I'm trying to do the following exercise:
Let $S_1$, $S_2$ be two regular surfaces that contain a common curve $\gamma$. Prove that if $S_1$ and $S_2$ make a constant angle along $\gamma$, then $\gamma$ is a line of curvature in $S_1$ if and only if it is a line of curvature in $S_2$.
What I understand from the surfaces making a constant angle along $\gamma$ is that
$$<\mathcal{N}^{S_1} \circ \gamma, \ \mathcal{N}^{S_2} \circ \gamma> \; = a$$ for some constant $a \in \mathbb{R}$, where $\mathcal{N}^{S_i} \circ \gamma$ is the normal unitary vector of the surface $S^i$ at each point of the curve $\gamma$. Derivating the prior expression, we have that
$$<d\mathcal{N}_{\gamma(s)}^{S_1}(\gamma'(s)), \mathcal{N}^{S_2}(\gamma(s))>+<\mathcal{N}^{S_1}(\gamma(s)), d\mathcal{N}_{\gamma(s)}^{S_2}(\gamma'(s))> \; = 0$$
Let's suppose that $\gamma$ is a line of curvature in $S_1$. Then
$$d\mathcal{N}_{\gamma(s)}^{S_1}(\gamma'(s)) = \lambda(s)\gamma'(s)$$
for some scalar function $\lambda$. Consequently,
$$< \lambda(s)\gamma'(s), \mathcal{N}^{S_2}(\gamma(s))>+<\mathcal{N}^{S_1}(\gamma(s)), d\mathcal{N}_{\gamma(s)}^{S_2}(\gamma'(s))> \; = 0$$
The first term is zero because $\gamma'(s)$ is a vector of the tangent plane of $S^2$ at the point $\gamma(s)$, and $\mathcal{N}^{S_2}(\gamma(s))$ is normal to that plane. Therefore,
$$<\mathcal{N}^{S_1}(\gamma(s)), d\mathcal{N}_{\gamma(s)}^{S_2}(\gamma'(s))> = 0$$
We now have that $\mathcal{N}^{S_1}(\gamma(s))$ is normal to both $d\mathcal{N}_{\gamma(s)}^{S_2}(\gamma'(s))$ and $d\mathcal{N}_{\gamma(s)}^{S_1}(\gamma'(s)) = \lambda(s)\gamma'(s)$. This is where I'm stuck: the proof would be done if $d\mathcal{N}_{\gamma(s)}^{S_1}(\gamma'(s))$ and $d\mathcal{N}_{\gamma(s)}^{S_2}(\gamma'(s))$ had the same direction, but couldn't they be orthogonal?