problem 3.39, Kristopher Tapp- Differential geometry of curves and surfaces. Let $S$ be a regular surface, $p \in S$, $v \in T_p S$ a nonzero vector, and $N$ a unit normal vector to $S$ at $p$. Prove that there exists a neighborhood, $V$, of $p$ in $S$ such that the intersection of $V$ with the plane $p+\operatorname{span}\{v,N\}$ is the trace of a regular surface.
I don't have an idea about this statement, Any hint? I need this proof to finish another problem in this context.
 A: Let me denote the (unit) normal to $S$ by $\mathbf n$.
After a rigid motion we may assume that $p=(0,0,0)$ and $\mathbf n(p)=(0,0,1)$. In particular
$$
   T_p(S)=\{z=0\},
$$
which implies that $\mathbf v=(a,b,0)$.
Take a parameterization $\xi\colon U\to\mathbb R^3$ of $S$ around $p$ of the form $\xi(u,v) = (u,v,\phi(u,v))$. Note that $$
  \nabla\phi(\mathbf0)=\mathbf 0.
$$
Let $\Pi$ be the plane generated by $\{\mathbf v,\mathbf n\}$. For $t\in(-\epsilon,\epsilon)$, with $\epsilon>0$ small enough, the map
\begin{equation}\label{eq119}
    \gamma(t) = t\mathbf v + \phi(ta,tb)\mathbf n(p)
\end{equation}
defines a regular curve because $\gamma'(0)=\mathbf v$ and therefore we are entitled to assume that $\gamma'(t)\ne\mathbf0$ in a nbh of $0$. In addition, $\textrm{im}(\gamma)\subseteq\Pi\cap S$ since
$$
              \gamma(t)=\xi(ta,tb).
$$
Now, using that $\xi$ covers a nbh of $p$, you should be able to prove that, given a point $q\in\Pi\cap S$ near $p$, there must exist $\tau$ such that
$$
   q = \xi(\tau a,\tau b)=\gamma(\tau).
$$
