Why is "Treating $d$ as the value from the origin or forgetting to compute $\mathbf{a}\cdot\mathbf{n}$" a common mistake in Equations of Planes?

Why it happens: Confusing the constant d with a coordinate; not realising d depends on which point is on the plane. How to avoid it: Always compute d = a·n using a KNOWN point on the plane, then write r·n = d. Source: 9231 Paper 1 Examiner Reports

Why is "Using $\cos$ instead of $\sin$ for the angle between a line and a plane" a common mistake in Equations of Planes?

Why it happens: Applying the standard = a·b|a||b| without adjusting for the normal. How to avoid it: The angle to the plane is the COMPLEMENT of the angle to the normal, so use = |d·n||d||n|. Source: 9231 Paper 1 Examiner Reports

Why is "Reporting the obtuse angle between two planes instead of the acute angle" a common mistake in Equations of Planes?

Why it happens: A negative dot product gives an obtuse , which is left unchanged. How to avoid it: Take the modulus of the dot product, or subtract from 180^ to get the acute angle when asked for it. Source: 9231 Paper 1 Examiner Reports

Why is "Choosing two parallel direction vectors for the parametric form" a common mistake in Equations of Planes?

Why it happens: Picking b and c that are scalar multiples, which collapse to a line, not a plane. How to avoid it: Ensure b and c are non-parallel; their cross product must be non-zero. Source: 9231 Paper 1 — parametric form questions

Why is "Failing to read the normal straight off the Cartesian coefficients" a common mistake in Equations of Planes?

Why it happens: Not knowing that (a, b, c) in ax+by+cz=d IS the normal vector. How to avoid it: Memorise: in ax+by+cz=d, the normal is (a, b, c) — no derivation needed. Source: 9231 Paper 1 mark schemes

Cambridge International A Levels Further Mathematics (9231)

Equations of Planes

Q: Why is "Sign or component errors in the cross product $\overrightarrow{AB}\times\overrightarrow{AC}$" a common mistake in Equations of Planes?

Why it happens: The middle component of the cross product carries a sign flip that is easy to drop. How to avoid it: Use the formula carefully and VERIFY by checking a second point satisfies the final equation. Source: 9231 Paper 1 mark schemes

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Detailed Study Notes

Detailed notes on Vectors for Cambridge International A Levels Further Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Equations of Planes — Cambridge International AS & A Level Further Mathematics 9231 Paper 1 (2026-2027 syllabus)

A plane is fixed by a point on it and a direction perpendicular to it (the normal). Master the three forms — scalar product $\mathbf{r}\cdot\mathbf{n}=d$ , Cartesian $ax+by+cz=d$ , and parametric $\mathbf{r}=\mathbf{a}+\lambda\mathbf{b}+\mu\mathbf{c}$ — and switch fluently between them.

At a glance

A plane needs a point and a normal $\mathbf{n}$ (a direction perpendicular to the whole plane).
Scalar product form: $\mathbf{r}\cdot\mathbf{n} = d$ , where $d = \mathbf{a}\cdot\mathbf{n}$ for any known point $\mathbf{a}$ .
Cartesian form: $ax + by + cz = d$ — the coefficients $(a, b, c)$ ARE the normal vector $\mathbf{n}$ .
Parametric form: $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b} + \mu\mathbf{c}$ uses a point and two non-parallel direction vectors lying in the plane.
Three points define a plane: take two difference vectors, cross them to get $\mathbf{n}$ , then build $\mathbf{r}\cdot\mathbf{n}=d$ .

What you’ll learn

Mapped to the Cambridge International A Level 9231 syllabus (2026-2027).

1.6a — Understand the equation of a plane in scalar product form $\mathbf{r}\cdot\mathbf{n}=d$ , Cartesian form $ax+by+cz=d$ , and parametric form $\mathbf{r}=\mathbf{a}+\lambda\mathbf{b}+\mu\mathbf{c}$ .
1.6b — Convert a plane between these three forms, recognising that the Cartesian coefficients are the components of the normal vector.
1.6c — Find the equation of a plane through three given points, or through a given point with a given normal direction.

The big idea — a point plus a normal pins down a plane

Choose a normal direction and one point; every point of the plane satisfies r·n = a·n.

The key idea. A line in 2D needs a point and a direction. A plane in 3D needs a point and a normal — a vector $\mathbf{n}$ perpendicular to the entire plane. Once $\mathbf{n}$ is fixed, the plane is the set of all points whose position vector $\mathbf{r}$ makes a right angle with $\mathbf{n}$ when measured from a known point $\mathbf{a}$ on the plane.

