# How do I find the dot product of two three-dimensional vectors?

Oct 17, 2014

3-D vector means it encompasses all the three co-ordinate axes, i.e. , the x , y and z axes.

We represent the unit vectors along these three axes by $\hat{i}$ , $\hat{j}$ and $\hat{k}$ respectively.

Unit vectors are vectors that have a direction and their magnitude is 1.

Now, we know that in order to find the dot product of two vectors, we multiply their magnitude by the cosine of the angle included between the vectors.

Now consider this:

$\hat{i}$.$\hat{i}$ = |$\hat{i}$| |$\hat{i}$| cos 0
=>$\hat{i}$ . $\hat{i}$= 1 1 1 = 1

Again:

$\hat{i}$.$\hat{j}$= |$\hat{i}$| |$\hat{j}$| cos 90
=> $\hat{i}$.$\hat{j}$= 1 1 0 = 0

Similarly, when the third co-ordinate axis is considered, we find the scalar product using the same rules that are used for finding out the scalar product of 2-D vectors.

Therefore:

$\hat{i}$.$\hat{k}$ = |$\hat{i}$| * |$\hat{k}$| * cos 90
=> $\hat{i}$. $\hat{k}$= 1 * 1* 0 = 0

Thus, for two vectors

$\vec{A}$ = a1 $\hat{i}$ + a2 $\hat{j}$ + a3 $\hat{k}$ and
$\vec{B}$ = b1 $\hat{i}$ + b2 $\hat{j}$ + b3 $\hat{k}$

we have:

$\vec{A}$. $\vec{B}$= (a1 $\hat{i}$ + a2 $\hat{j}$ + a3 $\hat{k}$) * ( b1 $\hat{i}$ + b2 $\hat{j}$ + b3 $\hat{k}$)

=> $\vec{A}$.$\vec{B}$= a1b1 + a2b2 + a3*b3