This whole subtopic is higher level. Nothing in it is on an SL paper.
Educerie · IB Diploma · Mathematics: analysis and approaches
Topic 3 Geometry and trigonometry · 3.18 Intersections and angles with planes
What you must be able to do
| You must be able to | Level | What it looks like in the exam |
|---|---|---|
| Find where a line meets a plane, or show it does not | HL | "Find the coordinates of the point where L meets Π" (3 to 4 marks) |
| Recognise a line parallel to a plane, or lying in it, from b · n = 0 | HL | "Show that L is parallel to Π"; "find k so that L lies in Π" (2 to 4 marks) |
| Find the line of intersection of two planes | HL | "Find a vector equation of the line of intersection" (4 to 6 marks) |
| Find where three planes meet, and interpret the solution geometrically | HL | "Solve the system and interpret it geometrically" (5 to 8 marks); "find k for which the planes meet in a line" |
| Find the angle between a line and a plane | HL | "Find the acute angle between L and Π" (4 marks), sin θ exact on Paper 1 |
| Find the angle between two planes | HL | "Find the angle between Π₁ and Π₂" (3 to 4 marks) |
| Use these ideas in context: a path hitting a surface, the slope of a roof or hillside, the distance from a point to a plane | HL | Paper 2 Section B |
Before you start
You need the vector, parametric and Cartesian forms of a line (3.14) and a plane (3.17), the scalar product and the angle between two vectors (3.13), and the vector product (3.16). You need to solve systems of linear equations, including systems with no solution or infinitely many, which is AHL 1.16; this page is its geometric picture. The formula booklet gives the equations of lines and planes and the angle between two vectors, but not the line-and-plane angle rule in its sine form; learn it here.
A column vector is printed on one line here: (1, 2, 2) means the column with entries 1, 2, 2. A point is written with its letter, P(7, 1, 7).
1The idea in one paragraph
Every intersection question is the same question: which points satisfy all the equations at once? For a line and a plane, write the line's general point in terms of λ, put it into the plane's equation, and solve for λ: one value means one point, no value means the line is parallel to the plane, and every value means the line lies in it. For two planes, two equations in three unknowns leave one free variable, so non-parallel planes meet in a line, whose direction is perpendicular to both normals, n₁ × n₂. For three planes, three equations in three unknowns usually have one solution, a single point, but the system can also give a line, a whole plane or nothing, and each of those has a picture. Angles come from normals and directions: the angle between two planes is the angle between their normals, and the angle between a line and a plane is 90° minus the angle between the line and the normal, which is why it uses sine.
2A line meets a plane
Substitute the parametric form of the line into the Cartesian equation of the plane, and solve for λ.
Take L: r = (2, 0, 1) + λ(1, 2, 2) and Π: 2x − y + 2z = 18.
When λ disappears. The coefficient of λ after substituting is exactly b · n, the scalar product of the line's direction with the plane's normal (here (1, 2, 2) · (2, −1, 2) = 4, the 4λ above). If b · n = 0, the line is perpendicular to the normal, which means it runs parallel to the plane. Then λ cancels completely, and one of two things is left. Figure 1 shows all three outcomes.
So: b · n ≠ 0 gives exactly one point; b · n = 0 gives none or infinitely many, and testing one point of the line in the plane tells you which.
The shortest distance from a point to a plane. The nearest point of a plane to a point P is the foot of the perpendicular F, where the line through P along the normal meets the plane. Every other point Q of the plane is further away, because PQ is the hypotenuse of the right-angled triangle PFQ. For P(7, 1, 7) and Π: 2x − y + 2z = 18:
Figure 2 shows the result. Going on to t = −2 gives (3, 3, 3), the reflection of P in the plane, a common last part.
3The angle between a line and a plane
The angle between a line and a plane is the angle between the line and its "shadow" in the plane, its projection, as Figure 3 shows. It is between 0° and 90°.
The normal is easier to work with than the shadow. If φ is the angle between the line and the normal, then θ = 90° − φ, and cos φ = sin θ. So
sin θ = |b · n| ÷ (|b||n|), where b is the line's direction and n the plane's normal. The modulus makes θ acute.
For L and Π of section 2:
The equivalent route: cos φ = 4/9 gives φ = 63.6°, so θ = 90° − 63.6° = 26.4°. Either is fine, but 63.6° on its own is the angle with the normal and loses the final mark. Paper 1 usually asks for sin θ exactly, here 4/9.
A line parallel to the plane has θ = 0; a line along the normal has θ = 90°.
