$\{(X,Y)\in \Bbb R^3: |X|=|Y|=1, \langle X,Y \rangle=0\}$ is homeomorphic to $\text{SO}(3, \Bbb R)$ 
Let $R=\{(X,Y)\in \Bbb R^3: |X|=|Y|=1, \langle X,Y \rangle=0\}$ with $\langle\,\cdot\,,\,\cdot\,\rangle$ the Euclidean scalar product.
Prove that $R$ is homeomorphic to $\text{SO}(3, \Bbb R)$.

I don't know where to start. My intuition is to build a map from $R$ to some  space $X$ in $\mathcal{F}(X,Y)$  , that preserves the scalar product which would mean that this space $X$ is $O(3)$. but I don't know how.
Thanks for your hints, comments.
 A: Hint Expand the defining equation $A^{\top} A = I$ in the usual definition $$SO(n, \Bbb R) := \{A : A^{\top} A = I, \det A = 1\}$$ in terms of the column vectors of $A$.

Writing $A = \pmatrix{{\bf A}_1 & \cdots & {\bf A}_n}$, the condition becomes $\langle {\bf A}_i , {\bf A}_j \rangle = \delta_{ij}$, $1 \leq i, j, \leq n$, where $\delta_{ij}$ is the Kronecker delta symbol.

A: My intuition would be to show that there for any $\vec X, \vec Y \in \mathbb R^3$ such that $|\vec X  | = | \vec Y | = 1$ and $\vec X . \vec Y = 0$, there exists a unique $M \in \rm{SO}(3)$ that sends $(1,0,0) \mapsto \vec X$ and $(0,1,0) \mapsto \vec Y$.
(Indeed, such an $M$ must necessarily send $(0,0,1) \mapsto \vec X \times \vec Y$, where $\times$ denotes the vector cross product. This is because $M$, being special orthogonal, maps right-handed orthonormal bases to right-handed orthonormal bases, and $\vec X \times \vec Y$ is the only vector that extends $\{ \vec X, \vec Y \}$ to form a right-handed basis. Having specified how $M$ acts on the standard basis of $\mathbb R^3$, $M$ is now fully determined.)
