Where $\theta$ is the angle between two vectors $\vec u$ and $\vec v$. What’s the fastest way to check that both vectors are approximately orthogonal. I know when $\vec u$ and $\vec v$ are exactly orthogonal $\vec u \cdot \vec v = 0$ and $\theta(\vec u, \vec v) = 90$.
However, say $$\vec u = [-214748.359375, 0.6362000107765198, 1.399999976158142, 1]$$ and $$\vec v = [8.385956100787163\times10^{-8}, 0.9103664774626016, 0.41380294430118253, 0]$$ Here $\vec u$ and $\vec v$ are not exactly but approximately orthogonal thus $\theta(\vec u, \vec v) = 89.9996957122442$ and their dot product $\vec u \cdot \vec v = 1.140490571783766$
Since comparing angles between vectors is a bit more expensive than computing a dot product. Is there any other way to quickly check that $\vec u$ and $\vec v$ are approximately orthogonal?
I find it a bit strange that their dot product isn’t approximately $0$ since the angle between them is approximately $90$.
$$(\cos\theta)^2=\frac{(u.v)^2}{(u.u)(v.v)}$$ so check that $(u.v)^2$ is much smaller than $(u.u)(v.v)$