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“c02VectorsInThePlane_WS_print” — 2018/10/16 — 8:45 — page 18 — #1 ii

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18 CHAPTER 2 Vectors in the plane • EXERCISE 2.2

Chapter 2 — Vectors in the plane

Exercise 2.2 — Vectors and scalars

1 a i r~ + s~

r~

s~

r + s~ ~s~

ii r~ − s~

r~ s~

r – s~ ~

s~–

iii s~ − r~r~

s – r~ ~

s~–

–

b i 2r~ + 2s~

2r~

s~

2r + 2s~ ~2s~

Same as 1 a i except scaled by a factor of 2.ii 2r~ − 2s~

2r~ s~–2s~

Same as 1 a ii except scaled by a factor of 2.iii 3s~ − 4r~

3s

–4r~

~~ ~3s – 4r

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

4

2

N

S

EW

Displacement = (4 − 2) north= 2 km north

The answer is C.4 a u~ + v~

= A toCb u~ − v~

= D toBc v~ − u~

= B toDd 3u~ + 2v~ − 2u~ − v~

= u~ + v~= A toC

5 u~ = 2v~ + w~w~ = v~ − u~⇒ u~ = 2v~ + v~ − u~

= 3v~ − u~2u~ = 3v~

u~ = 32v~

The answer is D.

6 a CH = CG + GH

= r~ + s~b CJ = CG + GJ

= s~ + t~c GD = GH + HD

= r~ − s~d FI = FE + EI

= r~ + s~e HE = HI + IE

= t~ − s~f DJ = DH + HG + GJ

= s~ − r~ + t~g CI = CD + DE + EI

= r~ + t~ + s~h JC = JG + GC

= −t~ − s~7 OD = 2 × a~ + 4 × b~

= 2a~ + 4b~8 EO = −3 × a~ + − 4 × b~

= −3a~ − 4b~9 Displacement, velocity, force

10 Speed, time, length11 1 magnitude and 2 angles (N–S) and (E–W).

Jacaranda Maths Quest 11 Specialist Mathematics Units 1 & 2 for Queensland Solutions Manual

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CHAPTER 2 Vectors in the plane • EXERCISE 2.2 19

12 a and b

Start

Finish

N

S

EW

R = N + E~�

~ ~

300 km

400 km

N~

E~

c R~ =√3002 + 4002

=√250 000

= 500 km

d tan 𝜃 = 400300

𝜃 = tan−1(43)

𝜃 = 53.1°clockwise from north

13

300 km

Start

300 km

T ES

45°

Finish

F

θ

N

S

EW

EF = TE = 300 cos 45°

= 300 ×√22

= 150√2 kmTotal distance east of the starting

point is 300 + 150√2 = 512.1 kmResultant bearing:

tan 𝜃 =150√2

(300 + 150√2)≈ 0.414

𝜃 = 22.5°Resultant bearing is 90° − 22.5°= 67.5° clockwise from north

14

600 km

400 kmθ

N

S

EW

R~

R~ =√6002 + 4002

=√520 000

= 721.1 km

tan 𝜃 = 600400

𝜃 = tan−1(32)

𝜃 = 56.3°Resultant bearing = 270° + 56.3°

= 326.3°15 a a~ + b~

8 east and 8 northb a~ + 3b~

18 east and 14 northc a~ − b~

2 west and 2 northd b~ − a~

2 east and 2 southe 3b~ − 4a~

3 east and 11 southf 0.5a~ + 2.5b~14 east and 10 north

g a~ − 2.5b~9.5 west and 2.5 south

h 4a~12 east and 20 north

i 2.5a~ − 1.5b~8 north

j b~ − 2.5a~2.5 west and 9.5 south

5

15

–15

–5–5–15 5 150

y

x

a~a

c

ged

bh

b~

f

j

i

16

10 m

4 m

A

B

θ

N

S

EW

AB =√102 + 42

=√116

= 10.77

𝜃 = tan−1(

410)

