OCR A Level Mathematics A: Vectors in two and three dimensions (H240) – 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 follow section 1.10 Vectors (1.10a to 1.10h) of the OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026). Questions on three dimensions and on kinematics are labelled A Level only. Direction angles are measured from the positive x-axis, in the interval [0°, 360°). Do the working by hand first, then check it.

If a topic is unfamiliar, begin with the study guide; for a quick refresher use the revision notes for a quick refresh. Course hub: /boards/ocr/a-level/mathematics/. Printable checklist: /checklists/ocr/a-level/mathematics/. Use the free 10-minute diagnostics to decide which topics to work on first.

Questions

1. Find the magnitude and the direction, to 1 decimal place, of the vector −5i − 2j. [3]

2. A force of 18 N acts in the direction 125°. Write it in the form xi + yj, giving x and y to 3 significant figures. [3]

3. (A Level only) Find the vector of magnitude 45 that has the same direction as 2i − 10j + 11k. [4]

4. (A Level only) The vectors a = 2i + pj − 3k and b = −6i + 9j + qk are parallel.

(a) Find p and q. [3] (b) Find |b|, giving your answer to 3 significant figures. [1]

5. (A Level only) The points A, B and C have coordinates (1, −1, 2), (3, 1, 3) and (2, 0, 7).

(a) Show that angle ABC is a right angle. [4] (b) Find the area of triangle ABC. [1]

6. A, B, C and D are the vertices of a quadrilateral with position vectors a, b, c and d. The midpoints of AB, BC, CD and DA are P, Q, R and S. Show that PQRS is a parallelogram. [5]

7. Three forces act on a particle: F₁ = (4i + 3j) N, F₂ = (−10i + 5j) N and F₃. The resultant of the three forces is (−3i + 4j) N.

(a) Find F₃. [2] (b) Find the magnitude and direction of F₃, giving the direction to 1 decimal place. [3] (c) The particle has mass 2.5 kg. Find its acceleration as a vector and the magnitude of that acceleration. [2]

8. (A Level only) A particle P has position vector (5i − 4j) m at time t = 0 s. It has velocity (−6i + 2j) m s⁻¹ and constant acceleration (2i − j) m s⁻².

(a) Find the velocity of P at t = 3 and show that P is then moving parallel to j. [2] (b) Find the position vector of P at t = 3. [2] (c) Find the distance of P from the origin at t = 3, to 3 significant figures. [2]

9. Two ships move at constant velocity. At time t hours, ship P has position (2i + 10j) + t(3i − 4j) km and ship Q has position (−4i − 2j) + t(6i + 2j) km.

(a) Write down the position vector of each ship when t = 1. [2] (b) Show that the ships collide, and find when and where. [3] (c) Find the distance between the ships when t = 0. [2] (d) Find the speed of each ship. [2]

10. (A Level only) ABCD is a parallelogram. Relative to an origin O, the vertices A, B and C are at a = i + 2k, b = 4i + 4j + 2k and c = 4i + 7j + 6k.

(a) Find →AB and |→AB|. [2] (b) Find the position vector of D. [2] (c) Show that ABCD is a rhombus. [2] (d) Find the position vector of the midpoint of AC and show that it is also the midpoint of BD. [2]

11. Two forces act on a particle: one of magnitude 12 N in the direction 30°, the other of magnitude 8 N in the direction 150°. Find the exact resultant in the form xi + yj, then its magnitude and direction to 1 decimal place. [6]

Answers

1. |v| = √(25 + 4) = √29 ≈ 5.4 [1]. The vector is in the third quadrant, with acute angle tan⁻¹(2/5) = 21.8° [1]. Direction = 180° + 21.8° = 201.8° [1]. Magnitude √29 ≈ 5.4 [3] Examiner insight: A sketch settles the quadrant. An answer that stops at the calculator’s 21.8° shows the vector has not been placed, and is likely to cost you credit.

2. x = 18 cos 125° [1] = −10.3 [1]. y = 18 sin 125° = 14.7 [1] [3] Examiner insight: Use the full angle in cos and sin and the signs look after themselves. Writing the answer as 10.3i + 14.7j loses the sign of x.

3. |2i − 10j + 11k| = √(4 + 100 + 121) = √225 = 15 [1]. Unit vector = (2i − 10j + 11k)/15 [1]. Multiply by 45 [1]: 6i − 30j + 33k [1] [4] Examiner insight: Dividing by the magnitude is the step that matters. Show the magnitude line before using it, so that one arithmetic slip still leaves the method visible.

4. (a) b = ka, so −6 = 2k and k = −3 [1]. Then 9 = k × p gives p = −3 [1], and q = k × (−3) = 9 [1]. (b) |b| = √(36 + 81 + 81) = √198 = 14.1 [1] [4] Examiner insight: Find the scale factor from the component with no unknown first. Checking that all three components fit b = −3a is a quick safeguard.

5. (a) →AB = 2i + 2j + k, so AB² = 9 [1]. →BC = −i − j + 4k, so BC² = 18 [1]. →AC = i + j + 5k, so AC² = 27 [1]. AB² + BC² = AC², so angle ABC is 90° by the converse of Pythagoras [1]. (b) Area = ½ × 3 × 3√2 = 9√2/2 ≈ 6.36 [1] [5] Examiner insight: On a “show that” question, give all three squared lengths and the sum. The conclusion in words, naming the right-angled vertex, completes the argument.

