Linear Algebra in $\mathbb R^3$: Projection onto a plane

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When I reviewed my linear algebra notes, there was comment like the below.

Let's say $n$ is a unit normal vector of the plane $P$ which contains $0=(0,0,0)\in\mathbb R^3$.

Then for $v \in\mathbb R^3$ the mapping $T(v) = v - \langle v,n\rangle\, n$ is a projection on the plane $P$.

Suddenly I've got a curious about those. What if the plane $P$ does not contain the point $0$? Does the above statement hold?

Well... Still, I couldn't find any counterexample. So at least my thought, it looks like it holds whatever the plane is. (If it is true, it could be a powerful tool for solving the projection.) What do you think about that?

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If the plane $P$ does not contain the origin, then it is an affine plane, but it is not a linear subspace in $\mathbb R^3$.

Hence you cannot expect there to be a projector-like linear map onto $P$ because the image of that linear map would be a subspace.