= 21.8°Bearing is 90° − 21.8°

= 068.2° True17

a

A

B C

D

~

b~

AC = a~ − b~BD = a~ + b~AC + BD = a~ − b~ + a~ + b~

= 2a~


ii


ii

ii


18 3u + 3v~ ~3(u + v)~ ~

u + v~ ~u~

v~3u~

3v~

3 (u~ + v~) = 3u~ + 3v~19

a + (b + c)~ ~

(a + b) + c ~~ ~

b + c~ ~

c~

b~

a~a + b~ ~

(a~ + b~) + c~ = a~ + (b~ + c~)20

s~s – 3r~ ~

–s~ ~~ ~–3r

~3r

3r – s

3r~ − s~ = − (s~ − 3r~)21

500 m

200 m

600 m

200 m

300 m

400 m

Start

N

S

EW

Net displacement vector is O~22

1

2

3

4

5

6

7

8

2 310 4 5 6

y

x

v~

v~

w~

v~w~ +

The horizontal component of w~ is 4The vertical component of w~ is 5The horizontal component of v~ is 2The vertical component of v~ is 3w~ + v~ = (6, 8)

23

1

2

3

4

5

6

3 521 4 60

y

x

v~

–v ~

w~

v~w~ –

w~ − v~ = (2, 2)24 a

5

10

15

20

12 2084 160

y

x

w~

4 w ~

4w~ = (16, 20)b

234

1

–2–3

–1

–4–5–6

–2–4 2 31–5 –3 –1 x

y

O

~–2v

~v

−2v~ = (−4, −6)25 One can deduce that x and y components can be added,

subtracted and multiplied separately.

Exercise 2.3 — Position vectors in the plane

1 a 3 𝚤 + 4 𝚥x = 3, y = 4

b 6 𝚤 − 3 𝚥x = 6, y = −3

c 3.4 𝚤 + √2 𝚥x = 3.4, y = √2

2 a i || v~ || =√62 + 62

=√36 + 36

=√72

= 6√2

ii 𝜃 = tan−1(66)

= tan−1 1

= 45°


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ii

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b i ||w~ || =√(−4)2 + 72

=√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~ || =√(−3.4)2 + (−3.5)2

=√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~ || = √3202 + (−10)2

= √102 400 + 100

= √102 500

= 50√41

(≈ 320.16)

ii tan−1(

−10320 )

= tan−1 (−0.031 25)= −1.8°

or 𝜃 = 360° − 1.8°= 358.2° (fourth quadrant)

3 i a 045°Tb 270° + 60.3° = 330.3°Tc 270° − 45.8° = 224.2°Td 90° + 1.8° = 091.8° T

ii a [6√2, 45°]b [√65, − 60.3°]c [4.88, 225.8°]d [320.16, 358.2°]

4

x

y

�

210°100w~

𝜃 = 90° + (360 − 210)°

= 240°

x = ||w~ || cos 240°= 100 cos 240°

= −50y = ||w~ || sin 240°

= 100 sin 240°

= −100 sin 60°

= −50√3

w~ = −50 𝚤 − 50√3 𝚤

5

x

y

60°30°

10

x = 10 cos 30°,= 5√3

y = 10 sin 30°

= 5The answer is C.

6

�

u~

x

y

457 km

147°

𝜃 = 90° − 147°

= −57°x = 457 cos (−57°)

= 248.9y = 457 sin (−57°)

= −383.3⇒ u~ = 248.9 𝚤 − 383.3 𝚥

7

�b~

x

y

125 km

331°

𝜃 = 90° + (360 − 331°)= 119°

x = 125 cos 119°

= −60.6y = 125 sin 119°

= 109.3⇒ b~ = −60.6 𝚤 + 109.3 𝚥

8�

a + b

45°

60°

a

b

a~ = 420 cos 45° 𝚤 − 420 sin 45° 𝚥= 210√2 𝚤 − 210√2 𝚥

b~ = 200 cos 60° 𝚤 − 200 sin 60° 𝚥= 100 𝚤 − 100√3 𝚥

a~ + b~ = (210√2 𝚤 − 210√2 𝚥) + (100 𝚤 − 100√3 𝚥)= (210√2 + 100) 𝚤 − (210√2 + 100√3) 𝚥


ii


ii

ii


|| a~ + b~ || =√(210√2 + 100)2

+ (210√2 + 100√3)2

= 615.4

𝜃 = tan−1

(−210√2 + 100√3

210√2 + 100 )= −49.8°

The resultant displacement is 615 km at 49.8° south of east.In polar form [615.4, −49.8°].

