Lesson 02

Equations of Planes

A plane has two descriptions that look nothing alike — 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 really the question “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.

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 — or, equivalently, by a point and a single normal vector \(\vv{n}\) perpendicular to it. The cross product converts between the two views: \(\vv{n} = \vv{u} \times \vv{v}\).

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.

The words non-parallel are doing real work. If \(\vv{v}\) were a multiple of \(\vv{u}\), every combination \(s\vv{u} + t\vv{v}\) would still be a multiple of \(\vv{u}\), and you would sweep out a line, not a plane. The cross product notices 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 \(s\vv{u}+t\vv{v}\).

The three forms

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 it comes from is what stops the sign going astray.

The normal vector and the cross product

The cross product of two vectors in R³ produces a third vector perpendicular to both. That is exactly what a plane needs: feed it two directions lying in the plane, and it hands back a normal.

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 then it is faster, because it 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 — but it is the same plane.
Cross product, expanded Interactive

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

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

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 the first job in any plane question is to say 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. Nothing to convert — 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 4 below does exactly 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. One extra piece of information is always needed.

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 — and it has a three-step recipe.

  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 is a multiple of the other — and assemble: \(\vv{r} = (3,0,0) + s(1,2,0) + t(0,3,1)\).

Verify by taking the cross product of your two directions: it should come back parallel to the original normal. Here \((1,2,0)\times(0,3,1) = (2,-1,3)\), which is the normal exactly.

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.

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

Practice

Worked examples

Example 1 · routine

A point and two directions, to a Cartesian equation

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

Show solution
  1. \(\vv{n} = \vv{u}\times\vv{v} = \begin{vmatrix}\hat{\imath}&\hat{\jmath}&\hat{k}\\1&1&0\\0&2&1\end{vmatrix}\)
  2. \(= \bigl((1)(1)-(0)(2),\ (0)(0)-(1)(1),\ (1)(2)-(1)(0)\bigr) = (1,\,-1,\,2)\)
  3. So the equation is \(x - y + 2z + D = 0\). Substitute \(P_0\): \(1 - 0 + 4 + D = 0 \Rightarrow D = -5\).
  4. \(\therefore\ x - y + 2z - 5 = 0\). Check a second point of the plane, \(P_0+\vv{u} = (2,1,2)\): \(2-1+4-5=0\ \checkmark\)
Example 2 · standard

A plane through three points

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

Show solution
  1. Two in-plane directions: \(\vv{u} = \overrightarrow{AB} = (1,-1,3),\quad \vv{v} = \overrightarrow{AC} = (-1,3,1)\).
  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 \(-2\) — note the sign change on every component: \(\vv{n} = (5,\,2,\,-1)\).
  4. \(5x + 2y - z + D = 0\); substitute \(A\): \(5 + 2 - 0 + D = 0 \Rightarrow D = -7\).
  5. \(\therefore\ 5x + 2y - z - 7 = 0\). Check both remaining points: \(B:\ 10+0-3-7=0\ \checkmark\qquad C:\ 0+8-1-7=0\ \checkmark\)
Example 3 · standard

Cartesian to vector form

Write \(2x-y+3z=6\) in vector form.

Show solution
  1. A point: set \(y=z=0\), so \(2x=6\) and \(\vv{r}_0 = (3,0,0)\).
  2. Directions solve the homogeneous equation \(2x-y+3z=0\). Take \(z=0,\ x=1 \Rightarrow y=2\): \(\vv{u} = (1,2,0)\).
  3. Take \(x=0,\ z=1 \Rightarrow y=3\): \(\vv{v} = (0,3,1)\). Neither is a multiple of the other, so they are independent.
  4. \(\therefore\ \vv{r} = (3,0,0) + s(1,2,0) + t(0,3,1)\).
  5. Check by going back: \((1,2,0)\times(0,3,1) = (2,-1,3)\), the original normal. ✓
Example 4 · demanding

The plane through a line and an external point

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

Show solution
  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 — otherwise no unique plane exists. 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: \(\vv{v} = \overrightarrow{PA} = (2,2,1)\).
  4. \(\vv{n} = \vv{u}\times\vv{v} = \bigl((1)(1)-(1)(2),\ (1)(2)-(2)(1),\ (2)(2)-(1)(2)\bigr) = (-1,\,0,\,2)\)
  5. Multiply by \(-1\) for tidiness: \(\vv{n}=(1,0,-2)\). Then \(x - 2z + D = 0\); substituting \(P\) gives \(1 + 2 + D = 0 \Rightarrow D = -3\).
  6. \(\therefore\ x - 2z - 3 = 0\). Check \(A:\ 3-0-3=0\ \checkmark\) and a second point of the line, \(t=1 \to (3,1,0):\ 3-0-3=0\ \checkmark\)

Watch out

Three ways to lose a mark here
  • The cross product is not commutative. \(\vv{u}\times\vv{v} \neq \vv{v}\times\vv{u}\) — the sign flips. The plane does not care, so either normal earns full marks; but be consistent within one solution, and never “fix” a sign halfway through.
  • Three points must be non-collinear. If \(\vv{u}\times\vv{v} = \vv{0}\), you have not found a bad normal — you have discovered that the data does not determine a plane at all. Say so.
  • Verify with a point you have not used. One substitution catches almost every cross-product slip. If you found the plane from three points, check it with the two you did not use to find \(D\).