The question comes from another question, I am asked to calculate the dimension and check orientability of the manifold
$$ V_2(\mathbb{R}^4) = \{(v_1,v_2) \in \mathbb{R}^4\times \mathbb{R^4} \mid |v_1|=|v_2|=1, v_1\cdot v_2=0\} $$
And I prove via the projection $\pi(v_1,v_2)=v_1$ that $V_2(\mathbb{R}^4)=S^3\times S^2$ ($\text{Im}(\pi) = S^3$ and $\forall x\in S^3, \pi^{-1}(x)=S^2$). Following that, the dimension is $5$ but what about orientability?
$S^3\times S^2$ is orientable since both manifolds are orientable, just take the product of the orientations. But what you showed is that $V_2(\mathbb R^4)$ is an $S^2$ bundle over $S^3$, you want to say that it is the trivial bundle.