# I Why is this Isometry a rotation?

1. Jun 27, 2016

Hello,

i need a little help. Did someone have an idea how to prove this?

Be $\Phi$ an direct isometry of the euclidean Space $\mathbb{R}^3$ with

$\Phi (\begin{pmatrix} 2\\0 \\1 \end{pmatrix})$=$\begin{pmatrix} 2\\1 \\0 \end{pmatrix}$ and $\Phi (\begin{pmatrix} 1\\1 \\0 \end{pmatrix})$=$\frac{1}{3} \begin{pmatrix} 1\\4 \\1 \end{pmatrix}$

Why is $\Phi$ a rotation?

2. Jun 27, 2016

### Staff: Mentor

All schoolwork-type question should be posted in the Homework Help forums, and you need to show your efforts toward solving the question. Advanced schoolwork questions are sometimes allowed in the technical forums, but only if the poster shows substantial work.

Can you say what level class this is for, and please show your thoughts for how to proceed?

3. Jun 27, 2016

### Staff: Mentor

Is it? Have you calculated the angles between the vectors before and after? Or the matrix wrt the basis (1,0,0), (0,1,0) and (0,0,1)?

4. Jun 27, 2016

### micromass

Staff Emeritus
It might also be a good idea to start with defining what you understand under "rotation" and "direct isometry".

5. Jun 28, 2016

### mathwonk

you might ask yourself whether every linear isometry might actually be an isometry, and if not why not. If you know the basic theory there is only one number to compute, (and that only up to sign!).

6. Jul 6, 2016