This is a worked example problem that shows how to find the angle between two vectors. The angle between vectors is used when finding the scalar product and vector product.

### About the Scalar Product

The scalar product is also called the dot product or the inner product. It's found by finding the component of one vector in the same direction as the other and then multiplying it by the magnitude of the other vector.

### Vector Problem

Find the angle between the two vectors:

A = 2i + 3j + 4k

B = i - 2j + 3k

### Solution

Write the components of each vector.

A_{x} = 2; B_{x} = 1

A_{y} = 3; B_{y} = -2

A_{z} = 4; B_{z} = 3

The scalar product of two vectors is given by:

A · B = A B cos θ = |A||B| cos θ

or by:

A · B = A_{x}B_{x} + A_{y}B_{y} + A_{z}B_{z}

When you set the two equations equal and rearrange the terms you find:

cos θ = (A_{x}B_{x} + A_{y}B_{y} + A_{z}B_{z}) / AB

For this problem:

A_{x}B_{x} + A_{y}B_{y} + A_{z}B_{z} = (2)(1) + (3)(-2) + (4)(3) = 8

A = (2^{2} + 3^{2} + 4^{2})^{1/2} = (29)^{1/2}

B = (1^{2} + (-2)^{2} + 3^{2})^{1/2} = (14)^{1/2}

cos θ = 8 / [(29)^{1/2} * (14)^{1/2}] = 0.397

θ = 66.6°