I need help in interconverting between these two definitions of a spherical cap in $\mathbb{R}^n$:
Figure 9 (Left): For a fixed unit vector $v$ and $\epsilon \in [0,1)$, the set $\theta\in S^{n-1}$ for which $$\langle\theta,v\rangle \ge \epsilon$$ is called the $\epsilon$-spherical cap about $v$. It is denoted by $C(\epsilon,v)$. Here $S^{n-1}$ is the sphere of radius $1$, i.e. $$S^{n-1} = \{x\in\mathbb{R}^n: \|x\| = 1\}$$ Figure 9 (Right): The cap of radius $r$ about $v$ is given by $$\{\theta \in S^{n-1}: \| \theta - v\|\le r\}$$
So how do I express $\epsilon$ in terms of $r$? This will help me switch between the two definitions of the spherical cap. Note that we are in $\mathbb{R}^n$ and not $\mathbb{R}^3$.
My try:
I started with the second definition and saw that $$\| \theta - v\|\le r \implies \langle \theta - v,\theta-v\rangle \leq r^2 \implies \|\theta\|^2 - \|v\|^2 - 2\langle\theta,v\rangle \leq r^2$$
Since $\theta,v\in S^{n-1}$ we have $\|\theta\| = \|v\| = 1$ so that $$\| \theta - v\|\le r\implies \langle\theta,v\rangle \geq 1-\frac{r^2}{2} $$
Now, looking at the first definition, we see that $\epsilon =-\frac{r^2}{2}$ but we had assumed $\epsilon\in[0,1)$ so this does not make sense!
Where am I going wrong, what am I missing?
Addendum:
It was a silly blunder. The right inequality is:
$$\| \theta - v\|\le r \implies \|\theta\|^2 + \|v\|^2 - 2\langle\theta,v\rangle \leq r^2 \implies \langle\theta,v\rangle \geq 1 - \frac{r^2}{2}$$
So we get $$\epsilon = 1 - \frac{r^2}{2}$$
It should be
$$\langle \theta - v,\theta-v\rangle \leq r^2 \implies \|\theta\|^2 + \|v\|^2 - 2\langle\theta,v\rangle \leq r^2.$$