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I n t r o d u c t i o n t o v e c t o r s M C 1 1 Q l d - 7 133
Exercise 7A — Vectors and scalars 1 a i r s+
% %
ii r s−
% %
iii s r−
%%
b i 2 2r s+% %
Same as 1 a i except scaled by a factor of 2. ii 2 2r s−
% %
Same as 1 a ii except scaled by a factor of 2. iii 3 4s r−
%%
2 a A to D = s t+
% %
b A to B = s t u v+ + +
% % % %
c D to A = s t− −
% %
d B to E = v u t− − −
% % %
e C to A = u t s− − −
% % %
3
Displacement = (4 − 2) north = 2 km north The answer is C. 4 a u v+
% %
= A to C b u v−
% %
= D to B c v u−
% %
= B to D d 3 2 2u v u v+ − −
% % % %
= u v+% %
= A to C
5 u%
= 2v w+% %
w
% = v u−% %
⇒ u%
= 2v v u+ −% % %
= 3v u−
% %
2u%
= 3v%
u%
= 32 v%
The answer is D. 6 a CH
uuur = CG GH+
uuur uuur
= r s+% %
b CJuur
= CG GJ+uuur uur
= s t+
% %
c GDuuur
= GH HD+uuur uuur
= r s−
% %
d FIuur
= FE EI+uuur uur
= r s+
% %
e HEuuur
= HI IE+uur uur
= t s−
% %
f DJuur
= DH HG GJ+ +uuur uuur uur
= s r t− +
%% %
g CIuur
= CD DE EI+ +uuur uuur uur
= r t s+ +
% % %
h JCuur
= JG GC+uur uuur
= t s− −
% %
7 a and b
c R%
= 2 2300 400+
= 250 000 = 500 km d tan θ = 400
300
θ = tan−1 ( )43
θ = 53.1° clockwise from north. 8
EF = TE = 300 cos 45° = 300 × 2
2
= 150 2 km Total distance east of the starting point is 300 + 150 2 = 512.1 km
Chapter 7 — Introduction to vectors
M C 1 1 Q l d - 7 134 I n t r o d u c t i o n t o v e c t o r s
Resultant bearing:
tan θ = 150 2(300 150 2)+
≈ 0.414 θ = 22.5° Resultant bearing is 90° − 22.5° = 67.5° clockwise from north 9
R
% = 2 2600 400+
= 520 000 = 721.1 km tan θ = 600
400
θ = tan−1 ( )32
θ = 56.3° Resultant bearing = 270° + 56.3° = 326.3° 10 a a b+
% %
8 east and 8 north b 3a b+
% %
18 east and 14 north c a b−
% %
2 west and 2 north d b a−
% %
2 east and 2 south e 3 4b a−
% %
3 east and 11 south f 0.5 2.5a b+
% %
14 east and 10 north g 2.5a b−
% %
9.5 west and 2.5 south h 4a
%
12 east and 20 north i 2.5 1.5a b−
% %
8 north j 2.5b a−
% %
2.5 west and 9.5 south
11
AB
uuur = 2 210 4+
= 116 = 10.77
θ = tan−1 410⎛ ⎞⎜ ⎟⎝ ⎠
= 21.8° Bearing is 90° − 21.8° = 068.2° True
12
AC
uuur = a b−
% %
BDuuur
= a b+% %
ACuuur
+ BDuuur
= b a ba + +−% % %%
= 2a
%
13
3( )u v+
% % = 3 3u v+
% %
14
( )a b c+ +
% % % = ( )a b c+ +
% % %
15
3r s−
% % = ( 3 )s r− −
%%
16
The horizontal component of w
%is 4
The vertical component of w%
is 5 The horizontal component of v
%is 2
The vertical component of v%
is 3 w v+
% % = (6, 8)
17
w v−
% % = (2, 2)
I n t r o d u c t i o n t o v e c t o r s M C 1 1 Q l d - 7 135
18 a
4w
% = (16, 20)
b
2v−
% = (− 4, − 6)
19 One can deduce that x and y components can be added, subtracted and multiplied separately.
20 ODuuur
= 2 4a b× + ×% %
= 2 4a b+
% %
The answer is B. 21 EO
uuur = 3 4a b− × + − ×
% %
= 3 4a b− −% %
The answer is D.
22
Net displacement vector is O
%
23 Displacement, velocity, force 24 Speed, time, length 25 1 magnitude and 2 angles (N–S) and (E–W).
Exercise 7B — Position vectors in two and three dimensions 1 a 3 4 2i j k+ −
% %%
x = 3, y = 4, z = −2 b 6 3i k−
% %
x = 6, y = 0, z = −3 c 1
23.4 2i j k+ +% %%
x = 3.4, y = 2 z = 12
2 a i | |v%
= 2 26 6+
= 36 36+
= 72 = 6 2 ii θ = tan−1 ( )6
6
= tan−1 1 = 45°
b i | |w%
= 2 2( 4) 7− +
= 16 49+
= 65 ii tan−1 ( )7
4−
= tan−1 (−1.75) = −60.3° θ = 180° − 60.3° = 119.7° (second quadrant)
c i | |a%
= 2 2( 3.4) ( 3.5)− + −
= 11.56 12.25+
= 23.81 ≈ 4.88 ii tan−1 ( )3.5
3.4−−
= tan−1 (1.0294) = 45.8° θ = 180° + 45.8° = 225.8° (third quadrant)
d i | |b%
= 2 2320 ( 10)+ −
= 102 400 100+
= 102 500
= 50 41 (≈ 320.16) ii tan−1 ( )10
320−
= tan−1 (− 0.031 25) = -1.8° or θ = 360° − 1.8° = 358.2° (fourth quadrant)
3 a 045° T b 270° + 60.3° = 330.3° T c 270° − 45.8° = 224.2° T d 90° + 1.8° = 091.8° Τ
4
θ = 90° + (360 − 210)° = 240° x = | |w
%cos 240°
= 100 cos 240° = −50 y = | |w
%sin 240°
= 100 sin 240° = −100 sin 60° = −50 3 w
% = 50 50 3i i− −
% %
5
x = 10 cos 30, y = 10 sin 30° = 5 3 = 5 The answer is C.
M C 1 1 Q l d - 7 136 I n t r o d u c t i o n t o v e c t o r s
6
θ = 90° − 147° = −57° x = 457 cos (−57°) = 248.9 y = 457 sin (−57°) = −383.3 ⇒ u
% = 248.9 i
% − 383.3 j
%
7
θ = 90° + (360 − 331°) = 119° x = 125 cos 119° = − 60.6 y = 125 sin 119° = 109.3 ⇒ b
% = −60.6 i
% + 109.3 j
%
8
420cos45 420sin 45
210 2 210 2
a i j
i j
= ° − °
= −% % %
% %
200cos60 200sin 60
100 100 3
b i j
i j
= −
= −
o o
% % %
% %
( ) ( )( ) ( )210 2 210 2 100 100 3
210 2 100 210 2 100 3
a b i j i j
i j
+ = − + −
= + − +% % % %% %
% %
( ) ( )2 2210 2 100 210 2 100 3
615.4
a b+ = + + +
=% %
1 210 2 100 3tan
210 2 10049.8
θ − ⎛ ⎞+= −⎜ ⎟⎜ ⎟+⎝ ⎠= −
The resultant displacement is 615 km at 49.8° south of east.
