Proving the vector identity $u\cdot (v \times w) = w \ \cdot (u \times v) = -v \ \cdot (u \times w)$ I'm stuck on proving the following identity:
$$u \ \cdot (v \times w) = w \ \cdot (u \times v) = -v \ \cdot (u \times w)$$
Geometrically I see how this works but computationally I have only been able to get this far:
$u \ \cdot (v \times w) = \left [\left \| u - proj_{v \times w} u \right \|  \right ]\left [\left \| v \times w \right \|  \right ] $
$= \left[ u-\frac{u \cdot (v \times w) }{\left \| v \times w \right \|} (v \times w)\right]\left[ \left\| v \times w \right\| \right]$
Where do I go from here? Or am I on the wrong track?
 A: Let $a,b,c$ be 3-vectors. The mixed product $a\cdot (b\times c)$ is equal to the determinant of the $3\times 3$-matrix, whose first row consists of coordinates of $a$, the second row consists of coordinates of $b$ and the third row consists of coordinates of $c$. Now your equalities follow from the standard property of determinants: if we switch two rows of a matrix, the determinant changes its sign.
A: Results of direct calculations of determinant for cross-product.
$$u\cdot (v \times w) =  u \cdot 
\begin{vmatrix} 
i & j & k \\
v_1 & v_2 & v_3 \\
w_1 & w_2 & w_3
\end{vmatrix}
= 
\begin{vmatrix} 
u_1 & u_2 & u_3 \\
v_1 & v_2 & v_3 \\
w_1 & w_2 & w_3
\end{vmatrix} 
=
-\begin{vmatrix} 
v_1 & v_2 & v_3 \\
u_1 & u_2 & u_3 \\
w_1 & w_2 & w_3
\end{vmatrix}
=-v\cdot(u\times w) 
$$
In analogous way you will get $w\cdot (u \times v )$.
A: The Levi-Civita tensor $\epsilon_{ijk}$ with Einstein summation notation is really handy in these sorts of proofs.  $\epsilon_{123}=1$ (this defines our coordinates as right-handed), and the entries change sign under the swap of any two indices.  This leaves $\epsilon_{123}=\epsilon_{231}=\epsilon_{312}=1$ and $\epsilon_{321}=\epsilon_{213}=\epsilon_{132}=-1$.  Einstein summation convention is that any repeated indices are summed over.  So, $A \times B = \epsilon_{ijk}A_jB_k$, which you can see by just writing out entries and comparing to the standard cross product.  With Einstein summation, then $A \cdot B = A_iB_i$, and $A\cdot (B \times C) = A_i \epsilon_{ijk} B_j C_k$.  The magic of the Levi-Civita tensor is that because these are now all scalar multiplications, you can re-order as you like, so $A\cdot (B \times C) = \epsilon_{ijk} A_i B_j C_k$.  Let's say you now did $B \cdot (A \times C)$ then this will be equal to $\epsilon_{ijk} B_i A_j C_k$.  There had better be nothing special about index names, so let's swap $i$ and $j$ to get $\epsilon_{jik} B_jA_iC_k = \epsilon_{jik} A_iB_jC_k$.  From the definition of the Levi-Civita tensor, though, $\epsilon_{ijk}=-\epsilon_{jik}$, so $A\cdot (B \times C)=-B \cdot (A \times C)$.  You can repeat this with any ordering of $A,B,C$, and you can see the if the sign has flipped by the ensuing order of the Levi-Civita indices.
In my experience, this is by far the easiest way of proving vector identities (once you've done a couple), especially if you remember $\epsilon_{ijk}\epsilon_{klm}=\delta_{il} \delta_{jm}-\delta_{im}\delta_{jl}$ (which you need to work out $A \times B \times C$. You can also use it for all the vector calculus identities as long as you don't reorder  differential operators/fields.
