IB DP Mathematics: Analysis and Approaches – Vectors, lines and planes (HL) Revision Notes

These revision notes condense the vectors, lines and planes unit for the final weeks before the exam. For full explanations and longer worked examples, go back to the vectors, lines and planes study guide.

The notes cover IB Diploma Programme Mathematics: Analysis and Approaches, aligned to the IB Mathematics: analysis and approaches guide, first assessment 2021, syllabus sections 3.12–3.18. All of it is HL only (AHL). It follows the IB guide for first assessment 2021, which remains the examined syllabus until the new course is first assessed in May 2029, so it applies to the May and November 2026, 2027 and 2028 HL sessions.

When you are ready, try the vectors practice questions. The IB DP Maths AA course hub and the printable syllabus checklist show the rest of the course.

Vectors are written in a row, such as (2, −1, 3), to fit a phone screen. Write them as columns in the exam.

Definitions (3.12)

• Position vector of A: OA = a, from the origin to A.
• Displacement vector: AB = b − a (“end minus start”).
• Base vectors: i, j, k are unit vectors along the axes; (v₁, v₂, v₃) = v₁i + v₂j + v₃k.
• Zero vector 0: all components 0. −v: same length as v, opposite direction.
• Parallel: w = kv for some scalar k (non-zero vectors).
• Magnitude: |v| = √(v₁² + v₂² + v₃²). Distance AB = |AB|.
• Unit vector in the direction of v: v/|v|.
• Skew lines: non-parallel lines that do not intersect (three dimensions only).
• Normal to a plane: a vector perpendicular to every direction in the plane.

Formulas

Result	Formula	Section
Scalar product	v · w = v₁w₁ + v₂w₂ + v₃w₃ = |v||w| cos θ	3.13
Angle between vectors	cos θ = (v · w)/(|v||w|)	3.13
Perpendicular test	v · w = 0 (non-zero vectors)	3.13
Parallel vectors	|v · w| = |v||w|	3.13
Line, vector form	r = a + λb	3.14
Line, parametric	x = x₀ + λl, y = y₀ + λm, z = z₀ + λn	3.14
Line, Cartesian	(x − x₀)/l = (y − y₀)/m = (z − z₀)/n	3.14
Vector product	v × w = (v₂w₃ − v₃w₂, v₃w₁ − v₁w₃, v₁w₂ − v₂w₁)	3.16
Magnitude of v × w	|v × w| = |v||w| sin θ	3.16
Parallelogram area	|v × w|	3.16
Triangle area	½|AB × AC|	3.16
Plane, vector form	r = a + λb + μc	3.17
Plane, normal form	r · n = a · n	3.17
Plane, Cartesian	ax + by + cz = d, normal (a, b, c)	3.17
Line–plane angle	sin φ = |b · n|/(|b||n|)	3.18
Plane–plane angle	cos θ = |n₁ · n₂|/(|n₁||n₂|)	3.18

Properties to quote: v · w = w · v and v · v = |v|²; v × w = −w × v and v × v = 0; both products distribute over addition and (kv) · w = k(v · w), (kv) × w = k(v × w).

Method in steps

Line through two points A and B

1. Direction b = AB = b − a.
2. Write r = a + λb.
3. For Cartesian form, divide each (coordinate − x₀) by its direction component.

Two lines: coincident, parallel, intersecting or skew?

1. Compare the directions. Multiples → parallel or coincident; test one point of L₁ in L₂ to decide.
2. Not multiples → equate a + λb = c + μd with different parameters.
3. Solve two component equations for λ and μ.
4. Substitute into the third. True → intersecting (find the point). False → skew.

Plane through three points A, B, C

1. Find AB and AC.
2. n = AB × AC. Simplify by a common factor if you can.
3. d = a · n. Write ax + by + cz = d.
4. Check the other two points.

Line meets plane

1. Write the line in parametric form.
2. Substitute x, y, z into the plane’s equation.
3. Solve for the parameter and substitute back for the point.

Line of intersection of two planes

1. Direction = n₁ × n₂.
2. Set one coordinate to 0 and solve the other two equations for a point.
3. Write r = point + t(direction). Check the point in both planes.

Three planes

1. Solve the 3 × 3 system (elimination by hand on Paper 1, GDC on Papers 2 and 3).
2. Read the geometry from the solution set: a point, a line, a plane, or none.

Vector proof (3.12)

1. Name the given vectors, for example OA = a and OC = c.
2. Write every other vector you need in terms of them, using AB = b − a.
3. Show the required fact: a scalar multiple for parallel, a zero scalar product for perpendicular, equal magnitudes for equal lengths.
4. Finish with a sentence that states the result in words.

