AQA A-Level Mathematics: J: Vectors (7357) – 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 Section J: Vectors (J1 to J5) of the AQA A-level Mathematics (7357) specification, version 1.3 (31 January 2018), for A-level exams from June 2018 onwards. The specification assesses Section J on Paper 2. A calculator is required in every 7357 paper, but “show that” parts need every step written out. Give answers to 3 significant figures unless a question asks for an exact value.

Learn the content first with the Vectors study guide and the Vectors revision notes. Course hub: /boards/aqa/a-level/mathematics/. Printable checklist: /checklists/aqa/a-level/mathematics/. Not sure where to start? Try the free 10-minute diagnostics.

Questions

1. a = 8i − 15j. Find |a| and the unit vector in the direction of a. [2]

2. v = −6i + 6√3 j. Find the magnitude of v and the angle v makes with the positive x-direction, measured anticlockwise. [3]

3. A force of magnitude 18 N acts at 210° anticlockwise from i. Write the force in the form (pi + qj) N, giving p and q to 3 significant figures. [2]

4. p = 4i − 4j + 7k and q = i + 2j − 2k.

(a) Find |p|. [1] (b) Find the unit vector in the direction of q. [2] (c) Find |2p − 3q|, giving your answer to 3 significant figures. [3]

5. The vectors a = 3i − 2j and b = (t + 2)i + (t − 8)j are parallel. Find the value of t and express b in terms of a. [3]

6. The points A and B have coordinates (2, 1, −3) and (8, −2, 3).

(a) Find →AB and the distance AB. [3] (b) C lies on AB with AC : CB = 2 : 1. Find the position vector of C. [2] (c) Find the exact distance of C from the origin O. [1]

7. OABC is a parallelogram with →OA = a and →OC = c. M is the midpoint of AB. P lies on the diagonal OB with OP : PB = 2 : 1.

(a) Show that C, P and M are collinear. [5] (b) State the ratio CP : PM. [1]

8. In triangle OAB, →OA = a and →OB = b, where a and b are not parallel. C is the midpoint of OA. D lies on AB with AD : DB = 1 : 3. The lines OD and BC meet at X.

(a) Find →OD in terms of a and b. [2] (b) Find →OX in terms of a and b, and find the ratio OX : XD. [5]

9. Forces F₁ = (6i − j) N and F₂ = (−2i + 4j) N act on a particle.

(a) A third force F₃ is added so that the particle is in equilibrium. Find F₃. [2] (b) Instead, the third force is (−i + qj) N, and the resultant of the three forces has magnitude 5 N. Find the possible values of q. [4] (c) Take q = 1. The particle has mass 0.5 kg. Find its acceleration as a vector, the magnitude of the acceleration, and the angle the acceleration makes with i. [3]

10. Unit vectors i and j point east and north. Distances are in km and t is the time in hours after 12:00. At 12:00 ship A is at (3i + 10j) and ship B is at (−5i + j). A moves with constant velocity (4i − 2j) km h⁻¹ and B with constant velocity (8i + 2j) km h⁻¹.

(a) Find the speed of A. [2] (b) Write down r_A and r_B, and show that the vector from A to B at time t is (4t − 8)i + (4t − 9)j. [3] (c) Show that at 14:00 ship B is due south of ship A, and state how far apart they are. [2] (d) Find the time at which the ships are closest together and the least distance between them. [3]

11. The points A, B and C have position vectors a = i + 2k, b = 3i + 4j − 2k and c = 7i + 4j.

(a) Find →AB, →BC and →AC, and their exact magnitudes. [4] (b) Show that angle ABC is a right angle. [2] (c) Find the exact area of triangle ABC. [2] (d) D is the point such that ABCD is a parallelogram. Find the position vector of D. [2]

Answers

1. |a| = √(64 + 225) = √289 = 17 [1] â = (8i − 15j)/17 [1] [2] Examiner insight: the unit vector must be divided by the magnitude in both components; dividing only one component loses the accuracy mark.

2. |v| = √(36 + 108) = √144 = 12 [1] Acute angle with the x-axis: tan α = 6√3/6 = √3, so α = 60° [1] v points left and up (second quadrant), so the angle is 180° − 60° = 120° [1] [3] Examiner insight: a calculator’s arctan(−√3) gives −60°; without a sketch or quadrant statement, that answer usually earns the method mark only.

3. Components 18 cos 210° and 18 sin 210° [1] = −9√3 and −9, so the force is (−15.6i − 9j) N [1] [2] Examiner insight: both components are negative because 210° is in the third quadrant; a sign error on either one loses the accuracy mark.

4. (a) |p| = √(16 + 16 + 49) = √81 = 9 [1] (b) |q| = √(1 + 4 + 4) = 3 [1] q̂ = (i + 2j − 2k)/3 [1] (c) 2p − 3q = (8 − 3)i + (−8 − 6)j + (14 + 6)k = 5i − 14j + 20k [1] |2p − 3q| = √(25 + 196 + 400) = √621 [1] = 24.9 (3 s.f.) [1] Examiner insight: find the vector 2p − 3q first; 2|p| − 3|q| is a different quantity and earns nothing in (c).

5. Parallel means b = λa, so (t + 2)/3 = (t − 8)/(−2) [1] −2(t + 2) = 3(t − 8), so −2t − 4 = 3t − 24 [1] t = 4, giving b = 6i − 4j = 2a [1] [3] Examiner insight: cross-multiplying with the negative denominator is where the sign usually goes wrong; checking b = 2a confirms the answer.

