You may or may not know this, but any vector $v$ in a finite-dimensional inner product space determines (uniquely) a *hyperplane* (a subspace of dimension: $\dim(V) - 1$):
$E_v = \{w \in V: \langle v,w\rangle = 0\}$.
The vector $\dfrac{1}{\|v\|}v$ serves as a (unit) *normal* to $E_v$, if we choose an orientation for $V$, we have that either:
$\dfrac{1}{\|v\|}v$ or $-\dfrac{1}{\|v\|}v$ can be called *the* normal to the hyperplane $E_v$ (the "usual" orientation for $\Bbb R^3$ is set so that $\mathbf{i} \times \mathbf{j} = \mathbf{k}$, often called the "right-hand" orientation, since it corresponds to forming the (positive) $x$-axis with the right-hand index finger, the (positive) $y$-axis with the right-hand middle finger, and the thumb (pointing up), is the (positive) $z$-axis. This is a purely arbitrary convention, which is why cross-products are often called "pseudo-vectors", their sign isn't independent of axis orientation).
So...where was I?
Pick any vector in $E_v$, say, $w$, and consider $w + v$.
Then $\langle w+v,v\rangle = \langle w,v\rangle + \langle v,v\rangle = 0 + \|v\|^2 \neq 0$ (unless $v = 0$).
Thus $\langle\dfrac{1}{\|v\|^2}(w+v),v\rangle = 1$, and it is clear we have just as many such vectors as we have elements of $E_v$.
Why does your result not hold when $\dim(V) = 1$?
