I would like to find the generalized formula of the identity $$(A\times B).(C\times D)=(A\cdot C)(B\cdot D)-(A\cdot D)(B\cdot C)$$ which holds in an Euclidian metric, within a general metric $g$ on $\Bbb R^3$.
Thanks for any help, I wish a nice evening.
This has a metric independent generalization in Geometric algebra (or Clifford Algebra) as follows
$$-(A\wedge B)\cdot(C\wedge D)=(A\cdot C)(B\cdot D)-(A\cdot D)(B\cdot C),$$
where the dot product of the bivectors indicates the scalar grade selection. This identity holds in $\Bbb R^n$, or in non-Eucliean generalizations.
The metric dependencies are all in the vector dot products. For example, with a basis and reciprocal basis, a vector $A$ can be written
$$A = \gamma_\mu a^\mu = \gamma^\mu a_\mu.$$
The dot product is:
$$A^2 = A \cdot A = a_\mu a_\nu g^{\mu\nu},$$
where
$$g^{\mu\nu} = \gamma^\mu \cdot \gamma^\nu.$$
In the distribution identity above, the wedge product is an antisymmetric sum of vectors:
$$A \wedge B = \frac{1}{2} ( A B - B A ).$$
This quantity is referred to as a bivector in Geometric Algebra (and in an Euclidean $\Bbb R^3$ space is the dual of the cross product).