9

�45°

45°

a + b

ab

a~ = 20 cos 45° 𝚤 + 20 sin 45° 𝚥= 10√2 𝚤 + 10√2 𝚥

b~ = 30 cos 45° 𝚤 − 30 sin 45° 𝚥= 15√2 𝚤 − 15√2 𝚥

a~ + b~ = (10√2 𝚤 + 10√2 𝚥) + (15√2 𝚤 − 15√2 𝚥)= (10√2 + 15√2) 𝚤 + (10√2 − 15√2) 𝚥

= 25√2 𝚤 − 5√2 𝚥

|| a~ + b~ || =√(25√2)2

+ (5√2)2

=√1300

= 10√13

≈ 36

𝜃 = tan−1

(−5√2

25√2 )= −11.3°

Take 36 steps in a direction 11.3° south of east.10

40°30°

a – b

ab

a~ = 15 cos 30° 𝚤 + 15 sin 30° 𝚥

=15√32

𝚤 + 152

𝚥

b~ = −12 sin 40° 𝚤 + 12 cos 40° 𝚥

a~ − b~ =(15√32

𝚤 + 152

𝚥)

− (−12 sin 40° 𝚤 + 12 cos 40° 𝚥)

=(15√32

+ 12 sin 40°)

𝚤 + (152

− 12 cos 40°) 𝚥

= 20.7038 𝚤 − 1.6925 𝚥|| a~ − b~ || =√20.70382 + 1.69252

= 20.8The scouts are 20.8 km apart.

11 a a~ = 3 𝚤 + 4 𝚥|| a~ || =√32 + 42

= 5

a~ = 35

𝚤 + 45

𝚥

b d~ = 3 𝚤 − 4 𝚥|| d~ || =√32 + (−4)2

= 5

d~ = 35

𝚤 − 45

𝚥

c b~ = 4 𝚤 + 3 𝚥|| b~ || =√42 + 32

= 5

b~ = 45

𝚤 + 35

𝚥

d e~ = −4 𝚤 + 3 𝚥|| e~ || =√(−4)2 + 32

= 5

e~ = −45

𝚤 + 35

𝚥

e c~ = 𝚤 +√2 𝚥

|| c~ || =√12 + (√2)2

=√3

c~ = 1

√3𝚤 +

√2

√3𝚥

12 v~ = 3 𝚤 − 4 𝚥|| v~ || =√32 + (−4)2

= 5

v~ = 35

𝚤 − 45

𝚥

The answer is B.

