}×\stackrel{→}{a}=0$)

The magnitude of $\left(\stackrel{→}{a}×\stackrel{→}{b}\right)$is 12.00 . Therefore,

$\left|\stackrel{→}{a}×\stackrel{→}{c}\right|=12.00$

Therefore, the magnitude of$\left|\stackrel{→}{a}×\stackrel{→}{c}\right|$ is 12.00 units .

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

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

$$\begin{gathered} \left|\stackrel{→}{b}×\stackrel{→}{c}\right|=\left|\stackrel{→}{b}×\left(-\stackrel{→}{a}-\stackrel{→}{b}\right)\right| \\ =\left|-\left(\stackrel{→}{b}×\stackrel{→}{a}\right)\right| \\ =\left|\stackrel{→}{a}×\stackrel{→}{b}\right| \\ =12 \end{gathered}$$

Therefore, the magnitude of cross product of$\stackrel{→}{b}×\stackrel{→}{c}$ is 12 units .

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