- #1
Doradus
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Hello,
i need a little help. Did someone have an idea how to prove this?
Thanks in advance.
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?
i need a little help. Did someone have an idea how to prove this?
Thanks in advance.
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?