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0580/41 Oct/Nov 2023, Question 10: Vector Proofs

Cambridge IGCSE Mathematics 0580 · Paper 41 (Extended) · Calculator · 13 marks

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The question, in short

Triangle ABC with B (1, −10), A (4, 14) and vector CA = (−11, 8); work out point C, vector BA, length of CA, then use triangle OMN with vectors a and b to find OR, k, c and MT.

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

Part (ai) · 2 marks

Using B (1, −10), A (4, 14) and CA = (−11, 8), determine the coordinates of point C.

  1. 1→CA = A − C, so C = A − →CA
  2. 2C = (4 − (−11), 14 − 8) = (15, 6)

Answer(15, 6)

Part (aii) · 1 mark

Calculate the vector BA from the coordinates of A and B.

  1. 1→BA = (4 − 1, 14 − (−10)) = column vector (3, 24)

Answercolumn vector (3, 24)

Part (aiii) · 2 marks

Work out the magnitude of the vector CA.

  1. 1|→CA| = √((−11)² + 8²) = √185
  2. 2= 13.60… ≈ 13.6

Answer13.6 or 13.60…

Part (bi) · 3 marks

With MR : RN = 3 : 2 on MN, prove that vector OR equals 2/5 a + 3/5 b.

  1. 1→MN = −a + b = b − a
  2. 2→MR = 3/5(b − a)
  3. 3→OR = a + 3/5(b − a) = 2/5 a + 3/5 b

Answera + 3/5(b − a) or b + 2/5(a − b) leading to 2/5 a + 3/5 b with no errors

Part (biia) · 4 marks

Given NT = 4a + kb and OT = c times OR, with ORT straight, find k and c.

  1. 1→OT = →ON + →NT = b + 4a + kb = 4a + (1 + k)b
  2. 2Also →OT = c(2/5 a + 3/5 b)
  3. 3a terms: 2/5 c = 4, so c = 10
  4. 4b terms: 3/5 × 10 = 1 + k, 6 = 1 + k, k = 5

Answerk = 5, c = 10

Part (biib) · 1 mark

Using your results, determine the vector MT.

  1. 1→MT = →MO + →OT = −a + 4a + 6b
  2. 2= 3a + 6b

Answer3a + 6b

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Also searched as: 0580/41/O/N/23 · 0580_w23_qp_41 · 0580 w23 41 · question 10.
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