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 · 3.17 Equations of a plane
What you must be able to do
| You must be able to | Level | What it looks like in the exam |
|---|---|---|
| Write the vector equation r = a + λb + μc from a point and two non-parallel directions, or from three points | HL | "Write down a vector equation of the plane" (2 to 3 marks) |
| Find a normal vector to a plane, usually by a vector product | HL | "Find a vector perpendicular to the plane Π" (2 to 3 marks) |
| Write the plane as r · n = a · n | HL | "Find an equation of the plane in the form r · n = d" (2 marks) |
| Find and use the Cartesian equation ax + by + cz = d, and read a normal from it | HL | "Find the Cartesian equation of Π" (3 to 5 marks); "write down a normal" (1 mark) |
| Convert between the vector and Cartesian forms in both directions | HL | "Hence find a vector equation of Π" (3 marks) |
| Build a plane from other data: a point and a normal, a parallel plane, a line and a point, two intersecting lines | HL | Part of a Section B question, 4 to 6 marks |
| Test whether a point lies in a plane | HL | "Show that D lies in Π" (2 marks) |
Before you start
You need the vector equation of a line (3.14), the scalar product (3.13) with the fact that perpendicular vectors have scalar product 0, and the vector product (3.16), which is the standard tool for finding a normal. The formula booklet gives all three equations of a plane: r = a + λb + μc, r · n = a · n, and ax + by + cz = d. It does not say that (a, b, c) is a normal to ax + by + cz = d; learn that.
A column vector is printed on one line here: (4, −3, 5) means the column with entries 4, −3, 5. A point is written with its letter, A(1, 0, 2).
1The idea in one paragraph
A line is fixed by a point and one direction. A plane is flat and two-dimensional, so it needs a point and two directions that are not parallel: from the point you can walk any amount λ along the first direction and any amount μ along the second, and the points you can reach this way fill the plane. That gives the vector equation r = a + λb + μc. There is a second, shorter description. A plane has one special direction at right angles to all of its own directions, its normal. A point R is in the plane exactly when the step from a known point A to R is perpendicular to the normal, which the scalar product turns into r · n = a · n. Multiply that out and you get one plain equation, ax + by + cz = d, in which the coefficients a, b, c are the components of the normal. The vector product links the two descriptions: the normal is b × c.
2A point and two directions: r = a + λb + μc
r = a + λb + μc, where a is the position vector of a point in the plane and b, c are two non-parallel vectors lying in the plane. From the formula booklet.
Figure 1 shows how it works. Starting at A, λ steps of b and μ steps of c reach one point for each pair (λ, μ). Together they cover the whole plane, like the squares on a (slanted) sheet of graph paper.
Why must b and c be non-parallel? If c is a multiple of b, then λb + μc is just a multiple of b, and you only ever move along one line. Two genuinely different directions are what make the surface two-dimensional.
The plane through three points. Use one point for a and two displacement vectors from it. For A(1, 0, 2), B(3, 1, 1) and C(0, 2, 4):
The three points must not lie on one line, or AB and AC are parallel. As with lines, the equation is not unique: any point of the plane and any two non-parallel directions in it will do. The parametric form reads it row by row: x = 1 + 2λ − μ, y = λ + 2μ, z = 2 − λ + 2μ.
Is a point in the plane? With this form you need λ and μ. For D(2, 3, 3), the x- and y-rows give 2λ − μ = 1 and λ + 2μ = 3, so λ = 1 and μ = 1. The z-row then gives 2 − 1 + 2 = 3 ✓, so D is in the plane. That works, but it is slow. The Cartesian form in section 4 does it in one line.
3The normal: r · n = a · n
A normal to a plane is a non-zero vector perpendicular to every direction in the plane. There is only one normal direction, although any non-zero multiple of a normal (including its negative) is also a normal.
Figure 2 shows why the normal gives an equation. Let A be a known point with position vector a, and R any point with position vector r. R is in the plane exactly when the step AR = r − a lies in the plane, and a vector lies in the plane exactly when it is perpendicular to n. So
(r − a) · n = 0, which rearranges to r · n = a · n. From the formula booklet.
Finding a normal. You need a vector perpendicular to two non-parallel directions in the plane, and that is exactly what the vector product produces (3.16). For the plane through A, B and C:
The number on the right is the same whichever point of the plane you use: B gives 12 − 3 + 5 = 14 and C gives 0 − 6 + 20 = 14. That is a free check that your normal is right.
4The Cartesian equation ax + by + cz = d
Write r = (x, y, z) and multiply out the scalar product in r · n = a · n:
ax + by + cz = d, and (a, b, c) is a normal to the plane.
This is the form you will use most. It makes three jobs one line long:
- Reading a normal: 4x − 3y + 5z = 14 has normal (4, −3, 5). Just read the coefficients.
- Testing a point: D(2, 3, 3) gives 8 − 9 + 15 = 14 ✓, so D is in the plane. E(4, 3, 1) gives 16 − 9 + 5 = 12 ≠ 14, so E is not.
- Comparing planes: planes with parallel normals are parallel (3.18).