Deriving the scalar product form. Let $\mathbf{a}$ be the position vector of a fixed point on the plane and $\mathbf{r}$ the position vector of any point on it. The displacement $\mathbf{r} - \mathbf{a}$ lies in the plane, so it is perpendicular to $\mathbf{n}$ :

$(\mathbf{r} - \mathbf{a}) \cdot \mathbf{n} = 0 \;\Rightarrow\; \mathbf{r}\cdot\mathbf{n} = \mathbf{a}\cdot\mathbf{n}$

The right-hand side is a fixed number, usually written $d$ :

$\boxed{\mathbf{r}\cdot\mathbf{n} = d, \qquad d = \mathbf{a}\cdot\mathbf{n}}$

The normal n stands perpendicular to the plane. Any displacement r − a within the plane is perpendicular to n, giving r·n = a·n = d.

Cambridge tip. The right angle between $\mathbf{n}$ and the plane is the whole point — if you can find ANY vector lying in the plane, dotting it with your candidate normal must give $0$ . Use this as a 10-second self-check.

A plane = one point $\mathbf{a}$ + a normal $\mathbf{n}$ perpendicular to it.
$(\mathbf{r}-\mathbf{a})\cdot\mathbf{n}=0$ collapses to $\mathbf{r}\cdot\mathbf{n}=d$ with $d=\mathbf{a}\cdot\mathbf{n}$ .
Any in-plane vector dotted with $\mathbf{n}$ must give $0$ .

See the full worked example for equations of planes →

The three forms and how they connect

Scalar product, Cartesian and parametric describe the same plane — the Cartesian coefficients ARE the normal.

A single plane can be written three ways. The exam expects you to recognise all three and move between them.

1. Scalar product form. $\mathbf{r}\cdot\mathbf{n} = d$ . Compact and ideal for finding distances and angles. Here $\mathbf{n}$ is the normal and $d = \mathbf{a}\cdot\mathbf{n}$ .

2. Cartesian form. Write $\mathbf{r} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$ and $\mathbf{n} = \begin{pmatrix} a \\ b \\ c \end{pmatrix}$ . Then $\mathbf{r}\cdot\mathbf{n} = d$ becomes:

$ax + by + cz = d$

Read this carefully: the numbers $a, b, c$ multiplying $x, y, z$ are exactly the components of the normal vector. So the plane $2x - y + 3z = 7$ has normal $\mathbf{n} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$ — you read it straight off.

3. Parametric (vector) form. $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b} + \mu\mathbf{c}$ , where $\mathbf{a}$ is a point on the plane and $\mathbf{b}, \mathbf{c}$ are two non-parallel direction vectors lying in the plane. As $\lambda$ and $\mu$ range over all real numbers, $\mathbf{r}$ sweeps out the whole plane.

Parametric form uses a point a and two in-plane directions b, c. Their cross product b × c gives the normal — the bridge to the scalar-product and Cartesian forms.

Converting parametric → scalar/Cartesian. Compute $\mathbf{n} = \mathbf{b} \times \mathbf{c}$ (the cross product of the two direction vectors — see subtopic 6.2), then $d = \mathbf{a}\cdot\mathbf{n}$ . Converting Cartesian → scalar is immediate: read off $\mathbf{n} = (a,b,c)$ and the constant is $d$ .

Cambridge tip. When a plane is given in Cartesian form, you almost never need to convert it — just lift the normal straight off the $x, y, z$ coefficients. Most marks are thrown away by students who needlessly re-derive what is sitting in front of them.

$ax+by+cz=d$ — the coefficients $(a,b,c)$ ARE the normal $\mathbf{n}$ .
Parametric $\mathbf{r}=\mathbf{a}+\lambda\mathbf{b}+\mu\mathbf{c}$ needs two non-parallel in-plane directions.
Parametric → scalar: $\mathbf{n}=\mathbf{b}\times\mathbf{c}$ , then $d=\mathbf{a}\cdot\mathbf{n}$ .

See the full worked example for equations of planes →

Finding a plane through three points

Two difference vectors, cross them for the normal, then dot with one point for d.

This is the single most common "find the plane" task in Paper 1. Given three points $A$ , $B$ , $C$ (not collinear), follow a fixed routine.

Step 1 — make two in-plane vectors. Both $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ and $\overrightarrow{AC} = \mathbf{c} - \mathbf{a}$ lie in the plane.

