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coolkat4

  • 3 years ago

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

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  1. Hoa
    • 3 years ago
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    i think it is 0+2kpi

  2. coolkat4
    • 3 years ago
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    thats no possiable as one of the given solutions..

  3. amistre64
    • 3 years ago
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    theres a relationship between cos and dot products, do you recall it?

  4. coolkat4
    • 3 years ago
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    yes.

  5. amistre64
    • 3 years ago
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    then the angle is simple to be found knowing:\[|u||v|cos~\alpha=u \cdot v\]

  6. amistre64
    • 3 years ago
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    solve for alpha :)

  7. coolkat4
    • 3 years ago
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    i remember the formula how do i plug the numbers into the formula?

  8. amistre64
    • 3 years ago
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    you know how to do a dot product right? and to find lengths of vectors?

  9. coolkat4
    • 3 years ago
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    yeah

  10. amistre64
    • 3 years ago
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    those are the numbers you need then, what do you get for: u.v and the lengths of u and v ?

  11. coolkat4
    • 3 years ago
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    38

  12. amistre64
    • 3 years ago
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    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
    • 3 years ago
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    \[\cos~\alpha=\frac{u.v}{|u||v|}\] \[\cos~\alpha=\frac{49}{\sqrt{37*65}}\] \[\alpha=cos^{-1}\frac{49}{\sqrt{37*65}}\]

  14. coolkat4
    • 3 years ago
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    the 38 came from 42+(-4)

  15. amistre64
    • 3 years ago
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    lol ... i do tend to add when i need to multiply when doing that

  16. amistre64
    • 3 years ago
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    -1*-4 is +4, and i did a +5 :)

  17. coolkat4
    • 3 years ago
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    okay thank you i got 20.28?

  18. amistre64
    • 3 years ago
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    i got something closer to 10

  19. amistre64
    • 3 years ago
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    http://www.wolframalpha.com/input/?i=arccos%2848%2Fsqrt%2837*65%29%29

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