Many application in mathematics, physics, engineering and geometry involve finding a vector in space that is orthogonal to two given vectors.

In this section we will learn a product that will yield such a vector, called cross product.

A product of two vectors $a$ and $b$ can be defined in such a way that the result is a vector. The result is written as $a \times b$ and called the cross product of $a$ and $b$.
The name vector product is also used in place of the term cross product
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If we have any two skew vectors $a$ and $b$, the right-hand rule is used to determine the sense of third vector $c$ that is required to be normal to the plane containing vectors $a$ and $b$.

Right Hand Rule

Since the cross product is a vector, its definition must specify both its magnitude and direction. The magnitude of $a \times b$ is $|a| |b| \sin \theta$, where $\theta$ is the angle between the two vectors $a$ and $b$. The direction of $a \times b$ is perpendicular to both $a$ and $b$ in a right-handed sense, ie., a right-handed screw rotated from $a$ towards $b$ moves in the direction of $a \times b$. we may therefore write
$a \times b$ = $|a| |b| \sin \theta$ $\hat{n}$, 

where $\hat{n}$ is a unit vector perpendicular to $a$ and $b$ in a right-handed sense.

Cross Product

Let $a$ and $b$ be arbitrary vectors, with $\hat{n}$ a unit vector normal to the plane of $a$ and $b$ chosen so that $a$, $b$ and $\hat{n}$, in this order, obey the right-hand rule.
Then the cross product of $\vec{a}$ and $\vec{b}$  is given as 

 $\vec{a} \times \vec{b}$ = $|a|.|b| \sin \theta \hat{n}$
The cross product has following properties:

1. The cross product is not commutative. Because of the right hand rule, $a \times b$ and $b \times a$ point in opposite directions, so

$a \times b$ = $-b \times a$.

2. If the two vectors $a$ and $b$ are parallel then
 
$a \times b$ = $0$.

3. $a \times a$ = $0$

4. The magnitude of the cross product of $a$ and $b$ is the area of the parallelogram made by the two vectors $a$ and $b$. Similarly the area of the triangle made by $a$ and $b$ is $\frac{|a \times b|}{2}$.

Cross Product Properties

The area of the parallelogram is the length of its base, $|a|$, multiplied by its height, $|b| sin \theta$   

5. The cross product of $a$ and $b$ only depends on the components of $b$ perpendicular to $a$. 

6. The cross product is distributive over addition, i.e.,

$a \times (b+c)$ = $a \times b$ + $a \times c$       
We know that the vector product $a \times b$ of two vectors $a$ and $b$ is itself a vector quantity. Therefore we can multiply it by another vector $c$. The product $(a \times b).c$ is called scalar triple product, which is a pure number. On the other hand the product $a \times (b \times c)$ is called vector triple product, which is again a vector quantity.
The vector product of two vectors one of which is itself the vector product of two vectors is a vector quantity called a “triple cross product”.

$a \times (b \times c)$ = $(a . c)b – (a . b)c$
From the definition the formula for cross product is given as,

$\vec{a} \times \vec{b}$ = $|a|.|b| sin \theta \hat{n}$

If the vectors $\vec{a}$ and $\vec{b}$ are perpendicular, then $\theta = 90$ and $\sin \theta = 0$

i.e., $|\vec{a} \times \vec{b}|$ = $|\vec{a}| |\vec{b}| \sin \theta$ = $|a|.|b|$


The cross product of parallel vectors are zero.
Let $A$ and $B$ be two matrices of the same size. Then the expression $AB -BA$ is called the cross product of $A$ and $B$ and is written as $A \times B$

Let $\vec{A}$ = $x \hat{i}+y \hat{j}+z \hat{k}$ and $\vec{B}$ = $a \hat{i}+b \hat{j}+c \hat{k}$

so the cross product of these two vectors can be defined by matrices form.

$A \times B$ = $\bigl(\begin{smallmatrix} i & j & k\\ x & y & z\\ a & b & c \end{smallmatrix}\bigr)$

= $A \times B$ = $\hat{i}(yc-zb)-\hat{j}(xc-za)+\hat{k}(xb-ya)$
The following are the example for cross product.

Solved Examples

Question 1: Find the cross product of following vectors

$\vec{A}$ = $5 \hat{i}+6 \hat{j}+2 \hat{k}$ and $\vec{B}$ = $ \hat{i}+ \hat{j}+ \hat{k}$
Solution:
 
Given $\vec{A}$ = $5 \hat{i}+6 \hat{j}+2 \hat{k}$ and $\vec{B}$ = $ \hat{i}+ \hat{j}+ \hat{k}$

$A \times B$ = $\bigl(\begin{smallmatrix} i & j & k\\ x & y & z\\ a & b & c \end{smallmatrix}\bigr)$

= $A \times B$ = $\hat{i}(yc-zb)-\hat{j}(xc-za)+\hat{k}(xb-ya)$

$A \times B$ = $\bigl(\begin{smallmatrix} i & j & k\\ 5 & 6 & 3\\ 1 & 1 & 1 \end{smallmatrix}\bigr)$

= $(6-2) \hat{i}-(5-2) \hat{j}+(5-6) \hat{k}$

= $4 \hat{i}-3 \hat{j}-1 \hat{k}$
 

Question 2: If $\vec{A}$ = $4\hat{i} + 5\hat{j} + \hat{k}$ and $\vec{B}$ = $-\hat{j}- \hat{k}$, Find the cross product.
Solution:
 
Given $\vec{A}$ = $4\hat{i} + 5\hat{j}+ \hat{k}$ and $\vec{B}$ = $-\hat{j}- \hat{k}$

$A \times B$ = $\bigl(\begin{smallmatrix} i & j & k\\ x & y & z\\ a & b & c \end{smallmatrix}\bigr)$

= $A \times B$ = $\hat{i}(yc-zb)-\hat{j}(xc-za)+\hat{k}(xb-ya)$

$A \times B$ = $\bigl(\begin{smallmatrix} i & j & k\\ 4 & 5 & 1\\ 0 & -1 & -1 \end{smallmatrix}\bigr)$

= $(-5+1) \hat{i}+(0+4) \hat{j}+(-4+0) \hat{k}$ = $-4 \hat{i}+4 \hat{j}-4 \hat{k}$