9
20cos45 20sin 45
10 2 10 2
a i j
i j
= ° + °
= +% % %
% %
30cos45 30sin 45
15 2 15 2
b i j
i j
= ° − °
= −% % %
% %
( ) ( )( ) ( )10 2 10 2 15 2 15 2
10 2 15 2 10 2 15 2
25 2 5 2
a b i j i j
i j
i j
+ = + + −
= + + −
= −
% % % %% %
% %
% %
( ) ( )2 225 2 5 2
1300
10 1336
a b+ = +
=
=≈
% %
1 5 2tan
25 211.3
θ − ⎛ ⎞−= ⎜ ⎟⎜ ⎟⎝ ⎠
= −
Take 36 steps in a direction 11.3° south of east.
10
15cos30 15sin30
15 3 152 2
a i j
i j
= ° + °
= +
% % %
% %
12sin 40 12cos40b i j= − ° + °% % %
( )15 3 15 12sin 40 12cos402 2
15 3 1512sin 40 12cos402 2
20.7038 1.6925
a b i j i j
i j
i j
⎛ ⎞− = + − − ° + °⎜ ⎟⎜ ⎟
⎝ ⎠⎛ ⎞ ⎛ ⎞= + ° + − °⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠
= −
% % % %% %
% %
% %
2 220.7038 1.6925
20.8a b− = +
=% %
The scouts are 20.8 km apart. 11 a a
% = 3 4i j+
% %
| |a%
= 2 23 4+ = 5 a
% = 3
5 i%
+ 45 j%
b d%
= 3 4i j−% %
| |d%
= 2 23 ( 4)+ − = 5 d
% = 3
5 i%
− 45 j%
c b%
= 4 3i j+% %
| |b%
= 2 24 3+ = 5 b
% = 4
5 i%
+ 35 j%
d e%
= 4 3i j− +% %
| |e%
= 2 2( 4) 3− + = 5 e
% = 4
5 i−%
+ 35 j%
e c%
= i%
+ 2 j%
| |c%
2 21 ( 2)
3
= +
=
c%
= 213 3
i j+% %
I n t r o d u c t i o n t o v e c t o r s M C 1 1 Q l d - 7 137
f f%
= 3.5 2.7i j− +% %
| |f%
2 2( 3.5) (2.7)4.42
= − +≈
f%
= 3.5 2.74.42 4.42
i j− +% %
= 0.792 0.611i j− +% %
12 v%
= 3 4i j−% %
| |v%
= 2 23 ( 4)+ − = 5 v
% = 3 4
5 5i j−% %
The answer is B. 13 v
% = 0.3 0.4i j+
% %
| |v%
= 2 20.3 0.4+ = 0.5
v%
= 0.3 0.40.5 0.5
i j+% %
= 0.6 0.8i j+% %
= 2(0.3 0.4 )i j+% %
= 2v%
14 w
% = 0.1 0.02i j− −
% %
| |w%
= 2 2( 0.1) ( 0.02)− + − = 0.102
w%
= 0.1 0.020.102 0.102
i j− −% %
= 0.98 0.20i j− −% %
15 w%
= 50 50 3i j− −% %
| |w%
= 2 2( 50) ( 50 3)− + −
= 2500 7500+
= 10 000 = 100
w%
= 50 50 3100 100
i j− −% %
= 1 32 2
i j− −% %
16 a ABuuur
= (4 0) (5 1)i j− + −% %
= 4 4i j+% %
b BAuuur
= (0 4) (1 5)i j− + −% %
= 4 4i j− −% %
17 a i ABuuur
= (4 0) ( 5 2)i j− + − −% %
= 4 7i j−% %
ii |AB|uuur
= 2 24 ( 7)+ −
= 65 b i AB
uuur = (5 2) (4 3)i j− + −
% %
= 3i j+% %
ii |AB|uuur
= 2 23 1+
= 10 c i AB
uuur = (0 4) (2 5)i j− + − −
% %
= 4 7i j− +% %
ii |AB|uuur
= 2 2( 4) 7− +
= 65 d i AB
uuur = (2 5) (3 4)i j− + −
% %
= 3i j− −% %
ii |AB|uuur
= 2 2( 3) ( 1)− + −
= 10 e i AB
uuur = (5 − 3) i
% + (7 − 7) j
%
= 2i%
ii |AB|uuur
= 22 = 2 f i AB
uuur = (3 7) ( 3 3)i j− + − − −
% %
= 4i−%
ii |AB|uuur
= 2( 4)− = 4
18 a BAuuur
= − ABuuur
= 4 7i j− +
% %
b BAuuur
= 3i j− −% %
c BAuuur
= 4 7i j−% %
d BAuuur
= 3i j+% %
e BAuuur
= 2i−%
f BAuuur
= 4i%
19 a ABuuur
= 4 7i j−% %
|AB|uuur
= 65
ˆ
ABuuur
= 1 (4 7 )65
i j−% %
= 4 765 65
i j−% %
b ABuuur
= 3i j+% %
|AB|uuur
= 10
ˆ
ABuuur
= 3 110 10
i j+% %
c ABuuur
= 4 7i j− +% %
|AB|uuur
= 65
ˆ
ABuuur
= 4 765 65
i j− +% %
d ABuuur
= 3i j− −% %
|AB|uuur
= 10
ˆ
ABuuur
= 3 110 10
i j− −% %
e ABuuur
= 2i%
|AB|uuur
= 2
ˆ
ABuuur
= 22
i%
= i%
f ABuuur
= 4i−%
|AB|uuur
= 4
M C 1 1 Q l d - 7 138 I n t r o d u c t i o n t o v e c t o r s
ABuuur
= 44
i−%
= i−%
20 u
% = 5 2i j−
% % and e
% = 2 3i j− +
% %
a i | |u%
= 2 25 ( 2)+ −
= 29
ii | |e%
= 2 2( 2) 3− +
= 13
iii u%
= 5 229 29
i j−% %
iv e%
= 2 313 13
i j− +% %
v u e+% %
= 5 2 2 3i j i j− − +% %% %
= 3i j+% %
vi | |u e+% %
= 2 23 1+
= 10 b | | | |u e+
% % = 29 13 | |u e+ > +
% % = 10
Therefore, reject the statement as the magnitudes are different.
21 u%
= 3 4i j− +% %
and e%
= 5i j−% %
a i | |u%
= 2 2( 3) 4− +
= 25 = 5
ii | |e%
= 2 25 ( 1)+ −
= 26
iii u%
= 3 45 5
i j− +% %
iv e%
= 5 126 26
i j−% %
v u e+% %
= ( 3 5) (4 1)i j− + + −% %
= 2 3i j+% %
vi | |u e+% %
= 2 22 3+
= 13 b | | | |u e+
% % = 5 26 | |u e+ > +
% % = 13
Therefore, reject the statement as the magnitudes are different.
22 a a b−% %
= (3 2) (2 3)i j− + −% %
= i j−% %
| |a b−% %
= 2 21 1+
= 2 b a b−
% % = (5 2) ( 2 5)i j− + − −
% %
= 3 7i j−% %
| |a b−% %
= 2 23 7+
= 58 23
a r
% = 3i
%
b b%
= 5 j%
c v%
= 3 5i j+% %
d tanθ = 53
θ = tan−1 53
= 59.0° True bearing = 90° − 59.0° = 031.0°
e | |v%
= 2 25 3+
= 34 km/h
24
θ = tan−1 35
⎛ ⎞⎜ ⎟⎝ ⎠
= 31.0° True bearing = 360° − 31.0° = 329.0° The boat should travel on a bearing of 329.0° to arrive at
the opposite bank due north of the starting position.
