### Cross Product

• Evaluate cross product.

• Understand physical interpretation of cross product and its applications.

• Understand properties and identities related to cross product.

When two vectors are multiplied in such a way that their multiplication is also a vector, then it is referred as cross product.
Cross products are mostly useful when we are studying phenomenons involving rotational effects, such as moment of a force.
In order to evaluate cross product, create a 3X3 determinant with first row as i,j,k and then write down i,j,k component values of individual vectors.

$\vec{a} \times \vec{b} = \begin{vmatrix} \overline{i} & \overline{j} & \overline{k}\\ a_{1} & a_{2} & a_{3} \\ b_{1}&b_{2} & b_{3} \end{vmatrix}$

Let us take the same example vectors again,
e.g. if $\bar{a}$ = 2$\overline{i}$ + 3$\overline{j}$ + 4$\overline{k}$ and
$\bar{b}$ = 3$\overline{i}$ - 8$\overline{j}$ + $\overline{k}$ then,
$\vec{a} \times \vec{b} = \begin{vmatrix} \overline{i} & \overline{j} & \overline{k}\\ 2 & 3 & 4 \\ 3& -8 & 1 \end{vmatrix}$ Then, $\vec{a} \times \vec{b}$ = 3-(-32)$\overline{i}$ - (2-12)$\overline{j}$ + (-16-9)$\overline{k}$ = 35$\overline{i}$ +10$\overline{j}$ -25$\overline{k}$

## Properties of Cross Product:

$\left|\vec{a} \times \vec{b}\right| = \left|\bar{a}\right| \left|\bar{b}\right|\sin \theta$

The order of vectors while taking cross product is important.

$\large \displaystyle \large \overline{\mathbf{a}}\times \overline{\mathbf{b}} = - (\large \overline{\mathbf{b}}\times \overline{\mathbf{a}})$ Now consider the cross product of two similar unit vectors,

$\textbf{i} \times \textbf{i} = 0,\ \textbf{j} \times \textbf{j} = 0,\ \textbf{k} \times \textbf{k} = 0$ This is because the angle between two similar unit vectors is zero, hence $\sin \theta$ becomes zero.
Alternatively, there is another way to look at this property. For cross product of two same vectors, the two rows of the determinant (which was mentioned above) will become zero. And from the properties of determinants, if two rows (or even two columns) are identical, then the value of that determinant becomes zero.
$\textbf{i} \times \textbf{j} = \textbf{k},\ \textbf{j} \times \textbf{k} = \textbf{i},\ \textbf{k} \times \textbf{i} = \textbf{j}$ The results are negative if the order of vectors is reversed.

### Solved Example:

#### 4-7-01

Cross product of two unit vectors has a magnitude = 1. The $\angle$ between the vectors will be:

Solution:
$\left | \vec{a} \times \vec{b}\right | = a b \sin \theta$ $1 = (1) (1) \sin \theta$ $\theta = 90^\circ$

### Solved Example:

#### 4-7-02

What is the cross product $\overline{A}$ x $\overline{B}$ of the vectors, $\overline{A}$ = $\overline{i}$ + 4$\overline{j}$ + 6$\overline{k}$ and $\overline{B}$ = 2$\overline{i}$ + 3$\overline{j}$ + 5$\overline{k}$ ?

Solution:
$\overline {A}=\overline{i}+4\overline {j}+6\overline {k}$ $\overline {B}=2\overline {i}+3\overline {j}+5\overline {k}$ \begin{align*} \overline {A} \times \overline {B} &= \begin{vmatrix} \overline{i} & \overline{j} & \overline{k}\\ 1 & 4 & 6 \\ 2& 3 & 5 \end{vmatrix}\\ &= \overline {i}\left( 20-18\right) -\overline {j}\left( 5-12\right) +\overline {k}\left( 3-8\right)\\ &=2\overline {i}+7\overline {j}-5\overline {k} \end{align*}

Find the magnitude of the following vector: $\bar{A} \times \bar{B}$ where, $\bar{A}$ = (-2, -5, 2) $\bar{B}$ = (-5, -2, -3)
$[(-2, -5, 2) \times (-5, -2, -3)] =(19, -16, -21)$ $\mathrm{Magnitude} = 23 \sqrt(2) = 32.5269$