$\langle X(t), Y(\alpha(t))\rangle=\pm 1.$

Question

I have a surface $S\subset \mathbb{R}^3$ containing a curve $\alpha:[a,b]\to S$ and vector unit continuous fields $X,Y$ s.t.:

• $Y:S\to T\mathbb{R}^3$ is normal;
• $X:[a,b]\to T\mathbb{R}^3 s.t. X(t)\in (T_{\alpha(t)}S)^{\perp}\forall t\in[a,b].$

One professor claims that

$\langle X(t), Y(\alpha(t))\rangle=\pm 1.$

I could not understand this. Why could not I have $X(t), Y(\alpha(t))$ parallel (both perp to $\alpha(t)$)?

Many thanks.

Answer 1

For each $t$ the vectors $X(t)$ and $Y(\alpha(t))$ are elements of the line $(T_{\alpha(t)} S)^\perp$, so they are parallel (or antiparallel). Since both have unit length, $\langle X(t), Y(\alpha(t)) \rangle = |X(t)| |Y(\alpha(t))| \cos \theta = \pm 1,$ where $\theta \in \{0, \pi\}$ is the angle between the two vectors.