## coolkat4 Group Title Find the angle between the given vectors to the nearest tenth of a degree. u = <6, -1>, v = <7, -4>? one year ago one year ago

1. Hoa Group Title

i think it is 0+2kpi

2. coolkat4 Group Title

thats no possiable as one of the given solutions..

3. amistre64 Group Title

theres a relationship between cos and dot products, do you recall it?

4. coolkat4 Group Title

yes.

5. amistre64 Group Title

then the angle is simple to be found knowing:$|u||v|cos~\alpha=u \cdot v$

6. amistre64 Group Title

solve for alpha :)

7. coolkat4 Group Title

i remember the formula how do i plug the numbers into the formula?

8. amistre64 Group Title

you know how to do a dot product right? and to find lengths of vectors?

9. coolkat4 Group Title

yeah

10. amistre64 Group Title

those are the numbers you need then, what do you get for: u.v and the lengths of u and v ?

11. coolkat4 Group Title

38

12. amistre64 Group Title

u = <6, -1>; sqrt(37) v = <7, -4>; sqrt(65) --------------- u.v 42+5 = 49 |u||v| = sqrt(37*65) .... what does 38 mean?

13. amistre64 Group Title

$\cos~\alpha=\frac{u.v}{|u||v|}$ $\cos~\alpha=\frac{49}{\sqrt{37*65}}$ $\alpha=cos^{-1}\frac{49}{\sqrt{37*65}}$

14. coolkat4 Group Title

the 38 came from 42+(-4)

15. amistre64 Group Title

lol ... i do tend to add when i need to multiply when doing that

16. amistre64 Group Title

-1*-4 is +4, and i did a +5 :)

17. coolkat4 Group Title

okay thank you i got 20.28?

18. amistre64 Group Title

i got something closer to 10

19. amistre64 Group Title