To expand on Victork Protsak's comment, if V is an n-dimensional real vector space with inner-product, the inner-product gives an isomorphism $V\to V^*$ and hence $V\otimes V \to \mathrm{End}(V)$. Under this isomorphism, $\Lambda^2(V)$ is identified with skew-adjoint endomorphisms of $V$, which is precisely the Lie algebra $\mathfrak{so}(V)$.
In the case dim V =3, the Hodge star gives an isomorphism $\Lambda^2(V) \to V$ and so in total we see that $V$ is canonically isomorphic to $\mathfrak{so}(V)$. A more direct way to see this isomorphism is to send the vector $v \in V$ to the generator of the right-handed rotation about the axis in the direction of $v$ with speed $|v|$.
The use of the phrase "right-handed" makes it clear that in order to identify $V$ and $\mathfrak{so}(V)$ we have used an orientation on $V$. IndeedV$; indeed, you need that for the Hodge star. What is interesting is that if you reverse the orientation on $V$, the map to $\mathfrak{so}(V)$ changes sign. This means that what ever orientation you choose on$V$, chose on $V$, the push-forward to $\mathfrak{so}(V)$ is the same. Conclusion: $\mathfrak{so}(3)$ is naturally oriented. This is analogous to the natural orientation on $\mathbb{C}$. A more prosaic way to describe the orientation is to pick two independent elements $x,y \in \mathfrak{so}(3)$ and then use $[x,y]$ to complete them to an oriented basis. (Of course, you then need to check that this doesn't depend on your choice of $x,y$.)