Step 2 — cross them for the normal. Since both lie in the plane, their cross product is perpendicular to the plane:

$\mathbf{n} = \overrightarrow{AB} \times \overrightarrow{AC}$

Step 3 — find $d$ . Substitute any one of the three points (say $A$ ) into $\mathbf{r}\cdot\mathbf{n} = d$ :

$d = \mathbf{a}\cdot\mathbf{n}$

Step 4 — write the answer in the form requested (scalar product or Cartesian).

Three non-collinear points define a plane. Vectors AB and AC lie in it; their cross product is the normal n, and d = a·n.

Special case — point with given normal. If you are simply given a point $\mathbf{a}$ and the normal $\mathbf{n}$ directly, skip straight to $d = \mathbf{a}\cdot\mathbf{n}$ and write $\mathbf{r}\cdot\mathbf{n} = d$ .

Cambridge tip. Always verify by checking that the other two points also satisfy your equation. If $A$ gives $d=14$ but $B$ gives $13$ , you have an arithmetic slip in the cross product — catch it before it costs you.

$\mathbf{n} = \overrightarrow{AB}\times\overrightarrow{AC}$ from two difference vectors.
$d = \mathbf{a}\cdot\mathbf{n}$ using any one of the three points.
Check all three points satisfy the final equation.

See the full worked example for equations of planes →

Quick recap

A plane is fixed by one point $\mathbf{a}$ and a normal $\mathbf{n}$ .
Scalar product form: $\mathbf{r}\cdot\mathbf{n} = d$ with $d = \mathbf{a}\cdot\mathbf{n}$ .
Cartesian form $ax+by+cz=d$ — read the normal $(a,b,c)$ straight off the coefficients.
Parametric form $\mathbf{r}=\mathbf{a}+\lambda\mathbf{b}+\mu\mathbf{c}$ uses two non-parallel in-plane directions.
Three points: $\mathbf{n}=\overrightarrow{AB}\times\overrightarrow{AC}$ , then $d=\mathbf{a}\cdot\mathbf{n}$ .
Always check the other points satisfy your plane equation.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

$\mathbf{r}\cdot\mathbf{n} = d$ with $d = \mathbf{a}\cdot\mathbf{n}$ (scalar product form).
$ax+by+cz=d$ — coefficients $(a,b,c)$ are the normal $\mathbf{n}$ .
$\mathbf{r}=\mathbf{a}+\lambda\mathbf{b}+\mu\mathbf{c}$ (parametric form).
$\mathbf{n}=\overrightarrow{AB}\times\overrightarrow{AC}$ for a plane through three points.
$(\mathbf{r}-\mathbf{a})\cdot\mathbf{n}=0$ — the underlying perpendicularity condition.

How it’s examined

Equations of planes (1.6) appear in most Paper 1 vector questions (5-10 marks). A typical question gives three points and asks for the plane in scalar product or Cartesian form, then flows into distances, angles, or line-plane intersections. The cross product $\overrightarrow{AB}\times\overrightarrow{AC}$ is where most method marks live — and where most arithmetic slips occur. Examiners reward candidates who quote the form correctly, evaluate $d=\mathbf{a}\cdot\mathbf{n}$ explicitly, and verify with a second point.

Sources: Cambridge International AS & A Level Further Mathematics 9231 syllabus (2026-2027); 9231 Examiner Reports 2022-2024 (Paper 1); 9231/11 & 9231/13 Further Pure Mathematics 1 — past papers and mark schemes. Last reviewed 2026-06-23.

Take this whole topic with you

Step-by-step worked examples — Equations of Planes

Step-by-step solutions to past-paper-style questions on equations of planes, written exactly the way a tutor would explain them at the board.

Question type:

1Plane through a point with a given normal
Getting startedDirect calculation• scalar-product, normal
Question
Find, in scalar product form, the equation of the plane through the point $A(2, -1, 4)$ with normal vector $\mathbf{n} = \begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix}$ .
Step-by-step solution
1. Step 1
  Recall the form $\mathbf{r}\cdot\mathbf{n} = d$ where $d = \mathbf{a}\cdot\mathbf{n}$ and $\mathbf{a}$ is the position vector of $A$ .
2. Step 2
  Compute $d = \mathbf{a}\cdot\mathbf{n}$ component by component.
  $d = (2)(3) + (-1)(1) + (4)(-2) = 6 - 1 - 8 = -3$
3. Step 3
  Write the plane in scalar product form.
  $\mathbf{r}\cdot\begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix} = -3$
Answer
$\mathbf{r}\cdot\begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix} = -3$ , equivalently $3x + y - 2z = -3$ .
Examiner tip
The whole task is one dot product. Beginners forget that $d$ is found by dotting the KNOWN point with the normal — not by guessing.
2Reading the normal off a Cartesian plane
Getting startedIdentify & classify• cartesian, normal
Question
A plane has Cartesian equation $4x - 2y + z = 9$ . (a) Write down a normal vector. (b) Verify that the point $(1, -1, 3)$ lies on the plane.
Step-by-step solution
1. Step 1
  Read the normal off the coefficients. In $ax+by+cz=d$ the normal is $(a, b, c)$ .
  $\mathbf{n} = \begin{pmatrix} 4 \\ -2 \\ 1 \end{pmatrix}$
2. Step 2
  Substitute the point into the left-hand side $4x - 2y + z$ .
  $4(1) - 2(-1) + (3) = 4 + 2 + 3 = 9$
3. Step 3
  Compare with the right-hand side. It equals $9$ , so the point satisfies the equation and lies on the plane.
Answer
(a) $\mathbf{n} = \begin{pmatrix} 4 \\ -2 \\ 1 \end{pmatrix}$ . (b) LHS $= 9 =$ RHS, so the point lies on the plane.
Examiner tip
The coefficients of $x, y, z$ ARE the normal — no working needed. Verification is a substitution, not a re-derivation.
3Plane through three points
Building confidenceMulti-step problem• three-points, cross-product
Question
Find, in scalar product form, the equation of the plane through $A(1, 0, 2)$ , $B(3, 1, 1)$ and $C(2, -1, 4)$ .
Step-by-step solution
1. Step 1
  Two in-plane vectors from $A$ .
  $\overrightarrow{AB} = \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}, \quad \overrightarrow{AC} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}$
2. Step 2
  Cross product gives the normal $\mathbf{n} = \overrightarrow{AB}\times\overrightarrow{AC}$ .
  $\mathbf{n} = \begin{pmatrix} (1)(2)-(-1)(-1) \\ (-1)(1)-(2)(2) \\ (2)(-1)-(1)(1) \end{pmatrix} = \begin{pmatrix} 1 \\ -5 \\ -3 \end{pmatrix}$
3. Step 3
  Find $d = \mathbf{a}\cdot\mathbf{n}$ using point $A$ .
  $d = (1)(1) + (0)(-5) + (2)(-3) = 1 + 0 - 6 = -5$
4. Step 4
  Check with $B$ : $(3)(1)+(1)(-5)+(1)(-3) = 3 - 5 - 3 = -5$ . ✓
Answer
$\mathbf{r}\cdot\begin{pmatrix} 1 \\ -5 \\ -3 \end{pmatrix} = -5$ .
Examiner tip
Verifying a second point catches cross-product slips for free. The cross product is where method marks are won or lost.
4Parametric form to Cartesian form
Building confidenceMulti-step problem• parametric, cartesian
Question
A plane is given parametrically by $\mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} + \mu\begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$ . Find its Cartesian equation.
Step-by-step solution
1. Step 1
  The two direction vectors lie in the plane, so the normal is their cross product.
  $\mathbf{n} = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} \times \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$
2. Step 2
  Evaluate the cross product component by component.
  $\mathbf{n} = \begin{pmatrix} (0)(-1)-(1)(1) \\ (1)(2)-(1)(-1) \\ (1)(1)-(0)(2) \end{pmatrix} = \begin{pmatrix} -1 \\ 3 \\ 1 \end{pmatrix}$
3. Step 3
  Find $d = \mathbf{a}\cdot\mathbf{n}$ with $\mathbf{a} = (1, 2, 0)$ .
  $d = (1)(-1) + (2)(3) + (0)(1) = -1 + 6 + 0 = 5$
4. Step 4
  Write the Cartesian equation $ax + by + cz = d$ .
  $-x + 3y + z = 5$
Answer
$-x + 3y + z = 5$ (equivalently $x - 3y - z = -5$ ).
Examiner tip
Either sign of the normal is acceptable provided $d$ matches it. The cross product converts the two directions into a single normal.
5Where a line meets a plane
StretchMulti-step problem• intersection, line-plane
Question
Find the point where the line $\mathbf{r} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} + t\begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$ meets the plane $3x - y + 2z = 8$ .
Step-by-step solution
1. Step 1
  Write the line's coordinates in terms of $t$ : $x = 1+2t$ , $y = -1+t$ , $z = 2-t$ .
2. Step 2
  Substitute into the plane equation $3x - y + 2z = 8$ .
  $3(1+2t) - (-1+t) + 2(2-t) = 8$
3. Step 3
  Expand and solve for $t$ .
  $3 + 6t + 1 - t + 4 - 2t = 8 \;\Rightarrow\; 8 + 3t = 8 \;\Rightarrow\; t = 0$
4. Step 4
  Substitute $t = 0$ back into the line to find the point.
  $\mathbf{r} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}$
Answer
The line meets the plane at $(1, -1, 2)$ .
Examiner tip
Substituting the parametric coordinates into the Cartesian plane gives one equation in $t$ — the cleanest route. Here the starting point already lies on the plane.
6Cartesian form to parametric form
StretchMulti-step problem• parametric, cartesian
Question
Express the plane $2x + y - z = 4$ in the parametric form $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b} + \mu\mathbf{c}$ .
Step-by-step solution
1. Step 1
  Choose two free parameters. Let $y = \lambda$ and $z = \mu$ , then solve for $x$ .
  $x = \frac{4 - y + z}{2} = 2 - \tfrac{1}{2}\lambda + \tfrac{1}{2}\mu$
2. Step 2
  Write the coordinate vector in terms of $\lambda, \mu$ .
  $\mathbf{r} = \begin{pmatrix} 2 - \tfrac12\lambda + \tfrac12\mu \\ \lambda \\ \mu \end{pmatrix}$
3. Step 3
  Group the constant, $\lambda$ and $\mu$ parts to read off $\mathbf{a}, \mathbf{b}, \mathbf{c}$ .
  $\mathbf{r} = \begin{pmatrix} 2 \\ 0 \\ 0 \end{pmatrix} + \lambda\begin{pmatrix} -\tfrac12 \\ 1 \\ 0 \end{pmatrix} + \mu\begin{pmatrix} \tfrac12 \\ 0 \\ 1 \end{pmatrix}$
4. Step 4
  Optional tidy-up: scale the direction vectors to clear fractions, e.g. $\mathbf{b} = (-1, 2, 0)$ and $\mathbf{c} = (1, 0, 2)$ .
Answer
$\mathbf{r} = \begin{pmatrix} 2 \\ 0 \\ 0 \end{pmatrix} + \lambda\begin{pmatrix} -1 \\ 2 \\ 0 \end{pmatrix} + \mu\begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}$ .
Examiner tip
Any valid point and any two non-parallel in-plane directions are accepted. A quick check: each direction vector must satisfy $\mathbf{n}\cdot\mathbf{b} = 0$ — here $(2,1,-1)\cdot(-1,2,0) = 0$ . ✓