Always simplify a normal before building the equation if it has a common factor: (−12, 0, 48) is a normal, but so is (−1, 0, 4), and the smaller numbers make every later step safer.
From Cartesian to vector form. You need a point and two directions in the plane. The quickest way: let two of the variables be the parameters and solve for the third. For 2x + 3y + 4z = 12, let x = λ and y = μ:
Another way is to find three points of the plane and proceed as in section 2. The intercepts are the easiest three: put two variables equal to 0 each time. For 2x + 3y + 4z = 12 they are (6, 0, 0), (0, 4, 0) and (0, 0, 3). Figure 3 shows how they also give you a sketch of the plane.
From vector to Cartesian form. Cross the two direction vectors to get n, then find d = a · n, exactly as in section 3. The alternative, eliminating λ and μ from the three parametric equations, works but takes longer and invites slips.
5Planes with a missing variable
When a variable is missing from the Cartesian equation, its coefficient is 0, so the normal has no component in that direction and the plane runs parallel to that axis. Figure 4 shows three.
- z = 2 is the horizontal plane two units above the xy-plane; its normal is (0, 0, 1). The xy-plane itself is z = 0.
- x = 3 is parallel to the yz-plane; normal (1, 0, 0).
- x + y = 4 has normal (1, 1, 0), which is horizontal, so the plane is vertical: it contains every vertical line through the line x + y = 4 in the xy-plane.
A common slip is to think x + y = 4 is a line. In the plane (2D) it is; in space it is a plane.
6Building a plane from other information
Every plane question comes down to one thing: find a point in the plane and a normal to it. Here are the common starting points.
A point and a normal. The plane through P(2, −1, 4) perpendicular to a line with direction (3, 1, −2). The line's direction is the normal.
A parallel plane. Parallel planes share a normal, so only d changes. The plane through (1, 1, 1) parallel to 4x − 3y + 5z = 14 is 4x − 3y + 5z = 4 − 3 + 5 = 6.
Two intersecting lines. In 3.15 the lines L₁: r = (1, −2, 3) + λ(2, 1, −1) and L₂: r = (3, 4, −5) + μ(1, −2, 3) met at (5, 0, 1). Their directions both lie in the plane, so their vector product is its normal, as Figure 5 shows.
d = 0 here, which means the plane passes through the origin. That happens, and is not a mistake.
A line and a point not on it. One direction is the line's direction. The other is the vector from a point on the line to the given point. Cross them for the normal.
Two parallel lines. Their shared direction is one direction in the plane; the vector joining a point on one to a point on the other is the second. (Two skew lines have no plane containing both, 3.15.)
7Where marks are lost
Using parallel directions in r = a + λb + μc. If c is a multiple of b the equation describes a line, not a plane. Check the two directions are not parallel, which for three points means they are not collinear.
Using position vectors as directions. For the plane through A, B, C the directions are AB and AC, not a, b and c themselves.
Reading a point off the Cartesian equation as if it were a normal. In 4x − 3y + 5z = 14 the normal is (4, −3, 5); the 14 is not a coordinate, and (4, −3, 5) is not (usually) a point of the plane.
Arithmetic slips in the normal. Check n against both directions with the scalar product, then check d with a second point. Both checks are quick; an error in n ruins every part that follows.
Stopping at r · n = a · n when the Cartesian form was asked for. Multiply the scalar product out and give ax + by + cz = d.
Treating x + y = 4 as a line in three dimensions. With no z in it, every value of z is allowed: it is a vertical plane.
Forgetting that normals and equations are not unique. (−4, 3, −5) is as good a normal as (4, −3, 5), and −4x + 3y − 5z = −14 is the same plane. Do not "correct" a right answer to match a scheme's sign.
8Work it right
- Name what you are finding: a point in the plane and a normal. Every method reduces to those two.
- Form direction vectors as displacements between points, and state that they are not parallel.
- Find the normal as a vector product, one component per line, and simplify any common factor.
- Check the normal with a scalar product against each direction: both must be 0.
- Find d as a · n, and check it with a second known point.
- Give the form asked for; if Cartesian, write ax + by + cz = d with the numbers in.
- To test a point, substitute into the Cartesian equation and compare with d.
9Try it
Marks in brackets. Q1 to Q4 are Paper 1 style, no calculator. Q5 is Paper 2 style.
Q1. The points A(2, 0, 1), B(1, 3, 0) and C(4, 1, 3) lie in the plane Π.
(a) Find AB and AC. 2 marks
(b) Find a vector normal to Π. 2 marks
(c) Find the Cartesian equation of Π. 2 marks
Q2. The plane Π has vector equation r = (1, 2, 0) + λ(1, 1, 1) + μ(2, −1, 0). Find the Cartesian equation of Π. 4 marks
Q3. The line L has equation r = (0, 1, 2) + t(1, −1, 3), and P is the point (2, 2, 1). Find the Cartesian equation of the plane that contains L and P. 5 marks
Q4. The plane Π has equation 3x − 2y + 6z = 12.