25 a 2 2 23 4 ( 5)+ + −
= 50 = 5 2
b 2 2 2( 3) ( 4) 5− + − +
= 50 = 5 2
c 2 2 20.5 ( 2) 3+ − + = 3.64
d 2 2 2( 2) ( 2 2) 1+ − +
= 11
e 2 2 2( 7) (14) ( 21)− + + −
= 686 = 7 14 or 26.19
f 2 2 21 1 1+ +
= 3 26 a a
% = 5 2 (4 3 2 )i j k i j k− + − + −
% % % %% %
= 5 3i j k− +% %%
| |a%
= 2 2 21 ( 5) 3+ − +
= 35 b b
% = 5 (2 )i j k i j k+ + − + −
% % % %% %
= 3 2i k+% %
| |b%
= 2 23 2+
= 13 c c
% = 3 ( 2 3 )i k i j k+ − − + +
% % % %%
= 4 2 2i j k− −% %%
| |c%
= 2 2 24 ( 2) ( 2)+ − + −
= 24 = 2 6 d d
% = 8 5 2 ( 3 )i j k i j k+ + − + −
% % % %% %
= 7 2 3i j k+ +% %%
I n t r o d u c t i o n t o v e c t o r s M C 1 1 Q l d - 7 139
| |d%
= 2 2 27 2 3+ +
= 62 27 C = (2, 6, 0), D = (3, −1, −2), E = (−4, 8, 10) and
F = (−2, −6, 6) CD
uuur = 3 2 (2 6 )i j k i j− − − +
% % %% %
= 7 2i j k− −%%
EFuur
= 2 6 6 ( 4 8 10 )i j k i j k− − + − − + +% % % %% %
= 2 14 4i j k− −% %%
= 2( 7 2 )i j k− −% %%
= 2 CDuuur
⇒ EF || CD
uur uuur
28 a
a
% = 20i
%, b%
= 15 j−%
a b−% %
= 20 15i j+% %
b | |a b−% %
= 2 220 15+ = 25 c tan θ = 15
20
θ = tan−1 ( )1520
= 36.9° True bearing = 90° − 36.9° = 053.1°
29 u%
= 3 4i j+% %
tan θ = 43
θ = tan−1 43
θ = 53.1° v
% = 4 3i j−
% %
tan θ = 34
−
θ = tan−1 ( )34−
θ = −36.9° Difference = 53.1 − −36.9 = 90° That is, the two vectors are perpendicular to each other. 3 × 4 + 4 × −3 = 12 − 12 = 0 To confirm that this is a pattern for all perpendicular
vectors, let 1 1u x i y j= +% % %
and let u%
make an angle of θ to
the positive direction of the x-axis. Let 2 2v x i y j= +% % %
be at
right angles to vector u%
. As the vectors are perpendicular, v%
will be at 90 – θ to the negative direction of the x axis,
1
1tan y
xθ = [1]
( )
( )( )
2
2
2
2
2
2
2
2
2
2
tan 90
sin 90cos 90
cossinsincos
tan [2]
yx
yx
yxx
yx
y
θ
θθθθθθ
θ
− =−
−=
− −
=−−=
−=
Equating [1] and [2]
1 2
1 2
1 2 1 2
1 2 1 20
y xx y
y y x xx x y y
−=
= −= +
30 a
v
% = − 4 12i j+
% %
b tan θ = 412
θ = 18.4° True bearing = 360° − 18.4° = 341.6°
c time = distancespeed
= 0.5 km12 km / h
= 0.041 667 hours Time taken is 2.5 minutes.
Exercise 7C — Multiplying two vectors — the dot product 1 If u
%= 3 3i j+
% % and v
% = 6 2i j+
% %
| |u%
= 2 23 3+ | |v%
= 2 26 2+
= 18 = 40 = 3 2 = 2 10
tan β = 2
6 tan α = 33
= 13 = 1
⇒ β = 18.4° ⇒ α = 45° θ = α β− = 45° − 18.4° = 26.6° cos θ = 0.8942 ∴ u v⋅
% % = | || |u v
% %cos θ
M C 1 1 Q l d - 7 140 I n t r o d u c t i o n t o v e c t o r s
= 3 2 2 10 0.8942× × = 23.99 2 3 3u i j= +
% % % and 6 2v i j= +
% % %
u v⋅% %
= 3 × 6 + 3 × 2 = 18 + 6 = 24 This is more accurate as no angle is required. 3 a u v⋅
% % = (2 3 5 ) (3 3 6 )i j k i j k+ + ⋅ + +
% % % %% %
= 2 × 3 + 3 × 3 + 5 × 6 = 6 + 9 + 30 = 45
b u v⋅% %
= (4 2 3 ) (5 2 )i j k i j k− + ⋅ + −% % % %% %
= 4 × 5 + −2 × 1 + 3 × −2 = 20 − 2 − 6 = 12 c u v⋅
% % = ( 4 5 ) (3 7 )i j k i j k− + − ⋅ − +
% % % %% %
= −1 × 3 + 4 × −7 + −5 × 1 = −3 − 28 − 5 = −36 d u v⋅
% % = (5 9 ) (2 4 )i j i j+ ⋅ −
% %% %
= 5 × 2 + 9 × − 4 = 10 − 36 = −26 e u v⋅
% % = ( 3 ) ( 4 )i j j k− + ⋅ +
% %% %
= −3 × 0 + 1 × 1 + 0 × 4 = 1 f u v⋅
% % = (10 ) ( 2 )i i⋅ −
% %
= 10 × −2 = −20 g u v⋅
% % = (3 5 ) ( )j k i+ ⋅
% %%
= 0 × 1 + 3 × 0 + 5 × 0 = 0 h u v⋅
% % = (6 2 2 ) ( 4 )i j k i j k− + ⋅ − − −
% % % %% %
= 6 × −1 + −2 × −4 + 2 × −1 = −6 + 8 − 2 = 0
4 u v⋅% %
= (3 3 3 ) ( 2 6 )i j k i j k− + ⋅ − +% % % %% %
= 3 × 1 + −3 × −2 + 3 × 6 = 3 + 6 + 18 = 27 The answer is E. 5
u v⋅
% % = | | | | cosu v θ
% %
= 6 × 5 cos 135° = −21.2 The answer is C. 6 u v⋅
% % = | | | | cosu v θ
% %
| |u%
= 7, | |v%
= 8 and θ = 180° − 50° = 130° u v⋅
% % = 7 × 8 cos 130°
= −35.996 ≈ −36
7 u u⋅% %
= ( ) ( )xi yj xi yj+ ⋅ +% %% %
= x × x + y × y = x2 + y2
8 u v⋅% %
= (2 5 ) ( 2 4 )i j k i j k− + ⋅ − − +% % % %% %
= 2 × −1 + −5 × −2 + 1 × 4 = −2 + 10 + 4 = 12 9 ( )w u v⋅ −
% % % = (5 2 ) [3 2 ( 2 )]i j i j i j− ⋅ + − −
% % %% % %
= (5 2 ) (2 4 )i j i j− ⋅ +% %% %
= 5 × 2 + −2 × 4 = 10 − 8 = 2 w u w v⋅ − ⋅
% % % % = (5 2 ) (3 2 ) (5 2 ) ( 2 )i j i j i j i j− ⋅ + − − ⋅ −
% % % %% % % %
= 5 × 3 + −2 × 2 − (5 × 1 + −2 × −2) = 15 − 4 − (5 + 4) = 15 − 4 − 9 = 2 ⇒ ( )w u v⋅ −
% % % = w u w v⋅ − ⋅
% % % %
10 ( )w u v⋅ +% % %
= (5 2 ) (3 2 2 )i j i j i j− ⋅ + + −% % %% % %
= (5 2 ) (4 )i j i− ⋅% %%
= 5 × 4 + −2 × 0 = 20 w u w v⋅ + ⋅
% % % % = (5 2 ) (3 2 ) (5 2 ) ( 2 )i j i j i j i j− ⋅ + + − ⋅ −
% % % %% % % %
= 5 × 3 + −2 × 2 + 5 × 1 + −2 × −2 = 15 − 4 + 5 + 4 = 20 ⇒ ( )w u v⋅ +
% % % = w u w v⋅ + ⋅
% % % %
11 Two vectors are perpendicular if their dot product is 0. A: (5 4 3 ) ( 5 4 3 )i j k i j k+ + ⋅ − − −
% % % %% %
= 5 × −5 + 4 × − 4 + 3 × −3 = −50 B: (5 4 3 ) (3 4 5 )i j k i j k+ + ⋅ + +
% % % %% %
= 15 + 16 + 15 = 46 C: (5 4 3 ) ( 5 )i j k i+ + ⋅ −
% % %%
= −25 D: (5 4 3 ) ( 3 5 )i j k i k+ + ⋅ − +
% % % %%
= −15 + 15 = 0 The answer is D.