13 v~ = 0.3 𝚤 + 0.4 𝚥|| v~ || =√0.32 + 0.42

= 0.5

v~ = 0.30.5

𝚤 + 0.40.5

𝚥

= 0.6 𝚤 + 0.8 𝚥= 2 (0.3 𝚤 + 0.4 𝚥)= 2v~

14 w~ = −0.1 𝚤 − 0.02 𝚥||w~ || =√(−0.1)2 + (−0.02)2

= 0.102

w~ = −0.10.102

𝚤 − 0.020.102

𝚥

= −0.98 𝚤 − 0.20 𝚥15 a AB = (4 − 0) 𝚤 + (5 − 1) 𝚥

= 4 𝚤 + 4 𝚥b BA = (0 − 4) 𝚤 + (1 − 5) 𝚥

= −4 𝚤 − 4 𝚥

16 a i AB = (4 − 0) 𝚤 + (−5 − 2) 𝚥= 4 𝚤 − 7 𝚥

ii || AB || =√42 + (−7)2

=√65


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ii

ii


b i AB = (5 − 2) 𝚤 + (4 − 3) 𝚥= 3 𝚤 + 𝚥

ii || AB|| =√32 + 12

=√10

c i AB = (0 − 4) 𝚤 + (2 − −5) 𝚥= −4 𝚤 + 7 𝚥

ii || AB || =√(−4)2 + 72

=√65

d i AB = (2 − 5) 𝚤 + (3 − 4) 𝚥= −3 𝚤 − 𝚥

ii ||AB|| =√(−3)2 + (−1)2

=√10

e i AB = (5 − 3) 𝚤 + (7 − 7) 𝚥= 2 𝚤

ii || AB || =√22

= 2

f i AB = (3 − 7) 𝚤 + (−3 − −3) 𝚥= −4 𝚤

ii || AB || =√(−4)2

= 4

17 a BA = −AB= −4 𝚤 + 7 𝚥

b BA = −3 𝚤 − 𝚥c BA = 4 𝚤 − 7 𝚥d BA = 3 𝚤 + 𝚥e BA = −2 𝚤f BA = 4 𝚤

18 a AB = 4 𝚤 − 7 𝚥|| AB || =√65

AB = 1

√65(4 𝚤 − 7 𝚥)

= 4

√65𝚤 − 7

√65𝚥

b AB = 3 𝚤 + 𝚥|| AB || =√10

AB = 3

√10𝚤 + 1

√10𝚥

c AB = −4 𝚤 + 7 𝚥|| AB || =√65

AB = − 4

√65𝚤 + 7

√65𝚥

d AB = −3 𝚤 + 𝚥|| AB || =√10

AB = − 3

√10𝚤 + 1

√10𝚥

e AB = 2 𝚤|| AB || = 2

AB = 22

𝚤

= 𝚤

f AB = −4 𝚤|| AB || = 4

AB = −44

𝚤

= − 𝚤19 u~ = 5 𝚤 − 2 𝚥 and e~ = −2 𝚤 + 3 𝚥

a i || u~ || =√52 + (−2)2

=√29

ii || e~ || =√(−2)2 + 32

=√13

iii u~ = 5

√29𝚤 − 2

√29𝚥

iv e~ = − 2

√13𝚤 + 3

√13𝚥

v u~ + e~ = 5 𝚤 − 2 𝚥 − 2 𝚤 + 3 𝚥= 3 𝚤 + 𝚥

vi || u~ + e~ || =√32 + 12

=√10

b || u~ || + || e~ || =√29 +√13 > || u~ + e~ || =√10Therefore, reject the statement as the magnitudes aredifferent.

20 u~ = −3 𝚤 + 4 𝚥 and e~ = 5 𝚤 − 𝚥

a i || u~ || =√(−3)2 + 42

=√25

= 5

ii || e~ || =√52 + (−1)2

=√26

iii u~ = −35

𝚤 + 45

𝚥

iv e~ = 5

√26𝚤 − 1

√26𝚥

v u~ + e~ = (−3 + 5) 𝚤 + (4 − 1) 𝚥= 2 𝚤 + 3 𝚥

vi || u~ + e~ || =√22 + 32

=√13

b || u~ || + || e~ || = 5 +√26 > || u~ + e~ || =√13Therefore, reject the statement as the magnitudes aredifferent.

21 a a~ − b~ = (3 − 2) 𝚤 + (2 − 3) 𝚥= 𝚤 − 𝚥

|| a~ − b~ || =√12 + 12

=√2

b a~ − b~ = (5 − 2) 𝚤 + (−2 − 5) 𝚥= 3 𝚤 − 7 𝚥

|| a~ − b~ || =√32 + 72

=√58


ii


ii

ii


22

i~

j~

�

b~v~

r~x

y

0 3 km/h

5 km/h

5

3

a r~ = 3 𝚤b b~ = 5 𝚥c v~ = 3 𝚤 + 5 𝚥

d tan 𝜃 = 53

𝜃 = tan−1 53

= 59.0°True bearing = 90° − 59.0°

= 031.0°

e || v~ || =√52 + 32

=√34 km/h23

�v~b~

x

y

= 5 km/h

r~ = 3 km/hN

S

EW

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 theopposite bank due north of the starting position.

