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Proving Cross Product distributivity, but

  1. Sep 2, 2010 #1
    ... with the cross product being only defined as: A X B = |A| |B| sin [tex]\theta[/tex] times a unit vector perpendicular to the plane of A&B (direction according to the right hand rule, in the usual way).
    where theta is the smallest angle between vectors A & B.

    A X ( B + C ) = A X B + A X C
    is the equation I have to prove without using a component-wise approach; I'm considering the case where one of the three vectors would be perpendicular to the two other as I have shown the coplanar case.

    I have tried using several approaches, graphical and algebraic, both unsuccessful so far.

    If you have any ideas please let me know, just to get started don't do anything more than suggesting something - I should be able to figure it out.

    Thanks!!
     
  2. jcsd
  3. Sep 2, 2010 #2
  4. Sep 2, 2010 #3
    Sadly I am not supposed to use anything that would deal with the operations applied to their components since I'm not supposed to have defined the cross product in such a way yet.
     
  5. Sep 2, 2010 #4

    gabbagabbahey

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    Well, with the definition you are given, you essentially have to prove

    [tex]|\textbf{B}+\textbf{C}|\sin \theta_{A,B+C} = |\textbf{B}|\sin\theta_{A,B}+|\textbf{C}|\sin\theta_{A,C}[/tex]

    Do you see why?
     
  6. Sep 2, 2010 #5
    but doesn't it contradict the triangular inequality? This (on my little drawing, at least) would be equivalent to showing that two sides of a triangle add up to the length of the third.

    I'll pause here and rethink about it for a minute.
     
  7. Sep 2, 2010 #6

    Office_Shredder

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    If you took away the sin(theta)s then it would be problematic, but those are all going to be different numbers
     
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