Relation S is $R^2-{\vec0}$ if angle between $\vec v$ and $\vec w$ is multiple of $\pi/2$. Is this an equivalence relation? What if $R^3-{\vec0}$

Question

Relation S is defined as $R^2-{\vec0}$ if vectors $\vec v$ and $\vec w$ have an angle that is multiple of $\pi/2$ is this an equivalence relation? What if $R^3-{\vec 0}$?

I wrote S={$\vec v, \vec w: cos^{-1}(\frac{\overline v * \overline w}{|\overline v |* |\overline w|})mod\frac{\pi}2$}. So we can clearly see it's reflexive because the angle between any vector and itself is $0$ so done. Then we can see it's symmetric because $\vec v$R$\vec w$ is going to be the same as $\vec w$R$\vec v$ in the above equation. Then I can see it's transitive for any vector that's symmetric, both those vectors will be reflexive so it'll be transitive. But I don't know about something like $(\vec w, \vec v)$ and $(\vec w, \vec t)$ being in the relation and how that shows $(\vec w, \vec t)$.

Also I don't know about $R^3-\vec 0$

Answer 1

1. If $(\vec v, \vec w)\angle=a\frac\pi2$ and $(\vec w,\vec t)\angle=b\frac\pi2$, then $(\vec v,\vec t)\angle=(a+b)\frac\pi2$.
2. In the 3d space, it doesn't work: you can find three vectors such that $\vec v\perp\vec t$ and $\vec w\perp\vec t$ with an arbitrary angle for $(\vec v,\vec w)\angle$.