A question about the angle of two vector field Let $S$ be an orientable surface and $\mathcal{N} : S \to S^2$ the Gauss map of $S$. Let $\alpha : I \to S$ be a curve on $S$ and $V$ and $W$ two unit vector fields on $\alpha$. We consider the another unit vector field $\overline{V} = (\mathcal{N} \circ \alpha) \wedge V$. I have proved that
$$
W(t) = a(t) V(t) + b(t) \overline{V}(t) \quad \mbox{ for all } \quad t \in I\mbox{,}
$$
being $a , b : I \to \mathbb{R}$ the two next differentiable functions:
$$
a(t) = \langle W(t) , V(t) \rangle \qquad \mbox{ and } \qquad b(t) = \langle W(t) , \overline{V}(t) \rangle
$$
for all $t \in I$. My question now is how prove that $a^2(t) + b^2(t) = 1$ for all $t \in I$. Thank you very much.

The field $W$ is a unit vector field as you said, which means that 
$$\forall t\in I,\langle W(t),W(t)\rangle=1.$$
Now all you need to do is to replace $W(t)$ with the expression you gave:
$$\forall t\in I,\langle a(t)V(t)+b(t)\bar{V}(t),a(t)V(t)+b(t)\bar{V}(t)\rangle=1.$$
Don't forget that, by definition of $\bar{V}$, $V(t)$ and $\bar{V}(t)$ are orthonormal for all $t\in I$.
