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Vectors question

  1. Jun 7, 2013 #1
    1. The problem statement, all variables and given/known data
    As in the diagram.


    2. Relevant equations
    Cosine law
    sine law


    3. The attempt at a solution
    if we do attempt the magnitudes of u and w and the angle between them 60 degrees, I can solve for the opposite side where v is supposed to be.
    I calculated
    [tex] |k \vec{v}|^2 = |u|^2 + |w|^2 - 2|u||w| \cos 60 [/tex]
    when I calculate the above I got the value of the right side as square root 52
    [tex] \sqrt{52} = \frac{\sqrt{52}}{7} |\vec{v}| [/tex]

    Would this mean that to get w as a linear combination of u and v, we simply can write

    [tex] \vec{w} = \vec{u} - \frac{\sqrt{52}}{7} \vec{v} [/tex]

    is that correct?

    Thanks for your help.
     

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  3. Jun 7, 2013 #2

    verty

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    Think of this as a trig question: what are the component vectors of ##\vec{w}##? Go from there.
     
  4. Jun 7, 2013 #3
    [tex] w_{x} = 8 \cos 60 \hat{i} [/tex]
    [tex] w_{y} = 8 \sin 60 \hat{j} [/tex]

    So of course x along the horizontal so
    [tex] w_{x} = |u| - |v| \cos \theta [/tex]
    [tex] w_{y} = - |v| \sin \theta [/tex]

    But this will not yield exact answers when solved
     
  5. Jun 7, 2013 #4
    PS are we to assume that u and v are perpendicular? That would remove the need for the above
     
  6. Jun 7, 2013 #5

    haruspex

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    The diagram is a little rough, but it looks as though it is supposed to be a closed triangle. If so, the given data (angle and magnitudes) are irrelevant.
     
  7. Jun 8, 2013 #6

    verty

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    You have assumed that ##\vec{u}## is parallel to ##\hat{j}## and has the same orientation, but you shouldn't make these assumptions. However, the assumption that ##\vec{u}## and ##\vec{v}## are perpendicular is necessary to make any sense of the question, I think.
     
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