Prove that there is no isometry from $\mathbb R^3$ to $\mathbb R^2$ How should I start of? I know if $T$ is an isometry then $T=A \vec x+\vec b$ where A is orthogonal matrix and $\vec b$ is a column vector. If I can prove that there is no orthogonal matrix from $\mathbb{R^3}$ to $\mathbb{R^2}$ then will it suffice?(which I don't think is true?)
(Also my understanding of isometries is quite basic, I know isometries preserve distance and dot product)

Consider a tetrahedron in $\mathbb R^3$. Its vertices are four points equidistant from each other.  
An isometry with $\mathbb R^2$ would create such a set in $\mathbb R^2$, but this is clearly impossible.
One way to see is that since distinct circles in $\mathbb R^2$ only intersect in at most two locations. If you pick two points $A,B$ distance $d$ apart, and you want to find two more points distance $d$ away from $A,B$, then you must select the intersections $C,D$ of circles radius $d$ around $A,B$. But unfortunately, $C$ and $D$ are necessarily $d\sqrt{3}$ apart, not $d$!
