The subspace in question:
$V=\{ \vec{u} \in \Bbb{R}^n : \vec{n}^T\vec{u}=\vec{0} \}$

I am assuming that $\vec{u} = \begin{bmatrix}x\_0 \\ x\_1 \\ \vdots \\ x\_2\end{bmatrix}$.

The dimension of a vector space/subspace is equal to the number of linearly independent vectors in its basis. So it's either 1 or n. Does $\vec{u}$ count as 1 or as n?