0.10 Planes in Space

Two points determine a line, and a point with a direction does too. Three points that are not in a line determine a plane — that is why a tripod never wobbles — but the description that will matter most in this book uses a point and a vector, as the line did, with one change: the vector stands perpendicular to the plane instead of lying along it. A nonzero vector $\mathbf{n}$ that is orthogonal to every vector lying in a plane is called a normal vector for the plane. The vector $\mathbf{e}_3$ is a normal for the floor; so is $5\mathbf{e}_3$, and so is $-\mathbf{e}_3$, since only the direction of a normal matters. Infinitely many planes share a given normal — a whole stack of parallel sheets — and picking one point $P_0$ selects a single sheet from the stack.

The figure draws the plane through $P_0 = (1, 2, 1)$ with normal $\mathbf{n} = (1, -2, 2)$, and puts the whole stack on a slider.

Explore in 3D (opens in a new tab)

Explore the figure

1. At the starting value $d = -1$, check that $P_0$ is on the sheet by substituting: $1 - 2(2) + 2(1) = -1$. Orbit until you see the sheet edge-on; the normal arrow then stands at a right angle to the line the sheet has become.
2. Drag $d$ to $2$. The sheet slides upward along the normal, leaving $P_0$ beneath it, and the normal does not turn. Every plane in the family $x - 2y + 2z = d$ has the same normal, because the normal is read from the coefficients on the left, which the slider does not touch.
3. Drag $d$ to $-4$. The sheet has passed back down through $P_0$ to the far side. Exactly one value of $d$ puts the sheet through $P_0$: a normal fixes the tilt of a plane, and one point fixes which plane.
4. Set $d$ back to $-1$ and reveal the hidden point $Q = (1, 0, -1)$ and the arrow from $P_0$ to $Q$. Substitution puts $Q$ on the sheet, $1 - 0 - 2 = -1$, and that arrow, $(0, -2, -2)$, lies inside it. Dot it with the normal: $0 + 4 - 4 = 0$. A vector in the plane is orthogonal to the normal — and that single fact is the plane's equation.

From a dot product to a linear equation

A point $P = (x, y, z)$ lies on the plane through $P_0 = (x_0, y_0, z_0)$ with normal $\mathbf{n} = (a, b, c)$ exactly when the displacement from $P_0$ to $P$ lies in the plane, that is, when it is orthogonal to $\mathbf{n}$:

$$\mathbf{n} \cdot (P - P_0) = a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$$

Multiplying out and moving the constants to the right gives $ax + by + cz = d$ with $d = ax_0 + by_0 + cz_0$. So every plane is the solution set of one linear equation in three unknowns, and — reading the argument backward — the solution set of any equation $ax + by + cz = d$ with $a$, $b$, $c$ not all zero is a plane with normal $(a, b, c)$. The coefficients are the normal. For the figure, $\mathbf{n} = (1, -2, 2)$ and $P_0 = (1, 2, 1)$ give $d = 1 - 4 + 2 = -1$: the equation $x - 2y + 2z = -1$, and the slider's other values are the parallel planes with the same left side.

Two consequences are immediate. Planes with proportional normals are parallel, since a normal is only a direction; and a plane through three points $P$, $Q$, $R$ has normal $(Q - P) \times (R - P)$, because Section 0.8 built the cross product to be orthogonal to both edge vectors, and therefore to the whole plane they span.

The same shape of equation exists in every dimension. In $\mathbb{R}^2$, $ax + by = d$ is a line, and $(a, b)$ is perpendicular to it — check it against the slope $-a/b$. In $\mathbb{R}^n$, $a_1 x_1 + \cdots + a_n x_n = d$ describes a hyperplane, an $(n - 1)$-dimensional flat sheet with normal $(a_1, \dots, a_n)$, invisible but governed by the same dot product. Chapter 1 begins by stacking several such equations and asking where their planes meet.

An original work of XYZ Homework, built around interactive XYZ 3D figures. Its chapter sequence is aligned to Interactive Linear Algebra (Margalit & Rabinoff, Georgia Tech, GNU FDL); this work is original, copies nothing from it, and is not affiliated with or endorsed by its authors. License: CC-BY-NC-SA-4.0.