The Chief Examiner reported that this question was quite popular among the candidates and majority of those who attempted it performed poorly. Teachers were encouraged to put in more effort at teaching this topic. Candidates expected response was as follows:
= + + = (4i – 2j) + (-5i – j) + (-i – 5j) = -2i - 8j
|| = = . Therefore unit vector in the direction of = (-2 – 8)
= ( +4).
In part (c) candidates were expected to apply the dot product as follows:
If was the required angle then cos = = (-5i - j).(-4i + 2j) = 20 - 2 = 18.
| = = | = = . Therefore, Cos = = 0.07894. Hence = 38o, correct to the nearest degree. Using the formula Area =½ absin C, the area of the triangle was obtained as ½ × × × sin 37.8705 = 7.0 square units, correct to one decimal place
