Vector Product and dot product identity: Levi-Civita symbols I want to prove that $\vec{a}\cdot(\vec{b}\times\vec{c}) = (\vec{c} \times \vec{a})\cdot\vec{b}$  using the  Levi-Civita symbols, however, I am not $100$% sure if my proof is correct.
Please see attached my proof, see the image 
Or see the (using MathJax) equations below
$$\vec{a}\cdot(\vec{b}\times \vec{c}) = a_i(\vec{b}\times\vec{c})_i = a_i\epsilon_{ijk}b_jc_ke_i = -\epsilon_{jik}a_ib_jc_ke_i = -(\vec{a}\times\vec{c})\cdot\vec{b} = (\vec{c}\times\vec{a})\cdot\vec{b}$$
My main concern is that when I change the indices for epsilon from $(i,j,k)$ to $(j,i,k)$, should I also change the index for $e$ vector from $i$ to $j$ as well? It's just in my proof I assume that $b_je_i$ gives vector $b$ and I do not know if I can state that given the different indices.
Thank you in advance and I hope this all does not sound too confusing.
 A: For the Levi-Civita symbols we have that, for two vectors $\vec{a}$ and $\vec{b}$,

$\vec{a}\cdot\vec{b} = \sum_ia_ib_i$ 

and using Einstein notation convention we can just wright $\vec{a}\cdot\vec{b} = a_ib^i$, or, we can just use that $(\vec{a}\cdot\vec{b} )_i = a_ib_i$. We also have that 

$(\vec{a}\times\vec{b})_i = \sum_j \sum_k \epsilon_{ijk}a_jb_k$

Or just $(\vec{a}\times\vec{b})_i = \epsilon_{ijk}a_jb_k$. So using that you can prove the relation as you did:
$$(\vec{a}\cdot(\vec{b}\times\vec{c})) = \sum_ia_i(\vec{b}\times\vec{c})_i = \sum_i\sum_j\sum_ka_i\epsilon_{ijk}b_jc_k = \sum_i\sum_j\sum_k\epsilon_{jki}c_ka_ib_j = \sum_j(\vec{c}\times\vec{a})_jb_j = (\vec{b}\cdot(\vec{c}\times\vec{a}))$$
Then this is what I think was your doubt.
A: That's almost right, but there are some inconsistencies in your notation.


*

*In the first step $a\cdot(b\times c)$, you have a scalar. Nothing wrong here. But note that since you begin with a scalar, you should have scalars in all the next steps.

*In the second step $a_{i}\cdot(b\times c)_{i}$ you use a dot ($\cdot$) between components. That is illegal. Components are numbers, and you can only use dot product between vectors. Hence, the second step should read just $a_{i}(b\times c)_{i}$

*Since $(b\times c)_{i} = \epsilon_{ijk}b_{j}c_{k}$, in step three you should have just $a_{i}\epsilon_{ijk}b_{j}c_{k}$, without the vectors $e_{i}$. This resonates with the note in (1), where I remarked you should have just scalars and not vector expressions. Also note that this goes against the summation convention, where it is only valid to sum over pairs of indices.

*The switch of indices and switch of sign is correct. Since, as mentioned in (3), you shouldn't write the vectors $e_i$, your concern about the index $i$ is just out of the question.


Steps 5 and 6 are indeed correct.
So, the correct derivation (with a pair of extra steps) is
$$a\cdot(b\times c) = a_{i}(b\times c)_{i} = a_{i}\epsilon_{ijk}b_{j}c_{k} = -a_{i}\epsilon_{jik}b_{j}c_{k} \\= -\epsilon_{jik}a_{i}c_{k}b_{j} = -(a\times c)_{j}b_{j} = -(a\times c)\cdot b = (c\times a)\cdot b$$
