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Cross product

  1. Dec 26, 2006 #1
    I didn't use the template, because I am not having difficulties with a problem.

    I am just starting to study rotational motion and there it appears the cross-product. I don't like to memorize formulae that I don't understand it's meaning.

    Why is [tex]\vec a \times \vec b}[/tex] defined mathematically the way it is. Is there some trick to memorize?

    Thanks in advance.
     
  2. jcsd
  3. Dec 26, 2006 #2

    Hurkyl

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    It's defined that way because it's useful.


    If it helps, cross products are almost just like ordinary multiplication: if we write the standard basis vectors as i, j, and k, then all you need to know to compute a cross product is that

    [tex]\begin{equation*}\begin{split}
    i \times i = j \times j = k \times k = 0 \\
    i \times j = k \\
    j \times k = i \\
    k \times i = j \\
    j \times i = -k \\
    k \times j = -i \\
    i \times k = -j
    \end{split}\end{equation*}[/tex]

    (which is pretty easy to memorize), and that you can apply the distributive rule. (but not the associative rule, or the commutative rule!!!)

    One geometric meaning to a cross products relates to perpendicularity -- you can already see that in the above identities. Another geometric meaning to the cross product of v and w is the "area" of the parallelogram with sides v and w, represented as a vector perpendicular to both v and w.
     
  4. Dec 26, 2006 #3
    Arg, I remember asking this question a while ago. It is NOT "defined that way because it's useful".....grrrrr. There is a GEOMETRICAL PROOF for why it works.

    Here, I found it. Ta- DAAA:

    http://www.math.oregonstate.edu/bridge/papers/dot+cross.pdf

    Enjoy.
     
    Last edited: Dec 26, 2006
  5. Dec 26, 2006 #4
    Thank you both for your help. It is much more clear now for me. :)
     
  6. Dec 26, 2006 #5
    What can you tell me about this unit vector [itex]\vec e_\theta[/itex]? Sorry, for the double post.
     
  7. Dec 26, 2006 #6

    cristo

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    Check out : http://mathworld.wolfram.com/SphericalCoordinates.html There's a good pic here, and the equations for the unit vectors. Note that [tex]
    \bold{e_\theta} \equiv \bold{\hat{\theta}} [/tex].
     
  8. Dec 26, 2006 #7
    Why is the direction of [tex]
    \bold{\hat{\theta}} [/tex] that one?
     
  9. Dec 26, 2006 #8

    cristo

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    Since [itex]\theta[/itex] is the azimuthal angle, then [itex]\bold{\hat{\theta}}[/itex] is the unit vector in the azimuthal direction. You can think of it in the same way as, say, [itex]\bold{\hat{x}}[/itex] is the unit vector in the direction of the x axis, then [itex]\bold{\hat{\theta}}[/itex] is the unit vector in the direction of the azimuthal "axis." Since [itex]\theta[/itex] is the azimuthal angle, the unit vector is thus the tangent vector in the direction of the azimuthal angle.
     
  10. Dec 26, 2006 #9
    Thank you cristo! I can see it now.

    Is your username some reference of The Count of Monte Cristo by Alexandre Dumas ?
     
  11. Dec 26, 2006 #10

    cristo

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    You're welcome. Haha, no my username is my nickname, derived from my surname. I prefer your version though- sounds more sophisticated!
     
  12. Dec 26, 2006 #11
    What is all this talk of memorization? Just use the right hand rule and determinants. Cross-products involve no memorization.
     
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