Quantcast

A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

coolkat4

  • 2 years ago

Find the angle between the given vectors to the nearest tenth of a degree. u = <6, -1>, v = <7, -4>?

  • This Question is Closed
  1. Hoa
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i think it is 0+2kpi

  2. coolkat4
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thats no possiable as one of the given solutions..

  3. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  4. coolkat4
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes.

  5. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  6. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    solve for alpha :)

  7. coolkat4
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  8. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  9. coolkat4
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah

  10. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  11. coolkat4
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    38

  12. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  14. coolkat4
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the 38 came from 42+(-4)

  15. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  16. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  17. coolkat4
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay thank you i got 20.28?

  18. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    i got something closer to 10

  19. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    http://www.wolframalpha.com/input/?i=arccos%2848%2Fsqrt%2837*65%29%29

  20. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.