12 ( ) ( )u v u v− ⋅ +% % % %
= 0 ⇒ u u v v⋅ − ⋅
% % % % = 0
u2 − v2 = 0 u2 = v2
The answer is B. 13 ( ) ( )u v u v− ⋅ +
% % % % = 2| |v
%
⇒ 2 2| | | |u v−% %
= 2| |v%
2| |u%
= 22 | |v%
or u%
= 2 v%
The answer is D. 14 a (4 3 ) (7 4 )i k j k− ⋅ +
% % %%
= 4 × 0 + 0 × 7 + −3 × 4 = −12
b ( 2 3 ) ( 9 4 )i j k i j k+ − ⋅ − + −% % % %% %
= −9 + 8 + 3 = 2
I n t r o d u c t i o n t o v e c t o r s M C 1 1 Q l d - 7 141
c (8 3 ) (2 3 4 )i j i j k+ ⋅ − +% % %% %
= 16 − 9 + 0 = 7 d (5 5 5 ) (5 5 5 )i j k i j k− + ⋅ + −
% % % %% %
= 25 − 25 − 25 = −25
15 a | |u%
= 2 24 ( 3)+ − , | |v%
= 2 27 4+
= 5 = 65
cos θ = u vuv⋅
% %
= 125 65
−×
≈ − 0.2977 θ = 107°
b | |u%
= 2 2 21 2 ( 3)+ + − ,
= 14
| |v%
= 2 2 2( 19) 4 ( 1)− + + −
= 98
cos θ = 214 98
= 0.053 99 θ = 87°
c | |u%
= 2 28 3+ , | |v%
= 2 2 22 ( 3) 4+ − +
= 73 = 29
cos θ = 773 29
= 0.152 14 θ = 81°
d | |u%
= 2 2 25 ( 5) 5+ − + , | |v%
= 2 2 25 5 ( 5)+ + −
= 75 = 75
cos θ = 2575 75−
= 13−
θ = 109° 16 u v⋅
% % = (2 3 ) (2 3 )i j i j+ ⋅ −
% %% %
= 4 − 9 = −5
| |u%
= 2 22 3+ , | |v%
= 2 22 ( 3)+ −
= 13 = 13
cos θ = 513 13
−
= 513−
= −0.384 62 θ = 112.6° ≈ 113° The answer is D. 17 u v⋅
% % = (2 3 ) ( 4 6 )i j i j− ⋅ − +
% %% %
= −8 − 18 = −26
| |u%
= 2 22 ( 3)+ − , | |v%
= 2 2( 4) 6− +
= 13 = 52
cos θ = 2613 52−
= 2626−
= −1 θ = 180° The answer is E. 18 v u⋅
% % = ( 3 ) (6 2 )ai j i j+ ⋅ −
% %% % = 0
6a − 6 = 0 6a = 6 a = 1 19 v u⋅
% % = ( 2 3 ) (4 3 2 )ai aj k i j k− + ⋅ − +
% % % %% % = 0
4a + 6a + 6 = 0 10a + 6 = 0 10a = − 6 a = 6
10−
a = 35−
20 2 4u i j= +% % %
Let v%
= (2 4 )k i j+% %
= 2 4ki kj+% %
u v⋅% %
= (2 4 ) (2 4 )i j ki kj+ ⋅ +% %% %
= 40
4 k + 16 k = 40 20 k = 40 k = 2 v
% = 4 8i j+
% %
21 4 3u i j= −% % %
Let v%
= (4 3 )k i j−% %
= 4 3ki kj−% %
u v⋅% %
= (4 3 ) (4 3 )i j ki kj− ⋅ −% %% %
= 80
16 k + 9 k = 80 25k = 80 k = 80
25
= 165
v%
= 64 485 5
i j−% %
Exercise 7D — Resolving vectors — scalar and vector resolutes 1 a u
% = 2 3i j+
% % and a
% = 4 5i j+
% %
i | |u%
= 2 22 3+
= 13
u%
= 1 (2 3 )13
i j+% %
u a⋅% %
= 1 (2 3 ) (4 5 )13
i j i j+ ⋅ +% %% %
= 8 1513 13
+
= 2313
= 23 1313
ii | |a%
= 2 24 5+
= 41
a%
= 1 (4 5 )41
i j+% %
a u⋅% %
= 1 (4 5 ) (2 3 )41
i j i j+ ⋅ +% %% %
M C 1 1 Q l d - 7 142 I n t r o d u c t i o n t o v e c t o r s
= 8 1541 41
+
= 2341
= 23 4141
b u%
= 5 2i j−% %
and a%
= 3i j−% %
i | |u%
= 2 25 ( 2)+ −
= 29
u%
= 1 (5 2 )29
i j−% %
u a⋅% %
= 1 (5 2 ) (3 )29
i j i j− ⋅ −% %% %
= 15 229 29
+
= 1729
= 17 2929
ii | |a%
= 2 23 ( 1)+ −
= 10
a%
= 1 (3 )10
i j−% %
ˆ .a u% %
= 1 (3 ) (5 2 )10
i j i j− ⋅ −% %% %
= 15 210 10
+
= 1710
= 17 1010
c u%
= 2 6i j− +% %
and 4a i j= −% % %
i | |u%
= 2 2( 2) 6− +
= 40 = 2 10
u%
= 1 ( 2 6 )2 10
i j− +% %
u a⋅% %
= 1 ( 2 6 ) ( 4 )2 10
i j i j− + ⋅ −% %% %
= 2 242 10 2 10
− −
= 262 10
−
= 13 1010
−
ii | |a%
= 2 21 ( 4)+ −
= 17
a%
= 1 ( 4 )17
i j−% %
a u⋅% %
= 1 ( 4 ) ( 2 6 )17
i j i j− ⋅ − +% %% %
= 2 2417 17
− −
= 2617
−
= 26 1717
−
d u%
= 3 2i j−% %
and a%
= − 4 3i j−% %
i | |u%
= 2 23 ( 2)+ −
= 13
u%
= 1 (3 2 )13
i j−% %
u a⋅% %
= 1 (3 2 ) ( 4 3 )13
i j i j− ⋅ − −% %% %
= 12 613 13
− +
= 613
−
= 6 1313
−
ii | |a%
= 2 2( 4) ( 3)− + − = 5
a%
= 1 ( 4 3 )5
i j− −% %
a u⋅% %
= 1 ( 4 3 ) (3 2 )5
i j i j− − ⋅ −% %% %
= 12 65 5
− +
= − 65
e u%
= 8 6i j−% %
and a%
= 5i j− +% %
i | |u%
= 2 28 ( 6)+ − = 10
u%
= 1 (8 6 )10
i j−% %
