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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?

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closed as off-topic by Peter, José Carlos Santos, kimchi lover, Namaste, Anubhav Mukherjee Jan 12 '18 at 4:39

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  • $\begingroup$ How do you connect with $(0,0,1)$? $\endgroup$ – Thomas Jan 11 '18 at 17:20
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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 $\int_\gamma x\,dy - \int_\Gamma x\,dy = Z-\frac12XY$?)

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