How to determine the foliation which integrates a 1-dimensional distribution? The question is as follows:
In the 3-sphere $S^3 \subset \mathbb{R}^4 $ consider the 1-dimensional distribution defined by
$$X = -y \frac{\partial}{\partial x} + x \frac{\partial}{\partial y } - \omega \frac{\partial}{\partial z } + z \frac{\partial}{\partial \omega }$$
Determine the foliation $F$ integrating this distribution.
$\textbf{Some attempt:}$
Let $\gamma (t) = (x(t ), y(t), z(t), \omega (t))$, with initial point $\gamma (0) = (x_0, y_0, z_0 , \omega_0)$. Hence, $\gamma'(t)=(x'(t ), y'(t), z'(t), \omega' (t))$, where 
$$ \begin{cases} \frac{dx}{dt} = -y \\ \frac{dy}{dt} = x \end{cases} ~~~~ \text{and}~~~~ \begin{cases} \frac{dz}{dt} = -\omega \\ \frac{d\omega}{dt} = z \end{cases}$$
From where this suggests we try $x(t) = A \cos t$ and plugging this into the second equation gives $y(t) = A \sin t$, and it's easy to check that this choice of $x(t)$ and $y(t)$ solves the second equation in the first system. And the same for second system, we have $z(t) = B \cos t$ and $\omega (t) = B \sin t$. 
So we have $\gamma (t) = (A \cos t , A \sin t , B \cos t, B \sin t)$ with $\gamma (0) = (A, 0 , B, 0)$. So we get $A = x_0$ and $B=z_0$.
Can some one please let me know if I am wrong?
Thanks!
 A: So the question seems rather to be: how do I understand the shape of the curve $\gamma(t)$ inside $\mathbb S^3$?
There are a number of ways. 
The easiest may be the following: choose a point $N$ such that: $\gamma(t)\ne N$ for any $t$. Now you can use the fact that $\mathbb S^3\setminus N$ is diffeomorphic to $\mathbb R^3$ (via stereographic projection from $N$). Composing $\gamma$ with the stereographic projection gives you a (closed) curve in $\mathbb R^3$ that you can visualize easily.
The previous one was not much specific of your curve and can be done for every curve in $\mathbb S^3$. For something more tailored on this $\gamma$ you should consider that your curve lies on the two hyperplanes $x_1/A-x_3/B=0$ and $x_2/A-x_4/B=0$ (here I'm assuming $A\ne0\ne B$), and both hyperplanes are through the origin. Therefore your curve sits inside a $2$-plane through the center intersecting the $3$-sphere and it should be then easy to convince you that this is nothing but a great circle on $\mathbb S^3$.
Then try and understand what happens when changing to various choices of $A$ and $B$...