4Two planes
Parallel or not? Two planes are parallel exactly when their normals are parallel. x + y + z = 6 and 2x + 2y + 2z = 5 have parallel normals (1, 1, 1) and (2, 2, 2); dividing the second by 2 gives x + y + z = 2.5, a different constant, so they are parallel and never meet. 2x + 2y + 2z = 12 would be the same plane as x + y + z = 6, coincident.
Otherwise they meet in a line, as Figure 4 shows. The line lies in both planes, so its direction is perpendicular to both normals, and the vector product gives it at once.
The line of intersection of two non-parallel planes has direction n₁ × n₂. Find one point on it by fixing one variable (often z = 0) and solving the two equations.
For Π₁: x + y + z = 6 and Π₂: 2x − y + z = 3:
A second method, all algebra. Let one variable be the parameter and solve for the other two. With z = t: x + y = 6 − t and 2x − y = 3 − t. Adding, 3x = 9 − 2t, so x = 3 − (2/3)t and y = 3 − (1/3)t. The line is r = (3, 3, 0) + t(−2/3, −1/3, 1), and (−2/3, −1/3, 1) is a multiple of (2, 1, −3). This is the general solution of AHL 1.16.
One trap. Putting z = 0 only works if the line actually crosses the plane z = 0, which fails when its direction has a z-component of 0. If z = 0 gives two inconsistent equations, set x = 0 or y = 0 instead.
The angle between two planes is the angle between their normals, as the edge-on view in Figure 5 shows: turning each plane through 90° turns it into its normal, and the angle between them is unchanged.
cos θ = |n₁ · n₂| ÷ (|n₁||n₂|), for the acute angle between the planes.
The angle of a roof or hillside with the horizontal is the angle between its plane and z = 0, whose normal is (0, 0, 1).
5Three planes
Three planes give a system of three linear equations in x, y and z, and AHL 1.16 tells you such a system has one solution, infinitely many, or none. Each outcome is a picture.
One solution: the planes meet at a single point, like the floor and two walls in the corner of a room (Figure 6).
Take Π₁ and Π₂ from section 4 with Π₃: x − y + 2z = −5.
A cross-check: (5, 4, −3) is t = 1 on the line of intersection of Π₁ and Π₂ from section 4.
Infinitely many solutions: a sheaf. Replace Π₃ by 3x + 2z = 9. Eliminating y from (1) and (2) gives 3x + 2z = 9, which is the third equation, so the elimination ends with 0 = 0. One equation adds nothing new: the three planes share the whole line r = (3, 3, 0) + t(2, 1, −3). Three planes through a common line are called a sheaf, like pages of a partly open book (Figure 7a).
No solution: a triangular prism. Replace Π₃ by 3x + 2z = 10. Now elimination gives 3x + 2z = 9 and 3x + 2z = 10 together, and subtracting gives 0 = 1, which is impossible. The system is inconsistent. No two of the normals (1, 1, 1), (2, −1, 1), (3, 0, 2) are parallel, so each pair of planes does meet in a line, but the three lines are parallel and separate. Seen end on, the planes form the sides of a triangular prism (Figure 7b).
Telling the cases apart. First compare normals, then solve.
| What elimination gives | Normals | Picture |
|---|---|---|
| One solution | not coplanar (no plane contains all three normals) | three planes meet at one point |
| A line of solutions (0 = 0 once) | no two parallel | a sheaf: three planes through one line |
| A line of solutions | two planes coincident | two identical planes cut by the third |
| A plane of solutions (0 = 0 twice) | all parallel, same plane | three coincident planes |
| No solution (0 = non-zero) | no two parallel | a triangular prism |
| No solution | two parallel, third not | two parallel planes, each cut by the third (Figure 7c) |
| No solution | all three parallel | three parallel planes, or two coincident and one parallel (Figure 7d) |
Finding a value that gives a line. Questions often put a letter k in the third equation and ask when the system has infinitely many solutions. Eliminate as usual: the left-hand sides reduce to 0, and the right-hand side becomes an expression in k, which must also be 0. For that k the planes form a sheaf; for any other k they form a prism (or, if some normals are parallel, one of the other no-solution pictures).
On Paper 2 the GDC's row-reduction (rref) shows a row of zeros for a line of solutions and a row 0 0 0 | 1 for none, but the interpretation in words is still yours to write.
6Where marks are lost
Using cos for the line–plane angle. cos θ = |b · n| ÷ (|b||n|) gives the angle with the normal. For the angle with the plane, use sin, or subtract from 90°.
Using a point instead of a direction. Every angle here uses direction vectors and normals only, never position vectors of points.
