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