Equivalence classes in $\mathbb{R}^{3}$ such that $Mv_{1} = v_{2}$ for orthogonal $M$ 
Let $ A = \mathbb{R}^{3}$, let $R$ be an equivalence relation such that $\left(v_{1},v_{2}\right) \in R$ if and only if $\exists P \in M_{3 \times 3}\left(\mathbb{R}\right)$ such that $P$ is orthogonal and $Pv_{1} = v_{2}$
Describe and sketch the equivalence classes of $R$ in $\mathbb{R}^{3}$

I am having difficulty attempting this question. I attempted to find the equivalence class, $C$, of $\left(1,0,0\right)^{T}$ for which I used: $$P = \begin{bmatrix}1 & 0 & 0 \\ 0 & \cos\left(\theta\right) & \sin\left(\theta\right)\\ 0 & -\sin\left(\theta\right) & \cos\left(\theta\right)\\ \end{bmatrix}$$ and said that $\displaystyle v = \left(\alpha,\beta,\gamma\right)^{T} \in C \implies \alpha = 1, \tan\left(\theta\right) = \frac{\gamma - \beta}{\gamma + \beta}$
But I feel that I am not following the right route with this approach. Thank you in advance.

An orthogonal matrix is an isometry when viewed as an operator. So for any orthogonal matrix $P$ this holds: $|v|=|Pv|$. In particular if $(v_1, v_2)\in R$ then $|v_1|=|v_2|$.
The other implication also holds, i.e. if $|v_1|=|v_2|$ then you can find an orthogonal matrix $P$ such that $Pv_1=v_2$ (for explicit construction in dimension 3 see this).
All in all, an equivalence class of $v$ in $R$ is a set of all vectors with the same norm, i.e. a sphere.
