In Andrew Pressley's "Elementary Differential Geometry" (second edition) on page 168 he gives a proof of Meusnier's Theorem, which is stated as follows:
Theorem: Let $\mathbf p$ be a point on the surface $\mathcal S$ and let $\mathbf v$ be a unit tangent vector to $\mathcal S$ at $\mathbf p$. Let $\Pi_\theta$ be the plane containing the line through $\mathbf p$ parallel to $\mathbf v$ and making an angle $\theta$ with the tangent plane $T_{\mathbf p}\mathcal S$, and assume that $\Pi_\theta$ is not parallel to $T_{\mathbf p}\mathcal S$. Suppose that $\Pi_\theta$ intersects $S$ in a curve with curvature $\kappa_\theta$. Then $\kappa_\theta\sin\theta$ is independent of $\theta$.
The proof is short, and begins as follows:
Assume that $\gamma_\theta$ is a unit-speed parametrization of the curve of intersection of $\Pi_\theta$ and $\mathcal S$. Then at $\mathbf p$, $\gamma_\theta' = \pm\mathbf v$, so $\gamma_\theta''$ is perpendicular to $\mathbf v$ $\underline{\text{and is parallel to $\Pi_\theta$}}$.
My question is about the underlined statement. How do we know that $\gamma_\theta''$ lies on $\Pi_\theta$? Looking at the diagram given, it seems obvious, in the sense that $\gamma_\theta''$ is parallel to the principle normal to $\gamma_\theta$, which, along with $\gamma_\theta'$ forms the basis for the oscullating plane, which also "looks" like the plane $\Pi_\theta$, but other than that heuristic argument I can't formalize why $\gamma_\theta''$ should lie on $\Pi_\theta$. I'm sure its easy, I'm just missing the obvious. Can anyone help out?
This is a formal typed answer 5 years later. Notice that the curve $\gamma$ lies in a $\textbf{plane}$, say $S$, then the plane equation for $S$ can assume to be $ax+by+cz = d$, where $a,b,c$ not all 0.
Now, suppose $\gamma:(-\epsilon,\epsilon) \to S$ is parametrized by arc-length: $\gamma(t) = (x(t),y(t),z(t))$ .To show $\gamma ''$ is parallel to $S$, it suffices to show $<\gamma'', N> = 0$ where $N$ is the unit normal of $S$. Since in $S$, $ax(t)+by(t)+cz(t)=d$, differentiating both sides we have $ax'+by'+cz'=0$. This implies the unit normal is $\displaystyle{N = (\frac{a}{\sqrt{a^2+b^2+c^2}},\frac{b}{\sqrt{a^2+b^2+c^2}},\frac{c}{\sqrt{a^2+b^2+c^2}})}$, which is a constant. By normal is perpendicular to tangent plane, $<N,\gamma'> = 0$. Differentiate it we have $<N',\gamma'>+<N, \gamma''>=0$. Therefore $<N,\gamma''>=0$.