Take any vector in \(\mathbb R^3\) and apply the rotation transformation.
\(\begin{bmatrix}\cos\theta&-\sin\theta&0\\\sin\theta&\cos\theta&0\\0&0&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}x\cos\theta-y\sin\theta\\x\sin\theta+y\cos\theta\\z\end{bmatrix}\)
Take the norm of the transformed vector.thet
\(\sqrt{(x\cos\theta-y\sin\theta)^2+(x\sin\theta+y\cos\theta)^2+z^2}\)
\(=\sqrt{x^2\cos^2\theta-2xy\cos\theta\sin\theta+y^2\sin^2\theta+x^2\sin^2\theta+2xy\sin\theta\cos\theta+y^2\cos^2\theta+z^2}\)
\(=\sqrt{x^2(\cos^2\theta+\sin^2\theta)+y^2(\sin^2\theta+\cos^2\theta)+z^2}\)
\(=\sqrt{x^2+y^2+z^2}\)
which is exactly the norm of \(\begin{bmatrix}x\\y\\z\end{bmatrix}\).
