This question was attempted by majority of the candidates and they performed very well
in it. Candidates' expected responses were:
In part (a), 6p - 5q = 6(5i +3j) - 5(4i + 7j) = 30i + 18j - 20i - 35j = 10i -17j
/6p - 5q/ = √10²+ (-17)² = 19.723.
In part (b), mp + nq = 14i + 13j => m(5i + 3J) + n(4i + 7J) = 14i + 13j. Equating the like
terms we have: 5m + 4n = 15 and 3m + 7n = 13. Solving these two equations
simultaneously gave m = 2 and n = 1.
In part (c), candidates were expected to recall that if 0 is the acute angle between p and
q, then Cosθ =
p*g p*q = (5i + 3j)*(4i +7j) = 20 + 21 = 41.
/pl./q/
. /p/ =√5² + 3² =√34 , /q/ =√4² + 7² =√ 65.
Therefore, Cosθ =√41/√(34)√(65)= 0.872. This implied that θ = Cos-¹(0.872) = 29.29°.
