1. The position vectors of points P, Q and R are 5i + 3j, 8i – j and 11i – 5j respectively.
2. Show that P, Q and R are collinear.
3. Find the scalars k1 and k2 such that 37i – j = 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.

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 = 37i – j, 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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