**In this post, we will see how to find out the angle between two vectors. **

Say **A** and **B** are 2 vectors, where **A** =Ax** i **+ Ay** j** + Az**k** and **B** = Bx **i **+ By **j** + Bz**k**

If we take the cross product of these 2 vectors, then we can write:

**A **· **B** = (Ax** i **+ Ay** j** + Az**k**) · (Bx **i **+ By **j** + Bz**k**)

Using the definition of scalar product and by applying the distributive law we get nine terms:

since i · i = j · j = k · k =1, and i · j = j · k = j · k = 0, we get**A** · **B** = Ax Bx + Ay By + Az Bz …………………. **(1)**

Again, as per the definition of scalar product, we can also write:**A** · **B **= AB cos θ ………. **(2)** [ where θ is the angle between vector A and vector B]

from equations (1) and (2), we can write:

AB cos θ = Ax Bx + Ay By + Az Bz

cos θ =( Ax Bx + Ay By + Az Bz) / AB ……………… (3)

**Using the above equation of θ we can easily find out the angle between 2 vectors.**

## Sample numerical problem to find the **angle between 2 vectors**

**Example 1**

Two vectors A and B are given by A = i + 5j − 7k and B = 6i − 2j + 3k. Find the angle between them.

**Solution**

Say the angle between A and B is φ

A · B = AB cos φ = Ax Bx + Ay By + Az Bz

cos φ =( Ax Bx + Ay By + Az Bz) / AB ………… (1)

Now let’s find out the magnitude of the 2 vectors. Then we will use these in equation 1.