u a⋅% %
= 1 (8 6 ) ( 5 )10
i j i j− ⋅ − +% %% %
= 40 610 10
− −
= 4610
−
= 235
−
ii | |a%
= 2 2( 5) 1− +
= 26
a%
= 1 ( 5 )26
i j− +% %
a u⋅% %
= 1 ( 5 ) (8 6 )26
i j i j− + ⋅ −% %% %
= 40 626 26
− −
= 4626
−
= 46 2626
−
= 23 2613
−
I n t r o d u c t i o n t o v e c t o r s M C 1 1 Q l d - 7 143
2 a u%
= 3i j−% %
and v%
= 2 5i j+% %
i | |u%
= 2 23 ( 1)+ −
= 10
u%
= 1 (3 )10
i j−% %
u v⋅% %
= 1 (3 ) (2 5 )10
i j i j− ⋅ +% %% %
= 6 510 10
−
= 1 101010
⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠
ii ||v%
= ˆ ˆ( )u v u⋅% % %
= 1 1 (3 )10 10
i j× −% %
= 1 (3 )10
i j−% %
= 3 110 10
i j−% %
iii v⊥%
= ||v v−% %
= 3 12 510 10
i j i j⎛ ⎞+ − −⎜ ⎟⎝ ⎠% %% %
= 17 5110 10
i j+% %
b u%
= 4 5i j+% %
and v%
= 8 10i j+% %
i | |u%
= 2 24 5+
= 41
u%
= 1 (4 5 )41
i j+% %
u v⋅% %
= 1 (4 5 ) (8 10 )41
i j i j+ ⋅ +% %% %
= 32 5041 41
+
= 82 4141
= 2 41 ii ||v
% = ˆ ˆ( )u v u⋅
% % %
= 12 41 (4 5 )41
i j× +% %
= 8 10i j+% %
iii v⊥%
= ||v v−% %
= 8 10 (8 10 )i j i j+ − +% %% %
= 0%
c u
%= 4 3i j+
% % and v
% = − 3 4i j+
% %
i | |u%
= 2 24 3+ = 5
u%
= 1 (4 3 )5
i j+% %
u v⋅% %
= 1 (4 3 ) ( 3 4 )5
i j i j+ ⋅ − +% %% %
= 12 125 5
− +
= 0
ii ||v%
= ˆ ˆ( )u v u⋅% % %
= 10 (4 3 )5
i j× +% %
= 0%
iii v⊥
% = ||v v−
% %
= − 3 4 0i j+ −% %%
= − 3 4i j+% %
d u%
= i j k+ +% %%
and v%
= 2i j k+ −% %%
i | |u%
= 2 2 21 1 1+ +
= 3
u%
= 1 ( )3
i j k+ +% %%
u v⋅% %
= 1 ( )(2 )3
i j k i j k+ + + −% % % %% %
= 2 1 13 3 3
+ −
= 23
ii ||v%
= ˆ ˆ( )u v u⋅% % %
= 2 1 ( )3 3
i j k× + +% %%
= 2 2 23 3 3
i j k+ +% %%
iii v⊥%
= ||v v−% %
= 2 223 3 3
i j k i j k2⎛ ⎞+ − − + +⎜ ⎟⎝ ⎠% % % %% %
= 4 1 53 3 3
i j k+ +% %%
e u%
= 2 3 4i j k+ +% %%
and 2 3 4v i j k= − −% % %%
i | |u%
= 2 2 22 3 4+ +
= 29
u%
= 1 (2 3 4 )29
i j k+ +% %%
u v⋅% %
= 1 (2 3 4 ) (2 3 4 )29
i j k i j k+ + ⋅ − −% % % %% %
= 4 9 1629 29 29
− −
= − 2129
ii ||v%
= ˆ ˆ( )u v u⋅% % %
= 21 1 (2 3 4 )29 29
i j k− × + +% %%
= − 42 63 8429 29 29
i j k− −% %%
iii v⊥%
= ||v v−% %
= 42 63 842 3 429 29 29
i j k i j k− − + + +% % % %% %
= 100 24 3229 29 29
i j k− −% %%
= 4 (25 6 8 )29
i j k− −% %%
M C 1 1 Q l d - 7 144 I n t r o d u c t i o n t o v e c t o r s
f u%
= 3i j k+ −% %%
and v%
= 2 3j k−%%
i | |u%
= 2 2 23 1 ( 1)+ + −
= 11
u%
= 1 (3 )11
i j k+ −% %%
u v⋅% %
= 1 (3 ) (2 3 )11
i j k j k+ − ⋅ −% % %% %
= 2 311 11
+
= 511
ii ||v%
= ˆ ˆ( )u v u⋅% % %
= 5 1 (3 )11 11
i j k× + −% %%
= 15 5 511 11 11
i j k+ −% %%
iii v⊥%
= ||v v−% %
= 15 5 52 311 11 11
j k i j k⎛ ⎞− − + −⎜ ⎟⎝ ⎠% % %% %
= − 15 17 2811 11 11
i j k+ −% %%
3
a Let the injured bushwalker’s position be denoted by v
%.
Let the path of the searcher be denoted by u%
. v
% = 2 3i j+
% %
u%
= (3 4 )k i j+% %
| |u%
= 2 23 4k + = 5k
u%
= 1 (3 4 )5
k i jk
× +% %
= 3 45 5
i j+% %
u v⋅%
= 3 4 (2 3 )5 5
i j i j⎛ ⎞+ ⋅ +⎜ ⎟⎝ ⎠% %% %
= 6 125 5
+
= 185
||v%
= ˆ ˆ( )u v u⋅% % %
= 18 3 45 5 5
i j⎛ ⎞+⎜ ⎟⎝ ⎠% %
= 54 7225 25
i j+% %
||| |v%
= 2 254 72
25 25⎛ ⎞ ⎛ ⎞+⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= 3.6 The searcher is 3.6 km from the camp when closest to the
bushwalker. b | |v⊥
% represents the minimum distance between the
searcher and the bushwalker.
v⊥%
= ||v v−% %
= 54 722 325 25
i j i j⎛ ⎞+ − +⎜ ⎟⎝ ⎠% %% %
= 4 325 25
i j− +% %
= 1 ( 4 3 )25
i j− +% %
| |v⊥%
= 2 21 ( 4) 325
− +
= 1 525
×
= 0.2 km
4
Let the yacht’s position be denoted by v
%.