Kinematics with a line

1. Read r = a + tb as starting position a and velocity b.
2. Speed is |b|. Position at time t: substitute t.
3. To test for a collision, use the same t for both objects. To test whether paths cross, use different parameters.

Small worked reminders

• (3, −1, 2) × (1, 0, 4) = ((−1)(4) − (2)(0), (2)(1) − (3)(4), (3)(0) − (−1)(1)) = (−4, −10, 1). Check with the scalar product: (−4, −10, 1) · (3, −1, 2) = −12 + 10 + 2 = 0.
• Plane through (2, 0, −1) with normal (1, 4, −2): d = 2 + 0 + 2 = 4, so x + 4y − 2z = 4.
• r = (0, 3, 1) + t(2, 2, 1): speed |(2, 2, 1)| = 3 units per unit time.
• Lines with directions (1, 2, 2) and (2, −1, 2): cos θ = |2 − 2 + 4|/(3 × 3) = 4/9, so θ = 63.6°.

Must-know distinctions

• Scalar product vs vector product. v · w is a number; v × w is a vector. Zero scalar product means perpendicular; zero vector product means parallel.
• Angle between vectors vs angle between lines. Vectors can give an obtuse angle. For lines (and planes), take the acute angle with the modulus |b₁ · b₂|.
• Line–line (cos) vs line–plane (sin). A line direction and a plane normal give sin φ, because the normal is at 90° to the plane.
• Parallel vs skew. Both never meet. Parallel lines have parallel directions; skew lines do not.
• Position a vs direction b in r = a + λb. a is a point; b sets the direction. Any point on the line can replace a, and any non-zero multiple can replace b.
• Velocity vs speed. In r = a + tb, b is the velocity; |b| is the speed.
• No solution for three planes. Either at least two planes are parallel, or they form a triangular prism. Check the normals to tell which.

Quick self-test

1. Find |(2, −6, 3)|.
2. Find the unit vector in the direction of (2, −6, 3).
3. Find (1, 2, 3) · (4, −5, 6).
4. Find t so that (t, 2, −1) is perpendicular to (3, t, 4).
5. Find (1, 0, 2) × (0, 1, 3).
6. Find the area of the parallelogram with sides (1, 0, 2) and (0, 1, 3).
7. Write r = (3, −1, 0) + λ(2, 5, −4) in Cartesian form.
8. Find the Cartesian equation of the plane through (1, −2, 4) with normal (3, 1, −2).
9. A particle has position r = (4, 1) + t(6, −8). Find its speed.
10. Find the angle between planes with normals (1, 1, 0) and (0, 1, 1).
11. Two lines have directions (1, −2, 3) and (−2, 4, −6). Can they be skew?
12. Does the line r = (1, 1, 1) + t(1, −1, 0) meet the plane x + y + 2z = 5?

Answers

1. √(4 + 36 + 9) = 7.
2. (2/7, −6/7, 3/7).
3. 4 − 10 + 18 = 12.
4. 3t + 2t − 4 = 0, so t = 4/5.
5. (0×3 − 2×1, 2×0 − 1×3, 1×1 − 0×0) = (−2, −3, 1).
6. |(−2, −3, 1)| = √14.
7. (x − 3)/2 = (y + 1)/5 = z/(−4).
8. d = 3 − 2 − 8 = −7, so 3x + y − 2z = −7.
9. |(6, −8)| = 10.
10. cos θ = 1/(√2 × √2) = 1/2, so θ = 60°.
11. No. The second direction is −2 times the first, so the lines are parallel or coincident.
12. (1, −1, 0) · (1, 1, 2) = 0, so the line is parallel to the plane. The point (1, 1, 1) gives 1 + 1 + 2 = 4 ≠ 5, so the line does not meet the plane.

Where marks are usually lost

• Writing AB = a − b instead of b − a, which reverses every direction that follows.
• Testing for intersection with one parameter for both lines, which forces a false answer.
• Stopping after two component equations and never checking the third.
• Stating “skew” without first showing that the directions are not parallel.
• A sign slip in the middle component of a vector product. Check the result is perpendicular to both vectors.
• Using cos φ for the angle between a line and a plane, which gives 90° − φ.
• Leaving an obtuse angle between two lines or two planes.
• Forgetting the ½ for a triangle area from a vector product.
• Describing a “no solution” case for three planes without saying whether any planes are parallel.
• Decimals on Paper 1 where exact answers such as 5√3/2 or 2/√42 are expected.

Official syllabus

International Baccalaureate Organization, Diploma Programme, Mathematics: analysis and approaches guide, first assessment 2021 (published February 2019, updated November 2020).