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BAC CAB Rule

  1. Mar 1, 2005 #1
    Hi!

    Please can u prove the following rule!

    Ax(BxC)=B(A.C)-C(A.B)

    this is usually remembered as "BAC CAB Rule"

    i just couldnt solve this question.

    Thanks in advance
     
  2. jcsd
  3. Mar 1, 2005 #2

    Galileo

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    Tedious, but straightforward is just writing out the components.
    Try verifying the equation one component a time.
    For example, the x-component of A X (B X C) is:

    [tex]A_y(B_xC_y-B_yC_x)-A_z(B_zC_x-B_xC_z)[/tex]
    , which can be rewritten as:
    [tex]B_x(A_yC_y+A_zC_z)-C_x(A_yB_y+A_zB_z)[/tex]

    you almost have the x component of B(AC) in the left term and the x-component of C(AB) in the right term. They both miss the same term [itex]B_x(A_xC_x)[/itex] so you can just add it to both terms, since they will cancel. This establishes the relation for the x-component. Do the same for the other two (or try to be smart and use some symmetry argument).
     
  4. Mar 1, 2005 #3
    Well thanks for ur reply but can u plz tell me in some more detail so that i can understand it more properly!

    Thanks!
     
  5. Mar 1, 2005 #4

    Galileo

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    I`m not sure which part is unclear to you.

    What I meant was:
    Ax(BxC)=B(A.C)-C(A.B)
    is true if and only if all three components for the vector on the left and on the right are equal. Equating one component a time makes it less cluttery.

    Let:
    [tex]\vec A = A_x \vec i + A_y \vec j + A_z \vec k[/tex]
    [tex]\vec B = B_x \vec i + B_y \vec j + B_z \vec k[/tex]
    [tex]\vec C = C_x \vec i + C_y \vec j + C_z \vec k[/tex]

    Let D = B x C:
    The expression for B x C is:
    [tex] \vec D = (B_yC_z-B_zC_y)\vec i + (B_zC_x-B_xC_z)\vec j +(B_xC_y-B_yC_x)\vec k[/tex]
    The x-component of A x D is:
    [tex]A_yD_z-A_yD_x[/tex]
    The components of D can be gotten from the expression from B x C, so:

    [tex](\vec A \times \vec D)_x = A_yD_z-A_zD_y=A_y(B_xC_y-B_yC_x)-A_z(B_zC_x-B_xC_z)[/tex]
    [tex]=B_x(A_yC_y+A_zC_z)-C_x(A_yB_y+A_zB_z)=[/tex]
    [tex]B_x(A_xC_x+A_yC_y+A_zC_z)-C_x(A_xB_x+A_yB_y+A_zB_z)=\left[\vec B (\vec A \cdot \vec C)-\vec C(\vec A \cdot \vec B)\right]_x[/tex]
     
    Last edited: Mar 1, 2005
  6. Mar 1, 2005 #5

    dextercioby

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    I think using euclidean tensors is the elegant way to do it...

    Daniel.
     
  7. Mar 2, 2005 #6
    Thank u very much Galileo but would u please explain ur last step for me that how u convert the 2nd last step to the last step.
    PLzzzzzzzzzzzzzzzzzzzz
    Thanks a lot again
     
  8. Mar 2, 2005 #7

    HallsofIvy

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    Why don't YOU try working out the calculation yourself and see what you get?
     
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