# Prove that final position of vectors of length r rest on surface of sphere with radius r.

I'm trying to mathematically prove that two non-zero vectors in $R^3$ of length $r$ in standard position's tips rest on the surface of a sphere with radius $r$.

At first this seemed fairly straight-forward since the sphere represents every possible resting point of any vector in standard position of length $r$, but how do you prove that?

This is assuming $r > 0$ and $r$ is a valid real number.

If $S_r$ denotes the sphere with radius $r>0$ centered at $0$ what You actually want to prove is: $$S_r=\{x\in\mathbb{R}^3:|x|=r\},$$ where $|.|$ denotes the euclidean norm. If You define the sphere with radius $r$ by the above equation there is nothing to prove. If You define the sphere as the boundary of the ball with radius $r$ centered at $0$, i.e. $S_r=\partial B_r$, where $$B_r=\{x\in\mathbb{R}^3:|x|<r\}$$ then You argue like this: Every neighborhood of a vector $x\in\mathbb{R}^3$ with $|x|=r$ contains elements of $B_r$ and of $\mathbb{R}^3\backslash B_r$ (why?) and thus $$\{x\in\mathbb{R}^3:|x|=r\}\subseteq S_r=\partial B_r$$ and every element that has only neighborhoods that contain elements of both $B_r$ and $\mathbb{R}^3\backslash B_r$ has to be of length $r$ and thus $$S_r\subseteq\{x\in\mathbb{R}^3:|x|=r\}$$ Just fill in the details. $\textbf{Hint:}$ for $r'\neq r$ and $x\in\mathbb{R}^3$ with $|x|=r'$ consider neighborhoods $U(x)=\{y\in\mathbb{R}^3:|x-y|<\frac{|r-r'|}{2}\}$.