A plane $\Pi$ passes through the point $A(1, 2, -1)$ and has a normal vector $\mathbf{n} = \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix}$. Find the Cartesian equation of the plane.

1. Identify the standard scalar product form for the equation of a plane: $\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}$. 2. Substitute the position vector of point $A$ and the normal vector into the right-hand side: $\mathbf{a} \cdot \mathbf{n} = (1)(3) + (2)(-1) + (-1)(2)$. 3. Calculate the scalar product constant: $3 - 2 - 2 = -1$. 4. Expand the left-hand side $\mathbf{r} \cdot \mathbf{n}$ using $\mathbf{r} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$ to get $3x - y + 2z$. 5. State the final Cartesian equation: $\mathbf{3x - y + 2z = -1}$.

Find the exact coordinates of the point of intersection between the line $L: \mathbf{r} = \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix} + t \begin{pmatrix} 1 \\ -1 \\ 3 \end{pmatrix}$ and the plane $\Pi: 2x + y - z = 5$.

1. Express the line equations parametrically in terms of $t$: $x = t$, $y = 1 - t$, and $z = 2 + 3t$. 2. Substitute these parametric expressions for $x$, $y$, and $z$ directly into the Cartesian equation of the plane: $2(t) + (1 - t) - (2 + 3t) = 5$. 3. Expand the brackets carefully: $2t + 1 - t - 2 - 3t = 5$. 4. Simplify and collect like terms to solve for the parameter $t$: $-2t - 1 = 5 \implies -2t = 6 \implies t = -3$. 5. Substitute $t = -3$ back into the original parametric equations of the line: $x = -3$, $y = 1 - (-3) = 4$, $z = 2 + 3(-3) = -7$. 6. State the final exact coordinates of the intersection point: $\mathbf{(-3, 4, -7)}$.

Find the acute angle between the planes $\Pi_1: x + 2y - z = 4$ and $\Pi_2: 2x - y + 3z = 7$. Give your answer in degrees to one decimal place.

1. Extract the normal vectors for each plane from their Cartesian coefficients: $\mathbf{n}_1 = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$ and $\mathbf{n}_2 = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$. 2. Calculate the scalar product of the normal vectors: $\mathbf{n}_1 \cdot \mathbf{n}_2 = (1)(2) + (2)(-1) + (-1)(3) = 2 - 2 - 3 = -3$. 3. Find the magnitude of each normal vector: $|\mathbf{n}_1| = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{6}$ and $|\mathbf{n}_2| = \sqrt{2^2 + (-1)^2 + 3^2} = \sqrt{14}$. 4. Substitute the values into the rearranged scalar product angle formula, taking the absolute value of the dot product to guarantee an acute angle: $\cos \theta = \frac{|-3|}{\sqrt{6}\sqrt{14}} = \frac{3}{\sqrt{84}}$. 5. Evaluate the inverse cosine using your GDC: $\theta = \arccos\left(\frac{3}{\sqrt{84}}\right) = 70.893...^\circ$. 6. State the final angle rounded to one decimal place: $\mathbf{70.9^\circ}$.

Particle A starts at $(0, 2, 1)$ and travels with velocity $\mathbf{v}_A = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}$ m/s. Particle B starts at $(3, 0, 0)$ and travels with velocity $\mathbf{v}_B = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}$ m/s. Find the exact distance between the two particles after $t = 3$ seconds.

Position of A at time $t$: $\mathbf{r}_A = (0, 2, 1) + t(1, -1, 2)$; at $t = 3$: $\mathbf{r}_A = (3, -1, 7)$ M1A1 Position of B at $t = 3$: $\mathbf{r}_B = (3, 0, 0) + 3(0, 1, 1) = (3, 3, 3)$ A1 Vector between them: $\overrightarrow{AB} = \mathbf{r}_B - \mathbf{r}_A = (0, 4, -4)$, so distance $= \sqrt{0 + 16 + 16}$ M1 $= \sqrt{32} = 4\sqrt{2}$ m A1

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IB Maths AA HL · Unit 3: Geometry and Trigonometry

IB Maths AA HL Planes Questions

Updated October 2026

Exam-style IB Maths AA HL planes questions with worked solutions. Start with the 3 fully worked examples below — each is solved step by step, showing the working that earns the marks — then try the practice questions and check your working against the mark scheme.

47 questions
Paper 1: 36
Paper 2: 11
12 easy
23 medium
11 hard
1 very hard
3 worked examples

Practise Planes questions →

🖩 How do I do this on my calculator? Keystrokes for the Casio fx-CG50, TI-84 Plus CE and TI-Nspire.

What you need to know

The cross product a × b gives a vector perpendicular to both. HL AA Paper 1 tests the coordinate formula and the |a × b| = |a||b|sinθ identity. 3D vectors — dot product, cross product, and applications overview →

Vector equations of lines: r = a + λb. Planes: r·n = d. HL AA Paper 2 always asks for the intersection point OR the shortest distance between skew lines. Lines and planes in 3D — intersections and distances overview →

What's examined in AA HL planes

The question bank covers these planes question types (number of questions in brackets):

3D Geometry Problems (28)
Vector Definitions & Operations (12)
Plane Equations (7)

Key formulas

Vector magnitude: $| \vec{a} | = \sqrt{a_{1}^{2} + a_{2}^{2} + a_{3}^{2}}$
Vector equation of a line: $\vec{r} = \vec{a} + t \vec{d}$
Cartesian equation of a plane: $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$

In the same notation as the IB formula booklet. All AA HL formulas →

Where students lose marks on planes

Examiner's noteVectors: skew lines without the non-parallel check, and planes

Where marks are lost: November 2025: many did not show that the lines were not parallel before concluding they were skew, made errors solving the simultaneous equations (often with scattered working), and did not give a plane's equation with integer coefficients. May 2024: the relationship between planes was often misread; the report recommends diagrams.

