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Explain why cosine formula is always -0,5.

  1. Aug 10, 2009 #1
    1. The problem statement, all variables and given/known data
    Pick any numbers that add to:
    [tex] x + y + z = 0 [/tex]
    Find the angle between your vector [tex]\textbf{v} = (x, y, z)[/tex]
    and the vecor [tex]\textbf{w} = (z, x, y)[/tex]
    Explain why [tex] \textbf{v}\bullet\textbf{w} / ||\textbf{v}||||\textbf{w}|| [/tex] is always [tex] -\frac{1}{2} [/tex]

    2. Relevant equations

    Cosine Forumla:


    3. The attempt at a solution

    I pick:

    [tex] \textbf{v} = (2, -1, -1) [/tex]
    [tex] \textbf{w} = (-1, 2, -1) [/tex]

    I insert the integers into the cosine forumla and get [tex] -\frac{1}{2} [/tex]

    As for the question why it is always -(1/2), I am not sure where to start.

    If you guys could push me in the right direction I would really appreciate it.

  2. jcsd
  3. Aug 10, 2009 #2


    User Avatar
    Homework Helper

    Well I'd write out what v.w works out as and what |v||w| works out as.

    EDIT: Hint: remember what (x+y+z)2 equals
  4. Aug 10, 2009 #3
    [tex]\textbf{v} \bullet \textbf{w} = xz + yz + xy [/tex]

    I can write this as:
    [tex] \textbf{v} \bullet \textbf{w} = \frac{1}{2} (x + y + z)^2 - \frac{1}{2} (x^2 + y^2 + z^2) [/tex]

    [tex]||\textbf{v}|| ||\textbf{w}|| = x^2 + y^2 + z^2 [/tex]

    This leads me to:

    [tex]\frac{\textbf{v} \bullet \textbf{w}}{||\textbf{v}|| ||\textbf{w}||} = \frac{\frac{1}{2} (x + y + z)^2}{(x^2 + y^2 + z^2)} - \frac{1}{2}[/tex]


    [tex] x + y + z = 0 [/tex]

    the answer is,

    [tex] - \frac{1}{2} [/tex]

    Thank you rock.freak667!
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