How to Prove the Vector Triple Product Identity?

[robierob12]

Homework Statement

Prove that

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

Homework 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 I am 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.

Rob

The Attempt at a Solution

 

[slearch]

cross product isn't defined for dimensions higher than three, so you would just prove it for three space.

 

[Fredrik]

I usually do these problems like these using the \varepsilon_{ijk} thingy and the summation convention. If you're not familiar with that notation, this may just confuse you. The epsilon thingy is defined by \varepsilon_{ijk}=1 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 \varepsilon_{132}=-1 and that \varepsilon_{122}=0. 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 u*v is u_i v_i=v_1u_1+v_2u_2+v_3u_3 (I don't remember the LaTeX code for the scalar product) and the cross product u\times v is \varepsilon_{ijk}u_jv_ke_i, where the e_i are the basis vectors of \mathbb{R}^3.

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

(u\times (v\times w))_i=\varepsilon_{ijk} u_j (v\times w)_k=\varepsilon_{ijk}\varepsilon_{klm}u_j v_l w_m
=(\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

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. \varepsilon_{k12}\varepsilon_{k12} and \varepsilon_{k12}\varepsilon_{k13}. When you've done that, you'll probably understand.

 

[robierob12]

Thanks... I don't know why I was thinking of tring to prove it for vectors outside of three space.