How One Normal Vector Pins Down an Entire Plane

A wall, a tabletop or a sheet of glass is a plane — a flat surface stretching in two directions. One point on it and one vector perpendicular to it, called the normal, fix it completely, and its equation gives distances, angles and the point where a line pierces it.

This lesson covers equations of a plane, the normal and distances, angles with planes and lines, and intersections of lines with planes.

**A plane through the point with position vector $\vec{a}$ and normal $\vec{n}$ has vector equation $(\vec{r}-\vec{a})\cdot \vec{n}=0$ and Cartesian form $A(x-x_1)+B(y-y_1)+C(z-z_1)=0$; its normal form is $lx+my+nz=p$, where l, m, n are the direction cosines of the normal and p is the distance from the origin; and its intercept form is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$.

Worked example (one-point form).** The plane through $(1,-2,3)$ with normal $2\hat{i}-\hat{j}+2\hat{k}$ is $2(x-1)-(y+2)+2(z-3)=0$, that is $2x-y+2z=10$.

Worked example (normal form). Dividing $2x-y+2z=10$ by $\sqrt{4+1+4}=3$ gives

$$\frac{2}{3}x-\frac{1}{3}y+\frac{2}{3}z=\frac{10}{3}$$

so the plane is $\frac{10}{3}$ units from the origin.

Worked example (intercept form). $6x+3y+2z=12$ becomes $\frac{x}{2}+\frac{y}{4}+\frac{z}{6}=1$, so the plane cuts the axes at 2, 4 and 6.

An everyday example. A sloping solar panel on a rooftop lies in a plane fixed by one of its corners and the direction its face points — the normal.

The substance. p in the normal form must be positive — if the right-hand side comes out negative, change every sign.

**The coefficients A, B and C in $Ax+By+Cz+D=0$ are direction ratios of the normal, and the perpendicular distance of $(x_1,y_1,z_1)$ from the plane is $\frac{|A{x}_1+B{y}_1+C{z}_1+D|}{\sqrt{A^2+B^2+C^2}}$.

Worked example (normal).** For $3x-4y+12z=7$, the normal has direction ratios 3, $-4$ and 12, and direction cosines $\frac{3}{13}$, $-\frac{4}{13}$ and $\frac{12}{13}$.

Worked example (distance). The distance of $(2,3,-5)$ from $x+2y-2z=9$ is

$$d=\frac{|2+6+10-9|}{\sqrt{1+4+4}}=\frac{9}{3}=3\text{ units}$$

Worked example (parallel planes). The distance between $2x-y+2z=3$ and $2x-y+2z=12$ is $\frac{|12-3|}{3}=3$ units.

An everyday example. The shortest distance from a bird perched on a branch to a sloping hillside is its perpendicular distance from the plane of the slope.

The substance. **The distance of the origin from $Ax+By+Cz+D=0$ is $\frac{|D|}{\sqrt{A^2+B^2+C^2}}$** — the special case with every coordinate zero.

**The angle between two planes is the angle between their normals, with $\cos \theta =\frac{|\overrightarrow{n_1}\cdot \overrightarrow{n_2}|}{|\overrightarrow{n_1}||\overrightarrow{n_2}|}$, while the angle between a line with direction $\vec{b}$ and a plane with normal $\vec{n}$ satisfies $\sin \varphi =\frac{|\vec{b}\cdot \vec{n}|}{|\vec{b}||\vec{n}|}$.

Special cases:**

- Planes perpendicular when $\overrightarrow{n_1}\cdot \overrightarrow{n_2}=0$; parallel when the normals are proportional
- A line parallel to a plane when $\vec{b}\cdot \vec{n}=0$; perpendicular when $\vec{b}$ is proportional to $\vec{n}$

Worked example (two planes). For $2x-y+z=6$ and $x+y+2z=7$:

$$\cos \theta =\frac{|2-1+2|}{\sqrt{6}\sqrt{6}}=\frac{3}{6}=\frac{1}{2}\Rightarrow \theta =60°$$

Worked example (line and plane). For the line $\frac{x-1}{2}=\frac{y}{1}=\frac{z+1}{2}$ and the plane $x+2y+2z=5$:

$$\sin \varphi =\frac{|2+2+4|}{3\times 3}=\frac{8}{9}\Rightarrow \varphi \approx 62.7°$$

An everyday example. The angle between a wall and a sloping roof is the angle between two planes, measured through their normals.

The substance. Two lines or two planes use cosine, but a line and a plane use sine — because the normal is perpendicular to the plane, not along it.

**Write the line in parametric form, $x=x_1+a\lambda$, $y=y_1+b\lambda$, $z=z_1+c\lambda$, substitute these into the equation of the plane, solve for $\lambda$, and put that value back into the line.

Worked example.** Find where $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$ meets $x-y+z=5$.

- General point: $(2+3\lambda ,-1+4\lambda ,2+12\lambda )$
- Substituting: $(2+3\lambda )-(-1+4\lambda )+(2+12\lambda )=5$
- So $5+11\lambda =5$ and $\lambda =0$
- The point is $(2,-1,2)$; check: $2+1+2=5$

An everyday example. Finding where the beam from a classroom projector hits a slanted screen is exactly a line-plane intersection.

The substance. **If $\lambda$ disappears from the equation, the line is parallel to the plane** — it either lies in the plane or never meets it.