OxfordAQA International GCSE Mathematics: Vectors and vector geometry (9260) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover OxfordAQA International GCSE Mathematics (9260), specification Version 5.1 (for exams May/June 2018 onwards), statement G22 in section 3.3.3. Every question is Extension only. If you need the method first, read the study guide or the revision notes. Course pages: the OxfordAQA IGCSE Mathematics hub and the checklist. Want a quick check of your level? Try the free 10-minute diagnostics.

Column vectors are typed as (x, y), with x on top. Write them as true columns in your answers. Bold letters are vectors.

Questions

1. (Extension only) a = (3, −5) and b = (−4, 2). Work out:

(a) a + b [1] (b) a − b [1] (c) 3a − 2b [2]

2. (Extension only) P is the point (−2, 4) and Q is the point (5, −1).

(a) Write →PQ as a column vector. [2] (b) Write down →QP. [1]

3. (Extension only) u = (4, 1) and v = (1, 3). On squared paper, start at the origin.

(a) Describe how to draw u + v and write down the point where you finish. [2] (b) Describe how to draw u − v using −v, and write down the point where you finish. [2]

4. (Extension only) a = (3, 1) and b = (−2, 4). Given that ma + nb = (0, 14), find m and n. [4]

5. (Extension only) p = (1, 6), q = (−3, 2) and r = (4, −5). Show that (p + q) + r = p + (q + r). Name the property. [3]

6. (Extension only) ABCD is a parallelogram. →AB = x and →AD = y.

(a) Write →AC in terms of x and y. [1] (b) Write →BD in terms of x and y. [2]

7. (Extension only) ABCD is a trapezium. AB is parallel to DC. →AB = 4a, →AD = b and →DC = 3a. M is the midpoint of BC.

(a) Find →BC in terms of a and b. [2] (b) Find →AM in terms of a and b. [2] (c) Find →DM in terms of a and b. [1]

8. (Extension only) →AB = 2a − b and →BC = −6a + 3b.

(a) Show that A, B and C lie on a straight line. [3] (b) Find the ratio AB : BC. [1]

9. (Extension only) In triangle OAB, →OA = a and →OB = b. P is the midpoint of OA. Q is on AB with AQ : QB = 1 : 3.

(a) Express →AB using a and b. [1] (b) Find →OQ. [2] (c) Find →PQ, giving your answer as a single fraction of a + b. [2] (d) State what (c) shows about PQ and the vector a + b. [1]

10. (Extension only) A is the point (1, 2), B is (5, 4) and C is (k, 8). A, B and C lie on a straight line.

(a) Write →AB as a column vector. [1] (b) Find k. [3]

11. (Extension only) PQRS is a parallelogram. →PQ = a and →PS = c. The diagonals PR and QS cross at X. Use vectors to show that the diagonals bisect each other. You may find the midpoint of each diagonal. [6]

12. (Extension only) In triangle OAB, →OA = a and →OB = b. P is the midpoint of OA. Q is on AB with AQ : QB = 1 : 2. R is the point with →OR = −b.

(a) Find →AB. [1] (b) Find →OQ. [2] (c) Find →PQ. [2] (d) Prove that P, Q and R are collinear. [3]

Answers

1. (a) a + b = (3 + (−4), −5 + 2) = (−1, −3) [1] (b) a − b = (3 − (−4), −5 − 2) = (7, −7) [1] (c) 3a = (9, −15) and 2b = (−8, 4) [1]. 3a − 2b = (9 + 8, −15 − 4) = (17, −19) [1] Examiner insight: Writing the scalar multiples first, as in part (c), makes your method clear and makes a slip in a subtraction easier to spot.

2. (a) →PQ is Q minus P [1]: (5 − (−2), −1 − 4) = (7, −5) [1] (b) →QP = −→PQ = (−7, 5) [1] Examiner insight: A reversed subtraction gives the negative of the correct vector. Check by asking whether Q is to the right of P, which it is, so the top entry should be positive.

3. (a) Draw u as 4 right and 1 up from the origin, ending at (4, 1). From there draw v, 1 right and 3 up [1]. The finish is (5, 4) [1] (b) −v is 1 left and 3 down. Draw u, then −v from the end of u [1]. The finish is (4 − 1, 1 − 3) = (3, −2) [1] Examiner insight: A description that says where each arrow starts, tip to tail, shows you understand the construction. A sketch alone may leave the second arrow’s start unclear.

4. ma + nb = (3m − 2n, m + 4n) [1] Top: 3m − 2n = 0, so m = 2n/3. Bottom: m + 4n = 14 [1] Substitute: 2n/3 + 4n = 14, so 14n/3 = 14 and n = 3 [1] Then m = 2. Check: 2(3, 1) + 3(−2, 4) = (6 − 6, 2 + 12) = (0, 14). m = 2, n = 3 [1] Examiner insight: Two unknown scalars need two equations, one from each entry. Checking in the original expression is a quick way to catch a sign error.

