Edexcel International GCSE Maths Vectors Past Paper Questions and Solutions (4MA1 Higher)

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Edexcel International GCSE Maths A (4MA1) · Higher · Vectors and Transformation Geometry

Vectors: Edexcel 4MA1 Past Paper Questions and Worked Solutions

This page collects every vectors question from the Pearson Edexcel International GCSE Mathematics A (4MA1) Higher tier Paper 1H and Paper 1HR, January 2020 to June 2025: 11 questions worth 56 marks. Questions run from easier to harder. Try each one first, then click “Show solution” for a step-by-step method with the reason for every step. All answers are checked against the official Pearson mark schemes.

12 marksHigherJanuary 2021 · Paper 1HR (4MA1/1HR) · Q13

Here are two vectors: AB = (5, 3) and CB = (−2, 4), written as column vectors (across, up). Find, as a column vector, AC.

Show solution
  1. AC = AB + BC, and BC = −CB = (2, −4)
  2. (5 + 2, 3 + (−4)) = (7, −1)

Answer: AC = (7, −1), i.e. 7 across and 1 down

28 marksHigherNovember 2021 · Paper 1H (4MA1/1H) · Q14

ABCDEF and GHIJKL are regular hexagons each with centre O. GHIJKL is an enlargement of ABCDEF, with centre O and scale factor 2. OA = a and OB = b.

ABCDEFGHIJKLOab

(a) Write the following vectors, in terms of a and b. Simplify your answers. (i) AB (ii) KI (iii) LD (5)

The triangle OAB has an area of 5 cm2.

(b) Calculate the area of the shaded region. (3)

Show solution
  1. Useful facts for a regular hexagon centred at O: OC = b − a, OD = −a, OE = −b, OF = a − b. The big hexagon’s vertices are twice as far out: OK = 2OE, OI = 2OC, OL = 2OF.

Part (a)(i)

  1. AB = AO + OB = −a + b

Answer: b − a

Part (a)(ii)

  1. OK = 2(−b) = −2b and OI = 2(b − a) = 2b − 2a
  2. KI = KO + OI = 2b + 2b − 2a

Answer: 4b − 2a

Part (a)(iii)

  1. OL = 2(a − b) = 2a − 2b and OD = −a
  2. LD = LO + OD = −2a + 2b − a

Answer: 2b − 3a

Part (b)

  1. The small hexagon is made of 6 triangles like OAB: area = 6 × 5 = 30 cm2
  2. The enlargement has scale factor 2, so areas multiply by 22 = 4: big hexagon = 4 × 30 = 120 cm2
  3. Shaded = 120 − 30

Answer: 90 cm2

35 marksHigherJune 2023 · Paper 1H (4MA1/1H) · Q19

OMA, ONB, MPB and NPA are straight lines. M is the midpoint of OA. ON : NB = 1 : 5. OM = a and ON = b.

OABMNPab

(a) Find in terms of a and b the vector AN (1)

(b) Use a vector method to find the ratio AP : PN (4)

Show solution

Part (a)

  1. OA = 2a because M is the midpoint. AN = AO + ON

Answer: −2a + b

Part (b)

  1. ON : NB = 1 : 5 so OB = 6 × ON and OB = 6b. Then MB = MO + OB = −a + 6b
  2. P is on AN: OP = OA + λAN = 2a + λ(b − 2a) = (2 − 2λ)a + λb
  3. P is on MB: OP = OM + μMB = a + μ(6b − a) = (1 − μ)a + 6μb
  4. Compare the b parts: λ = 6μ. Compare the a parts: 2 − 2λ = 1 − μ
  5. Substitute λ = 6μ: 2 − 12μ = 1 − μ, so 11μ = 1, μ = 1⁄11 and λ = 6⁄11
  6. AP = 6⁄11 of AN, so PN = 5⁄11 of AN.

Answer: AP : PN = 6 : 5

45 marksHigherJune 2023 · Paper 1HR (4MA1/1HR) · Q19

The diagram shows a triangle ABC where A, B and C represent the positions of three towns. AB = (7, −2) and BC = (−3, 5), written as column vectors (across, up). Pru travels directly from A to B and then directly from B to C. Yang travels directly from A to C. Given that the values for AB and BC are in kilometres, work out how much further Pru travels than Yang travels. Give your answer in km, correct to one decimal place.

