# How do you find the angle between the vectors #u=cos(pi/3)i+sin(pi/3)j# and #v=cos((3pi)/4)i+sin((3pi)/4)j#?

##### 2 Answers

#### 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=cos(pi/3)ulhati+sin(pi/3)ulhatj=1/2 ulhati+ sqrt(3)/2ulhatj = ((1/2),(sqrt(3)/2))#

#vec v=cos((3pi)/4)ulhati+sin((3pi)/4)ulhatj=-sqrt(2)/2ulhati + sqrt(2)/2ulhatj = ((-sqrt(2)/2),(sqrt(2)/2))#

The modulus is given by;

# |vec u| = |((1/2),(sqrt(3)/2))| \ \ \ = sqrt((1/2)^2+(sqrt(3)/2)^2)=sqrt(1/4+3/4)=sqrt(1)=1 #

# |vec v| = |((-sqrt(2)/2),(sqrt(2)/2))| = sqrt((-sqrt(2)/2)^2+(sqrt(2)/2)^2)=sqrt(2/4+2/4)=sqrt(1)=1 #

And the scaler product is:

# vec u * vec v = ((1/2),(sqrt(3)/2)) * ((-sqrt(2)/2),(sqrt(2)/2))#

# \ \ \ \ \ \ \ \ \ \ = (1/2)(-sqrt(2)/2) + (sqrt(3)/2)(sqrt(2)/2)#

# \ \ \ \ \ \ \ \ \ \ = -sqrt(2)/4 + (sqrt(2)sqrt(3))/4#

# \ \ \ \ \ \ \ \ \ \ = (sqrt(2)(sqrt(3)-1))/4#

And so using

# (sqrt(2)(sqrt(3)-1))/4 = 1 * 1 * cos theta #

# :. cos theta = (sqrt(2)(sqrt(3)-1))/4#

# :. cos theta = 0.258819 ... #

# :. theta = (7pi)/12#

So the acute angle between the vectors is

We can confirm these results graphically:

The vector

is in the direction

If a = 1, it is a unit vector, in the direction.

Here the unit vectors,

are in the directions

So, the angle

In the sense