Products of Two Vectors

# Vector Product of Two Vectors

When two vectors are multiplied in such a way that their product is a vector quantity then it is called vector product or cross product of two vectors.

Let $\overrightarrow{a}=(a_1,a_2)$ and $\overrightarrow{b}=(b_1,b_2)$ be two vectors in the Cartesian plane (i.e. xy plane) then the vector product of the two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$, denoted by $\overrightarrow{a}×\overrightarrow{b}$ (read $\overrightarrow{a}$ cross $\overrightarrow{b}$) is defined by

$\overrightarrow{a}×\overrightarrow{b}=(a_1b_2-a_2b_1)\:\overrightarrow{k}$ where $\overrightarrow{k}$ is the standard unit vector along the positive z-axis.

In other words, $\overrightarrow{a}×\overrightarrow{b}=(a_1b_2-a_2b_1)\:\overrightarrow{k}$ $=(a_1b_2-a_2b_1)(0,0,1)$ $=(0,0,a_1b_2-a_2b_1)$

Let $\overrightarrow{a}=(a_1,a_2,a_3)$ and $\overrightarrow{b}=(b_1,b_2,b_3)$ be two space vectors. Then, the vector product of $\overrightarrow{a}$ and $\overrightarrow{b}$ is defined by, $\overrightarrow{a}×\overrightarrow{b}=(a_1,a_2,a_3)×(b_1,b_2,b_3)$ $=(a_2b_3-a_3b_2, a_3b_1-a_1b_3, a_1b_2-a_2b_1)$

This vector $\overrightarrow{a}×\overrightarrow{b}$ is perpendicular to the plane of the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$. The components of the vector $\overrightarrow{a}×\overrightarrow{b}$ can be obtained by the following rule.

### Alternative Definition of Vector Product

The vector product or cross product of two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ denoted by $\overrightarrow{a}×\overrightarrow{b}$ is a vector which is perpendicular to the plane of the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ and whose magnitude is equal to the product of the magnitudes of the two vectors and sine of the angle between them.

If $\theta$ is the angle between the two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$, then $\overrightarrow{a}×\overrightarrow{b}=|\overrightarrow{a}||\overrightarrow{b}|\sin\theta\: \overrightarrow{n}$ where, $\overrightarrow{n}$ is the unit vector normal to $\overrightarrow{a}$ and $\overrightarrow{b}$.

The direction of $\overrightarrow{n}$ or $\hat{n}$ is perpendicular to the plane containing $\overrightarrow{a}$ and $\overrightarrow{b}$ and it points in the direction as given by right handed screw rule or right hand thumb rule.

If the cross product of one vector $(\overrightarrow{a})$ to another $(\overrightarrow{b})$ is made turning in anticlockwise, the direction of product i.e. of $\overrightarrow{n}$ will be upward and positive. But if the product be made turning in clockwise, the direction of product will be downward and negative.

## Vector Product of Two Vectors in Determinant Form

Let two space vectors be $\overrightarrow{a}=(a_1,a_2,a_3)=a_1\:\overrightarrow{i}+ a_2\:\overrightarrow{j}+ a_3\:\overrightarrow{k}$ $\overrightarrow{b}=(b_1,b_2,b_3)=b_1\:\overrightarrow{i}+ b_2\:\overrightarrow{j}+ b_3\:\overrightarrow{k}$

Then, $\overrightarrow{a}×\overrightarrow{b}=(a_1,a_2,a_3)×(b_1,b_2,b_3)$ $=(a_2b_3-a_3b_2, a_3b_1-a_1b_3, a_1b_2-a_2b_1)$ $=(a_2b_3-a_3b_2)\:\overrightarrow{i}+(a_3b_1-a_1b_3)\:\overrightarrow{j}+ (a_1b_2-a_2b_1)\:\overrightarrow{k}$ $=(a_2b_3-a_3b_2)\:\overrightarrow{i}-(a_1b_3-a_3b_1)\:\overrightarrow{j}+ (a_1b_2-a_2b_1)\:\overrightarrow{k}$ $=\left|\begin{array}{c} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{array} \right|$

## Expression for $\sin\theta$

Let $\overrightarrow{a}=(a_1,a_2,a_3)$ and $\overrightarrow{b}=(b_1,b_2,b_3)$ be two space vectors. Let $\theta$ be the angle between them. Then,

$\overrightarrow{a}×\overrightarrow{b}=(a_1,a_2,a_3)×(b_1,b_2,b_3)$ $=(a_2b_3-a_3b_2, a_3b_1-a_1b_3, a_1b_2-a_2b_1)$ and, $\overrightarrow{a}\cdot\overrightarrow{b}=a_1b_1+a_2b_2+a_3b_3=ab\cos \theta$