Concluding "parallel" from λ cancelling without looking at what is left. If λ cancels, 3 = 18 means parallel and separate, but 18 = 18 means the line lies in the plane. Say which.
Taking n₁ × n₂ as a point on the line of intersection. It is the direction. You still need one point, found by fixing a variable.
Setting z = 0 when the line never reaches z = 0. If the two equations become inconsistent, fix x or y instead.
Giving the obtuse angle between planes. Use the modulus in cos θ = |n₁ · n₂| ÷ (|n₁||n₂|), or subtract from 180°.
Saying "no solution, so the planes are parallel". Three planes can have no common point without any two being parallel: the triangular prism. Compare the normals before describing the picture.
Stopping at the algebra. "0 = 0, so infinitely many solutions" needs "the planes meet in a line (a sheaf)" when the question says interpret.
7Work it right
- For a line and a plane, write x, y, z in terms of λ, substitute into the plane, and solve; then substitute λ back and check the point in the plane.
- If λ cancels, state b · n = 0 and test a point of the line in the plane to decide between "parallel" and "lies in the plane".
- For two planes, check the normals are not parallel, find the direction as n₁ × n₂, find a point by fixing a variable, and check the line in both planes.
- For three planes, eliminate one variable twice, then finish; check the answer in all three equations.
- When elimination gives 0 = 0 or 0 = k, compare the normals, then name the picture: sheaf, prism, parallel planes.
- For angles, write which vectors you are using (direction and normal, or two normals), include the modulus, and say sin for a line and a plane, cos for two planes.
- On Paper 2, record the GDC's input and output (the system entered, the rref shown), then interpret in words.
8Try it
Marks in brackets. Q1 to Q3 are Paper 1 style, no calculator. Q4 and Q5 are Paper 2 style, with a GDC.
Q1. The line L has equation r = (1, 0, −2) + λ(2, −1, 3) and the plane Π has equation 3x + y − z = 15.
(a) Find the coordinates of the point where L meets Π. 4 marks
(b) Find the exact value of sin θ, where θ is the acute angle between L and Π. 3 marks
Q2. The line L has equation r = (2, 1, 0) + t(1, 1, 1) and the plane Π has equation x − 2y + z = k.
(a) Show that L is parallel to Π. 2 marks
(b) Find the value of k for which L lies in Π. 2 marks
Q3. The planes Π₁: x − y + 2z = 1 and Π₂: 2x + y − z = 5 meet in the line L.
(a) Find a vector equation of L. 5 marks
(b) Find the exact value of the cosine of the acute angle between Π₁ and Π₂. 2 marks
Q4. Three planes have equations x + 2y − z = 3, 2x − y + z = 1 and 4x + 3y − z = k.
(a) Find the value of k for which the three planes meet in a line. 3 marks
(b) For this value of k, find a vector equation of the line. 4 marks
(c) Describe the arrangement of the three planes when k takes any other value, giving a reason. 2 marks
Q5. A hillside is modelled by the plane x + 2y + 4z = 40, where distances are in metres and z is the height above a horizontal reference plane. A drone starts at (2, 3, 20) and flies in a straight line with velocity (1, 1, −2) m s⁻¹.
(a) Find the time at which the drone reaches the hillside, and the point where it lands. 4 marks
(b) Find the acute angle between the drone's path and the hillside. 3 marks
(c) Find the angle that the hillside makes with the horizontal. 3 marks
9In one breath
To find where things meet, find the points that satisfy every equation at once. A line meets a plane where its parametric form, substituted into the plane's Cartesian equation, gives a value of λ; if λ cancels, b · n = 0 and the line is parallel to the plane, lying in it if the leftover statement is true and missing it if it is false. The foot of the perpendicular from a point to a plane, and so the distance, comes from a line along the normal. Two planes with parallel normals are parallel or coincident; otherwise they meet in a line with direction n₁ × n₂ through a point found by fixing one variable. Three planes usually meet at one point, found by elimination, but 0 = 0 means a line (a sheaf) and 0 = non-zero means no common point, where the normals decide the picture: a triangular prism if no two are parallel, parallel planes otherwise. The angle between two planes is the acute angle between their normals, cos θ = |n₁ · n₂| ÷ (|n₁||n₂|), and the angle between a line and a plane uses sine, sin θ = |b · n| ÷ (|b||n|), because the normal is 90° away from the plane.
