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How is î x jcap = kcap? Please help!

  1. May 22, 2013 #1
    how is î x jcap = kcap? Please help!
     
  2. jcsd
  3. May 22, 2013 #2

    Fredrik

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    Do you know the definition of the cross product?
     
  4. May 22, 2013 #3
    yes. a x b = absinθ
     
  5. May 22, 2013 #4

    pwsnafu

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    Do you realise that the RHS of what you wrote is a scalar?
     
  6. May 22, 2013 #5

    Fredrik

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    That's not the definition, and that equality isn't correct. You may be thinking of the result ##\left|\mathbf a\times\mathbf b\right|=|\mathbf a||\mathbf b|\sin\theta##, where ##\theta## is the angle between the two vectors.

    There are many equivalent ways to define the cross product. One of them is
    $$(a_1,a_2,a_3)\times(b_1,b_2,b_3)=(a_2b_3-a_3b_2,a_3b_1-a_1b_3,a_1b_2-a_2b_1).$$ You should check what definition your book uses, and then try to use it to prove that it implies that
    $$\mathbf i\times\mathbf j=\mathbf k.$$
     
    Last edited: May 22, 2013
  7. May 22, 2013 #6

    pwsnafu

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    Again LHS is a vector, RHS is a scalar. :wink:
     
  8. May 22, 2013 #7

    Fredrik

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    LOL, yes I know. That's why I started typing that. Somehow I forgot to type the absolute value symbols on the left. I will edit my post.
     
  9. May 22, 2013 #8
    Oh I understood. (ab sin theta) was only the magnitude.

    Thank you very very much!
     
  10. May 22, 2013 #9

    HallsofIvy

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    One good way of defining the cross product is to start with
    1)[itex]\vec{i}\times\vec{j}= \vec{k}[/itex]
    2)[itex]\vec{j}\times\vec{k}=\vec{i}[/itex]
    3)[itex]\vec{k}\times\vec{i}= \vec{j}[/itex]
    Then extend it to all other vectors by "linearity" in the first component:
    [itex](\vec{u}+ \vec{v})\times \vec{w}= \vec{u}\times \vec{w}+ \vec{v}\times \vec{w}[/itex]
    and by "anti- commutativity":
    [itex]\vec{u}\times\vec{v}= -\vec{v}\times\vec{u}[/itex]

    What you are asking about is (1) above.

    Another, equivalent but less "sophisticated", way to define the cross product is to simply say that
    [itex](A\vec{i}+ B\vec{j}+ C\vec{k})\times(P\vec{i}+ Q\vec{j}+ R\vec{k})= (BR- CQ)\vec{i}- (AR- CP)\vec{j}+ (AQ- BP)\vec{k}[/itex]

    and then, [itex]\vec{i}\times\vec{j}[/itex] has A=1, B= 0, C= 0, P= 0, Q= 1, R= 0
    so the product is [itex](0(0)- 0(1))\vec{i}- (1(0)- 0(0))\vec{j}+ (1(1)- 0(0))\vec{k}= \vec{k}[/itex]
     
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