The path of the rescue boat be denoted by u%
. v
% = 5 2i j−
% %
u%
= (3 )k i j−% %
| |v⊥%
represents the closest distance the rescue boat gets to the yacht.
| |u%
= 2 23 ( 1)k + −
= 10k
u%
= 1 (3 )10
k i jk
−% %
= 1 (3 )10
i j−% %
u v⋅% %
= 1 (3 ) (5 2 )10
i j i j− ⋅ −% %% %
= 15 210 10
+
= 1710
||v%
= ˆ ˆ( )u v u⋅% % %
= 17 1 (3 )10 10
i j× −% %
= 17 (3 )10
i j−% %
= 51 1710 10
i j−% %
v⊥%
= ||v v−% %
= 51 175 210 10
i j i j⎛ ⎞− − −⎜ ⎟⎝ ⎠% %% %
= − 1 310 10
i j−% %
= 1 ( 3 )10
i j− −% %
| |v⊥%
= 2 21 ( 1) ( 3)10
− + −
= 1 1010
= 1010
km
or ≈ 0.316 km
I n t r o d u c t i o n t o v e c t o r s M C 1 1 Q l d - 7 145
Exercise 7E — Time-varying vectors 1 a u
% = 2ti tj−
% %
x = 2t
or t = 2x
y = − t
⇒ y = 2x−
b u%
= ( 1) 3t i t j− −% %
x = t − 1 t = x + 1 y = −3t = − 3(x + 1) = − 3x − 3 c u
% = 2( 3) 4t i t j+ +
% %
x = t + 3 t = x − 3 y = 4t 2
y = 4(x − 3)2
d u%
= 32ti t j+% %
x = 2t
t = 2x
y = t 3
y = 3
2x⎛ ⎞
⎜ ⎟⎝ ⎠
y = 3
8x
2 u%
= 2( 3) 4t i t j+ +% %
when t = 0,
u%
= 3i + 0 j The answer is B.
3 a v%
= 2( ) ,2t i t t j+ +%
t ≥ 0
x = ,2t x ≥ 0
or t = 2x y = t 2 + t = (2x)2 + 2x y = 4x2 + 2x
b When x = 0, y = 0 x = 1, y = 4 + 2 = 6
4 a v
% = 2( 1) ( 2 ) ,t i t t j+ + −
% % t ≥ 0
x = t + 1, x ≥ 1 (since t ≥ 0) or t = x − 1 y = t 2 − 2t = (x − 1)2 − 2(x − 1) = x2 − 2x + 1 − 2x + 2 y = x2 − 4x + 3
b y = x2 − 4x + 3, x ≥ 1 = (x − 2)2 + 3 − 4 = (x − 2)2 − 1 Turning point is (2, −1)
x-intercepts: x2 − 4x + 3 = 0 (x − 3)(x − 1) = 0 x = 1 or x = 3
5 a u
% = cos (2t) i
% + sin (2t) ,j
% t ≥ 0
x = cos 2t or x2 = cos2 2t y = sin 2t or y2 = sin2 2t ⇒ x2 + y2 = cos2 2t + sin2 2t x2 + y2 = 1
b Circle with centre (0, 0) and radius 1. When t = 0, u
% = .i
%
When t = 4π , u
% = .j
%
So the particle is moving from (1, 0) in an anticlockwise direction.
c Period = 22π
= π 6 a u
% = 3 cos (2t) i
% + 3 sin (2t) ,j
% t ≥ 0
x = 3 cos 2t or x2 = 9 cos2 2t y = 3 sin 2t or y2 = 9 sin2 2t ⇒ x2 + y2 = 9 cos2 2t + 9 sin2 2t x2 + y2 = 9
b Circle with centre (0, 0) and radius 3. When t = 0, u
% = 3 .i
%
When t = 4π , u
% = 3 .j
%
The particle is moving from (3, 0) in an anticlockwise direction.
c Period = 22π
= π 7 a u
% = (1 + cos t) i
% + (−2 + sin t) ,j
% t ≥ 0
x = 1 + cos t or cos t = x − 1 cos2 t = (x − 1)2
y = − 2 + sin t or sin t = y + 2 sin2 t = (y + 2)2
⇒ (x − 1)2 + (y + 2)2 = cos2 t + sin2 t (x − 1)2 + (y + 2)2 = 1
M C 1 1 Q l d - 7 146 I n t r o d u c t i o n t o v e c t o r s
b Circle with centre (1, −2) and radius 1. When t = 0, u
% = 2 2 .i j−
% %
When t = 2π , u
% = .i j−
% %
So the particle is moving from (2, −2) in an anticlockwise direction.
c Period = 21π
= 2π 8 u
% = 3 cos 2t i
% + sin 2t ,j
% t ≥ 0
x = 3 cos 2t
or cos 2t = 3x
cos2 2t = 2
9x
y = sin 2t or sin2 2t = y2
⇒ 2
9x + y2 = cos2 2t + sin2 2t
2
9x + y2 = 1
Graph is an ellipse with centre (0, 0) with a = 3 and b = 1. When t = 0, u
% = 3i
%
When t = 4π , u
% = j
%
The particle is moving in an anticlockwise direction from (3, 0).
9 u
% = 2 cos t i
% − 4 sin t ,j
% t ≥ 0
x = 2 cos t
or cos t = 2x
cos2 t = 2
4x
y = − 4 sin t
or sin t = 4y
−
sin2 t = 2
16y
⇒ 2
4x +
2
16y = cos2 t + sin2 t
2
4x +
2
16y = 1
Graph is an ellipse with centre (0, 0) with a = 2 and b = 4. When t = 0, u
% = 2 .i
%
When t = 2π , u
% = 4 .j−
%
The particle is moving in a clockwise direction from (2, 0).
10 a u%
= 2
2 24 4
( 2) ( 2)t ti jt t
+ ++ +% %
x = 2
24
( 2)tt
++
y = 24
( 2)t
t +
x + y = 2
24 4
( 2)t t
t+ +
+
= 2
2( 2)( 2)tt
++
x + y = 1 or y = 1 − x
When t = 0, ( )0u%
= 4 04 4
i j+% %
= i%
x = y
When 2
24
( 2)tt
++
= 24
( 2)t
t +
or t2 + 4 = 4t t2 − 4t + 4 = 0 (t − 2)2 = 0 t = 2 So x = y when t = 2.
and (2)u%
= 4 4 816 16
i j+ +% %
= 12 2
i j1+% %
or = 12 ( )i j+% %
b u%
= 2( 2) ( 1)t i t j+ + +% %
x = t + 2, x ≥ 2 or t = x − 2 y = (t + 1)2
y = (x − 2 + 1)2
y = (x − 1)2, x ≥ 2 c u
% = (2 cos t + 3) i
% + (3 sin t − 1) j
%
x = 2 cos t + 3 x − 3 = 2 cos t
or cos t = 32
x −
y = 3 sin t − 1 y + 1 = 3 sin t
or sin t = 13
y +
⇒ 23
2x −⎛ ⎞
⎜ ⎟⎝ ⎠
+ 21
3y +⎛ ⎞
⎜ ⎟⎝ ⎠
= cos2 t + sin2 t
2 2( 3) ( 1)
4 9x y− ++ = 1
11 Ship A’s position vector is u%
= (3 1) (4 2)t i t j+ + −% %
Ship B’s position vector is v%
= (2 3) (5 1)t i t j+ + +% %
I n t r o d u c t i o n t o v e c t o r s M C 1 1 Q l d - 7 147
Ship A: x = 3t + 1 x − 1 = 3t
or t = 13
x −
y = 4t − 2
= 4 13
x −⎛ ⎞⎜ ⎟⎝ ⎠
− 2
y = 4 4 23 3x − −
y = 4 103 3x − [1]
Ship B: x = 2t + 3 x − 3 = 2t
t = 32
x −
y = 5t + 1
= 5 32
x −⎛ ⎞⎜ ⎟⎝ ⎠
+ 1
= 5 15 12 2x − +
= 5 132 2x − [2]
Equate [1] and [2] to determine where the ships’ path cross.
4 103 3x − = 5 13
2 2x −
8x − 20 = 15x − 39 7x = 19 x = 19
7
Substitute x = 197 into [1]
y = 1974 10
3 3×
−
= 76 1021 3−
= 76 7021 21−
= 621
= 27
The ships paths cross at x = 197 and y = 2
7 .
