I was wondering: if $X$ is a non-vanishing smooth vector field defined on an open subset $U \subset \mathbb{R}^3$, there are two smooth vector fields $Y$ and $Z$ on $U$ such that $X(p) = Y(p) \wedge Z(p)$ (wedge product of vectors) for each $p \in U$?
The answer seems not to be unique. For instance, if $\partial/\partial z$ is the constant field in the $z$-direction, obviously $\partial/\partial z = \partial/\partial x \wedge \partial/\partial y$, so the answer is yes. But the field $X = x\partial/\partial x + y\partial/\partial y + z \partial/\partial z$ is normal to the unit sphere, which does not admit non-vanishing tangent vector fields. So the answer seems to be no in this last case.
Every smooth non-vanishing vector field $X$ in $U$ can be seen as a non-vanishing 1-differential form $\omega$. Now $\omega$ generates a two-dimensional distribution $\Delta$ (a subbundle of the tangent bundle $TU$) defined by $\Delta_p = \ker \omega_p \subset T_PU$.
My initial question now becomes: when $\Delta$ (as a vector bundle over $U$) is trivial?
What if we assume that $\Delta$ is completely integrable (i.e. satisfies Frobenius condition $\omega \wedge d\omega = 0$)? Could the answer be related to the integral surfaces of $\Delta$?
Thanks