Model Answers — Equations of Planes

High-scoring sample answers for equations of planes on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
3 marks
Find the equation of the plane through the point $(4, 1, -2)$ with normal $\mathbf{n} = \begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix}$ , giving your answer in the form $ax + by + cz = d$ .
Model answer
Find $d = \mathbf{a}\cdot\mathbf{n}$ using the given point: $d = (4)(1) + (1)(-3) + (-2)(2) = 4 - 3 - 4 = -3$

The normal supplies the coefficients, so the plane is: $x - 3y + 2z = -3$
Why this scores
M1 for $d = \mathbf{a}\cdot\mathbf{n}$ attempted, A1 for $d = -3$ , A1 for the correct Cartesian equation. The coefficients of $x, y, z$ are read straight from $\mathbf{n}$ .
Question 2
3 marks
The plane $\Pi$ has equation $5x + 2y - 3z = 1$ . (a) State a vector normal to $\Pi$ . (b) Determine whether the point $P(2, -1, 1)$ lies on $\Pi$ .
Model answer
(a) The coefficients of $x, y, z$ are the components of the normal: $\mathbf{n} = \begin{pmatrix} 5 \\ 2 \\ -3 \end{pmatrix}$

(b) Substitute $P(2, -1, 1)$ into the left-hand side: $5(2) + 2(-1) - 3(1) = 10 - 2 - 3 = 5$ Since $5 \neq 1$ , the point $P$ does not lie on $\Pi$ .
Why this scores
Part (a): B1 for the normal. Part (b): M1 for substituting $P$ into the equation, A1 for the correct conclusion with justification ( $5 \neq 1$ ). A bare 'no' without the value scores no A mark.
Question 3
6 marks
The points $A(2, 0, 1)$ , $B(1, 3, 2)$ and $C(4, 1, -1)$ lie on a plane $\Pi$ . Find the equation of $\Pi$ in the form $\mathbf{r}\cdot\mathbf{n} = d$ .
Model answer
Two direction vectors in the plane: $\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} -1 \\ 3 \\ 1 \end{pmatrix}, \qquad \overrightarrow{AC} = \mathbf{c} - \mathbf{a} = \begin{pmatrix} 2 \\ 1 \\ -2 \end{pmatrix}$

