What is the spherical line through the points $(0,-1,0)$ and $\left(0,\frac{1}{2},\frac{\sqrt{3}}{2}\right)$?
I solved:
$G = \{(x,y,z)\in S^2 \mid \exists\ a,b,c \in \mathbb{R}, ax+by+cz = 0\}$
$\left\{\begin{matrix} a(0)+ b(\frac{1}{2})+ c(\frac{\sqrt{3}}{2})=0\\ a(0)-b+c(0)=0 \end{matrix}\right.$
=> $\left\{\begin{matrix} b+ (\sqrt{3})c=0\\ -b=0 \end{matrix}\right.$
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I think my solution is wrong.
Please help me to solve this question.
You're not quite on the right track, here. The way you've written it $G$ is all of $S^2$, since if we simply take $a=b=c=0,$ then $ax+by+cz=0.$ What you want is to find some $a,b,c\in\Bbb R$ not all zero such that the set $$\{(x,y,z)\in S^2\mid ax+by+cz=0\}$$ contains your two points.
You're doing fine solving the system, so far. From $-b=0$ we conclude that $b=0$, whence $0=b+c\sqrt3=c\sqrt3,$ and so $c=0$. Now $a$ is a free variable of the system, but since $b=c=0$, then we must take $a\ne0$. Our set then becomes $$\{(x,y,z)\in S^2\mid ax=0\},$$ or equivalently $$\{(x,y,z)\in S^2\mid x=0\}$$ (since $a\ne0$).