This note shows how to describe a flat plane in three-dimensional space using a vector equation, a normal-vector equation and a Cartesian equation. You will learn to move between the three forms, because most plane problems in 3D geometry begin by getting the plane into the form that suits the question.
Key ideas
A plane is a flat surface that goes on for ever. You can fix a plane in space in two main ways: by giving a point on it together with two different directions that lie in it, or by giving a point on it together with a vector at right angles to it (the normal).
- Vector equation. Start at a known point with position vector \(\mathbf{a}\), then move any amount along two non-parallel direction vectors \(\mathbf{b}\) and \(\mathbf{c}\) that lie in the plane: \[ \mathbf{r}=\mathbf{a}+\lambda\mathbf{b}+\mu\mathbf{c},\qquad \lambda,\mu\in\mathbb{R}. \] The two parameters \(\lambda\) and \(\mu\) let you reach every point of the plane.
- Normal (scalar product) form. If \(\mathbf{n}\) is perpendicular to the plane and \(\mathbf{a}\) is a point on it, then for any point \(\mathbf{r}\) on the plane the vector \(\mathbf{r}-\mathbf{a}\) lies in the plane, so it is perpendicular to \(\mathbf{n}\): \[ (\mathbf{r}-\mathbf{a})\cdot\mathbf{n}=0 \quad\Longleftrightarrow\quad \mathbf{r}\cdot\mathbf{n}=\mathbf{a}\cdot\mathbf{n}. \]
- Cartesian equation. Writing \(\mathbf{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}\) and \(\mathbf{r}=\begin{pmatrix}x\\y\\z\end{pmatrix}\), the normal form becomes \[ ax+by+cz=d,\qquad d=\mathbf{a}\cdot\mathbf{n}. \] The coefficients of \(x\), \(y\), \(z\) are the components of a normal vector. This is the quickest way to read off the normal.
Finding a normal. If you know two direction vectors \(\mathbf{b}\) and \(\mathbf{c}\) in the plane, a normal is their vector product, \(\mathbf{n}=\mathbf{b}\times\mathbf{c}\). Any non-zero multiple of a normal is also a normal, so you can divide by a common factor to keep numbers small.
Direction vectors from points. If you are given three points \(A\), \(B\), \(C\) that are not on one line, use \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) as the two directions.
Ways to fix a plane. A plane is determined by any one of these: three points not in a line; a point and a normal; a point and two non-parallel directions; two intersecting lines; two parallel lines; a line and a point not on it.
Parallel planes have parallel normals, so they have the same left-hand side in the Cartesian form, up to a multiple, and different values of \(d\).
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(\mathbf{r}=\mathbf{a}+\lambda\mathbf{b}+\mu\mathbf{c}\) | Vector equation of a plane. \(\mathbf{a}\) is a point on it; \(\mathbf{b}\), \(\mathbf{c}\) are non-parallel vectors lying in it. |
| \(\mathbf{r}\cdot\mathbf{n}=\mathbf{a}\cdot\mathbf{n}\) | Normal form. Use when you know a point and a normal. |
| \(ax+by+cz=d\) | Cartesian equation. \(\begin{pmatrix}a\\b\\c\end{pmatrix}\) is a normal and \(d=\mathbf{a}\cdot\mathbf{n}\). |
| \(\mathbf{n}=\mathbf{b}\times\mathbf{c}\) | Normal found from two direction vectors in the plane. |
| \(\mathbf{b}\times\mathbf{c}=\begin{pmatrix}b_2c_3-b_3c_2\\ b_3c_1-b_1c_3\\ b_1c_2-b_2c_1\end{pmatrix}\) | Component form of the vector product. |
| Point on plane check | Substitute the coordinates into \(ax+by+cz\). The point is on the plane only if the result equals \(d\). |
Worked examples
Example 1 A drone flies in a flat sheet of airspace through the points \(A(1,2,3)\), \(B(2,0,5)\) and \(C(0,1,1)\). Find the Cartesian equation of the plane.
Find two direction vectors in the plane: \(\overrightarrow{AB}=\begin{pmatrix}2-1\\0-2\\5-3\end{pmatrix}=\begin{pmatrix}1\\-2\\2\end{pmatrix}\), \(\overrightarrow{AC}=\begin{pmatrix}0-1\\1-2\\1-3\end{pmatrix}=\begin{pmatrix}-1\\-1\\-2\end{pmatrix}\).
Normal: \(\mathbf{n}=\overrightarrow{AB}\times\overrightarrow{AC}=\begin{pmatrix}(-2)(-2)-(2)(-1)\\ (2)(-1)-(1)(-2)\\ (1)(-1)-(-2)(-1)\end{pmatrix}=\begin{pmatrix}6\\0\\-3\end{pmatrix}\).
Divide by 3 to get a simpler normal \(\begin{pmatrix}2\\0\\-1\end{pmatrix}\). So the plane is \(2x-z=d\).
Use point \(A\): \(d=2(1)-3=-1\).
Check with \(B\): \(2(2)-5=-1\). Check with \(C\): \(2(0)-1=-1\). Both work.
