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For a nonzero normal vector $\mathbf{n}$ and the plane equation $\mathbf{n}\cdot \mathbf{x}=d$, the closest point on a plane to $\mathbf{p}$ isIt subtracts only the normal displacement. Substituting gives $\mathbf{n}\cdot \mathbf{q}=d$, so $\mathbf{q}$ lies on the plane; the difference $\mathbf{p}-\mathbf{q}$ is a scalar multiple of its normal vector. These two checks identify the orthogonal projection onto a plane without solving a three-variable minimization problem.

For $\mathbf{n}=(2,-1,2)$, $d=3$, and $\mathbf{p}=(4,1,2)$, the residual $8$ givesIndeed, $(40-17+4)/9=3$, and $\Vert \mathbf{p}-\mathbf{q}\Vert =(8/9)\cdot 3=8/3$. Choosing the plane point $(1,1,1)$ instead gives distance $\sqrt{10}>8/3$: being on the plane does not make a point the nearest one.

As a function of $\mathbf{p}$, this projection isIt is an affine map, becoming a linear map when $d=0$. Projecting a second time changes nothing. It preserves displacements along the plane and removes the normal displacement, so it is a concrete example of an idempotent map.

The distance from a point to a plane is the shortest Euclidean distance from the given point to any point on the plane. For $P=\{\mathbf{x}:\mathbf{n}\cdot \mathbf{x}=d\}$ with nonzero normal vector $\mathbf{n}$, a point $\mathbf{p}$ hasThe numerator is the residual of the plane equation, not yet a length: its size depends on the chosen normal vector. Dividing by that vector's Euclidean norm removes this arbitrary scale.

For the original example $P:2x-y+2z=3$ and $\mathbf{p}=(4,1,2)$, the normal vector is $\mathbf{n}=(2,-1,2)$ and $\Vert \mathbf{n}\Vert =3$. The residual is $8-1+4-3=8$, so the distance is $8/3$. Using the equivalent plane equation $-6x+3y-6z=-9$ instead gives numerator $24$ and Euclidean norm $9$, hence the same $8/3$. A point already on the plane has zero residual and zero distance.

Why is the perpendicular route shortest? Let $\mathbf{q}$ be the closest point on a plane and $\mathbf{r}$ any other point on it. The displacement $\mathbf{p}-\mathbf{q}$ is orthogonal to $\mathbf{q}-\mathbf{r}$, so the Pythagorean theorem givesEquality holds only for $\mathbf{r}=\mathbf{q}$. This is the closest-point principle of orthogonal decomposition, applied after a translation of an affine plane to its direction linear subspace.^[ref]

The formula is useful whenever the deviation from a flat reference should be measured geometrically. A raw linear equation residual cannot be compared across arbitrary equation scalings; a perpendicular distance can. The same reasoning works for a hyperplane in any finite-dimensional Euclidean space, with the appropriate number of coordinates.

Use the distance from a point to a plane formula with the unit normal $\hat{\mathbf{n}}$. Since the plane passes through the origin, the distance isFor a direct check, the closest point on a plane is $\mathbf{p}-(\mathbf{p}\cdot \hat{\mathbf{n}})\hat{\mathbf{n}}=(1,\sqrt{2},-1{)}^T$, whose displacement from $\mathbf{p}$ has Euclidean norm $\sqrt{2}$.

Orient the plane $\mathbf{n}\cdot \mathbf{r}=d$ with a unit normal $\mathbf{n}$. A point $\mathbf{p}$ then has signed distance from a point to a planeIts absolute value is the ordinary distance from a point to a plane. The closest point on a plane is $\mathbf{p}-s\mathbf{n}$, so the displacement from the point to its orthogonal projection onto a plane is $-s\mathbf{n}$. Reversing the unit normal and $d$ reverses $s$ without changing the plane or the distance from a point to a plane. A sphere of radius $q$ centred at $\mathbf{p}$ has sphere-plane intersection radius $\sqrt{q^2-s^2}$ when $|s|<q$.

For a general, not necessarily unit, normal vector ${\mathbf{n}}_0$ in ${\mathbf{n}}_0\cdot \mathbf{x}=d_0$, the oriented distance is $({\mathbf{n}}_0\cdot \mathbf{p}-d_0)/\Vert {\mathbf{n}}_0\Vert$. In the worked example of distance from a point to a plane, orientation toward $(2,-1,2)$ assigns $+8/3$ to $(4,1,2)$; reversing that unit normal assigns $-8/3$. The closest point on a plane and the unsigned distance are unaffected.