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Vector Triple Product

  1. Apr 27, 2007 #1
    1. The problem statement, all variables and given/known data

    Prove that

    u x (v x w) = (u*w)v - (u*v)w

    2. Relevant equations

    I've been trying to get this one but keep ending up no where.

    I've tried the normal algebraic properties of the cross product but they lead me to a dead end.

    What im trying right now is just proving it in three space. Assigning each vector to a general form like (u1, u2, u3) and busting it out to see if I can get the right side identity.

    Is there an easyier way to start this out.
    Ideas are much appreciated.


    3. The attempt at a solution
  2. jcsd
  3. Apr 27, 2007 #2
    cross product isn't defined for dimensions higher than three, so you would just prove it for three space.
  4. Apr 28, 2007 #3


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    I usually do these problems like these using the [itex]\varepsilon_{ijk}[/itex] thingy and the summation convention. If you're not familiar with that notation, this may just confuse you. The epsilon thingy is defined by [itex]\varepsilon_{ijk}=1[/itex] and the requirement that it's totally antisymmetric, i.e. that if you swap two indices, it will change sign. This implies for example that [itex]\varepsilon_{132}=-1[/itex] and that [itex]\varepsilon_{122}=0[/itex]. The "summation convention" is that I don't bother writing out the sigmas for summation, since all the indices that we need to sum over always occur exactly twice.

    For example the scalar product [itex]u*v[/itex] is [itex]u_i v_i=v_1u_1+v_2u_2+v_3u_3[/itex] (I don't remember the LaTeX code for the scalar product) and the cross product [itex]u\times v[/itex] is [itex]\varepsilon_{ijk}u_jv_ke_i[/itex], where the [itex]e_i[/itex] are the basis vectors of [itex]\mathbb{R}^3[/itex].

    The ith component of the left hand side of the equation you're trying to prove is by definition of the cross product

    [tex](u\times (v\times w))_i=\varepsilon_{ijk} u_j (v\times w)_k=\varepsilon_{ijk}\varepsilon_{klm}u_j v_l w_m[/tex]
    [tex]=(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})u_j v_l w_m=u_j v_i w_j-u_j v_j w_i=(u*w)v_i-(u*v)w_i[/tex]

    The tricky step is the one where I replaced two epsilons with some Kronecker deltas. The easiest way to see that this identity must hold is to explicitly calculate e.g. [itex]\varepsilon_{k12}\varepsilon_{k12}[/itex] and [itex]\varepsilon_{k12}\varepsilon_{k13}[/itex]. When you've done that, you'll probably understand.
    Last edited: Apr 28, 2007
  5. Apr 28, 2007 #4
    Thanks... I don't know why I was thinking of tring to prove it for vectors outside of three space.
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