12 u%
= 2ti t j+% %
x = t y = t 2
or y = x 2
v%
= 2t te i e j+% %
x = e t
y = e 2t
= 2( )te or y = x 2
To coincide: (x-components) t = e t and (y-components) t 2 = 2te or t = e t
Paths coincide if t = e t. But e t > t for all t ≥ 0. Therefore the paths can never coincide. Furthermore, v
% is always ahead of u
% (Since e t > t).
Chapter review 1 u
% = 4 i
% − 3 j
% + 0.2 k
% and v
% = 2 i
% + 4 j
% − k
%
4 u%
− 2.5 v%
= 4(4 i%
− 3 j%
+ 0.2 k%
) − 2.5(2 i%
+ 4 j%
− k%
)
= 16 i%
− 12 j%
+ 0.8 k%
− 5 i%
− 10 j%
+ 2.5 k%
= 11 i%
− 22 j%
+ 3.3 k%
The answer is A. 2 v
% = x i
% + y j
% + z k
%
Where x = 400 sin 60°
= 34002
×
= 200 3 y = 400 cos 60°
= 14002
×
= 200 z = 40 v
% = 200 3 i
% + 200 j
% + 40 k
%
3 A: 2 2 23 1 1+ + = 11 5≠
B: 2 2 22 ( 5) 4+ + = 25 = 5
C: 2 2 25 5 5 5 3 5+ + = ≠
D: 2 2 26 3 4 61 5+ + = ≠
E: 225 25 5= ≠ The answer is B. 4 PQ
uuur = i−
% + 3 j
% − 4 k
% − (3 i
% + 4 j
% − 5 k
%)
= −4 i%
− j%
+ k%
|PQ|uuur
= 2 2 2( 4) ( 1) 1− + − +
= 18 = 3 2 The answer is A.
5
a HX
uuur = 5 i
% + 10 cos 45° i
% + 10 sin 45° j
% + 5 j
%
= 5 i%
+ 5 2 i%
+ 5 2 j%
+ 5 j%
= (5 + 5 2 ) i%
+ (5 + 5 2 ) j%
b |HX|uuur
= (5 + 5 2 ) 2 = 17.07 km
6 u%
= 3 i%
+ 2 j%
− 4 k%
and v%
= −4 i%
+ 5 j%
+ k%
u v⋅% %
= (3 2 4 ) ( 4 5 )i j k i j k+ − ⋅ − + +% % % %% %
= −12 + 10 − 4 = −6 The answer is D. 7 u
% = 3 i
% − 2 j
% and v
% = 16 i
% + 24 j
%
cos θ = | | | |u vu v
⋅% %
% %
= 48 4813 832
−
= 0 θ = cos−1(0) = 90° The answer is E.
M C 1 1 Q l d - 7 148 I n t r o d u c t i o n t o v e c t o r s
8 A: Magnitude = 2 24 3+
= 25 = 5 Not a unit vector.
B: Magnitude = 0 = 0 Not a unit vector.
C: Magnitude = 2 20.8 0.6+
= 1 = 1 A unit vector. and (3 4 ) (0.8 0.6 )i j i j− ⋅ +
% %% %
= 2.4 − 2.4 = 0
So the vectors are perpendicular. The answer is C. 9 u
% = 3 i
% + a j
% and v
% = 2a i
% − a j
%
(3 ) (2 )i aj ai aj+ ⋅ −% %% %
= 0
⇒ 6a − a2 = 0 a (6 − a) = 0 a = 0 or a = 6 Reject a = 0 as v
% = 0 in this case.
The answer is D.
10 cos θ = 2 2 2 2 2
(4 2 3 ) (2 2 )
4 ( 2) 3 2 ( 2)
i j k i j− + ⋅ −
+ − + + −% % %% %
= 8 429 8
+
= 122 29 2
= 658
⇒ θ = cos−1 ( )658
The answer is E. 11 u
% = 4 i
% +3 j
% and v
% = − i
% + 2 j
%
cos θ = | | | |u vu v
⋅% %
% %
= 2 2 2 2
(4 3 ) ( 2 )
4 3 ( 1) 2
i j i j+ ⋅ − +
+ − +% %% %
= 4 625 5
− +
= 25 5
= 2 525
θ = cos−12 525
⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠
= 1.3909 12 u
% = 3 5i j−
% % and v
% = 4i j− +
% %
a u v+% %
= 3 5i j−% %
+ 4i j− +% %
= 4i j− −% %
b u v−% %
= 3 5i j−% %
− ( 4 )i j− +% %
= 7 6i j−% %
c u v⋅% %
= (3 5 ) ( 4 )i j i j− ⋅ − +% %% %
= −12 − 5 = −17
d u%
= | |uu%%
= 2 2
3 5
3 ( 5)
i j−
+ −% %
= 1 (3 5 )34
i j−% %
= 3 534 34
i j−% %
e cos θ = | | | |u vu v
⋅% %
% %
= 2 2 2 2
17
3 ( 5) ( 4) 1
−
+ − − +
= 1734 17−
= 1717 2−
= 12
−
⇒ θ = 135° 13 v
% = 3 2 4i j k− +
% %%
| |v%
= 2 2 23 ( 2) 4+ − +
= 29
x-axis: cos θ = 329
θ = 56.1°
y-axis: cos θ = 229
−
θ = 111.8°
z-axis: cos θ = 429
θ = 42.0° 14 [ 2(1 3 ) ] (2 3 )pi p j pi j+ − ⋅ +
% %% % = 0
2p2 + 6(1 − 3p) = 0 2p2 + 6 − 18p = 0 p2 − 9p + 3 = 0
p = 29 ( 9) 4(1)(3)2
± − −
= 9 692
±
15 a%
= i%
− 2 j%
and b%
= 2 i%
+ 3 j%
| |b%
= 3 22 3+
= 13 b
% = 1
13(2 i%
+ 3 j%
)
b a⋅% %
= 113
(2 3 ) ( 2 )i j i j+ ⋅ −% %% %
= 113
(2 − 6)
= 413
−
The answer is B.
16 | |a%
= 2 21 ( 2)+ −
= 5 a
% = 1
5( i%
− 2 j%
)
I n t r o d u c t i o n t o v e c t o r s M C 1 1 Q l d - 7 149
ˆ ˆ( )a b a⋅% % %
= 1 1( 2 ) (2 3 ) ( 2 )5 5
i j i j i j− ⋅ + × −% % %% % %
= 15
(2 − 6) ( i%
− 2 j%
)
= 45− ( i
% − 2 j
%)
The answer is B. 17 a u
% = 2 i
% + 3 j
% − k
% and
v%
= i%
+ j%
− 2 k%
u%
= 1 ( )| |
uu %%
= 2 2 2
1 (2 3 )2 3 ( 1)
i j k+ −+ + − % %%
= 114
(2 i%
+ 3 j%
− k%
)
= 214
i%
+ 314
j%
− 114
k%
b ||v%
= ( ˆv u⋅% %
) u%
= 1( 2 ) (2 3 )14
i j k i j k⎡ ⎤+ − ⋅ + −⎢ ⎥⎣ ⎦% % % %% %
1 (2 3 )14
i j k+ −% %%
= 114 (2 + 3 + 2)(2 i
% + 3 j
% − k
%)
= 12 (2 i
% + 3 j
% − k
%)
v ⊥%
= v%
− ||v%
= i%
+ j%
− 2 k%
− 12 (2 i
% + 3 j
% − k
%)
= 1 32 2
j k− −%%
= 12
− ( j%
+ 3 k%
)
18 u%
= 3 sin t i%
+ 2 cos t j%
x = 3 sin t
sin t = 3x
sin2t = 2
9x
y = 2 cos t
cos t = 2y
cos2t = 2
4y
⇒ 2
9x +
2
4y = sin2t + cos2t = 1
The path is an ellipse. The answer is D.
