# To under the geometry of a space

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 $$\sin \theta?$$

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

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 $$\sin\theta$$.