# Isometries of $\mathbb{R}^3$

So I'm attempting a proof that isometries of $\mathbb{R}^3$ are the product of at most 4 reflections. Preliminarily, I needed to prove that any point in $\mathbb{R}^3$ is uniquely determined by its distances from 4 non-coplanar points, and then that an isometry sends non-coplanar points to non-coplanar points in $\mathbb{R}^3$. I've done the first preliminary step, and finished the proof assuming the second, but I can't find a simple way to prove the second...

Intuitively it makes a lot of sense that non-coplanar points be sent to non-coplanar points, but every method I've stumbled upon to prove such has been quite heavy computationally...

I know for example that any triangle chosen among the four points, A, B, C, D must be congruent to the triangles of their respective images, but what extra bit of information would allow me to say that the image of the whole configuration can't be contained in a single plane...

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Would this work? Construct the altitude from $D$ to the plane containing $A$, $B$, and $C$. Call the foot of this altitude $E$ (the point where the altitude meets the plane). Triangles $ADE$, $BDE$, and $CDE$ all have right angles at $E$ and you know that the isometry preserves angles, so triangles $A'D'E'$, $B'D'E'$, and $C'D'E'$ all have right angles at $E'$, which makes $A'$, $B'$, $C'$, and $E'$ coplanar and since $D'E'\ne 0$ and $\overline{D'E'}$ is perpendicular to that plane, $D'$ cannot be coplanar with the other points.

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Isometries are affine and preserve volume : so if $A$, $B$, $C$, $D$ are non-coplanar, their convex hull $ABCD$ has non-zero volume, and the convex hull $f(A)f(B)f(C)f(D)$, which is the image of the convex hull $ABCD$ under $f$ (since $f$ is affine) has non-zero volume, therefore $f(A)$, $f(B)$, $f(C)$ and $f(D)$ must be non coplanar.

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Isn't this essentially a stronger restatement of what I'm trying to prove? –  AsinglePANCAKE Dec 11 '11 at 22:50
actually, this is a result of measure theory when you construct Lebesgues measure. but I see what you mean, this is kind of heavy artillery for your problem... one other mean is to use barycentric coordinates consider the spanned vector space –  Glougloubarbaki Dec 11 '11 at 22:58
Hmmm, care to elaborate? There has to be a simpler explanation that I'm not seeing... –  AsinglePANCAKE Dec 12 '11 at 0:42

Well, how do you do it for collinear points? You may be able to generalize the idea.

But, I suspect, the easiest way is to find a condition equivalent to "coplanar" in terms of things you already know are preserved and reflected by isometries. (e.g. define "coplanar" in terms of "collinear")

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Well for non-collinear points I have the lovely SSS congruence axiom for triangles that ensures non-collinear points remain non-collinear. I don't know an easy way to prove SSSSSS for tetrahedra... –  AsinglePANCAKE Dec 11 '11 at 22:53