Answer:
(a) i) Vector BC = 3/2 a + 5b
ii) Vector AM = 15/4 a + 5/2 b
(b) Vector QP = -15/4 b where k = -15/4
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 + 5b
ii) ∵ 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