## anonymous one year ago Find the angle between the given vectors to the nearest tenth of a degree. u = <-5, -4>, v = <-4, -3>

1. anonymous

2. anonymous

Do you have any ideas about how to get started, or are you completely stuck? :)

3. johnweldon1993

Hint...dot product $\large \vec{u} \cdot \vec{v} = ||u|||v|| \cos(\theta)$ So $\large cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{||u|| ||v||}$

4. anonymous

5. anonymous

Okay, no problem! John's given you a clue here: we can use the dot product.

6. anonymous

Do you know how to find the dot product between two vectors? :)

7. anonymous

u * v = <-5,-4> , <-4,-3> =-5(-4) + -4(-3) = 20 + 12 = 32 |u| = √(-5)^2 + (-4)^2 =√25 +16 = √41

8. anonymous

|v| = √(-4)^2 + (-3)^2 = √16 + 9 = √25 = 5

9. anonymous

u * v = 32 |u| =√41 |v| = 5

10. anonymous

$\frac{ 32 }{ \sqrt{41}5}$

11. anonymous

Well done :) Carry on...

12. anonymous

im getting 0.02648....

13. anonymous

i calculated wrong

14. anonymous

now i got, 0.99951.. one of my answer choices is 0.9 degrees, is that correct?

15. anonymous

Well, try and remember what that 0.99951 means... :) It's not actually the angle!

16. johnweldon1993

Right, that leaves us with $\large cos(\theta) = 0.99951$ What do you get when you actually solve for the angle?

17. anonymous

1.8 degrees?

18. anonymous

Excellent, well done!

19. johnweldon1993

Correct indeed

20. anonymous

thank you guys so much for your help!

21. anonymous

You're most welcome :)