Vector product on any 3D euclidean vector space Take $(V, <\cdot , \cdot >)$ to be an euclidean vector space of dimension 3 with a choice of a basis $\mathcal{B}$ to be positively oriented.
Is there a sensible way to define a generalization of the vector product
$$\land :V\times V\rightarrow V $$
that satisfies the usual properties and only depends on the orientation of $V$ and not on the basis $\mathcal{B}$?
Possible solution:
If we define, for any positively oriented orthonormal basis $\mathcal{C} = (e_1,e_2,e_3)$:
$$\left(\sum_{i=1}^3x_ie_i\right)\land_{\mathcal{C}} \left(\sum_{i=1}^3y_ie_i\right) =(x_2y_3-x_3y_2)e_1+(x_3y_1-x_1y_3)e_2+(x_1y_2-x_2y_1)e_3$$
Is it true, then, that $\land_\mathcal{C}=\land_{\mathcal{C}'}$ for every $\mathcal{C},\mathcal{C}'$ positively oriented orthonormal bases?
 A: A usual definition of a vector product in three dimensional Euclidean space does not use a notion of a base. Namely, the vector product ${\bf a}\times {\bf b}$ is defined as a vector ${\bf c}$ that is perpendicular (orthogonal) to both ${\bf a}$ and ${\bf b}$, with a direction given by the right-hand rule and a magnitude equal to the area of the parallelogram that the vectors span, see Wikipedia for details.
Update. We can use that a Euclidean vector space $V$ of dimension three is isomorphic to $\Bbb R^3$, but there is a direct proof. Let $\mathcal C’=(e_1,e_2,e_3)$ and $\mathcal C’=(e’_1,e’_2,e’_3)$ be orthonormal basises of $V$. Suppose that for each $i=\{1,2,3\}$ let $e’_i=\sum m_{ij}e_j$ and a $3\times 3$ matrix $M=\|m_{ij}\|$ is orthogonal with positive determinant (that is $1$). Then its adjugate matrix $\operatorname{adj} M$ is equal to its transposal $M^T$, see here for a form of an adjugate of a $3\times 3$ generic matrix.
For each $(k,l,n)\in \{(1,2,3),(2,3,1),(3,1,2)\}$ we have $$e’_k \land_\mathcal{C’} e’_l= e’_n= m_{n1}e_1+ m_{n2}e_2+ m_{n3}e_3$$
and $$e’_k \land_\mathcal{C} e’_l=(\sum m_{ki}e_i) \land_\mathcal{C} (\sum m_{lj}e_j)=\sum_{i,j} m_{ki} m_{lj} e_i \land_\mathcal{C} e_j=$$ $$(m_{k2} m_{l3}- m_{k3} m_{l2})e_1+ (m_{k3} m_{l1}- m_{k1} m_{l3})e_2+(m_{k1} m_{l2}- m_{k2} m_{l1})e_3.$$  The equality  $$e’_k \land_\mathcal{C’} e’_l=e’_k \land_\mathcal{C} e’_l$$ now follows from the equality $\operatorname{adj} M=M^T$.
