Home / A-Level Further Maths 9231 / Vectors / Equations of planes
Cambridge International AS & A Level · Topic 1.6
A-Level Further Maths 9231 / Topic 1.6 / AS & A Level

One point and one normal fix a plane.

A plane in three dimensions is pinned down by a point on it and a vector at right angles to it. That one idea gives three equivalent equations, \(\mathbf{r}\cdot\mathbf{n} = p\), \(ax + by + cz = d\) and \(\mathbf{r} = \mathbf{a} + \lambda\mathbf{b} + \mu\mathbf{c}\), and the skill is moving between them to solve the problem in front of you.

Topic 1.6All vectors lessons →
The key idea

A plane through the point \(A\) with normal \(\mathbf{n}\) is \(\mathbf{r}\cdot\mathbf{n} = \mathbf{a}\cdot\mathbf{n}\). With \(\mathbf{n} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}\) this expands to \(ax + by + cz = d\): the coefficients of \(x\), \(y\), \(z\) are the normal. The parametric form \(\mathbf{r} = \mathbf{a} + \lambda\mathbf{b} + \mu\mathbf{c}\) uses a point and two non-parallel directions in the plane. To go from parametric to normal form, find \(\mathbf{n}\) with \(\mathbf{n}\cdot\mathbf{b} = 0\) and \(\mathbf{n}\cdot\mathbf{c} = 0\); to go from Cartesian to parametric, pick three points. The distance from \((x_{1}, y_{1}, z_{1})\) to \(ax + by + cz = d\) is \(\dfrac{|ax_{1} + by_{1} + cz_{1} - d|}{\sqrt{a^{2} + b^{2} + c^{2}}}\).

Π a r O n r − a A R (r − a) · n = 0
Fig. 1: the plane Π through the point A with normal n. For any point R in the plane, the vector r - a lies in the plane, so it is perpendicular to n: (r - a).n = 0, which rearranges to r.n = a.n
Section 01

A point and a normal give r.n = p.

A normal to a plane is a vector perpendicular to every line in the plane. Take a point \(A\) in the plane and a normal \(\mathbf{n}\). A point \(R\) lies in the plane exactly when \(\overrightarrow{AR} = \mathbf{r} - \mathbf{a}\) is perpendicular to \(\mathbf{n}\) (Fig. 1), so \((\mathbf{r} - \mathbf{a})\cdot\mathbf{n} = 0\).

Expanding the bracket gives the scalar product form:

\[\mathbf{r}\cdot\mathbf{n} = \mathbf{a}\cdot\mathbf{n}, \quad\text{written } \mathbf{r}\cdot\mathbf{n} = p.\]

Put \(\mathbf{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}\) and \(\mathbf{n} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}\): the scalar product is the Cartesian form

\[ax + by + cz = d.\]
Worked example From a point and a normal
Find, in the forms \(\mathbf{r}\cdot\mathbf{n} = p\) and \(ax + by + cz = d\), the equation of the plane through \(A(2, -1, 3)\) with normal \(\mathbf{i} + 4\mathbf{j} - 2\mathbf{k}\).
  1. \(p = \mathbf{a}\cdot\mathbf{n} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}\cdot\begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix} = 2 - 4 - 6 = -8\).
  2. Scalar product form: \(\mathbf{r}\cdot\begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix} = -8\).
  3. Cartesian form: \(x + 4y - 2z = -8\). Check with \(A\): \(2 - 4 - 6 = -8\).
Answer: \(\mathbf{r}\cdot(\mathbf{i} + 4\mathbf{j} - 2\mathbf{k}) = -8\), that is \(x + 4y - 2z = -8\)
Worked example Points and a parallel plane
The plane \(\Pi\) has equation \(\mathbf{r}\cdot\begin{pmatrix} 3 \\ -2 \\ 6 \end{pmatrix} = 12\). (a) Show that \(P(2, 3, 2)\) lies in \(\Pi\) and that \(Q(4, 0, 1)\) does not. (b) Find the equation of the plane parallel to \(\Pi\) through \((1, 1, 1)\).
  1. Cartesian form: \(3x - 2y + 6z = 12\).
  2. (a) \(P\): \(6 - 6 + 12 = 12\), so \(P\) lies in \(\Pi\). \(Q\): \(12 - 0 + 6 = 18 \ne 12\), so \(Q\) does not.
  3. (b) A parallel plane has the same normal: \(3x - 2y + 6z = d\) with \(d = 3 - 2 + 6 = 7\).
Answer: (a) \(P\) gives 12, \(Q\) gives 18; (b) \(3x - 2y + 6z = 7\)
Your turn
Find the Cartesian equation of the plane through \((0, 5, -2)\) perpendicular to the vector \(2\mathbf{i} + \mathbf{k}\).
Examiner trap

