Cambridge O Level Mathematics: Vectors in Two Dimensions: Column Vectors, Magnitude and Vector Geometry (4024) – 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 the Cambridge O Level Mathematics (Syllabus D) 4024 syllabus for 2025, 2026 and 2027, sections 7.2, 7.3 and 7.4. The syllabus is not tiered, so all candidates cover this content. Vectors are written as AB or a, and a column vector is written (x, y) with x above y in print. Read the study guide for full explanations, use the revision notes for quick recall, and check the course hub and the checklist. The final questions combine several skills. Give surds in exact form unless a decimal is asked for.

Questions

1. A translation maps the point A(−3, 4) onto the point A’(2, −1). Write down the column vector of the translation. [2]

2. p = (3, −2) and q = (−4, 6). Find, as column vectors,

(a) p + q [1] (b) 2p − q [2]

3. Calculate

(a) |(9, −12)| [2] (b) the positive value of k for which |(k, 5)| = 13. [2]

4. The position vectors of points A and B, relative to an origin O, are OA = (2, 5) and OB = (8, −3).

(a) Find AB as a column vector. [2] (b) Calculate |AB|. [1] (c) M is the midpoint of AB. Find the position vector of M. [2]

5. ABCD is a parallelogram. AB = a and AD = b. The point E is the midpoint of BC. Write each of these in terms of a and b, in its simplest form:

(a) AC [1] (b) AE [1] (c) DE [2]

6. OP = a + 2b, OQ = 3a + b and OR = 7a − b. Show that P, Q and R lie on a straight line. [4]

7. In triangle OAB, OA = a and OB = b. The point P lies on AB so that AP : PB = 3 : 1.

(a) Write AB in terms of a and b. [1] (b) Find OP in terms of a and b. [2] (c) Write PB in terms of a and b. [1]

8. In triangle OAB, OA = a and OB = b. The point M lies on OA with OM = (2/3)OA, and the point N lies on OB with ON = (2/3)OB.

(a) Find MN in terms of a and b. [2] (b) Write down two facts about MN and AB. [2] (c) The area of triangle OAB is 45 cm². Find the area of triangle OMN. [2]

9. The vectors (3, k − 1) and (6, 8) are parallel. Find the value of k. [3]

10. P is the point (−3, 2), Q is the point (5, 6) and R is the point (11, 9).

(a) Find the column vectors PQ and QR. [2] (b) Show that P, Q and R are collinear. [2] (c) Find the ratio PQ : QR in its simplest form. [1] (d) Calculate |PQ|, giving your answer as a decimal to 3 significant figures. [2]

11. OABC is a parallelogram with OA = a and OC = c. The point M is the midpoint of AB. The point N lies on BC with BN : NC = 1 : 3.

(a) Write OB and OM in terms of a and c. [2] (b) Find ON in terms of a and c. [2] (c) Show that MN = −(1/4)a + (1/2)c. [1] (d) Now a = (8, 4) and c = (2, 6). Find MN as a column vector and calculate |MN|. [2] (e) Show that MN is not parallel to OB. [1]

Answers

1. Image minus original: (2 − (−3), −1 − 4) [1] = (5, −5) [1] [2] Examiner insight: Take the start point away from the end point. Swapping them gives (−5, 5), the reverse translation, and a direction error can cost you the answer even when the method is clear.

2. (a) (3 + (−4), −2 + 6) = (−1, 4) [1] (b) 2p = (6, −4) [1]; 2p − q = (6 − (−4), −4 − 6) = (10, −10) [1] [3] Examiner insight: Write 2p on its own line first. The most common slip is 6 − 4 = 2 instead of 6 − (−4) = 10, and an intermediate line shows your method even if a final entry is wrong.

3. (a) √(81 + 144) [1] = √225 = 15 [1] (b) k² + 25 = 169 [1]; k² = 144, so k = 12 [1] [4] Examiner insight: Square the entries before adding them. Square (−12) as +144, not −144. For part (b), use the word “positive” to reject −12 explicitly.

4. (a) AB = OB − OA [1] = (8 − 2, −3 − 5) = (6, −8) [1] (b) √(36 + 64) = √100 = 10 [1] (c) OM = ½(OA + OB) = ½(10, 2) [1] = (5, 1) [1] [5] Examiner insight: Part (c) asks for a position vector, so give a column, not coordinates in brackets with the word “point”. Writing the formula first shows your method even if the arithmetic slips.

