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anonymous

  • 5 years ago

can some body help me with this vector problem? http://tinyurl.com/3wne3vx

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  1. amistre64
    • 5 years ago
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    ae is an addition of p+q; h is a fraction magnitude of AC

  2. amistre64
    • 5 years ago
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    and its scalar so AE = hp + hq AE = h(p+q)

  3. amistre64
    • 5 years ago
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    is that other symbol a gradient symbol?

  4. amistre64
    • 5 years ago
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    DE:EF is a ration of g:1...

  5. amistre64
    • 5 years ago
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    DE = g(EF)

  6. amistre64
    • 5 years ago
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    or just plain g = DE/EF

  7. amistre64
    • 5 years ago
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    AF = k(p)

  8. amistre64
    • 5 years ago
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    is g a magnitude?

  9. amistre64
    • 5 years ago
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    that seems to be the best I can pull out if it....

  10. amistre64
    • 5 years ago
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  11. anonymous
    • 5 years ago
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    i need help for part b

  12. amistre64
    • 5 years ago
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    p*q = cos(t) |p||q|

  13. amistre64
    • 5 years ago
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    (1/2)(2)(3) = 3 p*q = 3

  14. anonymous
    • 5 years ago
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    how do you find k?

  15. amistre64
    • 5 years ago
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    if you determined that g=1/k; then k = 1/g

  16. amistre64
    • 5 years ago
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    if DF is perp to AC then: DF * AF = 0

  17. amistre64
    • 5 years ago
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    DF*AC = 0

  18. amistre64
    • 5 years ago
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    given the magnitudes expressed as they are, that is helpful

  19. anonymous
    • 5 years ago
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    so (-p+kq)*(p+q) = 0

  20. amistre64
    • 5 years ago
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    |DF| = <kp>*<q> --------- |kp||q| and theres a place to put the cos(pi/3); just gotta recall it :)

  21. amistre64
    • 5 years ago
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    its the law of cosines... |DF|^2 = |k3|^2 + |2|^2 -2|k3||2| cos(pi/3)

  22. amistre64
    • 5 years ago
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    |DF|^2 = 9k^2 + 4 -12k(1/2) |DF|^2 = 9k^2 + 4 -6k

  23. amistre64
    • 5 years ago
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    |DF|^2 = 9k^2 -6k + 4 dont know how much this is gonna be useful :)

  24. amistre64
    • 5 years ago
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    angle b=120degrees or 2pi/3

  25. amistre64
    • 5 years ago
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    <kp>-<q> = <DF>

  26. amistre64
    • 5 years ago
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    k = p*q ----- 3

  27. amistre64
    • 5 years ago
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    p*q = 3 |3k||2| (1/2) = kp*q 3k = k(p*q) since DAB = pi/3, then q = <1,sqrt(3)>

  28. amistre64
    • 5 years ago
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    p = <0,3>

  29. amistre64
    • 5 years ago
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    right?

  30. anonymous
    • 5 years ago
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    yeah

  31. anonymous
    • 5 years ago
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    i get it now

  32. amistre64
    • 5 years ago
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  33. amistre64
    • 5 years ago
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    im starting to get it :)

  34. anonymous
    • 5 years ago
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    do you have any tips on how to solve vector because im going to have a very important test in 2 weeks?

  35. amistre64
    • 5 years ago
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    nothing that you probably dont already know, ive been reading up on them for the past week to get a feel for them.

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