Examiners flagged this in 2 exam series (May 2024, November 2025). Summarised in our own words from IB subject reports, IB Programme Resource Centre. Read the full note →

Planes worked examples

Worked example 1: Finding the Cartesian equation of a plane · easy

A plane $Π$ passes through the point $A (1, 2, - 1)$ and has a normal vector $n = (\begin{matrix} 3 \\ - 1 \\ 2 \end{matrix})$ . Find the Cartesian equation of the plane.

Solution

5 steps · 1 to fill in

Examiner tip: Don't forget that the coefficients of

x

y

, and

z

in a Cartesian plane equation are exactly the components of the normal vector; you can use this to quickly reverse-engineer a normal vector if given the plane equation.

Worked example 2: Intersection of a line and a plane · medium

Find the exact coordinates of the point of intersection between the line $L : r = (\begin{matrix} 0 \\ 1 \\ 2 \end{matrix}) + t (\begin{matrix} 1 \\ - 1 \\ 3 \end{matrix})$ and the plane $Π : 2 x + y - z = 5$ .

Solution

6 steps · 1 to fill in

Examiner tip: A frequent error is finding the value of the parameter

t

and stopping there; always remember to substitute

t

back into the line equation to find the actual 3D coordinates requested by the question.

Worked example 3: Finding the acute angle between planes · hard

Find the acute angle between the planes $Π_{1} : x + 2 y - z = 4$ and $Π_{2} : 2 x - y + 3 z = 7$ . Give your answer in degrees to one decimal place.

Solution

6 steps · 1 to fill in

Examiner tip: Always use the absolute value of the dot product (

| n_{1} \cdot n_{2} |

) when finding the acute angle between planes; omitting this step when the dot product is negative will result in you calculating the obtuse angle instead.

Try these IB Maths AA HL planes questions

Three questions from the bank, easiest first. Mark schemes and AI marking of your written working are in the practice area. One of them has its full worked solution here, open to everyone.

Question 1 · easy · 2 marks · Paper 1

Find the scalar (dot) product $a \cdot b$ where $a = (\begin{matrix} 2 \\ - 3 \\ 1 \end{matrix})$ and $b = (\begin{matrix} 1 \\ 4 \\ - 2 \end{matrix})$ .

Attempt it and see the mark scheme →

Question 2 · medium · 5 marks · Paper 2

Particle A starts at $(0, 2, 1)$ and travels with velocity $v_{A} = (\begin{matrix} 1 \\ - 1 \\ 2 \end{matrix})$ m/s. Particle B starts at $(3, 0, 0)$ and travels with velocity $v_{B} = (\begin{matrix} 0 \\ 1 \\ 1 \end{matrix})$ m/s. Find the exact distance between the two particles after $t = 3$ seconds.

Full worked solution Free, no sign-in

Position of A at time $t$ : $r_{A} = (0, 2, 1) + t (1, - 1, 2)$ ; at $t = 3$ : $r_{A} = (3, - 1, 7)$ M1A1
Position of B at $t = 3$ : $r_{B} = (3, 0, 0) + 3 (0, 1, 1) = (3, 3, 3)$ A1
Vector between them: $\vec{A B} = r_{B} - r_{A} = (0, 4, - 4)$ , so distance $= \sqrt{0 + 16 + 16}$ M1
$= \sqrt{32} = 4 \sqrt{2}$ m A1

Every question here has a full worked solution. See them all →

Question 3 · hard · 5 marks · Paper 2

Three planes are defined by the following Cartesian equations: $Π_{1} : x + y + z = 6$ $Π_{2} : 2 x - y + z = 3$ $Π_{3} : x + 2 y - z = 2$ Show that the three planes intersect at a single unique point, and find the coordinates of this point.

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FAQ

How many IB Maths AA HL planes questions are there?

There are 47 exam-style planes questions in the AA HL question bank (Paper 1: 36 · Paper 2: 11), graded 12 easy, 23 medium, 11 hard, 1 very hard. Every question has a full IB-style mark scheme (M, A and R marks).

Is planes on Paper 1 or Paper 2?

Both. In the bank, Paper 1: 36 · Paper 2: 11. Paper 1 is non-calculator, so practise the exact-value algebra as well as the GDC methods.

Where can I get the mark schemes?

Open the AA HL Unit 3 practice page: every question has a step-by-step IB-style mark scheme, and you can photograph your working for instant AI marking. The worked examples on this page are free.

What else is free here

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Teaching this topic?

Free ready-to-teach lesson (slides and teacher notes): Equations of planes; angles & intersections of planes.

IB Maths AA HL Planes Questions

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What you need to know

What's examined in AA HL planes

Key formulas

Where students lose marks on planes

Planes worked examples

Worked example 1: Finding the Cartesian equation of a plane · easy

Solution

Worked example 2: Intersection of a line and a plane · medium

Solution

Worked example 3: Finding the acute angle between planes · hard

Solution

Try these IB Maths AA HL planes questions

Question 1 · easy · 2 marks · Paper 1

Question 2 · medium · 5 marks · Paper 2

Full worked solution Free, no sign-in

Question 3 · hard · 5 marks · Paper 2

Revise planes for the exam

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