5. (a) AC = AB + BC = a + b [1] (since BC = AD = b) (b) BE = ½b, so AE = AB + BE = a + ½b [1] (c) DE = DA + AE = −b + a + ½b [1] = a − ½b [1] [4] Examiner insight: Show the route (DA then AE). Using DA = −b is the step many answers get wrong, and a visible route lets a marker see where a sign went astray.

6. PQ = OQ − OP = (3a + b) − (a + 2b) = 2a − b [1] QR = OR − OQ = (7a − b) − (3a + b) = 4a − 2b [1] QR = 2(2a − b) = 2 PQ, so PQ and QR are parallel [1] Q is common to both, so P, Q and R are collinear (a straight line) [1] [4] Examiner insight: A “show that” needs the conclusion in words. Parallel vectors alone do not prove collinearity, so name the shared point Q.

7. (a) AB = OB − OA = b − a [1] (b) AP = ¾ AB, as 3 of the 4 parts [1]; OP = a + ¾(b − a) = ¼a + ¾b [1] (c) PB = ¼ AB = ¼(b − a) [1] [4] Examiner insight: Convert the ratio 3 : 1 into the fraction 3/4 before you write anything else. The coefficients of a and b in OP add to 1, which is a quick check.

8. (a) OM = ⅔a and ON = ⅔b [1]; MN = ON − OM = ⅔b − ⅔a = ⅔(b − a) [1] (b) MN is parallel to AB [1]; MN is ⅔ the length of AB [1] (c) The triangles are similar with length ratio 2 : 3, so the area ratio is 4 : 9 [1]; area = (4/9) × 45 = 20 cm² [1] [6] Examiner insight: Use the square of the length ratio for area. Using 2/3 directly gives 30, a typical slip, so give the similarity reason in words.

9. (6, 8) = 2 × (3, 4) [1]; so k − 1 = 4 [1]; k = 5 [1] [3] Examiner insight: Find the scalar from the entries that are known, then apply it to the unknown entry. Check by substituting: (3, 4) and (6, 8) are indeed parallel.

10. (a) PQ = (5 − (−3), 6 − 2) = (8, 4) [1]; QR = (11 − 5, 9 − 6) = (6, 3) [1] (b) QR = ¾ PQ, since ¾ × 8 = 6 and ¾ × 4 = 3 [1]; PQ and QR are parallel and share Q, so P, Q and R are collinear [1] (c) PQ : QR = 8 : 6 = 4 : 3 [1] (d) |PQ| = √(64 + 16) = √80 [1] = 8.94 to 3 s.f. [1] [7] Examiner insight: In a multi-part question each part is separate, so state the scalar in part (b) in full. Give the decimal in part (d) only after writing the exact surd 4√5, because rounding early can shift the third figure.

11. (a) OB = OA + AB = a + c, because AB = OC [1]; OM = OA + ½AB = a + ½c [1] (b) BC = OC − OB = c − (a + c) = −a [1]; BN = ¼BC = −¼a, so ON = OB + BN = a + c − ¼a = ¾a + c [1] (c) MN = ON − OM = (¾a + c) − (a + ½c) [1] = −¼a + ½c, as required (d) MN = −¼(8, 4) + ½(2, 6) = (−2, −1) + (1, 3) = (−1, 2) [1]; |MN| = √(1 + 4) = √5 [1] (e) OB = (10, 10). MN would need to equal k × OB, which needs −1 = 10k and 2 = 10k, which give different k [1], so MN is not parallel to OB [8] Examiner insight: In a “show that” the printed answer is given, so the working is what you are showing. Write the full subtraction in part (c), not just the result.

Where marks are usually lost

• Giving OA − OB when AB is asked for, which reverses the answer.
• Subtracting a negative entry as if it were positive, such as 5 − (−2) = 3.
• Using ratio parts as the fraction (2/3 for a 2 : 3 split).
• Stopping a collinearity proof at “parallel” with no shared point and no conclusion.
• Scaling area by the length ratio instead of its square.
• Rounding a surd too early and carrying a rounded value forward.
• Leaving an answer as a number when a column vector is requested, and the reverse.
• Treating |a + b| as |a| + |b|.
• Not collecting like terms, so that an answer such as a + ½b − b is left unsimplified.

Next steps

• Return to the revision notes for quick recall.
• Re-read the study guide for any question you could not start.
• Visit the course hub and tick items off the checklist.
• Take one of our free 10-minute diagnostics to find your next weak area.
• Book a free trial class if you would like help from a teacher.

Official syllabus

Cambridge O Level Mathematics (Syllabus D) 4024 syllabus for 2025, 2026 and 2027, published by Cambridge International. Sections 7.2, 7.3 and 7.4.