The number \(p\) in \(\mathbf{r}\cdot\mathbf{n} = p\) is \(\mathbf{a}\cdot\mathbf{n}\), not \(\mathbf{n}\cdot\mathbf{n}\) and not \(|\mathbf{a}|\). And the normal is made only of the coefficients of \(x\), \(y\), \(z\): the constant \(d\) is not part of it, and a point in the plane, such as \((0, 0, \tfrac{d}{c})\), is not a normal.

Section 02

Two directions in the plane: the parametric form.

A plane also contains whole lines. Starting at the point \(A\), move any amount \(\lambda\) along a direction \(\mathbf{b}\) and any amount \(\mu\) along a second direction \(\mathbf{c}\), where \(\mathbf{b}\) and \(\mathbf{c}\) are not parallel. Every point of the plane is reached this way.

\[\mathbf{r} = \mathbf{a} + \lambda\mathbf{b} + \mu\mathbf{c}.\] \[\mathbf{b}\times\mathbf{c} = \begin{pmatrix} b_{2}c_{3} - b_{3}c_{2} \\ b_{3}c_{1} - b_{1}c_{3} \\ b_{1}c_{2} - b_{2}c_{1} \end{pmatrix}.\]
Worked example Parametric to Cartesian
Find a Cartesian equation of the plane \(\mathbf{r} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} + \mu\begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix}\).
  1. Let \(\mathbf{n} = \begin{pmatrix} p \\ q \\ s \end{pmatrix}\). Then \(p + 2q - s = 0\) and \(3p + q + s = 0\).
  2. Add: \(4p + 3q = 0\). Take \(p = 3\), \(q = -4\); then \(s = p + 2q = -5\). Check the second: \(9 - 4 - 5 = 0\).
  3. \(\mathbf{n} = \begin{pmatrix} 3 \\ -4 \\ -5 \end{pmatrix}\) (the vector product gives the same), and \(\mathbf{a}\cdot\mathbf{n} = 3 + 0 - 10 = -7\).
Answer: \(3x - 4y - 5z = -7\)
Worked example The plane through three points
Find the equation of the plane through \(A(1, 2, 0)\), \(B(3, 1, 2)\) and \(C(2, 4, 1)\), in parametric and in Cartesian form.
  1. \(\overrightarrow{AB} = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}\), \(\overrightarrow{AC} = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}\), so \(\mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix} + \mu\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}\).
  2. Normal \(\begin{pmatrix} p \\ q \\ s \end{pmatrix}\): \(2p - q + 2s = 0\) and \(p + 2q + s = 0\). Double the second and subtract: \(5q = 0\), so \(q = 0\) and \(s = -p\). Take \(\mathbf{n} = \mathbf{i} - \mathbf{k}\).
  3. \(\mathbf{a}\cdot\mathbf{n} = 1 - 0 = 1\), so \(x - z = 1\).
  4. Check: \(B\) gives \(3 - 2 = 1\) and \(C\) gives \(2 - 1 = 1\).
Answer: \(x - z = 1\), or \(\mathbf{r} = (\mathbf{i} + 2\mathbf{j}) + \lambda(2\mathbf{i} - \mathbf{j} + 2\mathbf{k}) + \mu(\mathbf{i} + 2\mathbf{j} + \mathbf{k})\)
Your turn
Find a Cartesian equation of the plane \(\mathbf{r} = (2\mathbf{i} + \mathbf{j}) + \lambda(\mathbf{i} + \mathbf{k}) + \mu(\mathbf{j} + 2\mathbf{k})\).
Examiner trap

For a plane through \(A\), \(B\), \(C\), the directions are \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\) and \(\overrightarrow{AC} = \mathbf{c} - \mathbf{a}\), not the position vectors \(\mathbf{b}\) and \(\mathbf{c}\). Using \(\overrightarrow{OB}\) and \(\overrightarrow{OC}\) gives a plane through \(A\) parallel to the wrong pair of lines. Always check the final equation with all three points.

