How can we show that a nondegenerate "open" triangle is a $2$-dimensioal $C^\infty$-submanifold of $\mathbb R^3$?

Question

How can we show that a nondegenerate triangle $\Delta$ (without the edges) is a $2$-dimensioal $C^\infty$-submanifold of $\mathbb R^3$?

Intuitively, the claim is obvious to me: For any point $p$ of $\Delta$, we could take an open ball $B$ around $p$ and rotate and translate $B$ such that the open circle given by the intersection of $B$ with $\Delta$ lies in the $(x,y)$-plane.

But how do we need to write this down rigorously?

Answer 1

Let $\vec r_i$, $1\le i\le 3$, be the vertices. Then consider $$\vec r \mapsto \bigl(\vec r\cdot(\vec r_2-\vec r_1),\vec r\cdot (\vec r_3-\vec r_1), (\vec r-\vec r_1)\cdot ((\vec r_3-\vec r_1)\times(\vec r_2-\vec r_1))\bigr ) .$$ It maps the plane containing $\Delta$ to the $xy$-plane.

Answer 2

The $\Sigma_0:=\ (x,y)$ plane is clearly a submanifold. By rotation and translation - which are smooth diffeomorphisms - the space is mapped to itself, $\Sigma_0$ is mapped to an arbitrary affine plane.
So, the plane $\Sigma$, which contains the triangle, is a submanifold, and $\Delta$ is an open subset of $\Sigma$, so it's a submanifold of $\Sigma$, hence also of $\Bbb R^3$.