Show that any element $g$ in $\operatorname{SO}(3)$ is an isometry in $\mathbb R^3$
$\operatorname{SO}(3) = \{ g \in M(3,\mathbb R) \mid gg^t = I, \ \det(g) =1 \}$
where $M(3,\mathbb R)$ is the set of matrices $3\times3$ in $\mathbb R$, and $I$ is the identity matrix.
My answer is:
$\|gx\|^2 =\langle gx,gx\rangle = g \langle x,gx \rangle = gg^t \langle x,x \rangle = \langle x,x \rangle$
Is my answer true?
We need to show that distances are preserved that is
$$|gx-gy|^2=|g(x-y)|^2=\langle g(x-y),g(x-y) \rangle=[g(x-y)]^Tg(x-y)=\\=(x-y)^Tg^Tg(x-y)=(x-y)^T(x-y)=\langle x-y,x-y \rangle=|x-y|^2$$
and thus we can conclude that every linear (or affine) transformation with orthogonal matrix is an isometry.