19 u%
= 21 2( 1)i t jt
+ −% %
x = 1t
, t > 0 ⇒ x > 0
or t = 1x
y = 2( 2 1t − )
= 221 1
x
⎡ ⎤⎛ ⎞ −⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦
= 2 21 1x
⎛ ⎞−⎜ ⎟⎝ ⎠
y = 22 2x
−
The path is hyperbolic. Asymptotes: x = 0, y = −2
Modelling and problem solving 1 a 1v
% = 8 i
% − 4 j
% + 13 k
% − (2 i
% + 8 j
% + k
%)
= 6 i%
− 12 j%
+ 12 k%
b 1v%
= 2 2 2
6 12 12
6 ( 12) 12
i j k− +
+ − +% %%
= 6 12 12
324
i j k− +% %%
= 118 (6 i
% − 12 j
% + 12 k
%)
= 13 ( i%
− 2 j%
+ 2 k%
)
c The fighter starts at 2 8i j k+ +% %%
and then continues along
the path m v%
. The equation of motion is therefore
( )ˆ2 8 2 8 2 23mi j k mv i j k i j k+ + + = + + + − +
% % % % % % %% % %
d
v%
= 1v%
= 13 ( i%
− 2 j%
+ 2 k%
)
||u%
= ˆ ˆ( )v u v⋅% % %
= 1 ( 2 2 ) ( 2 8 )3
i j k i j k⎡ ⎤− + ⋅ − − −⎢ ⎥⎣ ⎦% % % %% %
1 ( 2 2 )3
i j k− +% %%
= 19 (−2 + 16 − 2) ( i
% − 2 j
% + 2 k
%)
= 129 ( i
% − 2 j
% + 2 k
%)
= 43 ( i
% − 2 j
% + 2 k
%) from fighter’s starting position.
u⊥%
= u−%
+ ||u%
= 2 i%
+ 8 j%
+ k%
+ 43 ( i
% − 2 j
% + 2 k
%)
= 103 i%
+ 163 j%
+ 113 k%
u⊥%
= 13 (10 i
% + 16 j
% + 11 k
%) from station.
e | |u⊥%
= 2 2 21 10 16 113
+ +
= 1 4773
≈ 7.28 The distance is 7.28 km.
f Speed = 1160
| |v% km
h⎛ ⎞⎜ ⎟⎝ ⎠
= 60 1| |v%
= 60 2 2 26 ( 12) 12+ − +
M C 1 1 Q l d - 7 150 I n t r o d u c t i o n t o v e c t o r s
= 60 324 = 60 18× = 1080 The speed of the fighter is 1080 km/h.
g Vertical speed of the fighter = 160
12
= 720 km/h. 2 a C will have the same x coordinate as D, the same y
coordinate as E and the z coordinate will be 0. C is (3, 5.5, 0) b CE
uuur = OE OC−
uuur uuur
= 5.5 j%
+ 4 k%
− (3 i%
+ 5.5 j%
)
= −3 i%
+ 4 k%
c Other diagonal = 3 i
% + 5.5 j
% + 4 k
% − 5.5 j
%
= 3 i%
+ 4 k%
cos θ = 2 2 2 2
( 3 4 ) (3 4 )
( 3) 4 3 4
i k i k− + ⋅ +
− + +% % % %
= 9 165 5
− +×
= 725
θ = 73.7° d V = 3 5.5 4× × = 66 cm3
in litres V = 66 1000÷ = 0.066 litres. e d
% = 3 i
% + 5.5 j
% + 4 k
%
f | |d%
= 2 2 23 5.5 4+ +
= 55.25 = 7.43 cm g Other longest diagonal (connecting D to the y axis) is
given by: b
% = 3 5.5 4i j k− + −
% %%
| |b%
= 55.25 = 7.43
cos θ = | | | |d bd b
⋅% %
% %
= 29 30.25 16
(7.43)− + −
= 5.2555.25
= 0.095 02 θ = 84.5°
3 a
b ZYuuur
= 7 7 5i j i+ −% %%
= 2 7i j+% %
YXuuur
= 2 7 (7 7 )i j i j+ − +% %% %
= 5i−%
c OYuuur
= 7 7 0 0i j i j+ − +% %% %
= 7 7i j+% %
ZXuuur
= 2 7 5i j i+ −% %%
= 3 7i j− +% %
d cos θ = 2 2 2 2
( 3 7 ) (7 7 )
( 3) 7 7 7
i j i j− + ⋅ +
− + +% %% %
= 21 4958 98
− +
= 285684
= 2281421
= 282 7 29× ×
= 229
= 2 2929
e tan θ = 72
= 3.5 θ = 74.05 ≈ 74.1° f The vector resolute of OX on the x-axis is the
x-component of X. = 2 i
%
g Let P be the point such that OP
uuur = 7xi j+
% % (P lies on XY)
⇒ PZuuur
= 5 ( 7 )i xi j− +% % %
= (5 − x) 7i j−% %
OYuuur
= 7 7i j+% %
PZ OY⋅uuur uuur
= [(5 ) 7 ] (7 7 )x i j i j− − ⋅ +% %% %
= 0
⇒ 7(5 − x) − 49 = 0 35 − 7x − 49 = 0 7x = −14 x = −2 ⇒ OP
uuur = 2 7i j− +
% %
Coordinates of P are (−2, 7). h Area = bh where b = 5, h = 7 Area = 5 7× = 35 square units.
4 a
b Speed = 2 23 5+
= 34 = 5.83 m/s c tan θ = 5
3
= 1.6667 ⇒ θ = 59.0 The bearing is 059° Τ. d Distance = speed × time where speed = 3 m/s
I n t r o d u c t i o n t o v e c t o r s M C 1 1 Q l d - 7 151
time = 2 minutes or 120 seconds. Distance = 3 × 120 = 360 The width of the river is 360 metres. e Distance = 5 × 120 = 600 The swimmer is carried 600 metres downstream. f If the swimmer started on the north bank: Speed = 5.83 m/s (same as before) River width = 360 m Distance downstream = 600 m (same as before) Bearing:
tan θ = 53
θ = 59.0° Bearing is 180 − 59 = 121° T The results are the same, except the bearing is now 121° T. 5 a Assuming that Sally is at the origin, the vector would be: 25 sin 35° i
% + 25 cos 35° j
% + 8 k
%
= 14.34 i%
+ 20.48 j%
+ 8 k%
b Distance = 2 2 214.34 20.48 8+ +
= 689.066 = 26.25 m
c Total height is 8 + 1.75 = 9.75 m
θ = tan−1 9.7525
⎛ ⎞⎜ ⎟⎝ ⎠
= 21.3° The angle of elevation is 21.3°. 6 a Let v
% represent the skydiver’s path.
v%
= 3 5 4 (6 4 12 )i j k i j k− + − + +% % % %% %
= − 3 9 8i j k− −% %%
b | |v%
= 2 2 2( 3) ( 9) ( 8)− + − + −
= 154
v%
= 1 ( 3 9 8 )154
i j k− − −% %%
c 3 5 4i j k− +% %%
Distance = 2 2 23 ( 5) 4+ − +
= 50 = 5 2 km d Speed = | | 60v ×
% (km/h)
= 154 60× ≈ 744.6 km/h
