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0606/12 Feb/Mar 2025, Question 11: Vector Geometry

Cambridge IGCSE Additional Mathematics 0606 · Paper 12 (Non-calculator) · Non-calculator · 9 marks

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The question, in short

In triangle OAB with OA = a and OB = b, M is midpoint of OB and N satisfies ON = 3NA; lines BN and AM meet at X. Use BX = λBN and MX = μMA to find OX two ways, then find λ and μ.

Summarised in our own words. Use the official paper for the exact wording and the diagram.

Part (a) · 3 marks

Using BX = λBN, express OX in terms of a, b and λ.

  1. 1ON = (3/4)a
  2. 2BN = ON − OB = (3/4)a − b
  3. 3OX = OB + BX = b + λ((3/4)a − b)

AnswerOX = b + λ((3/4)a − b)

Part (b) · 2 marks

Using MX = μMA, express OX in terms of a, b and μ.

  1. 1OM = (1/2)b
  2. 2MA = a − (1/2)b
  3. 3OX = OM + MX = (1/2)b + μ(a − (1/2)b)

AnswerOX = (1/2)b + μ(a − (1/2)b)

Part (c) · 4 marks

Equate the two expressions for OX to find the values of λ and μ.

  1. 1b + λ((3/4)a − b) = (1/2)b + μ(a − (1/2)b)
  2. 2Compare a: (3/4)λ = μ
  3. 3Compare b: 1 − λ = 1/2 − (1/2)μ
  4. 41 − λ = 1/2 − (3/8)λ, so (5/8)λ = 1/2
  5. 5λ = 4/5, μ = 3/5

Answerλ = 4/5, μ = 3/5

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