Find perpendicular feet and distances

Check whether a point lies on the plane

For membership, substitute the point into the plane equation. For a perpendicular foot, move from the point along the normal until you reach the plane. For distance, take the length of that displacement; its parameter alone is not a distance.

Follow the source’s two steps: convert to scalar-product form $\mathbf r\cdot\mathbf n=d$ if needed, then substitute the point’s position vector $\mathbf p$. Equality $\mathbf p\cdot\mathbf n=d$ means the point lies on the plane.

Find the perpendicular foot

For a point P and plane $\Pi:\mathbf r\cdot\mathbf n=d$, the source constructs F in five steps:

1. Form the normal line $l_{PF}:\mathbf r=\mathbf p+\lambda\mathbf n$.
2. Identify F as the intersection of this line and the plane.
3. Write $\overrightarrow{OF}=\mathbf p+\lambda\mathbf n$ for its parameter.
4. Impose $(\mathbf p+\lambda\mathbf n)\cdot\mathbf n=d$ and solve for $\lambda$.
5. Substitute this parameter into the line to find F.

Check the foot and its boundary case

Since $\mathbf n\ne\mathbf0$, the denominator $\mathbf n\cdot\mathbf n$ is positive. Equivalently,

$$\lambda=\frac{d-\mathbf p\cdot\mathbf n}{|\mathbf n|^2},\qquad\mathbf f=\mathbf p+\lambda\mathbf n.$$

The source assumes P is off the plane. The same formula also covers P in the plane, when $\lambda=0$ and F=P.

Find the distance by projection or by its foot

Method 1: projection length. Choose any A in the plane and use the unit normal $\widehat{\mathbf n}=\mathbf n/|\mathbf n|$. The source takes the projection of $\overrightarrow{PA}$:

$$\begin{aligned}h&=\left|\overrightarrow{PA}\cdot\widehat{\mathbf n}\right|\\&=\frac{|(\mathbf a-\mathbf p)\cdot\mathbf n|}{|\mathbf n|}\\&=\frac{|d-\mathbf p\cdot\mathbf n|}{|\mathbf n|}.\end{aligned}$$

Its absolute value is a distance, so reversing PA to AP does not change h. The unit-normal hat and absolute-value bars are essential; native extraction can lose them.

When P is the origin O, the source obtains

$$\begin{aligned}h_O&=\left|\overrightarrow{OA}\cdot\widehat{\mathbf n}\right|\\&=\left|\overrightarrow{OA}\cdot\frac{\mathbf n}{|\mathbf n|}\right|\\&=\frac{|d|}{|\mathbf n|}.\end{aligned}$$

The signed perpendicular displacement from O toward the plane along the chosen normal is $d/|\mathbf n|$. The source separates $d>0$ (the plane is reached along $\mathbf n$) and $d<0$ (it is reached in the opposite direction). If $d=0$, the plane passes through O. Reversing the normal and d preserves the plane and distance but reverses this orientation-dependent signed value.

Method 2: perpendicular-foot length. Find F by the preceding five steps, then calculate $h=|\overrightarrow{PF}|$. This is the source’s alternative to projection.

The source remark recommends finding a convenient plane point by observation: $(0,2,0)$ lies on $\mathbf r\cdot(1,2,3)=4$. Any point used for projection must satisfy the plane equation; the base point of an unrelated line need not do so.