Mathematics 9709 · syllabus 3.7
Cambridge A Level Mathematics 9709 — Vectors topical past papers
Vectors in two and three dimensions: position vectors, magnitudes, the scalar product, and the vector equation of a straight line r = a + tb.
- Questions
- 107
- Years
- 2009–2026
- With mark scheme
- 100%
- Questions 2022–2026
- 32
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What Vectors covers
Vectors in 9709 are a Paper 3 topic (section 3.7). You work with column vectors and i, j, k notation, add and subtract vectors and multiply them by scalars, and interpret the results geometrically, as in a parallelogram or at the midpoint of a line segment. You find magnitudes and unit vectors, and use position and displacement vectors in two and three dimensions, often inside a cuboid or a pyramid.
The scalar product a·b = |a||b| cos θ gives the angle between two vectors and shows when they are perpendicular. A straight line is written r = a + tb, where a is the position vector of a point on it and b its direction. With two lines you decide whether they are parallel, intersect or are skew, and find the intersection when there is one; with a point and a line you find the foot of the perpendicular and so the distance from the point to the line.
Planes, the vector product and the shortest distance between skew lines are not part of 9709. Vectors draw on coordinate geometry for the idea of a line and on trigonometry for angles; the same thinking about components returns when you resolve forces in Mechanics, although Paper 4 does not use vector notation.
In the syllabus
The headings this topic has in the Cambridge International 2026–2027 syllabus, as the board prints them.
- 3.7 Vectors
What you should be able to do
- Add, subtract and scale vectors, and interpret the results geometrically
- Find magnitudes, unit vectors and displacement vectors between points
- Use the scalar product to find the angle between two vectors or two lines
- Write the vector equation of a line through two points, or through a point with a given direction
- Decide whether two lines are parallel, intersect or are skew, and find any point of intersection
- Find the foot of the perpendicular from a point to a line, and the perpendicular distance
Key ideas
- The displacement from A to B is b − a, where a and b are the position vectors of A and B
- |xi + yj + zk| = √(x² + y² + z²); a unit vector in the direction of a is a/|a|
- a·b = a₁b₁ + a₂b₂ + a₃b₃ = |a||b| cos θ, so a·b = 0 when a and b are perpendicular
- The line through the point with position vector a in the direction b is r = a + tb
- Lines are parallel if their direction vectors are multiples of each other; lines that are not parallel and do not meet are skew
How Vectors has been examined
From 2022 to 2026, Vectors was on Paper 3 · Pure Mathematics 3 in 11 of 11 exam series, worth about 13% of that paper’s marks.
| Paper | Questions | Share of marks | Series |
|---|---|---|---|
| Paper 3 · Pure Mathematics 3 | 32 | 13% | 11 of 11 |
Momentum’s own count, over the questions in its bank for these years: a series is one exam session, every zone included, and a question filed under two chapters counts in both. Median marks per question: 9.
Filed in Momentum under: Vectors.
A worked example
The line l passes through the point A(1, 2, −1) and has direction vector 2i − j + 2k. Find the position vector of the foot of the perpendicular from P(7, 2, 2) to l, and the perpendicular distance from P to l.
- A general point Q on l has position vector (1 + 2t)i + (2 − t)j + (−1 + 2t)k.
- So the vector PQ = (2t − 6)i − tj + (2t − 3)k.
- PQ is perpendicular to l, so PQ·(2i − j + 2k) = 0: 2(2t − 6) + t + 2(2t − 3) = 9t − 18 = 0, giving t = 2.
- The foot of the perpendicular is Q = 5i + 0j + 3k, and then PQ = −2i − 2j + k.
- The distance is |PQ| = √(4 + 4 + 1) = 3.
Answer: Foot of the perpendicular 5i + 3k; perpendicular distance 3 units
Written by Momentum for this page, with its own numbers; it is not taken from a past paper.
Mistakes that cost marks
- Finding the angle between two lines from their position vectors: only the direction vectors set the angle.
- Concluding that two lines intersect after solving two of the three component equations: the parameter values must also satisfy the third.
- Giving an obtuse angle when the acute angle between two lines is asked for: take the modulus of the scalar product, or subtract from 180°.
- Writing the vector from A to B as a − b: the displacement from A to B is b − a.
- Using the same parameter for both lines when testing for intersection, which wrongly forces them to meet at the same parameter value.
Exam technique
- Use different parameters, such as λ and μ, for two lines; solve two component equations and test the third.
- For the foot of the perpendicular from P to a line, write a general point Q on the line, form the vector PQ, and set its scalar product with the line's direction to zero.
- Draw a quick sketch of the points and lines, even in three dimensions; it helps you choose between position and displacement vectors.
Practising in Momentum
- Every question in a chapter as one list, from every paper and year in the bank, newest first or shuffled.
- Each question turns over to its own part of the mark scheme, where the board published one.
- Narrow by paper, year, session, zone and marks, and star the questions worth keeping.
- Pick chapters, a number of questions and a time, and the exam builder draws a mock to sit and mark.
Past-paper questions and mark schemes are the copyright of Cambridge International. Momentum shows them to signed-in learners for revision and does not publish them on these pages.