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Finding the angle of unit vectors

  1. Feb 8, 2010 #1
    If [itex]\vec{OA}[/itex] is the unit vector [itex]l_1i+m_1j+n_1k[/itex] and [itex]\vec{OB}[/itex] is the unit vector [itex]l_2i+m_2j+n_2k[/itex], by using the cosine formula in triangle OAB find the angle between [itex]\vec{OA}[/itex] &[itex]\vec{OB}[/itex]..

    I have tried expressing them as direction cosines , but none of that is working..can anyone point me in the right direction..
    Also I have not been introduced to scalar product ..is there an easy way of go about doing this without using the scalar product
    Thanks
     
    Last edited: Feb 8, 2010
  2. jcsd
  3. Feb 8, 2010 #2

    D H

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    Hint: Use the law of cosines.
     
  4. Feb 8, 2010 #3
    the question does state to use that ..a further hint maybe:(
     
  5. Feb 8, 2010 #4

    D H

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    What is the law of cosines?
     
  6. Feb 8, 2010 #5
    a^2=b^2+c^2-2bccosA ..

    and since the length is 1 b=c

    but what do I do with a unit vector?
     
  7. Feb 8, 2010 #6

    D H

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    You are trying to solve for cos(A), not a^2. You already know (or can know) a^2. Why don't you re-arrange the above in terms of solving for cos(A)?
     
  8. Feb 9, 2010 #7
    thanks ...I have solved it now
     
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