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Thread: Using Law of Cosines To Prove Vector Algebra

  1. #1
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    Using Law of Cosines To Prove Vector Algebra

    Use the law of cosines $\displaystyle c^2 = a^2 + b^2 = 2ab \cos \theta$, where a, b and c are the lengths of the sides of a triangle and theta is the angle between sides a and b. To show:-

    $\displaystyle U_x V_x + U_y V_y + U_z V_z = ||U|| ||V|| \cos \theta$

    Hint: Set $\displaystyle c^2 = ||U - V||, a^2 = ||U||^2$ and $\displaystyle b^2 = ||V||^2$. Use the dot product properties.

    I've had an attempt at this but I'm 99% certain I am going about this the wrong way.

    $\displaystyle ||U-V|| = ||U||^2 + ||V||^2 = 2||U|| ||V|| \cos \theta = U \cdot V = ||U|| ||V|| \cos \theta$
    Last edited by mr fantastic; Aug 22nd 2009 at 03:59 PM. Reason: Improved the latex for much better readability.
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  2. #2
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    Quote Originally Posted by RedKMan View Post
    Use the law of cosines $\displaystyle c^2 = a^2 + b^2 = 2ab \cos \theta$, where a, b and c are the lengths of the sides of a triangle and theta is the angle between sides a and b. To show:-

    $\displaystyle U_x V_x + U_y V_y + U_z V_z = ||U|| ||V|| \cos \theta$

    Hint: Set $\displaystyle c^2 = ||U - V||, a^2 = ||U||^2$ and $\displaystyle b^2 = ||V||^2$. Use the dot product properties.

    I've had an attempt at this but I'm 99% certain I am going about this the wrong way.

    $\displaystyle ||U-V|| = ||U||^2 + ||V||^2 = 2||U|| ||V|| \cos \theta = U \cdot V = ||U|| ||V|| \cos \theta$
    Using the hint:
    $\displaystyle c^2 = \left( {U - V} \right) \cdot \left( {U - V} \right) = U \cdot U - 2U \cdot V + V \cdot V = \left\| U \right\|^2 - 2U \cdot V + \left\| V \right\|^2$


    Brief latex tutorial: [tex]||U|| ||V|| \cos(\theta)[/tex] gives $\displaystyle ||U|| ||V|| \cos(\theta)$
    Last edited by mr fantastic; Aug 22nd 2009 at 04:01 PM. Reason: Edited quote, re-organised the latex advice
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    I've been trying to work out how you have arrived at this answer for a few days now and I'll be honest I don't get it or understand it at all.
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  4. #4
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    Quote Originally Posted by RedKMan View Post
    I've been trying to work out how you have arrived at this answer for a few days now and I'll be honest I don't get it or understand it at all.
    Are you saying that you do not understand $\displaystyle c^2 = \left( {U - V} \right) \cdot \left( {U - V} \right) = U \cdot U - 2U \cdot V + V \cdot V = \left\| U \right\|^2 - 2U \cdot V + \left\| V \right\|^2$?
    If that is the case, you need to have a good one on one with your instructor.

    One the other hand, if you are saying that you do not understand how to finish, ok.
    Note that $\displaystyle U \cdot V = \left\| U \right\|\left\| V \right\|\cos (\theta )$ where $\displaystyle \theta$ is the angle between $\displaystyle U~\&~V$.
    Then from $\displaystyle c^2 = \left\| U \right\|^2 - 2U \cdot V + \left\| V \right\|^2$ we get $\displaystyle c^2 = a^2+b^2 - 2ab\cos(\theta). $
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  5. #5
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    OK, now I get it. Thanks. I don't have an instructor.
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