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Math Help - 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 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:-

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

    Hint: Set  c^2 = ||U - V||, a^2 = ||U||^2 and 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.

    ||U-V|| = ||U||^2 + ||V||^2 = 2||U|| ||V|| \cos  \theta = U \cdot V = ||U|| ||V|| \cos  \theta
    Last edited by mr fantastic; August 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 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:-

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

    Hint: Set  c^2 = ||U - V||, a^2 = ||U||^2 and 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.

    ||U-V|| = ||U||^2 + ||V||^2 = 2||U|| ||V|| \cos  \theta = U \cdot V = ||U|| ||V|| \cos  \theta
    Using the hint:
    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 ||U|| ||V|| \cos(\theta)
    Last edited by mr fantastic; August 22nd 2009 at 04:01 PM. Reason: Edited quote, re-organised the latex advice
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  3. #3
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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 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 U \cdot V = \left\| U \right\|\left\| V \right\|\cos (\theta ) where \theta is the angle between U~\&~V.
    Then from c^2  = \left\| U \right\|^2  - 2U \cdot V + \left\| V \right\|^2 we get 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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