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Let $α = dz - ydx \in Ω^1 (\mathbb{R}^3)$.

Prove that $\forall p,q \in \mathbb{R}^3,\ \exists \gamma: [0,1] \rightarrow \mathbb{R}^3$ smooth, such that $γ(0)=p, γ(1) =q$ and $\gamma$ is tangent to $ker\ \alpha$ for all $t\in [0,1]$.

If we take $\gamma(t) = p(1-t) + qt$ we math the conditions $γ(0)=p, γ(1) =q$, but how to match the last condition : $\alpha (\gamma'(t)) = 0$?

Thank you for any insights.

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The third condition says that for each $t$, we must have $$ \alpha(\gamma(t)) [\gamma'(t)] = 0. \tag{*} $$ If we write $$ \gamma(t) = (x(t), y(t), z(t)) $$ then $$ \gamma'(t) = (x'(t), y'(t), z'(t)) $$ and equation (*) becomes $$ z'(t) - y(t) x'(t) = 0, $$ which you can, perhaps, solve (or can at least prove has a solution).

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  • $\begingroup$ Thank you! I will try to prove that it has a solution as I didn't manage to derive an explicit solution in terms of p and q $\endgroup$ – PerelMan Dec 10 '18 at 1:14

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