Question

What is the relation between dot product and cross product?

Answer 1

Hint: The dot product and the cross product are the two operations which act on the vectors. The dot product of two vectors gives a scalar quantity. And the cross product of two vectors gives a vector quantity.

Complete step by step solution:
There are two types of multiplication in vector algebra. They are dot product and cross product. We use both of these operations on the vectors. The dot product of two vectors gives us a scalar quantity and the cross product of two vectors gives us a vector quantity. Since the dot product produces a scalar quantity from the vectors, it is also called the scalar product. Since the cross product produces another vector when it acts on the vectors, it is also called the vector product.
Suppose there are two $3$-dimensional vectors $\overrightarrow{A}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$ and $\overrightarrow{B}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}.$
If we are asked to find the dot product of the vectors given above, what we have to do is to multiply the corresponding coordinate values and add them together to get the required scalar quantity.
That is, the dot product of the vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ is given by
$\Rightarrow \overrightarrow{A}\cdot \overrightarrow{B}=(a_1\hat{i}+a_2\hat{j}+a_3\hat{k})\cdot (b_1\hat{i}+b_2\hat{j}+b_3\hat{k})$
The multiplication is similar to the polynomial multiplication.
But here, we get the dot product as
$\Rightarrow \overrightarrow{A}\cdot \overrightarrow{B}=(a_1\cdot b_1)+(a_2\cdot b_2)+(a_3\cdot b_3)$
And the reason for this is $\hat{i}\cdot \hat{i}=\hat{j}\cdot \hat{j}=\hat{k}\cdot \hat{k}=1$ and $\hat{i}\cdot \hat{j}=\hat{j}\cdot \hat{k}=\hat{k}\cdot \hat{i}=\hat{j}\cdot \hat{i}=\hat{k}\cdot \hat{j}=\hat{i}\cdot \hat{k}=0.$
So, if we go back to the first step, we will get the second step as
$\begin{array}{rl}& \Rightarrow \overrightarrow{A}+\overrightarrow{B}=(a_1.b_1)(\hat{i}\cdot \hat{i})+(a_2\cdot b_2)(\hat{j}\cdot \hat{j})+(a_3.b_3)(\hat{k}\cdot \hat{k})+(a_1.b_2)(\hat{i}\cdot \hat{j})+(a_1\cdot b_3)(\hat{i}\cdot \hat{k})+(a_2.b_1)(\hat{j}\cdot \hat{i})\\ & +(a_2\cdot b_3)(\hat{j}\cdot \hat{k})+(a_3.b_1)(\hat{k}\cdot \hat{i})+(a_3\cdot b_2)(\hat{k}\cdot \hat{j})\end{array}$
So, we get
$\Rightarrow \overrightarrow{A}+\overrightarrow{B}=(a_1.b_1)1+(a_2\cdot b_2)1+(a_3.b_3)1+0.$
Our answer is
$\Rightarrow \overrightarrow{A}+\overrightarrow{B}=(a_1.b_1)+(a_2\cdot b_2)+(a_3.b_3).$
And this is a scalar quantity.
Now, let us try to find the cross product of the vectors $\overrightarrow{A}$ and $\overrightarrow{B}.$
The cross product of these two vectors is the following determinant:
$\Rightarrow \overrightarrow{A}\times \overrightarrow{B}=\left|\begin{array}{rl}& \begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\end{array}\\ & \begin{array}{ccc}{a}_1& a_2& a_3\end{array}\\ & \begin{array}{ccc}{b}_1& b_2& b_3\end{array}\end{array}\right|$
The point to be noted is $$\hat{i}\times \hat{j}=\hat{k},\hat{j}\times \hat{k}=\hat{i},\hat{k}\times \hat{i}=\hat{j},\hat{j}\times \hat{i}=-\hat{k},\hat{k}\times \hat{j}=-\hat{i},\hat{i}\times \hat{k}=-\hat{j}$$ and $\hat{i}\times \hat{i}=\hat{j}\times \hat{j}=\hat{k}\times \hat{k}=0.$
And the determinant is a vector quantity.
The relation between dot product and cross product is,
$\begin{array}{rl}& \Rightarrow (\overrightarrow{u}\times \overrightarrow{v})\cdot \overrightarrow{u}=0\\ & \Rightarrow (\overrightarrow{u}\times \overrightarrow{v})\cdot \overrightarrow{v}=0.\end{array}$

Note: The dot product of two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ can be defined in terms of the angle $\theta$ made by them as $\overrightarrow{A}\cdot \overrightarrow{B}=\left|A\right|\left|B\right|\cos \theta$ where $\left|A\right|=\sqrt{{\left(a_1\right)}^2+{\left(a_2\right)}^2+{\left(a_3\right)}^2}$ and $\left|B\right|=\sqrt{{\left(b_1\right)}^2+{\left(b_2\right)}^2+{\left(b_3\right)}^2}.$ Similarly, the cross product of two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ can be defined in terms of the angle $\theta$ made by them as $\overrightarrow{A}\cdot \overrightarrow{B}=\left|A\right|\left|B\right|\sin \theta \hat{n}$ where $\hat{n}$ is the unit vector, $\left|A\right|=\sqrt{{\left(a_1\right)}^2+{\left(a_2\right)}^2+{\left(a_3\right)}^2}$ and $\left|B\right|=\sqrt{{\left(b_1\right)}^2+{\left(b_2\right)}^2+{\left(b_3\right)}^2}.$