Let $\vec{x_1},\vec{x_2}\in{R^3}$.
$C=\{(\vec{x_1},\vec{x_2})\ \vert\ ||\vec{x_1}-\vec{x_2}||_2 = l\}$ where $l\in R$.
I would like to show whether the set $C$ is convex or not but am not sure where to start. The notion of a line segment between some $x,y\in C$ is confusing me. Are there any tips on how I should start or if I should reformulate this problem?
So you want to regard the convexity of $$C=\{(\vec{x_1},\vec{x_2})\ \vert\ ||\vec{x_1}-\vec{x_2}||_2 = l\}.$$ It's a subset of $\mathbb{R}^3 \times \mathbb{R}^3$. Take for example $x_1,y_1$ and $t\in ]0,1[$ such that $\|x_1\|_2=\|y_1\|_2=l$ and such that $t x_1$ and $(1-t)y_1$ are not colinear. Then $t(x_1,0)+(1-t)(y_1,0)=(tx_1+(1-t)y_1,0)\not\in C$ because $$\|tx_1+(1-t)y_1-0\|_2<t\|x_1\|_2+(1-t)\|y_1\|_2=l.$$
Let $(\vec{x}_1,\vec{x}_2),(\vec{x}'_1,\vec{x}'_2)\in C$. The question is whether $(\alpha\vec{x}_1+(1-\alpha)\vec{x}'_1,\alpha\vec{x}_2+(1-\alpha)\vec{x}'_2)\in C\;\forall \alpha\in(0,1).$ Compute
$$ ||\alpha(\vec{x}_1-\vec{x}_2)+(1-\alpha)(\vec{x}'_1-\vec{x}'_2)||_2\leq \alpha||(\vec{x}_1-\vec{x}_2)||_2+(1-\alpha)||(\vec{x}'_1-\vec{x}'_2)||_2=l_{1,2} $$
and, therefore, $(\alpha\vec{x}_1+(1-\alpha)\vec{x}'_1,\alpha\vec{x}_2+(1-\alpha)\vec{x}'_2)\not \in C\;\forall \alpha\in(0,1).$