Normal = cross product $\overrightarrow{AB}\times\overrightarrow{AC}$ : $\mathbf{n} = \begin{pmatrix} (3)(-2)-(1)(1) \\ (1)(2)-(-1)(-2) \\ (-1)(1)-(3)(2) \end{pmatrix} = \begin{pmatrix} -7 \\ 0 \\ -7 \end{pmatrix}$

Dividing by $-7$ gives the simpler normal $\mathbf{n} = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}$ .

Find $d = \mathbf{a}\cdot\mathbf{n}$ with $A(2, 0, 1)$ : $d = (2)(1) + (0)(0) + (1)(1) = 3$

Check with $B$ : $(1)(1) + (3)(0) + (2)(1) = 3$ . ✓

$\mathbf{r}\cdot\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = 3$
Why this scores
M1 for two correct difference vectors, M1 A1 for the cross product, M1 for $d = \mathbf{a}\cdot\mathbf{n}$ , A1 for $d$ , A1 for the fully correct plane. Simplifying the normal is good practice but not required for the marks.
Question 4
5 marks
A plane is given by $\mathbf{r} = \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} + \mu\begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix}$ . Find its Cartesian equation.
Model answer
The two direction vectors lie in the plane, so the normal is their cross product: $\mathbf{n} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \times \begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} (1)(1)-(0)(2) \\ (0)(0)-(1)(1) \\ (1)(2)-(1)(0) \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}$

Find $d = \mathbf{a}\cdot\mathbf{n}$ with $\mathbf{a} = (0, 1, 2)$ : $d = (0)(1) + (1)(-1) + (2)(2) = -1 + 4 = 3$

Cartesian equation: $x - y + 2z = 3$
Why this scores
M1 A1 for the cross product, M1 for $d = \mathbf{a}\cdot\mathbf{n}$ , A1 for $d = 3$ , A1 for the equation. A useful check: each direction vector dotted with $\mathbf{n}$ must give $0$ (e.g. $(1,1,0)\cdot(1,-1,2) = 0$ ).
Question 5
8 marks
The line $\ell$ has equation $\mathbf{r} = \begin{pmatrix} 2 \\ 0 \\ -1 \end{pmatrix} + t\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$ and the plane $\Pi$ has equation $2x - y + z = 5$ . (a) Find the point of intersection of $\ell$ and $\Pi$ . (b) Find the acute angle between $\ell$ and $\Pi$ , giving your answer to $1$ decimal place.
Model answer
(a) Line coordinates: $x = 2 + t$ , $y = 2t$ , $z = -1 + 3t$ . Substitute into $\Pi$ : $2(2+t) - (2t) + (-1+3t) = 5$ $4 + 2t - 2t - 1 + 3t = 5 \;\Rightarrow\; 3 + 3t = 5 \;\Rightarrow\; t = \tfrac{2}{3}$ Substituting $t = \tfrac23$ back into $\ell$ : $\mathbf{r} = \begin{pmatrix} 2 + \tfrac23 \\ \tfrac43 \\ -1 + 2 \end{pmatrix} = \begin{pmatrix} \tfrac83 \\ \tfrac43 \\ 1 \end{pmatrix}$

(b) The angle $\theta$ between the line direction $\mathbf{d} = (1, 2, 3)$ and the plane normal $\mathbf{n} = (2, -1, 1)$ satisfies $\sin\alpha = \frac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}$ where $\alpha$ is the angle between the line and the plane. $\mathbf{d}\cdot\mathbf{n} = (1)(2)+(2)(-1)+(3)(1) = 3$ $|\mathbf{d}| = \sqrt{1+4+9} = \sqrt{14}, \qquad |\mathbf{n}| = \sqrt{4+1+1} = \sqrt{6}$ $\sin\alpha = \frac{3}{\sqrt{14}\,\sqrt{6}} = \frac{3}{\sqrt{84}} = 0.3273\ldots$ $\alpha = 19.1^\circ \text{ (1 d.p.)}$
Why this scores
(a) M1 substitute into plane, A1 for $t = \tfrac23$ , A1 for the point. (b) B1 for using $\sin$ (not $\cos$ ) of the angle between line and plane, M1 for $\frac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}$ , A1 for $0.327$ , A1 for $19.1^\circ$ . The classic error is using $\cos$ and reporting the complement $70.9^\circ$ .
Question 6
5 marks
Two planes have equations $\Pi_1: x + 2y - 2z = 1$ and $\Pi_2: 2x - y + 2z = 4$ . Find the acute angle between the planes, giving your answer to $1$ decimal place.
Model answer
The angle between two planes equals the angle between their normals. $\mathbf{n_1} = \begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix}, \qquad \mathbf{n_2} = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$

