Vectors Practice Answers
Uploaded by currymuncher Β· 4 March 2025
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1 Worksheet 1 β Vectors 1. Answer (a) magnitude of π΄π΅#####β = &π΄π΅#####β& = β3! + 4! = 5 (b) magnitude of π΅πΆ#####β = &π΅πΆ#####β& = -(β5)! + (β12)! = 13 (c) magnitude of πΆπ·#####β = &πΆπ·#####β& = -3! + (β5)! = 5.83 (3.s.f) 2. Answer (a) ππ΄#####β = 7β4 6 9 ππ΅#####β = : β1 π¦ β 2< (b) ππΆ#####β = ππ΄#####β + π΄πΆ#####β = 7β4 6 9 + 7 4 β109 = 7 0 β49 C = (0, β4) (c) π΄π#####β = βππ΄#####β = 7 4 β69 (d) π΄π΅#####β = π΄π#####β + ππ΅#####β = 7 4 β69 + : β1 π¦ β 2< = : 3 π¦ β 8< π΄π΅#####β = ππ΄πΆ#####β : 3 π¦ β 8< = π 7 4 β109 : 3 π¦ β 8< = 7 4π β10π9 3 = 4π π = " # π¦ β 8 = β10 7 " #9 π¦ = β7.5 + 8 = $ ! 3. Answer (a) &ππ#####β& = β3! + 4! = 5 (b) ππ######β = ππ#####β + ππ#####β
2 ππ#####β = 70 39 ππ######β = 70 39 + 7β3 4 9 = 7β3 7 9 π = (β3, 7) (c) ππ #####β = 7π₯ 09 ππ #####β = ππ#####β + ππ #####β = β 70 39 + 7π₯ 09 = 7 π₯ β39 (d) ππ#####β = πππ #####β 7β3 4 9 = π 7 π₯ β39 7β3 4 9 = 7 π₯π β3π9 β3 = π₯π 4 = β3π π = β # " (e) β3 = π₯π β3 = π₯ 7β # "9 π₯ = % # ππ #####β = F2 $ # 0 G 4. Answer (a) ππ΄#####β = 75 29 and ππ΅#####β = 7 6 β39 π΄π΅#####β = π΄π#####β + ππ΅#####β = β 75 29 + 7 6 β39 = 7β5 β29 + 7 6 β39 = 7 1 β59 &π΄π΅#####β& = -1! + (β5)! = 5.10 units
3 (b) π of π΄π΅ = &"&! '&( = β5 π¦ = ππ₯ + π 2 = β5(5) + π π = 27 π¦ = β5π₯ + 27 (c) π΄π΅#####β = 2π΄πΆ#####β = 2π΅πΆ#####β $ ! π΄π΅#####β = π΄πΆ#####β = π΅πΆ#####β π΄πΆ#####β = $ ! 7 1 β59 = L $ ! β ( ! M ππΆ#####β = ππ΄#####β + π΄πΆ#####β = 75 29 + L $ ! β ( ! M = L 5 $ ! β $ ! M πΆ = (5.5, β0.5) 5. ππ#####β = ππ######β + ππ#####β ππ######β = βππ######β = βπ ππ#####β = " ( ππ#####β = " ( π ππ#####β = βπ + + " ( π 6. Answer (a) Answer (i) ππ #####β = ππ#####β + ππ #####β ππ #####β = 2ππ#####β = 2π ππ #####β = π + 2π (ii) ππ #####β = ππ#####β + ππ #####β ππ#####β = 3ππ#####β = β3ππ#####β = β3π ππ #####β = β3π + π + 2π = β2π + 2π (iii) ππ#####β = ππ #####β + π π#####β
4 ππ #####β = π (since Q is midpoint of PR, PQ = QR) π π#####β = $ ' π π#####β = β $ ' ππ #####β = β $ ' (β2π + 2π) ππ#####β = π β $ ' (β2π + 2π) = π + ! ' π β ! ' π = ! ' π + # ' π = $ " (π + 2π) (b) ππ#####β = πππ #####β ππ#####β = $ " ππ #####β Since they are a factor of one another, they are parallel. 7. Answer 8. Answer
5 9. Answer
6 Worksheet 2 β Vectors 1. Answer 2. Answer
7 3.
8 4. Answer (a) Answer (i) π΄πΆ#####β = π΄π·#####β + π·πΆ#####β = 4π + " ( π΄π΅#####β = 4π + " ( (5π) = 4π + 3π (ii) π΅π#####β = π΅π΄#####β + π΄π#####β = β5π + ( ) π΄πΆ#####β = β5π + ( ) (4π + 3π ) = β5π + !*+ ) + $(, ) = ( ! π β !(, ) = ( ) (4π β 5π) (iii) ππ·#####β = ππ΄#####β + π΄π·#####β = β ( ) π΄πΆ#####β + 4π
9 = β ( ) (4π + 3π ) + 4π = " ! π β $( ) π = " ) (4π β 5π) (b) π΅π#####β = πππ·#####β ( ) (4π β 5π) = " ) π(4π β 5π) ( ) = " ) π π = ( " π΅π#####β = ( " ππ·#####β Thus B, X and D lie on the same straight line. 5. Answer (a) Answer (i) π΄π΅#####β = π΄π#####β + ππ΅#####β = β 7β1 5 9 + 7 2 β39 = 7 3 β89 (ii) β3! + 8! = 8.54 units (3.s.f.) (iii) ππΆ#####β = ππ΄#####β + π΄πΆ#####β = 7β1
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