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anonymous

  • 5 years ago

If vector a=-8(vector b) and vector c=7(vector b) and what is angle between a and c ??

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  1. amistre64
    • 5 years ago
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    cos(t) = a*c/|a||c|

  2. amistre64
    • 5 years ago
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    b*b = |b|^2

  3. amistre64
    • 5 years ago
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    <-8xb, -8yb> < 7xb, 7yb> -------------- -xb + -yb = -1<b>

  4. amistre64
    • 5 years ago
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    thats messed up lol

  5. anonymous
    • 5 years ago
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    yeah.. i didnt really understand that..

  6. anonymous
    • 5 years ago
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    what do u mean in ur 3rd reply..??

  7. amistre64
    • 5 years ago
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    a*c = -8<xb,yb> * 7<xb,yb>

  8. amistre64
    • 5 years ago
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    -56x^2b + -56 y^b is the only values we can get from it right?

  9. amistre64
    • 5 years ago
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    -56|b|^2 is the top value

  10. amistre64
    • 5 years ago
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    the bottom is -8|b| * 7|b| = -56|b|^2 cos(t) = 1

  11. amistre64
    • 5 years ago
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    t = 90 begrees right?

  12. amistre64
    • 5 years ago
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    so hard to type in the dark lol

  13. amistre64
    • 5 years ago
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    or it might just be 180 degrees; since a and c are scalars of b and they are facing inopposite directions.

  14. anonymous
    • 5 years ago
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    I worked it similarly. I don't know if we got same place. \[-8b.-7b =\left| a \right|\left| c \right|\cos \theta\] \[-8b.-7b =8b7b \cos \theta\] \[-1=\cos \theta\] \[\cos^{-1} -1=\theta\] \[\theta = 3.14 (radians)\]

  15. anonymous
    • 5 years ago
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    I didn't dot the left hand side. May be that is a mistake that needs to be cleaned up. I should go to sleep now.

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