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A cross product question.

  1. Mar 13, 2005 #1
    I do not understand something about cross products. Say,

    [tex] \vec A\times \vec B=\vec C=(C_x, C_y, C_z) [/tex]


    [tex] \vec C=(A_x \hat x+A_y \hat y+A_z \hat z)\times (B_x \hat x+B_y \hat y+B_z \hat z)[/tex]

    but why is this equivalent to

    [tex] (A_x B_y - A_y B_x)\hat x \times \hat y + (A_x B_z - A_z B_x) \hat x \times \hat z + (A_y B_z - A_z B_y) \hat y \times \hat z [/tex]


    Can someone show me how do i get this? Preferbly an algebraic method instead of a geometric one, because I am poor at visualizing stuff.
  2. jcsd
  3. Mar 13, 2005 #2


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    [tex] \vec C=(A_x \hat x+A_y \hat y+A_z \hat z)\times (B_x \hat x+B_y \hat y+B_z \hat z)[/tex]

    can be written as (using the distributive property)

    [tex] \vec C= A_x \hat x \times (B_x \hat x+B_y \hat y+B_z \hat z)[/tex]

    + [tex] A_y \hat y \times (B_x \hat x+B_y \hat y+B_z \hat z)[/tex]

    + [tex] A_z \hat z \times (B_x \hat x+B_y \hat y+B_z \hat z)[/tex]

    then [tex] \hat x \times \hat x = \hat y \times \hat y = \hat z \times \hat z[/tex] = 0

    and [tex] \hat x \times \hat y[/tex] = - [tex] \hat y \times \hat x[/tex] and so on.
  4. Mar 13, 2005 #3
    Hey, thanks, that was helpful, i've gotten it already.....

    : )
  5. Mar 13, 2005 #4


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    If you're dealing with vector identites,i've got an advice:learn cartesian tensors.

  6. Mar 13, 2005 #5


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    another computational way is to use a definition of cross product such as: the cross product of A and B is the determinant of the 3 by 3 matrix with x,y,z (your notation) in the first rwo and the entries of A in the second row, and the entries of B in the third row.
  7. Mar 13, 2005 #6
    mathwonk: the matrix idea helped too.....

    dexter: I apparently havent even started on vector calc yet, so I don't think i can go into tensors yet.

    : )
  8. Mar 13, 2005 #7


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    do not be afraid of tensors. the dot product is the first tensor we meet. tensors are multilinear as opposed to merely linear. the dot product is bilinear, hence the simplest tensor.

    a derivative is a linear map that approximates a fucntion. the second derivative is a bilinear map that approximates the difference between function and its derivative, and so on....

    do not be misled by the confusing gobbledygook found here about upper and lower indices as being tensors. that is pablum for people who refuse to learn what tensors are.

    there is some truth in it, but it is like saying that a linear map is a matrix. i.e. the matrix is a notational representation of a linear map, not the linear map itself. similarly an array of upper and lower indices is a representation of a tensor and not the tensor itself.

    do not be seduced by the "dark side"!
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