Stage 1 · Learn

Check your understanding

Four quick checks. Each correct answer earns XP and records this skill on your profile.

Quick check+10 XP

Which vector is normal to the plane \(3x - y + 4z = 7\)?

Quick check+10 XP

The plane through \((1, 2, -1)\) with normal \(2\mathbf{i} - 3\mathbf{j} + \mathbf{k}\) is

Quick check+10 XP

Which point lies in the plane \(x + 2y - z = 3\)?

Quick check+10 XP

Which vector is normal to the plane \(\mathbf{r} = (\mathbf{i} + \mathbf{j}) + \lambda(\mathbf{i} + 2\mathbf{k}) + \mu(\mathbf{j} - \mathbf{k})\)?

Section 03

Cartesian to parametric, and the distance to a plane.

To write \(ax + by + cz = d\) in parametric form, find three points in it that do not lie on one line (set two coordinates to convenient values and solve for the third), then use one point and the two vectors joining it to the others. Equivalently, let two of \(x\), \(y\), \(z\) be the parameters.

For distances, the formula list gives the perpendicular distance from the point \((x_{1}, y_{1}, z_{1})\) to the plane \(ax + by + cz = d\):

\[\frac{|ax_{1} + by_{1} + cz_{1} - d|}{\sqrt{a^{2} + b^{2} + c^{2}}}.\]
Worked example Cartesian to parametric
Write the plane \(2x + y - z = 4\) in the form \(\mathbf{r} = \mathbf{a} + \lambda\mathbf{b} + \mu\mathbf{c}\).
  1. Three points: \(x = y = 0\) gives \(z = -4\), so \(A(0, 0, -4)\); \(B(2, 0, 0)\); \(C(0, 4, 0)\).
  2. \(\overrightarrow{AB} = \begin{pmatrix} 2 \\ 0 \\ 4 \end{pmatrix}\) and \(\overrightarrow{AC} = \begin{pmatrix} 0 \\ 4 \\ 4 \end{pmatrix}\); halve and quarter them to \(\begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}\) and \(\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\).
  3. Check against the normal \(2\mathbf{i} + \mathbf{j} - \mathbf{k}\): \(2 - 2 = 0\) and \(1 - 1 = 0\), so both directions lie in the plane.
Answer: \(\mathbf{r} = \begin{pmatrix} 0 \\ 0 \\ -4 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + \mu\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\) (many correct answers)
Worked example Distances from the origin and from a point
The plane \(\Pi\) has equation \(2x - y + 2z = 9\). Find the distance of \(\Pi\) from the origin, and the perpendicular distance from \(P(4, -1, 5)\) to \(\Pi\).
  1. \(|\mathbf{n}| = \sqrt{4 + 1 + 4} = 3\).
  2. From the origin: \(\dfrac{|9|}{3} = 3\).
  3. From \(P\): \(\dfrac{|2(4) - (-1) + 2(5) - 9|}{3} = \dfrac{|8 + 1 + 10 - 9|}{3} = \dfrac{10}{3}\).
  4. \(O\) gives \(0 - 9 < 0\) and \(P\) gives \(19 - 9 > 0\): they are on opposite sides of \(\Pi\).
Answer: 3 from the origin, \(\dfrac{10}{3}\) from \(P\)
Your turn
Find the distance between the parallel planes \(x + 2y + 2z = 3\) and \(x + 2y + 2z = 12\).
Examiner trap

\(\mathbf{r}\cdot(2\mathbf{i} - \mathbf{j} + 2\mathbf{k}) = 9\) is not 9 units from the origin: divide by \(|\mathbf{n}| = 3\). Divide by \(\sqrt{a^{2} + b^{2} + c^{2}}\), not by \(a^{2} + b^{2} + c^{2}\), and keep the modulus signs so a distance is never negative.

Stage 2 · Exam

Exam-style questions

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

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Find the perpendicular distance from \(P(2, -1, 3)\) to the plane \(2x + y - 2z = 5\).

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Find the distance from the origin to the plane \(\mathbf{r}\cdot\begin{pmatrix} 6 \\ 2 \\ -3 \end{pmatrix} = 14\).

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Which is a vector equation of the plane \(3x - y + z = 5\)?

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Planes pinned down, in every form.

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Six original Cambridge-style questions
A plane from a point and a normal, a plane through three points, a parametric plane tested with points, a Cartesian plane rewritten, distances and sides, and a proof of the distance formula.