Dot product and magnitudes: $\mathbf{n_1}\cdot\mathbf{n_2} = (1)(2)+(2)(-1)+(-2)(2) = 2 - 2 - 4 = -4$ $|\mathbf{n_1}| = \sqrt{1+4+4} = 3, \qquad |\mathbf{n_2}| = \sqrt{4+1+4} = 3$

Angle between normals: $\cos\theta = \frac{\mathbf{n_1}\cdot\mathbf{n_2}}{|\mathbf{n_1}||\mathbf{n_2}|} = \frac{-4}{9} = -0.4444\ldots \;\Rightarrow\; \theta = 116.4^\circ$

Take the acute angle (the planes meet at the supplement): $180^\circ - 116.4^\circ = 63.6^\circ \text{ (1 d.p.)}$
Why this scores
M1 for both normals, M1 for $\cos\theta = \frac{\mathbf{n_1}\cdot\mathbf{n_2}}{|\mathbf{n_1}||\mathbf{n_2}|}$ , A1 for $\pm\frac{4}{9}$ , M1 for taking the acute value, A1 for $63.6^\circ$ . Because the dot product is negative, the obtuse angle appears first — the acute angle is its supplement.

Key Formulae — Equations of Planes

The formulae you need to memorise for equations of planes on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Plane — scalar product form
$\mathbf{r}\cdot\mathbf{n} = d, \qquad d = \mathbf{a}\cdot\mathbf{n}$
$\mathbf{r}$
position vector of a general point on the plane
$\mathbf{n}$
a normal vector (perpendicular to the plane)
$\mathbf{a}$
position vector of a known point on the plane
$d$
constant equal to $\mathbf{a}\cdot\mathbf{n}$
When to use
Whenever you have a point and a normal — the most flexible form for angles and distances.
Plane — Cartesian form
$ax + by + cz = d$
$(a, b, c)$
components of the normal vector $\mathbf{n}$
$d$
constant $= \mathbf{a}\cdot\mathbf{n}$
When to use
Substituting coordinates to test points; the coefficients are read directly as the normal.
Plane — parametric (vector) form
$\mathbf{r} = \mathbf{a} + \lambda\mathbf{b} + \mu\mathbf{c}$
$\mathbf{a}$
position vector of a point on the plane
$\mathbf{b}, \mathbf{c}$
two non-parallel direction vectors lying in the plane
$\lambda, \mu$
real parameters ranging over all reals
When to use
When a plane is generated from a point and two in-plane directions.
Normal from three points
$\mathbf{n} = \overrightarrow{AB} \times \overrightarrow{AC}$
$\overrightarrow{AB}, \overrightarrow{AC}$
two difference vectors in the plane
$\mathbf{n}$
the resulting normal vector
When to use
Finding the plane through three non-collinear points.
Angle between a line and a plane
$\sin\alpha = \frac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}$
$\mathbf{d}$
direction vector of the line
$\mathbf{n}$
normal vector of the plane
$\alpha$
acute angle between the line and the plane
When to use
Use $\sin$ (not $\cos$ ) because $\alpha$ is measured to the plane, not its normal.
Angle between two planes
$\cos\theta = \frac{\mathbf{n_1}\cdot\mathbf{n_2}}{|\mathbf{n_1}||\mathbf{n_2}|}$
$\mathbf{n_1}, \mathbf{n_2}$
normals of the two planes
$\theta$
angle between the planes (take the acute value)
When to use
The angle between two planes equals the angle between their normals.

