ANSWERS 729
3 Paola is 25 km from his starting point at a bearing of 347 o.
4a 600 km/h due north b 400 km/h due north
c 510 km/h at a bearing of 011 : 3 o
5 The ship is travelling at a speed of 23 : 3 knots at a bearing of
134 o.
EXERCISE 24D
1a bcdef2a¡QS! b ¡PR! c 0 (zero vector) d ¡RQ! e¡QS!
f¡RQ!
3 The plane must fly 4 : 57 owest of north at 501 : 6 km/h.
4aThe boat must head 26 : 6 owest of north. b 28 : 3 km/h
EXERCISE 24E.1
1a b c d2a¡ 4
2¢
b¡ 0
¡ 3¢
c¡¡ 3
¡ 4¢
d¡¡ 6
0¢
e¡¡ 6
4¢
f¡ 2
¡ 4¢
3a¡ 9
3¢
b¡ 2
5¢
c¡ 4
1¢
d¡¡ 2
4¢
e¡ 11
¡ 1¢
f¡¡ 4
¡ 1¢
g¡¡ 9
¡ 3¢
h¡¡ 2
¡ 5¢
i¡¡ 11
1¢
4a¡ 3
4¢
b¡¡ 4
¡ 2¢
c¡¡ 2
1¢
d¡ 3
¡ 3¢
e¡ 1
5¢
EXERCISE 24E.2
1a¡ 5
¡ 4¢
b¡ 5
¡ 4¢
c¡ 1
¡ 4¢
d¡ 1
¡ 4¢
e¡ 0
¡ 6¢
f¡ 0
¡ 6¢
g¡ 4
¡ 6¢
h¡ 3
¡ 7¢
2a¡ 1
6¢
b¡¡ 5
1¢
c¡¡ 6
4¢
dp
10 units
e
p
26 units f 5
p
2 units
3a¡¡ 3
¡ 3¢
b¡¡ 4
9¢
4ap
17 units b 6 units cp
13 units
dp
26 units e 2p
5 units f 13 jajunits
5a i¡¡ 2
¡ 3¢
iip
13 units bi¡ 5
¡ 2¢
iip
29 units
ci¡¡ 3
¡ 4¢
ii 5 units di¡¡ 12
5¢
ii 13 units6a¡ 160
240¢
b¡¡ 80
120¢
c¡ 80
40¢
di¼ 431 m ii ¼ 288 m iii¼ 144 m iv ¼ 89 m
e¡ 160
400¢ This vector describes the position of the hole from
the tee.
Its length gives the length of the hole, i.e., ¼ 431 m.7a
¡!
SA=¡¡ 2
2¢
,
¡!
AB=¡ 2
5¢
,
¡!
BC=¡ 5
3¢
,
¡!
CD=¡ 4
¡ 2¢
,
¡DE!=¡ 0
¡ 5¢
, ¡!EF=¡¡ 4
0¢
,¡FG!=¡¡ 4
2¢
,¡GS!=¡¡ 1
¡ 5¢
b¡ 0
0¢
cThe finishing point is the same as the starting point, i.e., we
are back where we started.
8 k=§ 4 9ak=5 bjuj=jvj=5p
2 unitsEXERCISE 24F
1a bcdefghSo,a¡ 4
6¢
b¡¡ 12
6¢
cμ
1(^112)
¶
d¡¡ 6
7¢
e¡ 10
7¢
f¡¡ 8
12¢
g¡ 4
2¢
h¡ 4
2¢
3a b c- q p
pq- -q p
pq-- q p
pq-ppq- -qp- q
pq-
p- q
pq-rr
2 r- s
- s
-3s -s
- s
r
Qw_\r- s
- s
rrs-2srrr3 rs+
rr- s
- s
- s
2-3rsr
Qw_\sQw_\sr+
rr s2+rs
Qw_\(2+rs)p qpq=p
q
qpIB MYP_3 ANS
cyan magenta yellow black(^05255075950525507595)
100 100
(^05255075950525507595)
100 100
Y:\HAESE\IGCSE01\IG01_an\729IB_IGC1_an.CDR Thursday, 20 November 2008 10:20:41 AM PETER