Now, $(\overrightarrow{a}×\overrightarrow{b})^2=(\overrightarrow{a}×\overrightarrow{b})\cdot (\overrightarrow{a}×\overrightarrow{b})$ $=(a_2b_3-a_3b_2)^2+(a_3b_1-a_1b_3)^2+(a_1b_2-a_2b_1)^2$ $=(a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2)-(a_1b_1+a_2b_2+a_3b_3)^2$ $=a^2b^2-(\overrightarrow{a}×\overrightarrow{b})^2$

$\text{or,}\;|\overrightarrow{a}×\overrightarrow{b}|^2=|\overrightarrow{a}|^2|\overrightarrow{b}|^2- |\overrightarrow{a}|^2|\overrightarrow{b}|^2\cos^2\theta$ $=|\overrightarrow{a}|^2|\overrightarrow{b}|^2 (1-\cos^2\theta)$ $=|\overrightarrow{a}|^2|\overrightarrow{b}|^2\sin^2\theta$

$\therefore|\overrightarrow{a}×\overrightarrow{b}|=|\overrightarrow{a}||\overrightarrow{b}|\sin\theta$ $\therefore\sin\theta=\frac{|\overrightarrow{a}×\overrightarrow{b}|}{|\overrightarrow{a}| |\overrightarrow{b}|}$

## Geometrical Interpretation of Vector Product of Two Vectors

Let the two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ be represented by $\overrightarrow{OA}$ and $\overrightarrow{OB}$ respectively. Let $\theta$ be the angle between the vectors.

$\therefore\overrightarrow{OA}=\overrightarrow{a}\;\;\text{and}\;\;\overrightarrow{OB}= \overrightarrow{b}$ Draw a parallelogram $OACB$. Join $AB$ and from $B$, draw $BN\perp OA$.

Now, the magnitude of $\overrightarrow{a}×\overrightarrow{b}$ is given by, $|\overrightarrow{a}×\overrightarrow{b}|=ab\sin\theta$

Again, area of the parallelogram $OACB$ $=OA×BN$ $=OA×OB\sin\theta$ $=ab\sin\theta$

Hence, the magnitude of cross product of two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is equal to the area of the parallelogram with magnitudes of $\overrightarrow{a}$ and $\overrightarrow{b}$ as its adjacent sides.

Cor. Since the area of the parallelogram $OACB$ is twice the area of the triangle $OAB$, we have, $|\overrightarrow{a}×\overrightarrow{b}|=2\Delta OAB$

Hence, the magnitude of cross product of two vectors is equal to twice the area of a triangle whose sides are represented by $\overrightarrow{a}$ and $\overrightarrow{b}$.

## Unit Vectors

The three mutually perpendicular unit vectors $\overrightarrow{i}$, $\overrightarrow{j}$ and $\overrightarrow{k}$ in space are defined by $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$ respectively. Then,

$\overrightarrow{i}×\overrightarrow{i}=(1,0,0)×(1,0,0)=(0,0,0)=0$ Similarly, $\overrightarrow{j}×\overrightarrow{j}=\overrightarrow{k}×\overrightarrow{k}=0$

Again, $\overrightarrow{i}×\overrightarrow{j}=(1,0,0)×(0,1,0)=(0,0,1)=\overrightarrow{k}$ Similarly, $\overrightarrow{j}×\overrightarrow{k}=\overrightarrow{i}\;\;\text{and}\;\;\overrightarrow{k}×\overrightarrow{i}=\overrightarrow{j}$

Further, $\overrightarrow{j}×\overrightarrow{i}=(0,1,0)×(1,0,0)=(0,0,-1)$$=-(0,0,1)=-\overrightarrow{k}$ Similarly, $\overrightarrow{k}×\overrightarrow{j}=-\overrightarrow{i}\;\;\text{and}\;\;\overrightarrow{i}×\overrightarrow{k}=-\overrightarrow{j}$

The value of the cross product of two mutually perpendicular unit vectors can be remembered by using the diagram circle.

## Properties of Vector Product

1. The vector product of $\overrightarrow{a}×\overrightarrow{b}$ is perpendicular to both $\overrightarrow{a}$ and $\overrightarrow{b}$.

Let $\overrightarrow{a}=(a_1,a_2,a_3)$ and $\overrightarrow{b}=(b_1,b_2,b_3)$. Then, $\overrightarrow{a}×\overrightarrow{b}=(a_2b_3-a_3b_2,a_3b_1-a_1b_3,a_1b_2-a_2b_1)$

Now, $(\overrightarrow{a}×\overrightarrow{b})\cdot\overrightarrow{a}$ $=(a_2b_3-a_3b_2,a_3b_1-a_1b_3,a_1b_2-a_2b_1)\cdot(a_1,a_2,a_3)$ $=(a_2b_3-a_3b_2)a_1+(a_3b_1-a_1b_3)a_2+(a_1b_2-a_2b_1)a_3$ $=0$

$\therefore$ $\overrightarrow{a}×\overrightarrow{b}$ is perpendicular to $\overrightarrow{a}$.

Similarly, $\overrightarrow{a}×\overrightarrow{b}$ is perpendicular to $\overrightarrow{b}$.

