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Proof 3d vector cross products

  1. Apr 1, 2007 #1
    2 questions i have;

    1. proove that; p x ( q + r ) = p x q + p x r

    2. and p x ( q x r ) = ( p x q ) x r

    where;

    p = p1i + p2j + p3k
    q = q1i + q2j + q3k
    r = r1i + r2j + r3k
     
  2. jcsd
  3. Apr 1, 2007 #2

    cristo

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    Well, start with telling us what you know. What is the cross product between two vectors?
     
  4. Apr 1, 2007 #3

    robphy

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    Sounds like homework.

    Do you know the result of these expressions: i x i , i x j, etc... , ?
     
  5. Apr 1, 2007 #4
    yes well so far for Qn1.
    i have left handside p x (q + r) = (p1i + p2j + p3k)( (q1+ r1)i + (q2+ r2)j + (q3+ r3)k)

    am i on the right track?
     
  6. Apr 1, 2007 #5

    robphy

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    p x (q + r) = (p1i + p2j + p3k) x ( (q1+ r1)i + (q2+ r2)j + (q3+ r3)k)
    Don't forget these important symbols.
     
  7. Apr 1, 2007 #6
    thanks i dont know where to go from there
     
  8. Apr 1, 2007 #7

    robphy

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    Can you now explicitly calculate the cross-product of two vectors?
     
  9. Apr 2, 2007 #8
    Cross product is not associative, so i don't see how

    px(qxr) = (pxq)xr
     
  10. Apr 2, 2007 #9

    D H

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    SeReNiTy is correct. [itex]\vec p\times(\vec q \times\vec r) = \vec (p\times\vec q) \times\vect [/itex] only under some special circumstances. Tthe two forms are not equal in general. Google "vector triple product".
     
  11. Apr 3, 2007 #10
    true, i found out; px(qxr) = (pxq)xr
    if you let q=p

    then px(pxr)=(pxp)xr

    such that (pxp)=0

    therefore its not true statement
     
  12. Apr 3, 2007 #11
    Just grind through that. By that I mean set up the determinant and perform the algebra.
     
  13. Apr 4, 2007 #12

    Then wat is distributive law?:confused: :confused:

    Vinodh
     
  14. Apr 4, 2007 #13

    cristo

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    Number 1. in the original post is the distributivity (under addition) of the cross product: i.e. ax(b+c)=axb +axc
     
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