waecE-LEARNING
Further Mathematics Paper 2, Nov/Dec. 2011  
Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Main
General Comments
Weakness/Remedies
Strength
Question 16
  1. The position vectors of points P, Q and R are 5i + 3j, 8ij and 11i – 5j respectively.
  2. Show that P, Q and R are collinear.
  3. Find the scalars k1 and k2 such that 37ij = k1p + k2 r where p and r are position vectors of P and R respectively.
  4. Given that m = 3i – 4j and n = 6i + 4j, find the angle between the two vectors, correct to the nearest degree.
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Observation

It was reported that this question was attempted by majority of the candidates and their performance was described as fair. Candidates were expected to respond as follows:
In part (a), gradient of PQ =  = . Gradient of QR =  = . Since the gradients were the same, it implied that the points were collinear. K1p + k2r = k1(5i + 3j) + k2(11i – 5j) = (5k1 + 11k2)i + (3k1 – 5k2)j. If k1p + k2r = 37ij, then 5k1 + 11k2 = 37 and 3k1 – 5k2 = -1. Solving these equations simultaneously gave k1 = 3 and k2 = 2.
In part (b), candidates were expected to recall that if θ was the angle between two vectors m and n, then cos θ = . Here, m = 3i – 4j and n = 6i + 4j. |m| =  = 5. |n| =  = . Therefore, cos θ =  =  = 0.0555. hence, θ = cos-1(0.0555) = 86.82o = 87o to the nearest degree.

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