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?

2 Answers
Jun 14, 2018

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´´

Jun 14, 2018

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^@