Question

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`?

Answer 1

`(7pi)/12`

Explanation:

The angle `theta` between two vectors `vec A` and `vec B` is related to the modulus (or magnitude) and scaler (or dot) product of `vec A` and `vec A` by the relationship:

`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 `vecu` and `vecv` be `theta` then:

`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 `vec A * vec B = |A| |B| cos theta` we have:

`(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 `(7pi)/12`

We can confirm these results graphically:

Answer 2

The vector

`< a cos alpha, a sin alpha >`

is in the direction

`theta = alpha`.

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

Here the unit vectors,

`v = < cos (3/4pi), sin (3/4pi) >` and

`u = < cos pi/3, sin pi/3 >`

are in the directions

`theta = 3/4pi and pi/3`, respectively.

So, the angle `u to v`, in between, is `3/4pi-pi/3 = 5/12pi`.

In the sense `u to v`, it is the supplement `pi-5/12pi = 7/12 pi`.