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Determine the unit vector

  1. Aug 16, 2013 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    The vector -i+j+k bisects the angle between the vector c and 3i+4j. Determine a unit vector along c.

    2. Relevant equations

    3. The attempt at a solution

    Taking the dot product of the two vectors (other than c) gives me the cosine of the angle = 1/5√3.
    This is also equal to the angle between the angle bisector and c. But now I can't take the dot product anymore as I don't know anything about c.
     
  2. jcsd
  3. Aug 16, 2013 #2

    haruspex

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    If you have two vectors, u and v, can you write down in terms of u and v a vector that bisects the angle between them?
     
  4. Aug 16, 2013 #3

    utkarshakash

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    lu+mv where l and m are constants.
     
  5. Aug 16, 2013 #4

    haruspex

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    That can give you any angle in the plane formed by the vectors. To get the bisector there will be a relationship between l, m, |u| and |v|.
     
    Last edited: Aug 16, 2013
  6. Aug 17, 2013 #5

    utkarshakash

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    OK So here's what I did:
    I assumed c=xi+yj+zk

    [itex]-i+j+k= \lambda \left( \dfrac{3i+4j}{5} + \dfrac{ \vec{c}}{|\vec{c}|} \right) [/itex]

    Equating the respective components of both sides I get three equations

    [itex] \frac{3}{5} + \frac{x}{|\vec{c}|} = \frac{-1}{\lambda} \\
    \frac{4}{5} + \frac{y}{|\vec{c}|} = \frac{1}{\lambda} \\
    \lambda \frac{z}{|\vec{c}|} = 1 [/itex]

    Now taking the dot product of angle bisector with c gives me the value of |c|=5(-x+y+z). Using this in the above three equations gives me x=y=z=0 !:surprised
     
  7. Aug 17, 2013 #6

    haruspex

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    I don't see how it gives you that. How about taking the dot product of c with itself?
     
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