Let $x$ is a point on a surface $S$, we can write a vector $\boldsymbol{v}$ at $x$. We can write $\boldsymbol{v}$ as
$$ \boldsymbol{v} = \boldsymbol{v}_{||} + \boldsymbol{v}_{\perp} $$ Where the first and second terms on the RHS are the tangential and normal vectors.
The normal vector is $$ \boldsymbol{v}_{\perp} = (\boldsymbol{v}\cdot \hat{n})\hat{n} $$
where $\hat{n}$ is the unit normal to the surface. My question is, if this normal vector is zero and ALL of the components of the unit normal vector $\hat{n}$ are non-zero, then does this guarantee that $\boldsymbol{v}=0$?
Nope. Consider the vector $n=\frac{1}{\sqrt6}(1,1,-2)$. Then the vector $v=(1,1,1)$ satisfies $v\cdot n=0$ but $v\neq 0$.