6. (a) →AB = b − a [1] = 6i − 3j + 6k [1] AB = √(36 + 9 + 36) = √81 = 9 [1] (b) c = a + (2/3)→AB [1] = (2i + j − 3k) + (4i − 2j + 4k) = 6i − j + k [1] (c) OC = √(36 + 1 + 1) = √38 [1] Examiner insight: an answer of a − b in (a) can still earn follow-through credit for the distance, but the vector itself is marked wrong.

7. (a) →OB = a + c [1] →AB = →OC = c, so →OM = a + (1/2)c [1] →CM = →OM − →OC = a − (1/2)c [1] →OP = (2/3)(a + c), so →CP = (2/3)a + (2/3)c − c = (2/3)a − (1/3)c [1] →CP = (2/3)(a − (1/2)c) = (2/3)→CM. The vectors are parallel and share the point C, so C, P and M are collinear [1] (b) CP : PM = 2 : 1 [1] Examiner insight: this is a “show that”, so the final mark depends on saying both “parallel” and “common point C”; a scalar multiple with no conclusion is incomplete.

8. (a) →OD = a + (1/4)(b − a) [1] = (3/4)a + (1/4)b [1] (b) X on OD: →OX = λ((3/4)a + (1/4)b) [1] X on BC: →OX = b + μ((1/2)a − b) = (μ/2)a + (1 − μ)b [1] a and b are not parallel, so equate coefficients: (3/4)λ = μ/2 and (1/4)λ = 1 − μ [1] Solving: λ = 4/7, μ = 6/7 [1] →OX = (3/7)a + (1/7)b, and OX = (4/7)OD, so OX : XD = 4 : 3 [1] Examiner insight: using the same letter for both parameters makes the equations unsolvable; you need two different scalars, one for each line.

9. (a) F₁ + F₂ + F₃ = 0 [1] F₃ = −(4i + 3j) = (−4i − 3j) N [1] (b) R = (6 − 2 − 1)i + (−1 + 4 + q)j = 3i + (3 + q)j [1] 9 + (3 + q)² = 25 [1] (3 + q)² = 16, so 3 + q = ±4 [1] q = 1 or q = −7 [1] (c) R = 3i + 4j, so a = R/0.5 = (6i + 8j) m s⁻² [1] |a| = √(36 + 64) = 10 m s⁻² [1] tan θ = 8/6, so θ = 53.1° above i [1] Examiner insight: in (b) the negative square root gives a second valid answer; leaving out q = −7 loses the final mark.

10. (a) speed = |4i − 2j| = √(16 + 4) = √20 [1] = 4.47 km h⁻¹ [1] (b) r_A = (3 + 4t)i + (10 − 2t)j [1] r_B = (−5 + 8t)i + (1 + 2t)j [1] r_B − r_A = (−8 + 4t)i + (−9 + 4t)j = (4t − 8)i + (4t − 9)j [1] (c) At t = 2: r_B − r_A = 0i − 1j [1] The east component is 0 and the north component is negative, so B is due south of A, 1 km away [1] (d) d² = (4t − 8)² + (4t − 9)² = 32t² − 136t + 145; d(d²)/dt = 64t − 136 = 0, so t = 17/8 [1] d² = 32(17/8)² − 136(17/8) + 145 = 0.5, so d = √0.5 = 0.707 km [1] t = 2.125 h after noon, which is 14:07:30 [1] Examiner insight: in (b) the answer is given, so writing r_A − r_B and then “fixing” the signs will not earn the final mark; the subtraction must be in the right order.

11. (a) →AB = 2i + 4j − 4k, |→AB| = √36 = 6 [1] →BC = 4i + 0j + 2k, |→BC| = √20 = 2√5 [1] →AC = 6i + 4j − 2k [1] |→AC| = √56 = 2√14 [1] (b) AB² + BC² = 36 + 20 = 56 [1] = AC², so by the converse of Pythagoras angle ABC = 90° [1] (c) Area = (1/2)(6)(2√5) [1] = 6√5 [1] (d) For ABCD, →AD = →BC, so d = a + c − b [1] = (1 + 7 − 3)i + (0 + 4 − 4)j + (2 + 0 + 2)k = 5i + 4k [1] Examiner insight: “exact” means leaving √20 as 2√5 and the area as 6√5; a decimal such as 13.4 alone does not earn the accuracy mark in (c).

Where marks are usually lost

• Writing →AB as a − b. It is b − a: end minus start.
• Taking an angle straight from arctan without a sketch, so a second- or third-quadrant direction comes out wrong.
• Calculating 2|p| − 3|q| when the question wants |2p − 3q|.
• Showing two vectors are parallel but not naming the shared point before saying “collinear”.
• Using one parameter for two different lines when finding where they meet.
• Dropping the negative root when an unknown appears inside a magnitude.
• Writing the speed when the question asks for velocity, or leaving out units (N, m s⁻², km h⁻¹).
• In “show that” parts, skipping the line that turns a scalar multiple into a conclusion.
• Rounding decimals early in multi-step distance or time questions, so the final answer drifts.

Next steps

• Recap with the Vectors revision notes.
• Re-read any weak section in the Vectors study guide.
• Resolving forces with trigonometry: Trigonometry practice questions.
• Minimising a function, as in question 10(d): Differentiation practice questions.
• Course hub: /boards/aqa/a-level/mathematics/.
• Printable checklist: /checklists/aqa/a-level/mathematics/.
• All free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

AQA A-level Mathematics (7357) specification, version 1.3, 31 January 2018, for A-level exams June 2018 onwards: Section 3.11 J: Vectors (J1 to J5), with Appendix A: mathematical notation (vectors) and Appendix B: mathematical formulae and identities.