Let $\gamma: \mathbb{R} \to \mathbb{R^3}, \gamma(t) := (0,0,1)+\cos(t)v_1+\sin(t)v_2$,
where $v_1 =(\frac{2}{3},\frac{1}{3},-\frac{2}{3}),v_2=(\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}})$.
I know this curve is a circle centered at $(0,0,1)$ with radius 1 in the plane spanned by $v_1,v_2$. If it is in $\mathbb{R^2}$, I can calculate the determinant to find the orientation. However, I have no idea if it is in $\mathbb{R^3}$.
Any help would be appreciated.