Two-dimensional distributions in R^3

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

• I don't quite see the relation between the stuff with differential forms and your first question, which has an obvious positive answer: the orthogonal plane to $X$ contains a smooth vector field $Y$, so that can arrange $Z$ to be orthogonal to $Y$ and of length ensuring $X=Y\wedge Z$ everywhere. This question seems not adequate for MO, voting to close. – Benoît Kloeckner Feb 25 '13 at 20:39
• @Benoît: Oops! As the OP pointed out by example, there will not always exist a smooth nonvanishing vector field $Y$ orthogonal to $X$ if the domain $U$ on which the nonvanishing vector field $X$ is defined has a nontrivial second homology group. However, I can't see anything interesting to say, other than that the $2$-plane field in $U$ that is orthogonal to $X$ (which is clearly orientable and hence equivalent to a complex line bundle) has a nonzero section, which is equivalent to the nonvanishing of a Cech cohomology class. The integrability of this plane field is irrelevant, though. – Robert Bryant Feb 25 '13 at 23:08
• My question is equivalent to ask whenever $X$ has an orthogonal nonvanishing smooth field. – Yvoz Mar 1 '13 at 19:48