Proposition. Let $B$ be the point on the plane $q=\left\{ \mathbf{p}\in\mathbb{R}^{n}:\mathbf{p}\cdot\mathbf{n}=d\right\}$ that is closest to $A$, then the line $AB$ is perpendicular to the plane.
This Proposition is intuitively/geometrically "obvious", but I was wondering how we'd prove it from the usual definitions (distance, scalar product, etc.), including these of the plane and orthogonality/perpendicularity:
Definition 1. A plane in $\mathbb{R}^{n}$ is any set of points that can be written as $$\left\{ \mathbf{p}\in\mathbb{R}^{n}:\mathbf{p}\cdot\mathbf{n}=d\right\},$$ for some $\mathbf{n}\in\mathbb{R}^{n}\setminus\left\{ \mathbf{0}\right\} $ and some $d\in\mathbb{R}$.
Definition 2. Two vectors $\mathbf{u}$ and $\mathbf{v}$ are perpendicular if $\mathbf{u}\cdot \mathbf{v} = 0$.
If $AB$ is not perpendicular to the plane, pick a point $C$ in the plane such that $AC$ is. Then $ABC$ is a right triangle with hypotenuse $AB$, so $B$ is not the closest point by Pythagoras' theorem.
Note that Pythagoras' theorem can be expressed purely in terms of scalar product:
$$\|u+v\|^2=\|u\|^2+\|v\|^2-2\left<u,v\right> $$
implies that $$\|u+v\|^2=\|u\|^2+\|v\|^2$$ whenever $u$ and $v$ are orthogonal by your definition.