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I have this 1-form on $\mathbb R^3$ given by $\omega=dz+\frac{x}{2}dy-\frac{y}{2}dx$. If $p_0=(x_0,y_0,z_0)$ and $\vec v=(u_0,v_0,w_0)$, then find the set of tangent vectors $\vec v_{p_0}$ such that $\omega_{p_0}(\vec v_{p_0})=0$

I'm confused because I don't know how to deal with the base of the tangent vector $p_0$. I think the solution should be $(dz+\frac{x_0}{2}dy-\frac{y_0}{2}dx)$ applied to $\vec v_0$, but I'm not sure. Help would be appreciated.

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Yes, that's exactly what you need to look at. – Zhen Lin Apr 17 '12 at 7:27
So then $(dz+\frac{x_0}{2}dy-\frac{y_0}{2}dx)(w_0\vec e_3+v_0\vec e_2+u_0\vec e_1)$ where $e_i$ is the ith standard basis vector. Which should simplify to $w_0+\frac{x_0 v_0}{2}-\frac{u_0 y_0}{2}$? – AnalysisHelp Apr 17 '12 at 16:03

In terms of coordinates, the equation $\omega_{p_0}(\vec v_{p_0})=0$ becomes $$w_0+\frac{x_0}{2}v_0-\frac{y_0}{2}x_0=0 \tag1$$ The answer is the set of all tangent vectors $\vec v=(u_0,v_0,w_0)$ for which (1) holds.

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