smooth vector field on the boundary of unit ball

Question

Let C be the boundary of the unit ball in $\mathbb{R}^n$, and let $v$ be a smooth vector field on $C$. What does the condition $x\cdot v(x)>0$ for all $x$ in C mean?

Answer 1

It means your vector field is pointing outwards.

Obs: One can conclude from here that if $v$ is extended to the whole ball $B$ then it will be zero at at least one point inside.

Answer 2

Since $v$ is tangent to the sphere, if you consider $\phi_t$ its flow, $\|\phi_t\|^2=1$ by differentiating that you have $<x,v(x)>=0$. So $<x,v(x)>$ can't be >0.