Answer:(a) i) Vector BC = 3/2 a + 5bii) Vector AM = 15/4 a + 5/2 b(b) Vector QP = -15/4 b where k = -15/4Step-by-step explanation:* Lets explain how to solve this problem∵ ABCD is a trapezium∵ AB // DC∵ The vector AB = 3a∵ Vector DC = 3/2 vector AB∴ Vector DC = 3/2 × 3a = 9/2 a∵ Vector AD = 5b(a)i) ∵ Vector BC = vector BA + vector AD + vector DC∵ Vector AB = 3a , then vector BA = -3a∵ Vector AD = 5b , vector DC = 9/2 a∴ Vector BC = -3a + 5b + 9/2 a = (-3a + 9/2 a) + 5b∴ Vector BC = 3/2 a + 5bii) ∵ Vector AM = vector AB + vector BM∵ M is the mid-point of BC∴ Vector BM = 1/2 vector BC∵ Vector BC = 3/2 a + 5b∴ Vector BM = 1/2(3/2 a + 5b) = (1/2 × 3/2) a + (1/2 × 5) b∴ Vector BM = 3/4 a + 5/2 b∴ Vector AB = 3a∴ Vector AM = 3a + 3/4 a + 5/2 b = (3a + 3/4 a ) + 5/2 b∴ Vector AM = 15/4 a + 5/2 b(2)∵ 7 DQ = 5 QC ⇒ divide both sides by 7∴ DQ = 5/7 DC ∴ The line DC = 7 + 5 = 12 parts ⇒ DQ 5 parts and QC 7 parts∵ DQ = 5/12 DC∵ Vector DC = 9/2 a∴ Vector DQ = 5/12 (9/2 a) = 45/24 a ⇒ divide up and down by 3∴ Vector DQ = 15/8 a∵ P is the mid point of AM∴ Vector AP = 1/2(15/4 a + 5/2 b) = (1/2 × 15/4) a + (1/2 × 5/2) b∴ Vector AP = 15/8 a + 5/4 b∵ Vector QP = QD + DA + AP∵ Vector DQ = 15/8 , then vector QD = -15/8 a∵ Vector AD = 5b , then vector DA = -5b∴ Vector QP = -15/8 a + -5b + 15/8 a + 5/4 b∴ Vector QP = (-15/8 a + 15/8 a) + (-5b + 5/4 b)∴ Vector QP = -15/4 b∵ -15/4 is constant∴ Vector QP = k b ⇒ proved