Answers308P1^
Start with the vector OQ→ = a 1 i + a 2 j.Length = aa 12 + 22
Now look at the triangle OQP.OP^2 = OQ^2 + QP^2= (^) (a 1 2 + a 2 2 ) + a 3 2
⇒ OP = aaa^21 + + 22 32
Exercise 8A (Page 261)
1 (i) (^3) i + 2 j
(ii)^5 i − 4 j
(iii)^3 i
(iv)^ − 3 i − j
2 For all question 2:
(i)
(^13 , 56.3°)
(ii)^
( 13 , −33.7°)
(iii)
( 4 2, −135°)
(iv) (^)
( 5, 116.6°)
(v)^
(5, −53.1°)
3 (i) 3.74
(ii) 4.47
(iii) 4.90
(iv) 3.32
(v) 7
(vi) 2.24
4 (i) 2 i − 2 j
(ii)^2 i
(iii)^ − 4 j
(iv) 4 j
(v) 5 k
(vi) −i − 2 j + 3 k
(vii)^ i + 2 j − 3 k
(viii)^4 i − 2 j + 4 k
(ix) 2 i − 2 k
(x) − 8 i + 10 j + k
5 (i) (^) A: 2i + 3 j, C: − 2 i + j
(ii) A
→
B = − 2 i + j, C
→
B = 2 i + 3 j
(iii) (a) A
→
B = O
→
C
(b) C
→
B = O
→
A
(iv) A parallelogram
Activity 8.1 (Page 266)
(i) (a) (^) F
(b) C
(c) Q
(d) (^) T
(e) S
(ii) (a) O
→
F
(b) O
→
E, C
→
F
(c) O
→
G, P
→
S, A
→
F
(d) B
→
D
(e) Q
→
S, P
→
T
Exercise 8B (Page 269)
1 (i) (^) 86
(ii) (^) 11
(iii) (^) 00
(iv) (^) −^81
(v) –3j
2 (i) 2 i + 3j + k
(ii) i – k
(iii) j – k
(iv) 3 i + 2j – 5 k
(v) –6k
3 (i) (a) b
(b) a + b
(c) – a + b
(ii) (a) (^12) (a + b)
(b) 12 (–a + b)
(iii) (^) PQRS is any parallelogram
and P
→
M = 12 P
→
R, Q
→
M = 12 Q
→
S
O
x
y
a 2 4
a 1
3
O 4
a
j
i