Answer: \(2x-z=-1\)
Example 2 A plane has vector equation \(\mathbf{r}=\begin{pmatrix}1\\0\\2\end{pmatrix}+\lambda\begin{pmatrix}1\\1\\0\end{pmatrix}+\mu\begin{pmatrix}0\\2\\1\end{pmatrix}\). Write it in Cartesian form.
The directions are \(\mathbf{b}=\begin{pmatrix}1\\1\\0\end{pmatrix}\) and \(\mathbf{c}=\begin{pmatrix}0\\2\\1\end{pmatrix}\).
Normal: \(\mathbf{b}\times\mathbf{c}=\begin{pmatrix}(1)(1)-(0)(2)\\ (0)(0)-(1)(1)\\ (1)(2)-(1)(0)\end{pmatrix}=\begin{pmatrix}1\\-1\\2\end{pmatrix}\).
So the plane is \(x-y+2z=d\). The point \((1,0,2)\) is on the plane, so \(d=1-0+4=5\).
Check with another point: \(\lambda=1,\mu=0\) gives \((2,1,2)\), and \(2-1+4=5\). Correct.
Answer: \(x-y+2z=5\)
Example 3 Find a Cartesian equation of the plane that contains the line \(\mathbf{r}=\begin{pmatrix}1\\2\\0\end{pmatrix}+t\begin{pmatrix}2\\1\\-1\end{pmatrix}\) and the point \(P(3,0,1)\).
One direction in the plane is the line direction \(\mathbf{b}=\begin{pmatrix}2\\1\\-1\end{pmatrix}\).
A second direction joins the point \(A(1,2,0)\) on the line to \(P\): \(\overrightarrow{AP}=\begin{pmatrix}2\\-2\\1\end{pmatrix}\). It is not parallel to \(\mathbf{b}\), so the two vectors span the plane.
Normal: \(\mathbf{b}\times\overrightarrow{AP}=\begin{pmatrix}(1)(1)-(-1)(-2)\\ (-1)(2)-(2)(1)\\ (2)(-2)-(1)(2)\end{pmatrix}=\begin{pmatrix}-1\\-4\\-6\end{pmatrix}\).
Multiply by \(-1\): use \(\begin{pmatrix}1\\4\\6\end{pmatrix}\). The plane is \(x+4y+6z=d\).
Using \(A\): \(d=1+8+0=9\). Check with \(P\): \(3+0+6=9\).
Answer: \(x+4y+6z=9\)
Common errors
- Common error: using two vectors that are parallel (or one of them zero) as the directions of a plane. Check that neither is a multiple of the other, otherwise they describe only a line.
- Common error: writing the position vector of a point as the direction of the plane. The directions must be differences of points, such as \(\overrightarrow{AB}\), not the points themselves.
- Common error: getting \(d\) from the normal instead of from a point. Always substitute a point on the plane: \(d=\mathbf{a}\cdot\mathbf{n}\).
- Common error: slips in the vector product, especially the sign of the middle component. Check your answer by substituting every given point into the final equation.
- Common error: thinking a plane has one unique equation. Multiplying both sides of \(ax+by+cz=d\) by a constant gives the same plane, and a vector equation can use many different points and directions.
Exam tip: After finding a Cartesian equation, substitute each given point back in. It takes seconds and catches arithmetic slips in the vector product or in \(d\).
Check yourself
- Find the Cartesian equation of the plane with normal \(\begin{pmatrix}3\\-1\\2\end{pmatrix}\) passing through \((1,1,1)\).
Answer
\(d=3(1)-1+2(1)=4\), so \(3x-y+2z=4\). - Does the point \((2,3,1)\) lie on the plane \(3x-y+2z=4\)?
Answer
\(3(2)-3+2(1)=5\), which is not 4. The point is not on the plane. - Find the plane parallel to \(2x-3y+z=5\) that passes through \((1,2,-1)\).
Answer
Parallel planes share a normal, so the plane is \(2x-3y+z=d\). \(d=2-6-1=-5\). The plane is \(2x-3y+z=-5\). - Find a vector equation and the Cartesian equation of the plane through \(A(1,0,0)\), \(B(0,2,0)\) and \(C(0,0,3)\).
Answer
\(\overrightarrow{AB}=\begin{pmatrix}-1\\2\\0\end{pmatrix}\), \(\overrightarrow{AC}=\begin{pmatrix}-1\\0\\3\end{pmatrix}\). A vector equation is \(\mathbf{r}=\begin{pmatrix}1\\0\\0\end{pmatrix}+\lambda\begin{pmatrix}-1\\2\\0\end{pmatrix}+\mu\begin{pmatrix}-1\\0\\3\end{pmatrix}\). Normal: \(\begin{pmatrix}(2)(3)-0\\ 0-(-1)(3)\\ 0-(2)(-1)\end{pmatrix}=\begin{pmatrix}6\\3\\2\end{pmatrix}\). With point \(A\), \(d=6\), so \(6x+3y+2z=6\). Check \(B\): \(6\). Check \(C\): \(6\).