Intuition behind $\vec{e_i} \times \vec{e_j}=\epsilon_{ijk} \vec{e_k}$ (Levi Civita) Let $\vec{e_i}$ denote a unit vector.
Then we can write:
$\vec{e_i} \times \vec{e_j}=\epsilon_{ijk} \vec{e_k}$,
where $\epsilon_{ijk}$ is the Levi Civita symbol.
Can someone intuitively explain me why this is true and what the formula tells me?
 A: The formula is a compact representation of the nine equations for the cross products of the basis vectors. I'll call the basis vectors $e_1,e_2,e_3$ in this answer, i.e. I'm not going to draw arrows over them.
You might want to start by recognizing that "k" is a "dummy variable" and then summing on it.  That gives you:
  $e_i \times e_j = \epsilon_{ij1}e_1+\epsilon_{ij2}e_2+\epsilon_{ij3}e_3$
or, alternately:
  $e_i \times e_j = (\epsilon_{ij1},\epsilon_{ij2},\epsilon_{ij3})$
Now that "k" is gone, you're left with just "i" and "j" indices and it's a little more obvious that you can plug in values for each to get the cross product result for the two corresponding basis vectors.  Here's a list of the nine equations that result from all the combinations of i and j:
$e_1 \times e_1=(\epsilon_{111},\epsilon_{112},\epsilon_{113})=(0,0,0)$
 $e_1 \times e_2=(\epsilon_{121},\epsilon_{122},\epsilon_{123})=(0,0,$+$1) = $ +$e_3$
 $e_1 \times e_3=(\epsilon_{131},\epsilon_{132},\epsilon_{133})=(0,$-$1,0) = $ -$e_2$
 $e_2 \times e_1=(\epsilon_{211},\epsilon_{212},\epsilon_{213})=(0,0,$-$1) = $ -$e_3$
 $e_2 \times e_2=(\epsilon_{221},\epsilon_{222},\epsilon_{223})=(0,0,0)$
 $e_2 \times e_3=(\epsilon_{231},\epsilon_{232},\epsilon_{233})=($+$1,0,0) = $ +$e_1$
 $e_3 \times e_1=(\epsilon_{311},\epsilon_{312},\epsilon_{313})=(0,$+$1,0) = $ +$e_2$
 $e_3 \times e_2=(\epsilon_{321},\epsilon_{322},\epsilon_{323})=($-$1,0,0) = $ -$e_1$
 $e_3 \times e_3=(\epsilon_{331},\epsilon_{332},\epsilon_{333})=(0,0,0)$
A: Consider an orthonormal basis $e_1,e_2,e_3$ for $\mathbb{R}^3$,we can write $a\times b$ as a uniqUe linear combination of $e_i\times e_j$(i,j=1,2,3) also we can write $e_i\times e_j$ as a unique combination in the basis $e_1,e_2,e_3$ with the form ${c}^1_{i,j}e_1+{c}^2_{i,j}e_2+{c}^3_{i,j}e_3$.
Now we show ${c}^k_{i,j}$ with ${\epsilon}_{i,j,k}$ and called levi-civita symbol, ${\epsilon}_{i,j,k}$ is  projection with sign of the vector $e_i\times e_j$ on $e_k$.
${\epsilon}_{i,j,k}=0$ if there are coinciding values of the indices i, j, k.
${\epsilon}_{i,j,k}=1$ if the values of the indices i, j, k form an even permutation of the numbers 1, 2, 3.
${\epsilon}_{i,j,k}=-1$ if the values of the indices i, j, k form an odd permutation of the numbers 1, 2, 3.
