# How do you find the angle between the vectors #u=<3, 2># and #v=<4, 0>#?

##### 1 Answer

The angle is

#### Explanation:

The angle

# vec A * vec B = |A| |B| cos theta #

By convention when we refer to the angle between vectors we choose the acute angle.

So for this problem, let the angle betwen

#vec u=<<3, 2>># and#vec v=<<4, 0>>#

The modulus is given by;

# |vec u| = |<<3, 2>>| = sqrt(3^2+2^2)=sqrt(9+4)=sqrt(13) #

# |vec v| = |<<4, 0>>| = sqrt(4^2+0^2)=sqrt(16)=4 #

And the scaler product is:

# vec u * vec v = <<3, 2>> * <<4, 0>>#

# \ \ \ \ \ \ \ \ \ \ = (3)(4) + (2)(0)#

# \ \ \ \ \ \ \ \ \ \ = (12)#

And so using

# 12 = sqrt(13) * 4 * cos theta #

# :. cos theta = (12)/(4sqrt(13))#

# :. cos theta = 0.83205 ... #

# :. theta = 33.69006 °#

# :. theta = 33.7 °# (3sf)