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Homework Help: Question about cross product

  1. Sep 5, 2010 #1
    1. The problem statement, all variables and given/known data

    This is actually a concept question, but since its kind of elementary i post it here

    I understand the calculation of the cross product, what i do not understand is why the cross product that only involve in 2 dimension will have the result of 3rd dimension


    2. Relevant equations

    A cross B = AB sin

    A dot B = AB cos

    3. The attempt at a solution

    If i analyze the equation, i find that Asin is equal to A's y component, and if you times that with B, it will only result in a vector perpendicular to B and has the magnitude of A's y component times B, how does that end up with vector that is perpendicular to both A and B

    and the right hand rule doesn't explain it either, it just shows how to obtain the direction of the third vector

    besides, how does a difference in trigonometric function made cross product a vector, and scalar product a scalar?

    from my understanding, AB cos is just like AB sin, it only gives a number, so where does the direction comes from

    the concept of vector multiplication is really confusing, i hope someone can help explaining this

    Thanks in advance
     
    Last edited: Sep 5, 2010
  2. jcsd
  3. Sep 6, 2010 #2

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    You need to be a little more careful in distinguishing vectors from magnitudes. The precise equations are

    A x B = |A| |B| sin(x) (1)
    where x is the angle between the vectors A and B, and |A| and |B| are just the magnitude (without direction information) of A and B.
    Similarly,
    A · B = |A| |B| cos(x) (2)

    The difference is, that A · B is just a number, whereas A x B produces a new vector. So to calculate A · B you just have one formula, namely (2). To calculate A x B you actually have three formulas, one for each component, which go like
    (A x B)x = Ay Bz - Az By (3),
    etc. If you then calculate the magnitude of the new vector A x B, you will get formula (1) back.

    So the most important ingredient to clearing up your confusion, I think, is that you should see formula (3) as defining the cross product, and see formula (1) for the magnitude as a consequence of that.
     
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