Lesson 02

Equations of Planes

A plane has two descriptions that look nothing alike. You can give two directions lying in it, or one normal standing on it, and the cross product is the bridge between them. Almost every plane question is a version of “which of those two descriptions does the given information hand me, and which does the answer want?”

By the end of this lesson

  • Produce the vector, parametric and Cartesian forms of a plane.
  • Use \(\vv{n} = \vv{u}\times\vv{v}\) to move between the direction view and the normal view.
  • Build a plane from any of the five defining data sets.
  • Run a Cartesian equation backwards into vector form.

Section 1

A point and two directions

A plane is determined by a point \(P_0\) and two non-parallel direction vectors \(\vv{u}\) and \(\vv{v}\) lying in it. Compare that with a line, which needed one direction and one parameter. A plane is two-dimensional, so it needs two directions and two parameters.

From \(P_0\) you walk \(s\) steps along \(\vv{u}\) and then \(t\) steps along \(\vv{v}\), and between them those two moves reach every point of the plane and nothing else.

\[\vv{r} = \vv{r}_0 + s\vv{u} + t\vv{v}, \qquad s,\,t \in \R\]
(2.1)

The words non-parallel are doing real work in (2.1). If \(\vv{v}\) were a multiple of \(\vv{u}\), every combination \(s\vv{u} + t\vv{v}\) would still be a multiple of \(\vv{u}\), so you would sweep out a line and never fill a plane. The cross product catches this for you: it returns \(\vv{0}\) exactly when the two vectors are parallel.

A plane spanned by a point and two direction vectors Coordinate axes with a shaded plane. From the point P zero, two non-parallel vectors u and v are drawn inside the plane, and dashed construction lines complete the parallelogram to the point s u plus t v, showing how every point of the plane is reached. x y z u v su + tv P0
fig. 2.1Two non-parallel directions from one point sweep out the whole plane. The dashed lines complete the parallelogram to \(s\vv{u}+t\vv{v}\).
Cross product, expanded Interactive

Enter two vectors and watch the determinant expand term by term, with the perpendicularity check done for you.

\[(1,1,0)\times(0,2,1) = \begin{vmatrix}\hat{\imath}&\hat{\jmath}&\hat{k}\\1&1&0\\0&2&1\end{vmatrix} = (1,\,-1,\,2)\]

Section 2

The normal vector and the Cartesian form

The cross product of two vectors in R³ produces a third vector perpendicular to both, which is exactly what a plane needs. Feed it two directions lying in the plane and it hands back a normal, \(\vv{n} = \vv{u}\times\vv{v}\).

Set it out as a determinant every time. Writing the \(\hat{\imath},\,\hat{\jmath},\,\hat{k}\) row above the two vectors and expanding along it is slower than the memorised component formula for about a week, and after that it is faster. It also never produces the middle-component sign error that the memorised version invites.

The normal is unique only up to a scalar multiple, and its sign depends on the order of the factors: \(\vv{v}\times\vv{u} = -(\vv{u}\times\vv{v})\). Both answers describe the same plane, so either is acceptable, and it is always worth dividing out a common factor before finding \(D\).

The normal vector of a plane as a cross product Coordinate axes with a shaded plane. Two in-plane vectors u and v are drawn from a point, and their cross product n stands perpendicular to the plane, marked with a right-angle symbol. x y z π n = u × v u v
fig. 2.2The cross product of two in-plane directions stands perpendicular to the plane. Reverse the order of the factors and the arrow points the other way — it is still the same plane.

The four forms of a plane

Vector
\(\vv{r} = \vv{r}_0 + s\vv{u} + t\vv{v},\quad s,t \in \R\)
Parametric
\(x = x_0 + su_1 + tv_1,\quad y = y_0 + su_2 + tv_2,\quad z = z_0 + su_3 + tv_3\)
Cartesian
\(Ax + By + Cz + D = 0,\ \text{with normal } \vv{n} = (A,B,C)\)
Point–normal
\(\vv{n} \cdot (\vv{r} - \vv{r}_0) = 0\)

The point–normal form is the one to remember, because the Cartesian form is just it multiplied out. Saying “the vector from \(P_0\) to any point of the plane is perpendicular to \(\vv{n}\)” and expanding the dot product gives \(Ax+By+Cz-(Ax_0+By_0+Cz_0)=0\), so \(D = -\vv{n}\cdot\vv{r}_0\). In practice you rarely compute \(D\) that way — you write down \(Ax+By+Cz+D=0\) and substitute the known point — but knowing where the formula comes from keeps the sign straight.

Plane builder, five ways Interactive

Choose which data you have been given, type the numbers, and watch the same plane appear in all three forms.

\[\vv{r} = (1,0,2) + s(1,1,0) + t(0,2,1) \quad\Longleftrightarrow\quad x - y + 2z - 5 = 0\]

Section 3

Five ways to define a plane

A plane is pinned down the moment you have a point and two independent directions. Every one of these five data sets is that information in disguise, and your first job in any plane question is to work out which one you have been given.