6. p = (a + b)/2, q = (b + c)/2, r = (c + d)/2 and s = (d + a)/2 [1]. →PQ = q − p = (c − a)/2 [1]. →SR = r − s = (c − a)/2 [1]. So →PQ = →SR [1]. Opposite sides PQ and SR are equal and parallel, so PQRS is a parallelogram [1] [5] Examiner insight: Use position vectors for the midpoints rather than a diagram. State that equal vectors mean equal and parallel sides, because that sentence is what turns the algebra into a proof.

7. (a) F₃ = (−3i + 4j) − (4i + 3j) − (−10i + 5j) [1] = 3i − 4j N [1]. (b) |F₃| = √(9 + 16) = 5 N [1]. Fourth quadrant, acute angle tan⁻¹(4/3) = 53.1° [1]. Direction = 360° − 53.1° = 306.9° [1]. (c) a = (−3i + 4j)/2.5 [1] = −1.2i + 1.6j m s⁻², with magnitude √(1.44 + 2.56) = 2 m s⁻² [1] [7] Examiner insight: Subtract all the known forces from the resultant. Reversing the sign (adding them instead) is a common slip, and a check that the three forces sum to the resultant catches it.

8. (a) v = (−6i + 2j) + 3(2i − j) [1] = −j, which has no i-component, so P is moving parallel to j [1]. (b) r = (5i − 4j) + 3(−6i + 2j) + 4.5(2i − j) [1] = −4i − 2.5j m [1]. (c) Distance = √(16 + 6.25) = √22.25 [1] = 4.72 m [1] [6] Examiner insight: The position formula needs the starting position added on. Leaving it out gives the displacement instead, which is a different quantity.

9. (a) P: (2 + 3)i + (10 − 4)j = 5i + 6j km [1]. Q: (−4 + 6)i + (−2 + 2)j = 2i km [1]. (b) i-components: 2 + 3t = −4 + 6t, so t = 2 [1]. j-components at t = 2: 10 − 8 = 2 and −2 + 4 = 2, so both agree [1]. They collide 2 hours after t = 0 at 8i + 2j km [1]. (c) →PQ at t = 0 = (−4i − 2j) − (2i + 10j) = −6i − 12j [1]. Distance = √(36 + 144) = √180 = 6√5 ≈ 13.4 km [1]. (d) P: √(9 + 16) = 5 km h⁻¹ [1]. Q: √(36 + 4) = 2√10 ≈ 6.32 km h⁻¹ [1] [9] Examiner insight: A collision needs both components to match at the same time. Solving one equation and stopping leaves the argument unfinished, so always substitute into the other.

10. (a) →AB = b − a = 3i + 4j [1]. |→AB| = √(9 + 16) = 5 [1]. (b) d = a + c − b [1] = i + 3j + 6k [1]. (c) →BC = c − b = 3j + 4k, so |→BC| = 5 [1]. Adjacent sides are equal and ABCD is a parallelogram, so it is a rhombus [1]. (d) Midpoint of AC = (a + c)/2 = (5/2)i + (7/2)j + 4k [1]. Midpoint of BD = (b + d)/2 = (5i + 7j + 8k)/2, which is the same point [1] [8] Examiner insight: For D, think of a closed route round the parallelogram: →AD = →BC. Writing that statement first makes the working easy to follow and easy to credit.

11. F₁ = 12 cos 30° i + 12 sin 30° j = 6√3 i + 6j [1]. F₂ = 8 cos 150° i + 8 sin 150° j = −4√3 i + 4j [1]. Resultant = 2√3 i + 10j N [1]. Magnitude = √(12 + 100) = √112 = 4√7 ≈ 10.6 N [1]. Both components are positive, so the angle is tan⁻¹(10/(2√3)) [1] = 70.9° [1] [6] Examiner insight: Work in components. Adding the two magnitudes (20 N) is a typical error. If the question asks for an exact resultant, keep the surds.

Where marks are usually lost

• Giving the acute angle from tan⁻¹ and not the direction in [0°, 360°).
• Getting →AB the wrong way round, so every sign in the answer is reversed.
• Squaring a negative component and writing a negative result.
• Rounding in the middle of a multi-part answer and carrying an inaccurate value forward.
• Ending a proof (collinear, parallelogram, right angle) without a conclusion in words.
• Adding magnitudes of forces or velocities instead of adding the vectors.
• Mixing up velocity and speed, or position and displacement.
• Using s = ut + ½at² for the position and omitting the starting position.
• Writing a direction for a 3D vector, which the specification excludes.

Next steps

• Review the method in the vectors revision notes.
• Re-read the vectors study guide for the topics you got wrong.
• Return to the OCR A Level Mathematics hub or print the checklist.
• Take all the free 10-minute diagnostics to find the next weak spot.
• Book a free trial class if you would like help from a teacher.

Official syllabus

OCR, A Level in Mathematics A (H240), Version 3.1 (August 2026), for first assessment in 2018, section 1.10 (Vectors), with the constant acceleration formulae from section 3.02e.