# Thread: The unit sphere in R^3 is arcwise connected

1. ## The unit sphere in R^3 is arcwise connected

So I attempted to find the intersection of the unit sphere with a plane that contains any two arbitrary points $a, b$ on the sphere and the origin, but I only came up with a really convoluted expression that I am not even sure can be used to find a function $f: [0,1] \to \mathbb{R}^{3}$ such that $f(0) = a, f(1) = b$ and for all $t\in [0,1]$, $f(t)$ is on the unit sphere. Any help would be appreciated.

2. Originally Posted by Pinkk
So I attempted to find the intersection of the unit sphere with a plane that contains any two arbitrary points $a, b$ on the sphere and the origin, but I only came up with a really convoluted expression that I am not even sure can be used to find a function $f: [0,1] \to \mathbb{R}^{3}$ such that $f(0) = a, f(1) = b$ and for all $t\in [0,1]$, $f(t)$ is on the unit sphere. Any help would be appreciated.
You can parametrize the sphere with two angles $\vartheta, \varphi$ so that you get $x=r\sin\vartheta \cos\varphi, y = r\sin \vartheta \sin\varphi, z=r\cos\vartheta$.
This is a map $g:\, [0;2\pi)\times [-\pi/2;\pi/2]\to \in\mathbb{R}^3, (\varphi,\vartheta)\mapsto (x,y,z)$.
Assuming that you have coordinates $(\varphi_a,\vartheta_a), (\varphi_b,\vartheta_b)$ for two given points $a,b$ on the unit sphere, you next parametrize the line segment connecting the points $(\varphi_a,\vartheta_a), (\varphi_b,\vartheta_b)$ that lie in the rectangular area of the domain of these two parameters. This gives you the map $h:\, [0;1]\to [0;2\pi)\times [-\pi/2;\pi/2]$.
Now a parametrization of the arc connecting a and b on the sphere is $f\,: [0;1]\to \mathbb{R}^3, t\mapsto g(h(t))$, I think.

3. What would a parameterization of the line segment look like, simply $h(t) = (\varphi_a,\vartheta_a) + t(\varphi_b - \varphi_a,\vartheta_b - \vartheta_a)$?

4. Originally Posted by Pinkk
What would a parameterization of the line segment look like, simply $h(t) = (\varphi_a,\vartheta_a) + t(\varphi_b - \varphi_a,\vartheta_b - \vartheta_a)$?
Sure, why not? Since the domain of the parametrization of the sphere by $g:\, (\varphi,\vartheta)\to\mathbb{R}^3$ is a rectangle and thus convex, that parametrization h(t) of the line segment is sure to give you a line segment that remains completely within the domain of g.