This set is a manifold let $S$ be the set of pairs $(x,y)$ where x,y are orthogonal unit vectors in $\mathbb R^3$. i am trying to show this is a topological manifold. for starters one needs to define a suitable topology on it. i was thinking let a set $U$ be open in $S$ iff $U \cap S^2$ (intersection with sphere) is open in $S^2$ in the subspace topology? am i going in the right direction? I'd really appreicate some help. thank you 
this is really about topological manifold, sorry for not stating it earlier. 
here is the definition:

a manifold is a second countable Hausdorff space that is locally
  Euclidean.

 A: Your set $S$ is a subset of $\mathbb R^6$, so give it the subspace topology.  That ensures it's 2nd countable and Hausdorff.
To show it's a manifold, notice $S$ is the pairs $(x,y) \in \mathbb R^3 \times \mathbb R^3$ such that:
$$ |x|^2 =1,\ \  |y|^2=1, \ \ x\cdot y = 0 $$
This is the same as saying $S = f^{-1}(1,1,0)$ where $f(x,y) = (|x|^2, |y|^2, x \cdot y)$.  
So the idea would be to show $(1,1,0)$ is a regular-value of $f$, then apply the preimage theorem as Qiaochu cites. The pre-image theorem is basically just the implicit function theorem from calculus, but re-cast in a convenient formalism for saying things are manifolds. 
A: It is a smooth manifold of dimension 3. 
Indeed : let $F : S^2 \times S^2 \rightarrow \mathbb{R}$ be defined by $F(x,y)=\sum_i x_i y_i$ (the standard inner product). $F$ is a smooth map on the smooth manifold $S^2 \times S^2$ and its derivative does not vanish so it is a submersion. Bu the submersion property, the set of its zeroes is a manifold of dimension $2*2-1=3$.
