Why is this Isometry a rotation?

• I
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?

berkeman
Mentor
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?
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?

fresh_42
Mentor
2021 Award
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)?

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

mfb
mathwonk