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.