Further Mathematics Paper 2, Nov/Dec. 2014
 Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Main
Weakness/Remedies
Strength

Question 14

The position vectors of two points P and R are p = 4i – 2j and r = 2i + 3j respectively.  Find:

(a)   ê3p – 2rê;

(b)   the values of the scalars m and n such that 8i + 8j = mp + nr;

(c)   the cosine of the acute angle between p and r leaving the answer in surd form.

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Observation

This question was reported to be quite unpopular with the candidates as majority of them did not attempt it.  However, majority of those who attempted it performed well.

In part (a), candidates were expected to obtain 3p as 3(4i - 2j) = 12i - 6j.
2r = 2(2i + 3j) = 4i + 6j. Therefore,  ê3p – 2rê = ê12i – 6j) – (4i + 6j) ê = ê8i – 12jê = 2 =   = 4.

In part (b) candidates were expected to respond as follows:
mp = m(4i - 2j) = 4mi – 2mj; nr = n(2i + 3j) = 2ni + 3nj.  Therefore mp + nr = (4m + 2n)i + (3n – 2m)j. Since mp + nr = 8i + 8j, it implied that 4m + 2n = 8 and 3n – 2m = 8.  Solving these equations simultaneously gave n = 3 and m =  .

In part (c), candidates were expected to apply the dot product thus:
If θ was the acute angle between p and r, then  = . êpê =   = ;  êrê=  =  and p ٠r = (4i – 2j).(2i+3j) = 2 Therefore, cos θ =  = .

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