### 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°