Vector Triple Product Proof

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.