# Double cross product

1. Sep 22, 2015

### geoffrey159

1. The problem statement, all variables and given/known data
If $u,v,w\in\mathbb{R}^3$, show that $u\times(v\times w) = (u.w) v - (u.v) w$.

2. Relevant equations

3. The attempt at a solution
Since $u\times(v\times w)$, $v$ and $w$ are orthogonal to $v\times w$, these vectors are coplanar. Therefore, there must be reals $a,b$ such that $u\times(v\times w) = a v + b w$.
Then, $0 = u.(u\times(v\times w)) = a u.v + b u.w$ and $a = u.w$ and $b = - u.v$ work. But why not $a = -u.w$ and $b = u.v$ ? Thanks

2. Sep 22, 2015

### ehild

Write the vectors in Cartesian components and expand the double product.

3. Sep 22, 2015

### Ray Vickson

An alternative method is to write vectors as $/vec{v} = (v_1,v_2,v_3)$ for example, and use the so-called Levi-Civita symbol to express the cross-product:
$$(\vec{u} \times \vec{v}) _i = \sum_{j,k} \epsilon_{ijk} u_j v_k$$
Here the Levi-Civita symbol $\epsilon$ is
$$\epsilon_{ijk} = \begin{cases} +1 & \text{if} \:\; ijk \; \text{is an even permutation of} \: \; 123 \\ -1 & \text{if} \:\; ijk \; \text{is an odd permutation of} \:\; 123 \\ 0 & \text{otherwise} . \end{cases}$$
There are identities available for sums of products like $\sum_k \epsilon_{ijk} \epsilon_{klm}$ which would be the thing needed in the triple product. See, eg., http://www.ucl.ac.uk/~ucappgu/seminars/levi-civita.pdf or https://en.wikipedia.org/wiki/Levi-Civita_symbol .

4. Sep 23, 2015

### geoffrey159

Thank you for your answers, but is it possible to avoid lengthy calculations ?
In the OP, I am to the point where $(a,b) \in \{\pm (u.w, - u.v) \}$.
How should I decide for the sign ?

5. Sep 23, 2015

### Ray Vickson

Some detailed calculations are just unavoidable. However, if you pay attention to what I said in #3, that will lead to just about the shortest and most straightforward proof available. Besides, you have NOT reached the point where $(a,b) \in \{\pm (u.w, - u.v) \}$, because you had one relation between $a$ and $b$, and you said that your particular choice "works". However, in principle, there are infinitely many other choices of $a,b$ that also work for that single condition. Somehow, you need more conditions in order to pin down $a,b$ convincingly and uniquely.

Last edited: Sep 23, 2015
6. Sep 23, 2015

### geoffrey159

Oops you're right, I'm just at $\begin{pmatrix}a\\b\end{pmatrix} \in \mathbb{R} \begin{pmatrix}u.w \\ -u.v \end{pmatrix}$.
Thanks again, I will try your proof.