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Dot product

  1. Mar 15, 2013 #1
    find 2 unit vectors that make an angle of pi/3 with <3,4>

    <3,4>dot<a,b>=5/2=3a+4b
    b=5/8-3/4 a

    |<a,b>|=1
    such that
    a^2+25/64+15/16a+9/16 a^2=1
    25/16 a^2+15/16/ a=39/64
    100a^2+60a=39
    a^2+3/5a=39/100
    (a+3/10)^2=48/100
    a=(4sqrt(3)+-3)/10
    so
    b=5/8-(12sqrt(3)+9)/40
    =(200-96sqrt(3)-72)/320
    =(128-96sqrt(3))/320
    =(32-24sqrt(3))/80
    =(8-6sqrt(3))/20
    =(4-3sqrt(3))/10
    so far so good

    b=5/8-(12sqrt(3)-9)/40
    =(200-96sqrt(3)+72)/320
    =(272-96sqrt(3))/320
    =(68-24sqrt(3))/80
    =(34-12sqrt(3))/40
    =(17-6sqrt(3))/20 // correct me if im wrong
     
  2. jcsd
  3. Mar 16, 2013 #2

    tiny-tim

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    hi nameVoid! :smile:

    your method looks basically correct (but i haven't checked it)

    however, it's really complicated

    try first finding the vector perpendicular to (3,4) and with the same magnitude :wink:
     
  4. Mar 16, 2013 #3
    The solution for b2 is incorrect by the book
     
  5. Mar 16, 2013 #4

    tiny-tim

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    shouldn't it be minus 15/16 ? :wink:
     
  6. Mar 19, 2013 #5
    Ah yes my mistake
     
  7. Mar 20, 2013 #6

    HallsofIvy

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    Looking at <3, 4>, you should have seen that <4, -3> is perpendicular without needing to write anything!
     
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