Every vectors in two dimensions question from 8 Cambridge IGCSE Mathematics (0580) past papers (May/June 2025 to May/June 2026), syllabus section 7.2. 8 Extended-tier questions. Each question is labelled with its question paper code (for example 0580/21/M/J/26) and every final answer matches the official Cambridge mark scheme. Try each question first, then click “Show solution” to check your working. This page is part of our IGCSE Maths 0580 Transformations and Vectors past paper questions.
0580 Vectors in Two Dimensions past paper questions: Extended (Papers 2 and 4)
May/June 2026 · Paper 2 · 0580/21/M/J/26 · Q8(b)
P is the point (0, 4) and Q is the point (1, 8). Find the vector QP.
Show solution
- The vector QP goes from Q to P, so subtract Q from P.
- x: 0 − 1 = −1. y: 4 − 8 = −4.
- QP = (−1, −4) (as a column vector)
May/June 2026 · Paper 2 · 0580/23/M/J/26 · Q10
The position vector of A is (4, −1) and the vector AB = (−7, 5), written as column vectors. Find the position vector of B.
Show solution
- The position vector of B is OB = OA + AB.
- Add the top numbers: 4 + (−7) = −3. Add the bottom numbers: −1 + 5 = 4.
- OB = (−3, 4) (as a column vector)
May/June 2026 · Paper 4 · 0580/42/M/J/26 · Q5
P is the point (−3, 5) and PQ is the column vector (8, −3). Find the coordinates of the point Q.
Show solution
- The vector (8, −3) means 8 right and 3 down from P.
- Add to the coordinates of P: x = −3 + 8 = 5 and y = 5 + (−3) = 2.
- Q = (5, 2)
Feb/Mar 2026 · Paper 2 · 0580/22/F/M/26 · Q19(a)
Point A has coordinates (−4, 3) and point B has coordinates (4, 7). Find the vector BA.
Show solution
- The vector BA goes from B to A, so subtract B from A.
- x: −4 − 4 = −8. y: 3 − 7 = −4.
- BA = (−8, −4) (as a column vector)
Oct/Nov 2025 · Paper 2 · 0580/23/O/N/25 · Q8
DE is the column vector (3, −4). (a) Find 5DE. (b) Find |DE|. (c) D is the point (−2, −3). Find the coordinates of the point E.
Show solution
- Multiply each part of the vector by 5: 5 × 3 = 15 and 5 × (−4) = −20.
- 5DE = (15, −20)
- The magnitude (length) of a vector (a, b) is √(a² + b²).
- |DE| = √(3² + (−4)²) = √(9 + 16) = √25 = 5
- Start at D(−2, −3) and move by the vector (3, −4): 3 right and 4 down.
- E = (−2 + 3, −3 − 4) = (1, −7)
Oct/Nov 2025 · Paper 4 · 0580/41/O/N/25 · Q10
A is the point (2, 1) and AB is the column vector (2, 4). Find the coordinates of B.
Show solution
- A column vector (2, 4) means move 2 to the right and 4 up.
- Start at A(2, 1).
- x-coordinate: 2 + 2 = 4. y-coordinate: 1 + 4 = 5.
- B = (4, 5)
May/June 2025 · Paper 2 · 0580/21/M/J/25 · Q8
m = (11, 5) and n = (8, −3), written as column vectors.
(a) Find 2m − n.
(b) The vector (5, √y) has a magnitude of 7. Find the value of y.
Show solution
- Multiply each part of m by 2: 2m = (22, 10).
- Subtract n part by part: (22 − 8, 10 − (−3)) = (14, 13).
- 2m − n = (14, 13)
- The magnitude (length) of a vector (a, b) is √(a² + b²).
- √(5² + (√y)²) = 7. Square both sides: 25 + y = 49.
- y = 49 − 25 = 24
May/June 2025 · Paper 2 · 0580/22/M/J/25 · Q9
A is the point (3, −1). AB = (2, −4) as a column vector.
(a) AC = 2AB. Find the coordinates of the point C.
(b) The length of AB is k√5. Find the value of k.
(c) P is a point on AB. AP : PB = 1 : 3. Find the position vector of P.
Show solution
- AC = 2AB, so multiply each part of AB by 2: AC = (4, −8).
- Start at A(3, −1) and move by (4, −8): C = (3 + 4, −1 − 8).
- C = (7, −9)
- The length of a vector (a, b) is √(a² + b²).
- |AB| = √(2² + (−4)²) = √(4 + 16) = √20.
- Simplify: √20 = √4 × √5 = 2√5, so k = 2
- AP : PB = 1 : 3, so P is 1⁄4 of the way from A to B (1 part out of 1 + 3 = 4).
- AP = 1⁄4 × (2, −4) = (0.5, −1).
- OP = OA + AP = (3, −1) + (0.5, −1) = (3.5, −2)
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