Tangent space of manifold has two unit vectors orthogonal to tangent space of its boundary

I'm reading spivak calculus on manifolds and got stuck. Let M be a k-dimensional manifold with boundary in $\mathbb{R^{n}}$, and $M_{x}$ is the tangent space of M at x with dimension k, then $\partial M_{x}$ is the (k-1) dimensional subspace of $M_{x}$ Spivak says that this implies, there are only two unit vectors in $M_{x}$ that are orthogonal to $\partial M_{x}$. I am confused as to why only two unit vectors?

Given a finite dimensional inner product space $V$ and a codimension 1 subspace (that is, a hyperplane) $W \subset V$, there are precisely two unit vectors othogonal to $W$. Because $W$ is codimension 1, we automatically have that $\text{dim}(W^\perp) = 1$; and a 1-dimensional inner product space has precisely two unit vectors.
• Is there a linear algebra theorem that relates dimension of inner product space to the number of unit vectors in that space? Don't know how dim(W$^\perp$) =1 shows it has two unit vectors. Sorry for the trouble. – Kimi Lee Aug 9 '15 at 16:08
• @KimiLee: Let $U$ be 1-dimensional and pick a unit vector $u$. Then every other vector is of the form $\lambda u$; and $\|\lambda u\| = |\lambda|$. This is only 1 for $\lambda = \pm 1$. So the only unit vectors are $\pm 1$. In a higher-dimensional normed space there are infinitely many unit vectors. – user98602 Aug 9 '15 at 16:11