Key Definitions and Keywords — Equations of Planes

Definitions to memorise and the exact keywords mark schemes credit for equations of planes answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Plane
Examiner keyword
A flat, two-dimensional surface extending infinitely in 3D space, fixed by one point on it and a normal direction perpendicular to it.
Normal vector $\mathbf{n}$
Examiner keyword
A vector perpendicular to every line lying in the plane. In Cartesian form $ax+by+cz=d$ , the normal is $(a, b, c)$ .
Scalar product form
Examiner keyword
$\mathbf{r}\cdot\mathbf{n} = d$ , where $d = \mathbf{a}\cdot\mathbf{n}$ for any point $\mathbf{a}$ on the plane.
Parametric (vector) form
Examiner keyword
$\mathbf{r} = \mathbf{a} + \lambda\mathbf{b} + \mu\mathbf{c}$ , a point plus two non-parallel in-plane direction vectors scaled by parameters $\lambda, \mu$ .
Cartesian form of a plane
Examiner keyword
$ax + by + cz = d$ , obtained by writing the scalar product $\mathbf{r}\cdot\mathbf{n} = d$ in components.
Collinear / non-collinear points
Points are collinear if they lie on a single straight line. Three non-collinear points uniquely define a plane; three collinear points do not.

Common Mistakes and Misconceptions — Equations of Planes

The traps other students keep falling into on equations of planes questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Treating $d$ as the value from the origin or forgetting to compute $\mathbf{a}\cdot\mathbf{n}$
9231 Paper 1 Examiner Reports
Why it happens
Confusing the constant $d$ with a coordinate; not realising $d$ depends on which point is on the plane.
How to avoid it
Always compute $d = \mathbf{a}\cdot\mathbf{n}$ using a KNOWN point on the plane, then write $\mathbf{r}\cdot\mathbf{n} = d$ .
✕Sign or component errors in the cross product $\overrightarrow{AB}\times\overrightarrow{AC}$
9231 Paper 1 mark schemes
Why it happens
The middle component of the cross product carries a sign flip that is easy to drop.
How to avoid it
Use the formula carefully and VERIFY by checking a second point satisfies the final equation.
✕Using $\cos$ instead of $\sin$ for the angle between a line and a plane
9231 Paper 1 Examiner Reports
Why it happens
Applying the standard $\cos\theta = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$ without adjusting for the normal.
How to avoid it
The angle to the plane is the COMPLEMENT of the angle to the normal, so use $\sin\alpha = \frac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}$ .
✕Reporting the obtuse angle between two planes instead of the acute angle
9231 Paper 1 Examiner Reports
Why it happens
A negative dot product gives an obtuse $\theta$ , which is left unchanged.
How to avoid it
Take the modulus of the dot product, or subtract from $180^\circ$ to get the acute angle when asked for it.
✕Choosing two parallel direction vectors for the parametric form
9231 Paper 1 — parametric form questions
Why it happens
Picking $\mathbf{b}$ and $\mathbf{c}$ that are scalar multiples, which collapse to a line, not a plane.
How to avoid it
Ensure $\mathbf{b}$ and $\mathbf{c}$ are non-parallel; their cross product must be non-zero.
✕Failing to read the normal straight off the Cartesian coefficients
9231 Paper 1 mark schemes
Why it happens
Not knowing that $(a, b, c)$ in $ax+by+cz=d$ IS the normal vector.
How to avoid it
Memorise: in $ax+by+cz=d$ , the normal is $(a, b, c)$ — no derivation needed.

Equations of Planes

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Plane through a point with a given normal

2Reading the normal off a Cartesian plane

3Plane through three points

4Parametric form to Cartesian form

5Where a line meets a plane

6Cartesian form to parametric form

Plane — scalar product form

Plane — Cartesian form

Plane — parametric (vector) form

Normal from three points

Angle between a line and a plane

Angle between two planes

Plane

Normal vector n\mathbf{n}n

Scalar product form

Parametric (vector) form

Cartesian form of a plane

Collinear / non-collinear points

✕Treating ddd as the value from the origin or forgetting to compute a⋅n\mathbf{a}\cdot\mathbf{n}a⋅n

✕Sign or component errors in the cross product AB→×AC→\overrightarrow{AB}\times\overrightarrow{AC}AB×AC

✕Using cos⁡\coscos instead of sin⁡\sinsin for the angle between a line and a plane

✕Reporting the obtuse angle between two planes instead of the acute angle

✕Choosing two parallel direction vectors for the parametric form

✕Failing to read the normal straight off the Cartesian coefficients

Normal vector $\mathbf{n}$

✕Treating $d$ as the value from the origin or forgetting to compute $\mathbf{a}\cdot\mathbf{n}$

✕Sign or component errors in the cross product $\overrightarrow{AB}\times\overrightarrow{AC}$

✕Using $\cos$ instead of $\sin$ for the angle between a line and a plane