Question

If `vecu=(3,-1)` and `vecv=(-6,4)`, then how do you find the angle between `u` and `v` vectors?

If `vecu=(3,-1)` and `vecv=(-6,4)`, then how do you find the angle between `u` and `v` vectors?

Answer 1

See below

Explanation:

We know that dot product of two vectors is defined as

`vecv·vecu=abs(u)·abs(v)·costheta` (1)

with `theta` angle between both vectors

By other hand, we know also that `absu=sqrt(u_1^2+u_2^2`

Then `absu=sqrt(9+1)=sqrt10`
`absv=sqrt(36+16)=sqrt52`

And `vecv·vecu=v_1u_1+v_2u_2=abs(u)·abs(v)·costheta` (1)

Applying this to our case

`vecu·vecv=-18-4=-22` (2)

`vecu·vecv=sqrt10·sqrt52·costheta` (3)

Combining (2) and (3) `costheta=-22/(sqrt10·sqrt52)=-0.96476382`

`theta=arccos -0.96476382=164º 44´41.57´´`

Answer 2

Compute the dot product using `vecu*vecv=(u_x)(v_x)+(u_y)(v_y)`
Compute the magnitudes, `|vecu| and |vecv|`
Use the dot product formula, `vecu*vecv= |vecu||vecv|cos(theta)`, to find the angle.

Explanation:

Compute the dot product using `vecu*vecv=(u_x)(v_x)+(u_y)(v_y)`

`vecu*vecv=(3)(-6)+(-1)(4)`

`vecu*vecv=-22`

Compute the magnitudes, `|vecu| and |vecv|`

`|vecu| = sqrt(u_x^2+u_y^2)`

`|vecu| = sqrt(3^2+(-1)^2)`

`|vecu| = sqrt10`

`|vecv| = sqrt(v_x^2+v_y^2)`

`|vecv| = sqrt((-6)^2+4^2)`

`|vecv| = 2sqrt13`

Use the dot product formula, `vecu*vecv= |vecu||vecv|cos(theta)`, to find the angle.

`-22= (sqrt10)(2sqrt13)cos(theta)`

`theta = cos^-1(-22/((sqrt10)(2sqrt13)))`

`theta = cos^-1(-22/((sqrt10)(2sqrt13)))`

`theta = 164.7^@`