2. The vector product of two vectors is not commutative i.e. if $\overrightarrow{a}$ and $\overrightarrow{b}$ be two vectors then $\overrightarrow{a}×\overrightarrow{b}≠\overrightarrow{b}×\overrightarrow{a}$

Let $\overrightarrow{a}=(a_1,a_2,a_3)$ and $\overrightarrow{b}=(b_1,b_2,b_3)$.

Now, $\overrightarrow{a}×\overrightarrow{b}=\left|\begin{array}{c} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{array} \right|$ $=-\left|\begin{array}{c} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k} \\ b_1 & b_2 & b_3 \\ a_1 & a_2 & a_3 \end{array} \right| \;\; (R_1\leftrightarrow R_2)$ $=-\overrightarrow{b}×\overrightarrow{a}$

$\therefore\overrightarrow{a}×\overrightarrow{b}≠\overrightarrow{b}×\overrightarrow{a}$

### 3. Distributive Law

If $\overrightarrow{a}$, $\overrightarrow{b}$, $\overrightarrow{c}$ be any three vectors, then $\overrightarrow{a}×(\overrightarrow{b}+\overrightarrow{c})=\overrightarrow{a}×\overrightarrow{b}+\overrightarrow{a}×\overrightarrow{c}$

Let $\overrightarrow{a}=(a_1,a_2,a_3)$, $\overrightarrow{b}=(b_1,b_2,b_3)$ and $\overrightarrow{c}=(c_1,c_2,c_3)$.

Then, $\overrightarrow{b}+\overrightarrow{c}=(b_1,b_2,b_3)+(c_1,c_2,c_3)$ $=(b_1+c_1,b_2+c_2,b_3+c_3)$

Now, $\overrightarrow{a}×(\overrightarrow{b}+\overrightarrow{c})=\left|\begin{array}{c} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k} \\ a_1 & a_2 & a_3 \\ b_1+c_1 & b_2+c_2 & b_3+c_3 \end{array} \right|$

$=\left|\begin{array}{c} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{array} \right|+\left|\begin{array}{c} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k} \\ a_1 & a_2 & a_3 \\ c_1 & c_2 & c_3 \end{array} \right|$

[Each element of $R_3$ being the sum of two terms]

$=\overrightarrow{a}×\overrightarrow{b}+\overrightarrow{a}×(\overrightarrow{c}$ $\therefore\overrightarrow{a}×(\overrightarrow{b}+\overrightarrow{c})=\overrightarrow{a}×\overrightarrow{b}+\overrightarrow{a}×\overrightarrow{c}$

4. For any scalar $n$, $(n\:\overrightarrow{a})×\overrightarrow{b}=n(\overrightarrow{a}×\overrightarrow{b})=\overrightarrow{a}×(n\:\overrightarrow{b})$

Let $\overrightarrow{a}=(a_1,a_2,a_3)$ and $\overrightarrow{b}=(b_1,b_2,b_3)$. Then, $n\:\overrightarrow{a}=(na_1,na_2,na_3)$

Now, $(n\:\overrightarrow{a})×\overrightarrow{b}=\left|\begin{array}{c} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k} \\ na_1 & na_2 & na_3 \\ b_1 & b_2 & b_3 \end{array} \right|$ $=n \left|\begin{array}{c} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{array} \right|$ $=n(\overrightarrow{a}×\overrightarrow{b})$

Similarly, $\overrightarrow{a}×(n\:\overrightarrow{b})=n(\overrightarrow{a}×\overrightarrow{b})$ $\therefore(n\:\overrightarrow{a})×\overrightarrow{b}=n(\overrightarrow{a}×\overrightarrow{b})=\overrightarrow{a}×(n\:\overrightarrow{b})$

## Examples of Vector Product of Two Vectors

1. Torque $(\overrightarrow{\tau})$ acting on a particle is equal to the cross product of its position vector $(\overrightarrow{r})$ and force vector $(\overrightarrow{F})$ i.e. $\overrightarrow{\tau}= \overrightarrow{r}×\overrightarrow{F}$
2. Angular momentum $(\overrightarrow{L})$ of a particle is equal to the cross product of its position vector $(\overrightarrow{r})$ and its linear momentum $(\overrightarrow{P})$ i.e. $\overrightarrow{L}=\overrightarrow{r}×\overrightarrow{P}$
3. Linear velocity $(\overrightarrow{v})$ of a particle in rotational motion is equal to the cross product of its angular velocity $(\overrightarrow{\omega})$ and its displacement vector $(\overrightarrow{r})$ i.e. $\overrightarrow{v}=\overrightarrow{\omega}×\overrightarrow{r}$
4. Tangential acceleration $(\overrightarrow{a})$ of a particle in rotational motion is equal to the cross product of its angular acceleration $(\overrightarrow{\alpha})$ and its position vector $(\overrightarrow{r})$ i.e. $\overrightarrow{a}=\overrightarrow{\alpha}×\overrightarrow{r}$