$2$ dimensional foliations of space whose leaves contain the trajectories of a given vector field

Assume that $X$ is a non-vanishing vector field on $\mathbb{R}^3$.

Is there a $2$-dimensional foliation of space such that every trajectory of $X$ is contained in a leaf of the $2$-dimensional foliation?

As a related question:

Is there a classification of all $1$-dimensional foliations of space tangent to the unit speed vector field $t$ for which the distributions $\{t,n\}$ and $\{t,b\}$ are integrable? Here $\{t,n,b\}$ is the associated Frenet frame.

No. For a counterexample, start with the Hopf map $S^3\to S^2$, a fiber bundle with $S^1$ fibers. Its fibers are the leaves of a $1$-dimensional foliation of $S^3$ in which all leaves are closed and the space of leaves is $S^2$. Choose a vector field tangent to the leaves. Remove one point from $S^3$ to get $\mathbb R^3$. A $2$-dimensional foliation of the kind asked for would give a $1$-dimensional foliation of $S^2$.
• Thank you very much for this interesting answer. I understand the tangent vector field to the Hopf foliation is not contained in any 2 dimensional smooth distribution $D$ otherwise since any such distribution would be map to a $1$ dimensional foliation of $S^2$.Am I correct? – Ali Taghavi Oct 1 '17 at 11:16
• Yes, that was my reasoning. No, the Reeb foliation has a torus leaf, but no torus in $S^3$ can be transverse to the Hopf fibration because that would make it a covering space of $S^2$. In fact, no $2$-dimensional foliation of $S^3$ that has a compact leaf can be transverse to the Hopf fibration. – Tom Goodwillie Oct 1 '17 at 18:28
• My apology if my question is elementary. How the foliation on total space, which is not necessarily $S^1$ invariant, give us a foliation of the bases space? – Ali Taghavi Aug 4 at 21:55
• For each point in $S^2$ its preimage in $\mathbb R^3$ would be contained in one of the leaves. Therefore each leaf would be the preimage of some subset of $S^2$. In order for the preimages of these subsets to be the leaves of a foliation, the sets themselves would have to be the leaves of a foliation. – Tom Goodwillie Aug 5 at 23:02