Show solution
  1. The length of a vector (across, up) is √across2 + up2 (Pythagoras).
  2. AB = √72 + (−2)2 = √53 = 7.280… km
  3. BC = √(−3)2 + 52 = √34 = 5.830… km
  4. AC = AB + BC = (7 − 3, −2 + 5) = (4, 3), so AC = √16 + 9 = 5 km
  5. Pru travels 7.280… + 5.830… = 13.111… km
  6. Difference = 13.111… − 5

Answer: 8.1 km

55 marksHigherJanuary 2022 · Paper 1H (4MA1/1H) · Q22

The diagram shows triangle OAB. OA = 8a and OB = 6b. M is the point on OB such that OM : MB = 1 : 2. N is the midpoint of AB. P is the point of intersection of ON and AM. Using a vector method, find OP as a simplified expression in terms of a and b. Show your working clearly.

OABMNP
Show solution
  1. OM = 1⁄3 × 6b = 2b
  2. ON = OA + ½AB = 8a + ½(6b − 8a) = 4a + 3b
  3. P on ON: OP = λ(4a + 3b)
  4. P on AM: OP = OA + μAM = 8a + μ(2b − 8a) = (8 − 8μ)a + 2μb
  5. Compare: 4λ = 8 − 8μ and 3λ = 2μ, so μ = 1.5λ. Then 4λ = 8 − 12λ, giving λ = 0.5
  6. OP = 0.5(4a + 3b)

Answer: 2a + 1.5b

65 marksHigherJanuary 2023 · Paper 1HR (4MA1/1HR) · Q22

The diagram shows triangle OAB with OA extended to E. OA = a and OB = b. M is the point on OB such that OM : MB = 4 : 1. N is the point on AB such that AN : NB = 3 : 2. OA : AE = 5 : 3.

(a) Find an expression for ON in terms of a and b. Give your answer in its simplest form. (2)

(b) Use a vector method to show that MNE is a straight line. (3)

OAEBMNab
Show solution

Part (a)

  1. AB = b − a, and AN = 3⁄5 of AB
  2. ON = OA + 3⁄5AB = a + 3⁄5(b − a) = 2⁄5a + 3⁄5b

Answer: ON = 1⁄5(2a + 3b)

Part (b)

  1. OM = 4⁄5b and OE = 8⁄5a (because OE = 5 + 3 = 8 parts and OA = 5 parts)
  2. MN = ON − OM = 2⁄5a + 3⁄5b − 4⁄5b = 1⁄5(2a − b)
  3. ME = OE − OM = 8⁄5a − 4⁄5b = 4⁄5(2a − b)
  4. So ME = 4MN: the vectors are parallel, and they share the point M.

Answer: MNE is a straight line (shown)

76 marksHigherJanuary 2021 · Paper 1H (4MA1/1H) · Q23

OAB is a triangle. OA = 2a and OB = 2b. M is the midpoint of AB. N is the point on OB such that ON : NB = 2 : 1. P is the point on AN such that OPM is a straight line. Use a vector method to find OP : PM. Show your working clearly.

OABMNP
Show solution
  1. OM = OA + ½AB = 2a + ½(2b − 2a) = a + b
  2. ON = 2⁄3 × 2b = 4⁄3b, so AN = 4⁄3b − 2a
  3. P on OM: OP = λ(a + b)
  4. P on AN: OP = 2a + μ(4⁄3b − 2a) = (2 − 2μ)a + 4⁄3μb
  5. Compare: λ = 2 − 2μ and λ = 4⁄3μ. So 4⁄3μ = 2 − 2μ, giving 4μ = 6 − 6μ, μ = 0.6 and λ = 0.8
  6. OP = 0.8OM, so OP is 4⁄5 of OM and PM is 1⁄5.

Answer: OP : PM = 4 : 1

85 marksHigherJune 2025 · Paper 1H (4MA1/1H) · Q23

OAB is a triangle. P is the midpoint of OA. Q is a point on OB. ABR and PQR are straight lines. OA = 12a and OB = 8b.

