finding a differential curve tangent to a distribution I have the following distribution on $\mathbb{R}^3$
$${\cal{D}}_{(x,y,z)} = \langle\{\partial_x,\partial_y + x\partial_z\}\rangle$$
I want to show that for any $(x,y,z)$ in $\mathbb{R}^3$, there exists a path $\gamma$ from $0$ to $(x,y,z)$ tangent to $\cal{D}$ i.e. $\dot{\gamma}(t)\in{\cal{D}}_{\gamma(t)}$
Any hints?

HINT: It helps to translate into differential forms. The distribution is given by $\omega = x\,dy - dz = 0$. So, given $(X,Y,Z)$, your goal will be to choose a path $\gamma$ from $(0,0,0)$ to $(X,Y,0)$ so that $Z=\int_\gamma x\,dy$. How does this lift to a path tangent to your distribution?
(As a further hint, if needed, start by taking the line segment $\Gamma$ joining the points in the plane. How can you find $\gamma$ so that $\int_\gamma x\,dy - \int_\Gamma x\,dy = Z-\frac12XY$?)
