Lagrange's Identity in $\Bbb L^3$. I'm working in $\Bbb L^3 = (\Bbb R^3, \langle \cdot, \cdot \rangle_L)$, with $\langle {\bf x},{\bf y}\rangle_L = x_1y_1+x_2y_2-x_3y_3$. Let $\|\cdot\|_L = \sqrt{|\langle \cdot, \cdot \rangle_L|}$ be the "norm" and $\times_L$ be the vector product (it is the euclidean product reflected on the plane $z = 0$). I'm studying Lagrange's Identity: $$\|{\bf x}\times_L {\bf y}\|_L^2 = \|{\bf x}\|_L^2\|{\bf y}\|_L^2 - \langle {\bf x},{\bf y}\rangle_L^2.$$
I have proved that if ${\bf x},{\bf y} \in \Bbb L^3$ span a spacelike plane, then it is true. It is also true if they are lightlike and parallel. If ${\bf x}$ is spacelike, ${\bf y}$ is lightlike and ${\bf x}\times_L{\bf y}$ is also lightlike, then the identity is still true. There is only one case that I'm struggling to kill here: when both ${\bf x}$ and ${\bf y}$ are spacelike, with ${\bf x}\times_L{\bf y}$ lightlike. In every other case is false.
I don't know how exactly to go at it, and I'm having trouble finding a counter-example. If we had the situation $\langle {\bf x},{\bf y}\rangle_L = 0$ possible, then it would be easy, but this can't happen, otherwise the plane spanned would be spacelike and not lightlike. Can someone give me a hand? Thanks.
 A: Build a clifford algebra from this space.  Let $\epsilon = e_1 e_2 e_3$ with $e_3 e_3 = -1$ and $e_1 e_1 = e_2 e_2 = +1$.  Then see that multiplication with $\epsilon$ produces the following:
$$\epsilon e_1 e_2 = -e_3, \quad \epsilon e_2 e_3 = e_1, \quad \epsilon e_3 e_1 = e_2$$
This allows us to write the Lorentzian cross product as
$$a \times b = \epsilon (a \wedge b)$$
Now consider the product
$$\langle abba \rangle_0 = (a \cdot b)^2 - (a \wedge b)^2 = a^2 b^2$$
See that $\epsilon^2 = +1$, so we can substitute $a \wedge b = \epsilon(a \times b)$ and write this as
$$(a \cdot b)^2 - (a \times b)^2 = a^2 b^2$$
This identity is always true; note the sign difference from the identity you're considering.  This is the correct Lorentzian analogue of the Lagrange identity, as the sign on $(a \times b)^2$ depends on the sign of $\epsilon^2$--in Euclidean 3d space, $\epsilon^2 = -1$ and you get the "usual" Lagrange identity.

Edit: retract this section. While I admit I have not directly answered your question, it's my hope that I've illustrated why the usual Lagrange identity doesn't hold, but that it does have an analogue in Lorentzian space that does hold.
Indeed, in the case that $a\times b$ is lightlike, then this Lorentzian analogue suggests
$$(a \cdot b)^2 - (a \times b)^2 = (a \cdot b)^2 - 0 = a^2 b^2$$
And we can see that this is true for your example, as well.