OPQABR

(a) Express AB in terms of a and b. (1)

AB : BR = 1 : 2 and OQ = nb

(b) Use a vector method to find the value of n. (4)

Show solution

Part (a)

  1. Go from A to O then O to B: AB = AO + OB = −12a + 8b

Answer: 8b − 12a

Part (b)

  1. AB : BR = 1 : 2 means AR is 3 times AB: AR = 3(8b − 12a) = 24b − 36a
  2. P is the midpoint of OA, so PA = 6a. Then PR = PA + AR = 6a + 24b − 36a = 24b − 30a
  3. Q lies on PR, so PQ = λPR for some number λ. Then OQ = OP + λPR = 6a + λ(24b − 30a) = (6 − 30λ)a + 24λb
  4. Q also lies on OB, so OQ has no a part: 6 − 30λ = 0, so λ = 1⁄5
  5. OQ = 24 × 1⁄5b = 4.8b

Answer: n = 4.8 (or 24⁄5)

95 marksHigherJanuary 2022 · Paper 1HR (4MA1/1HR) · Q24

OXYZ is a trapezium. OX = a, XY = b and OZ = 3b. M is the midpoint of OX. N is the point such that MNZ and ONY are straight lines. Given that ON : OY = λ : 1, use a vector method to find the value of λ.

OXYZMN
Show solution
  1. OY = OX + XY = a + b, so ON = λ(a + b)
  2. OM = ½a and MZ = MO + OZ = −½a + 3b
  3. N on MZ: ON = ½a + μ(3b − ½a) = (½ − ½μ)a + 3μb
  4. Compare: λ = ½ − ½μ and λ = 3μ
  5. 3μ = ½ − ½μ, so 7⁄2μ = ½ and μ = 1⁄7

Answer: λ = 3⁄7

105 marksHigherJanuary 2023 · Paper 1H (4MA1/1H) · Q24

OAED is a quadrilateral. OA = 2a, OB = 2b and DE = 7a + 3b. B lies on AD with AB : BD = 1 : 2. The point C on AB is such that OCE is a straight line. Use a vector method to find the ratio of OC : CE.

AEDOBC2a2b7a + 3b
Show solution
  1. AB = 2b − 2a, and BD = 2 × AB, so BD = 4b − 4a
  2. OD = OB + BD = 2b + 4b − 4a = 6b − 4a
  3. OE = OD + DE = 6b − 4a + 7a + 3b = 3a + 9b
  4. C on OE: OC = λ(3a + 9b). C on AB: OC = 2a + μ(2b − 2a) = (2 − 2μ)a + 2μb
  5. Compare: 3λ = 2 − 2μ and 9λ = 2μ. Adding: 12λ = 2, so λ = 1⁄6
  6. OC = 1⁄6 of OE, so CE = 5⁄6 of OE.

Answer: OC : CE = 1 : 5

115 marksHigherJune 2021 · Paper 1H (4MA1/1H) · Q26

OACB is a trapezium. OA = 2a, OB = 5b and AC = 3b. The diagonals, OC and AB, of the trapezium intersect at the point P. Find and simplify an expression, in terms of a and b, for OP. Show your working clearly.

OACBP2a3b5b
Show solution
  1. OC = OA + AC = 2a + 3b, and AB = AO + OB = −2a + 5b
  2. P is on OC: OP = λ(2a + 3b) = 2λa + 3λb
  3. P is on AB: OP = 2a + μ(−2a + 5b) = (2 − 2μ)a + 5μb
  4. Compare a: 2λ = 2 − 2μ. Compare b: 3λ = 5μ, so λ = 5⁄3μ
  5. Substitute: 10⁄3μ = 2 − 2μ, so 10μ = 6 − 6μ, μ = 3⁄8 and λ = 5⁄8
  6. OP = 5⁄8(2a + 3b)

Answer: 5⁄4a + 15⁄8b

Tip: Quicker check: triangles APC and BPO are similar with ratio AC : OB = 3 : 5, so OP : PC = 5 : 3 and OP = 5⁄8OC.

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