Answers
Q1. (a) Substitute x = 1 + 2λ, y = −λ, z = −2 + 3λ: 3(1 + 2λ) − λ − (−2 + 3λ) = 15, so 3 + 6λ − λ + 2 − 3λ = 15, 5 + 2λ = 15 and λ = 5. The point is (11, −5, 13); check: 33 − 5 − 13 = 15. M1 for substituting the parametric form into the plane, A1 for λ = 5, M1 for substituting λ back, A1 for (11, −5, 13).
(b) b · n = (2)(3) + (−1)(1) + (3)(−1) = 2, |b| = √14 and |n| = √11, so sin θ = 2/√154. M1 for using the direction of L and the normal of Π, M1 for sin θ = |b · n| ÷ (|b||n|) or the 90° − φ route, A1 for 2/√154. Giving cos θ = 2/√154 scores M1 M0 A0.
Q2. (a) b · n = (1)(1) + (1)(−2) + (1)(1) = 0, so the direction of L is perpendicular to the normal of Π, and L is parallel to Π. M1 for the scalar product of the direction and the normal, R1 for the conclusion with the reason.
(b) L lies in Π when its point (2, 1, 0) is in Π: 2 − 2 + 0 = k, so k = 0. M1 for substituting a point of L into the plane, A1 for k = 0.
Q3. (a) Direction: (1, −1, 2) × (2, 1, −1) = ((−1)(−1) − (2)(1), (2)(2) − (1)(−1), (1)(1) − (−1)(2)) = (−1, 5, 3). Point: put z = 0, then x − y = 1 and 2x + y = 5, so 3x = 6, x = 2, y = 1, giving (2, 1, 0). So r = (2, 1, 0) + t(−1, 5, 3). Check: 2 − t − 1 − 5t + 6t = 1 ✓ and 4 − 2t + 1 + 5t − 3t = 5 ✓. M1 for the vector product of the normals (or solving with a parameter), A1 for (−1, 5, 3) or a multiple, M1 for fixing a variable to find a point, A1 for a correct point, A1 for a complete equation with "r =".
(b) n₁ · n₂ = 2 − 1 − 2 = −1, |n₁| = |n₂| = √6, so cos θ = 1/6. M1 for the scalar product of the normals over their magnitudes, A1 for 1/6. −1/6 scores A0 because the acute angle was asked for.
Q4. (a) 2 × (first) + (second) gives 4x + 3y − z = 2(3) + 1 = 7, which has the same left-hand side as the third equation. The planes share a line exactly when the third equation is this one, so k = 7. (Equivalently, eliminating reduces the third equation to 0 = k − 7.) M1 for a correct elimination or spotting the combination, A1 for reaching 0 = k − 7 or equivalent, A1 for k = 7.
(b) Adding the first two equations: 3x + y = 4. Let x = t, so y = 4 − 3t, and from the first equation z = x + 2y − 3 = t + 8 − 6t − 3 = 5 − 5t. So r = (0, 4, 5) + t(1, −3, −5). Check in the second: 2t − (4 − 3t) + (5 − 5t) = 1 ✓. M1 for eliminating a variable, M1 for introducing a parameter, A1 for the point, A1 for the direction (1, −3, −5) or a multiple. A GDC rref with the parameter introduced correctly earns full marks.
(c) For k ≠ 7 the system is inconsistent, so the planes have no common point. No two of the normals (1, 2, −1), (2, −1, 1), (4, 3, −1) are parallel, so each pair of planes meets in a line: the planes form a triangular prism. R1 for no common point with the reason, A1 for "triangular prism" justified by the normals not being parallel.
Q5. (a) Position r = (2 + t, 3 + t, 20 − 2t). Substituting: (2 + t) + 2(3 + t) + 4(20 − 2t) = 40, so 88 − 5t = 40 and t = 9.6 s. The drone lands at (11.6, 12.6, 0.8). M1 for writing the position at time t, M1 for substituting into the plane, A1 for t = 9.6, A1 for the point.
(b) b · n = 1 + 2 − 8 = −5, |b| = √6, |n| = √21, so sin θ = 5/√126 = 0.4454… and θ = 26.5° (3 s.f.). M1 for using the velocity and the normal, M1 for the sine relationship with the modulus, A1 for 26.5°. 63.5° (the angle with the normal) scores M1 M0 A0.
(c) The horizontal has normal (0, 0, 1). cos θ = |(1, 2, 4) · (0, 0, 1)| ÷ (√21 × 1) = 4/√21 = 0.8728…, so θ = 29.2° (3 s.f.). M1 for the normal (0, 0, 1) of the horizontal, M1 for the angle between the normals, A1 for 29.2°.
Educerie · written from the published IB Diploma Programme Mathematics: analysis and approaches guide, first assessment 2021, section 3.18 Intersections and angles with planes. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.