Show that there are at least two points on a manifold to which a vector is normal Let $M \subset \mathbb R^3$ be a $2$-dimensional manifold which is also a compact set. And let $v \in \mathbb R^3$ be a vector which satisfies $ ||v|| = 1$.
The task is to prove that there are at least two points $x,y \in M$ such that $v$ is a normal vector to their tangent spaces.
There was given a hint to look at the extremum points of the function $f(x) = <x,v>$ which means $f(x) = x_1v_1 + x_2v_2 + x_3v_3$. Since the functions is continuous and $M$ is compact then the function get a minimum and maximum on $M$. Let's say the $x$ is the maximum and $y$ is a the minimum. Moreover, I noticed that $\nabla f = (v_1,v_2,v_3) = v$.
However, I am not so sure how to continue from here. I am supposed to show that $v$ is normal to both $T_xM$ and $T_yM$ but I don't see how exactly. I still haven't used the fact that $M$ is a manifold so it probably uses that.
Help would be appreciated

Following the hint you propose, $f(x)=\langle x, v\rangle$ has a maximum and a minimum. If they both coincide, it turns out that your $2$-manifold is a compact subset of a plane in $\mathbb{R}^3$, which would imply that it is a surface with boundary, contradicting your definition of submanifold as something that is locally a graph of a function with open domain in $\mathbb{R}^2$ (Invariance of Domain implicitly used).
Then if $x_0$ is the maximum of $f$ and $\gamma: (-\epsilon, \epsilon)\rightarrow M$ is a curve with $\gamma(0)=x_0$, $t=0$ is a critical point of $f\circ \gamma$ (this is just a directional derivative of $f$ at $x_0$). This means that $0=\nabla f(x_0)\cdot \gamma'(0)=v\cdot \gamma'(0)$. This proves that $v\perp T_{x_0}M$, since $T_{x_0}M$ is exactly the plane containing all vectors tangent to curves through $x_0$.
The exact same argument proves that $v\perp T_{y_0}M$.