5. p + q = (−2, 8), so (p + q) + r = (2, 3) [1] q + r = (1, −3), so p + (q + r) = (2, 3) [1] Both give (2, 3). This is the associative property of vector addition. [1] Examiner insight: A “show that” with numbers needs both groupings written out in full. Naming the property is part of the answer when the question asks for it.

6. (a) →AC = →AB + →BC = x + y (since →BC = →AD) [1] (b) →BD = →BA + →AD [1] = −x + y = y − x [1] Examiner insight: In a parallelogram, opposite sides are equal vectors, so →BC can be replaced by →AD. State that fact when you use it.

7. (a) →BC = →BA + →AD + →DC [1] = −4a + b + 3a = b − a [1] (b) →AM = →AB + ½→BC [1] = 4a + ½(b − a) = 7a/2 + b/2, that is 3.5a + 0.5b [1] (c) →DM = →DA + →AM = −b + 3.5a + 0.5b = 3.5a − 0.5b [1] Examiner insight: A route that goes against an arrow needs the sign changed. Writing the route out, as in part (a), is safer than doing it in your head.

8. (a) →BC = −6a + 3b = −3(2a − b) [1]. So →BC = −3→AB [1]. The vectors are parallel and share the point B, so A, B and C are collinear. [1] (b) The lengths are in the ratio 1 : 3, so AB : BC = 1 : 3 [1] Examiner insight: Parallel is not enough for a collinearity proof. The sentence about the shared point is what turns “parallel” into “on one line”.

9. (a) →AB = −a + b = b − a [1] (b) →AQ = ¼→AB, so →OQ = a + ¼(b − a) [1] = ¾a + ¼b [1] (c) →OP = ½a. →PQ = →OQ − →OP = ¾a + ¼b − ½a [1] = ¼(a + b) [1] (d) →PQ is a scalar multiple of a + b, so PQ is parallel to a + b. [1] Examiner insight: Converting AQ : QB = 1 : 3 into one quarter of the whole line comes first. Using 1 : 3 directly as a fraction is a common slip.

10. (a) →AB = (5 − 1, 4 − 2) = (4, 2) [1] (b) →AC = (k − 1, 6) [1]. Collinear, so →AC is a multiple of →AB. The bottom entries give 6 = 3 × 2, so the multiple is 3 [1]. k − 1 = 12, so k = 13 [1] Examiner insight: Find the scalar from the entry that has no unknown, then use it for the other. State why the multiple exists, because A is shared by both vectors.

11. →PR = →PQ + →QR = a + c, since →QR = →PS [1] Midpoint of PR: →PM = ½(a + c) [1] →QS = →QP + →PS = −a + c [1] Midpoint of QS from P: →PN = →PQ + ½→QS = a + ½(c − a) [1] = ½a + ½c [1] Both midpoints are at the same vector ½(a + c) from P, so they are the same point, X. The diagonals bisect each other. [1] Examiner insight: A proof needs a concluding sentence that says what has been shown. Finding both midpoints from the same starting point makes the comparison clear.

12. (a) →AB = b − a [1] (b) →AQ = ⅓→AB, so →OQ = a + ⅓(b − a) [1] = ⅔a + ⅓b [1] (c) →PQ = →OQ − ½a [1] = ⅙a + ⅓b [1] (d) →PR = →OR − →OP = −b − ½a [1] = −½(a + 2b) = −3(⅙a + ⅓b) [1]. So →PR = −3→PQ. The vectors are parallel and share P, so P, Q and R are collinear. [1] Examiner insight: Take the common factor out in a way that matches the other vector, here ⅙a + ⅓b. Then the multiple, −3, can be read off directly.

Where marks are usually lost

• Using A minus B instead of B minus A for →AB.
• Dropping a sign when subtracting a vector with a negative entry.
• Multiplying only one entry of a column by a scalar.
• Using a ratio such as 1 : 3 as the fraction of the line, instead of one quarter.
• Not collecting terms in a and b into a single expression.
• Proving two vectors are parallel and then not saying the points are collinear.
• Giving one numerical example when a general proof was needed.
• Writing →AB when the working shows →BA.

Next steps

• Re-read the revision notes for the key facts and the proof table.
• Go back to the study guide for any step that felt unfamiliar.
• Visit the OxfordAQA IGCSE Mathematics hub and tick off statements on the checklist.
• Take the free 10-minute diagnostics to find your next weak spot.
• Book a free trial class if you would like a teacher to go through these questions with you.

Official syllabus

OxfordAQA International GCSE Mathematics (9260), specification Version 5.1 (for exams May/June 2018 onwards), published by OxfordAQA. Statement G22, section 3.3.3 Transformations, Matrices and Vectors.