# Vector Triple Product

1. Apr 27, 2007

### robierob12

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.

Rob

3. The attempt at a solution

2. Apr 27, 2007

### slearch

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

3. Apr 28, 2007

### Fredrik

Staff Emeritus
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.

Last edited: Apr 28, 2007
4. Apr 28, 2007

### robierob12

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