One Three non-collinear points Take one as \(P_0\) and the two vectors joining it to the others as \(\vv{u}\) and \(\vv{v}\). If the three points are collinear, \(\vv{u}\times\vv{v}=\vv{0}\) and no plane is determined.
Two A point and two directions The definition itself. There is nothing to convert, so write the vector form straight down.
Three A point and a normal The quickest route to a Cartesian equation. The normal supplies \(A,B,C\) and the point supplies \(D\).
Four A line and a point not on it The line gives one direction and one point; joining that point to the external one gives the second direction. Example 3 below does this.
Five Two intersecting or parallel-distinct lines Intersecting lines give a common point and two directions. Parallel distinct lines give one direction, and the vector joining a point of each gives the other.
Not enough A line on its own Infinitely many planes contain a given line; picture a page rotating about its spine. You always need one more piece of information.

Section 4

Cartesian back to vector form

Every textbook shows the forward direction. The reverse comes up just as often: a question gives you \(2x-y+3z=6\) and asks for a vector equation. There is a three-step recipe for it.

  1. Find one point. An intercept is easiest — set two variables to zero and solve for the third. Here \(y=z=0\) gives \((3,0,0)\).
  2. Find two independent directions. A direction lies in the plane exactly when it is perpendicular to the normal, so solve the homogeneous equation \(2x-y+3z=0\) twice, choosing convenient values. Setting \(z=0,\ x=1\) gives \((1,2,0)\); setting \(x=0,\ z=1\) gives \((0,3,1)\).
  3. Check they are independent — neither a multiple of the other — then assemble: \(\vv{r} = (3,0,0) + s(1,2,0) + t(0,3,1)\).

Practice

Worked examples

Example 1 · routine

Find the Cartesian equation of the plane through \(P_0(1,0,2)\) containing the directions \(\vv{u}=(1,1,0)\) and \(\vv{v}=(0,2,1)\).

  1. \(\vv{n} = \vv{u}\times\vv{v} = \bigl((1)(1)-(0)(2),\ (0)(0)-(1)(1),\ (1)(2)-(1)(0)\bigr) = (1,\,-1,\,2)\)
  2. So the equation is \(x - y + 2z + D = 0\). Substitute \(P_0\): \(1 - 0 + 4 + D = 0 \Rightarrow D = -5\).
  3. Check with a second point of the plane, \(P_0+\vv{u} = (2,1,2)\): \(2-1+4-5=0\ \checkmark\)
\(x - y + 2z - 5 = 0\)
Example 2 · standard

Find the Cartesian equation of the plane through \(A(1,1,0)\), \(B(2,0,3)\) and \(C(0,4,1)\).

  1. Two in-plane directions: \(\vv{u} = \overrightarrow{AB} = (1,-1,3),\qquad \vv{v} = \overrightarrow{AC} = (-1,3,1)\).
  2. \(\vv{n} = \vv{u}\times\vv{v} = \bigl((-1)(1)-(3)(3),\ (3)(-1)-(1)(1),\ (1)(3)-(-1)(-1)\bigr) = (-10,\,-4,\,2)\)
  3. Divide by \(-2\), watching the sign change on every component: \(\vv{n} = (5,\,2,\,-1)\). Then \(5x + 2y - z + D = 0\), and substituting \(A\) gives \(5 + 2 - 0 + D = 0 \Rightarrow D = -7\).
  4. Check the two points not used to find \(D\): \(B:\ 10+0-3-7=0\ \checkmark\qquad C:\ 0+8-1-7=0\ \checkmark\)
\(5x + 2y - z - 7 = 0\)
Example 3 · demanding

Find the Cartesian equation of the plane containing the line \(\vv{r} = (1,0,-1) + t(2,1,1)\) and the point \(A(3,2,0)\).

  1. The line supplies a point \(P(1,0,-1)\) and a direction \(\vv{u}=(2,1,1)\).
  2. First confirm \(A\) is not on the line, or there is no unique plane. From \(x:\ 1+2t=3 \Rightarrow t=1\), but then \(y = 1 \neq 2\), so \(A\) is off the line.
  3. The second direction joins the two: \(\vv{v} = \overrightarrow{PA} = (2,2,1)\). Then \(\vv{n} = \vv{u}\times\vv{v} = \bigl((1)(1)-(1)(2),\ (1)(2)-(2)(1),\ (2)(2)-(1)(2)\bigr) = (-1,\,0,\,2)\).
  4. Multiply by \(-1\) for tidiness: \(\vv{n}=(1,0,-2)\), so \(x - 2z + D = 0\). Substituting \(P\) gives \(1 + 2 + D = 0 \Rightarrow D = -3\). Check \(A:\ 3-0-3=0\ \checkmark\)
\(x - 2z - 3 = 0\)

Watch out

Common mistakes

  • Treating the cross product as commutative. \(\vv{u}\times\vv{v} \neq \vv{v}\times\vv{u}\), because the sign flips. The plane does not care, so either normal earns full marks — just be consistent within one solution, and never “fix” a sign halfway through.
  • Reporting a zero cross product as a bad normal. If \(\vv{u}\times\vv{v} = \vv{0}\,\), the three points are collinear or the two directions are parallel, and the data does not determine a plane at all. Say so; that is the answer.
  • Never checking the equation against a point you did not use. One substitution catches almost every cross-product slip. If you built the plane from three points, verify it with the two that did not go into finding \(D\).