= Solution
The <vector triple product> with $|u|=1$ gives
$$
f(r)=r-u(u\cdot r),\qquad\boxed{A=I-uu^T,\quad A_{ij}=\delta_{ij}-u_iu_j.}
$$
This is the <orthogonal projection onto a plane> perpendicular to $u$, so it is a <linear map>. Its <image> is the plane $u^\perp$ through the origin, while its <kernel> is the normal line $\operatorname{span}\{u\}$. In components,
$$
A=\begin{pmatrix}1-u_1^2&-u_1u_2&-u_1u_3\\-u_1u_2&1-u_2^2&-u_2u_3\\-u_1u_3&-u_2u_3&1-u_3^2\end{pmatrix}.
$$
The equation $Ar=r$ is equivalent to $u(u\cdot r)=0$, hence to $u\cdot r=0$. Its solution set is precisely the two-dimensional plane $u^\perp$, with infinitely many nonzero vectors, not merely the normal line. These are the <eigenvectors> with <eigenvalue> one; the normal direction has <eigenvalue> zero.
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