Let $x\in S^2$(unit sphere in $\mathbb{R}^3$) be fixed. We consider, for $0\leq \theta \leq \pi$, $$L_\theta=\{y\in S^2: x\cdot y=\cos \theta \}. $$ So $S^2=\displaystyle\cup_{0\leq \theta \leq \pi}=L_\theta.$

I want to understand the geometric interpretation of $L_\theta.$ Can we interprete $L_\theta$ as circle of radius $\sin \theta?$

  • $\begingroup$ Is $S^3$ a 3D spherical shell of radius $1$? $\endgroup$
    – Andrei
    Sep 22 at 17:23
  • $\begingroup$ Sorry that should be $S^2$. Let me correct that $\endgroup$ Sep 22 at 18:02
  • $\begingroup$ Take $x$ to be the north pole and draw the picture. $\endgroup$ Sep 22 at 18:23

1 Answer 1


Let's write $y$ in terms of two directions, one along $x$ and one perpendicular. $$y=y_{||}+y_\perp$$Then $$x\cdot y=x\cdot y_{||}+x\cdot y_\perp$$ Since $|x|=|y|=1$, $$x\cdot y=y_{||}$$ If $y\in \mathbb R^3$, then this would be a plane, perpendicular to $x$, at a distance $\cos\theta$ from the origin. Since $y\in S^2$, then $L_\theta$ is the intersection of the plane with the sphere, a circle at a distance $\cos\theta$ from the origin, with the center along the $x$ direction, and radius $\sin\theta$.


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