(a) Write down a vector normal to Π. 1 mark
(b) Find the coordinates of the points where Π meets the x-, y- and z-axes. 3 marks
(c) Hence find a vector equation of Π. 2 marks
(d) Find the Cartesian equation of the plane parallel to Π that passes through (1, 1, 1). 2 marks
Q5. A sloping roof is modelled as part of a plane. Three of its corners are A(0, 0, 3), B(8, 0, 5) and C(0, 6, 3), where distances are in metres and z is the height above the ground.
(a) Find a vector normal to the roof. 3 marks
(b) Find the Cartesian equation of the plane of the roof. 2 marks
(c) A solar panel is to be fixed at the point D(4, 3, h) on the roof. Find h. 2 marks
10In one breath
A plane is fixed by a point and two non-parallel directions, r = a + λb + μc, or by a point and a normal, a vector perpendicular to every direction in the plane: R is in the plane exactly when r − a is perpendicular to n, so (r − a) · n = 0 and r · n = a · n. Multiply that out and you have the Cartesian equation ax + by + cz = d, whose coefficients (a, b, c) are a normal and whose d is a · n for any point A of the plane. The vector product of two directions in the plane gives the normal, so for three points cross AB and AC, check the answer with two scalar products, and check d with a second point. Go from Cartesian to vector form by letting two variables be parameters, or by finding three points such as the intercepts. A variable missing from the equation means the plane is parallel to that axis. Every plane question, from a point and a normal, a parallel plane, two intersecting lines, or a line and a point, reduces to the same two things: a point in the plane and a normal to it.
Answers
Q1. (a) AB = (−1, 3, −1) and AC = (2, 1, 2). A1 for each.
(b) AB × AC = ((3)(2) − (−1)(1), (−1)(2) − (−1)(2), (−1)(1) − (3)(2)) = (7, 0, −7), so (1, 0, −1) is a normal (any non-zero multiple). M1 for a vector product of two directions in the plane, A1 for (7, 0, −7) or a multiple.
(c) d = (2, 0, 1) · (1, 0, −1) = 2 − 1 = 1, so x − z = 1. Check: B gives 1 − 0 = 1 and C gives 4 − 3 = 1. M1 for substituting a point into x − z = d, A1 for x − z = 1 or any multiple, such as 7x − 7z = 7.
Q2. n = (1, 1, 1) × (2, −1, 0) = ((1)(0) − (1)(−1), (1)(2) − (1)(0), (1)(−1) − (1)(2)) = (1, 2, −3). d = (1, 2, 0) · (1, 2, −3) = 1 + 4 + 0 = 5. So x + 2y − 3z = 5. M1 for crossing the two direction vectors, A1 for (1, 2, −3), M1 for finding d with the point, A1 for the equation.
Q3. The line's point Q(0, 1, 2) and P give QP = (2, 1, −1). n = (1, −1, 3) × (2, 1, −1) = ((−1)(−1) − (3)(1), (3)(2) − (1)(−1), (1)(1) − (−1)(2)) = (−2, 7, 3). d = (2, 2, 1) · (−2, 7, 3) = −4 + 14 + 3 = 13. So −2x + 7y + 3z = 13 (or 2x − 7y − 3z = −13). Check with Q: 0 + 7 + 6 = 13. M1 for a second direction from a point of L to P, A1 for (2, 1, −1), M1 for the vector product with L's direction, A1 for (−2, 7, 3) or a multiple, A1 for the equation.
Q4. (a) (3, −2, 6). A1.
(b) y = z = 0 gives 3x = 12, so (4, 0, 0); x = z = 0 gives −2y = 12, so (0, −6, 0); x = y = 0 gives 6z = 12, so (0, 0, 2). A1 for each point, written as coordinates.
(c) Using (4, 0, 0) and directions to the other two intercepts, (−4, −6, 0) and (−4, 0, 2), which simplify to (2, 3, 0) and (2, 0, −1): r = (4, 0, 0) + λ(2, 3, 0) + μ(2, 0, −1). M1 for two directions formed from their intercepts, A1 for a correct equation with "r =". Check: (2, 3, 0) × (2, 0, −1) = (−3, 2, −6), a multiple of (3, −2, 6).
(d) d = 3 − 2 + 6 = 7, so 3x − 2y + 6z = 7. M1 for keeping the normal and substituting (1, 1, 1), A1 for the equation.
Q5. (a) AB = (8, 0, 2) and AC = (0, 6, 0). AB × AC = ((0)(0) − (2)(6), (2)(0) − (8)(0), (8)(6) − (0)(0)) = (−12, 0, 48), so (−1, 0, 4) is a normal. A1 for both direction vectors, M1 for their vector product, A1 for (−12, 0, 48) or a multiple.
(b) d = (0, 0, 3) · (−1, 0, 4) = 12, so −x + 4z = 12. Check: B gives −8 + 20 = 12. M1 for finding d, A1 for the equation.
(c) −4 + 4h = 12, so h = 4: the panel is fixed 4 m above the ground. The equation has no y, so the roof's height does not depend on y, and D's y-coordinate of 3 plays no part. M1 for substituting D into their equation, A1 for h = 4.
Educerie · written from the published IB Diploma Programme Mathematics: analysis and approaches guide, first assessment 2021, section 3.17 Equations of a plane. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.