24 a

a~

b~

a – b

x

y

0 20 km/h

15 km/h

~~ b~ = 15j–~

a~ = 20 𝚤, b~ = −15 𝚥a~ − b~ = 20 𝚤 + 15 𝚥

b || a~ − b~ || =√202 + 152

= 25

c tan 𝜃 = 1520

𝜃 = tan−1(1520)

= 36.9°True bearing = 90° − 36.9°

= 053.1°

25 u~ = 3 𝚤 + 4 𝚥

tan 𝜃 = 43

𝜃 = tan−1 43

𝜃 = 53.1°v~ = 4 𝚤 − 3 𝚥

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

= 0To confirm that this is a pattern for all perpendicular vectors,let u~ = x1 𝚤 + y1 𝚥 and let u~ make an angle of 𝜃 to the positivedirection of the x-axis. Let v~ = x2 𝚤 + y2 𝚥 be at right angles tovector u~ . As the vectors are perpendicular, v~ will be at 90 − 𝜃to the negative direction of the x-axis.

�i

jy2

x1x2

y1v

a90 – �

tan 𝜃 =y1x1

[1]

tan (90 − 𝜃) =y2

−x2sin (90 − 𝜃)cos (90 − 𝜃)

=y2

−x2cos 𝜃sin 𝜃

=y2

−x2sin 𝜃cos 𝜃

= −x2y2

tan 𝜃 = −x2y2

[2]

Equating [1] and [2]

y1x1

= −x2y2

y1y2 = −x1x20 = x1x2 + y1y2

26 a

v~ �b~

r~

r~x

y

0

12 km/h

4 km/h

4 km/hv~ = −4 𝚤 + 12 𝚥


ii


ii

ii


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 hoursTime taken is 2.5 minutes.

Exercise 2.4 — Scalar multiplication of vectors

1 a~ = 4 𝚤 − 5 𝚥 b~ = 3 𝚤 + y 𝚥c~ = 3a~ + 2b~

= 3 (4 𝚤 − 5 𝚥) + 2 (3 𝚤 + y 𝚥)= 6 𝚤 − 15 𝚥 + 6 𝚤 + 2y 𝚥= 12 𝚤 + (2y − 15) 𝚥 // to x-axis

⇒ 2y − 15 = 0

y = 152

2 CD = 7 𝚤 − 5 𝚥 C (x, − 3) D (4, y)OD = 4 𝚤 + y 𝚥 OC = x 𝚤 − 3 𝚥CD = OD − OC = (4 − x) 𝚤 + (y + 3) 𝚥 = 7 𝚤 − 5 𝚥𝚤∶ 4 − x = 7

x = −3𝚥∶ y + 3 = −5

y = −8

3 OM = OA + OB2

OA = 3 𝚤 + 4 𝚥OB = 7 𝚤 + 8 𝚥

OM = 12

(10 𝚤 + 12 𝚥)

OM = 5 𝚤 + 6 𝚥4 a x 𝚤 + 3 𝚥 = 𝜆 (5 𝚤 − 6 𝚥)

⇒ 𝜆 = −12

⇒ x = −52

b −4 𝚤 + 5 𝚥 = 𝜆 (6 𝚤 + y 𝚥)

⇒ 5 = −2y3

y = −152

−4 = 6𝜆

𝜆 = −23

5 a 2a~ + 3b~ = 2 (−2 𝚤 + 3 𝚥) + 3 (4 𝚤 − 2 𝚥)= −4 𝚤 + 6 𝚥 + 12 𝚤 − 6 𝚥= 8 𝚤

⇒ //parallel to the x-axisb 4c~ − 3d~ = 4 (3 𝚤 − 5 𝚥) − 3 (4 𝚤 − 3 𝚥)

= 12 𝚤 − 20 𝚥 − 12 𝚤 + 9 𝚥= −11 𝚤

⇒ //parallel to the y-axis

c 3a~ + 4b~ = 3 (x 𝚤 − 5 𝚥) + 4 (5 𝚤 + 3 𝚥)= (3x + 20) 𝚤 + (12 − 15) 𝚥

parallel to the y-axis

⇒ 3x + 20 = 0 ⇒ x = −203

d 5c~ + 7d~ = 5 (5 𝚤 − 3 𝚥) + 7 (4 𝚤 + y 𝚥)= (25 + 28) 𝚤 + (7y − 15) 𝚥

parallel to the x-axis

⇒ 7y − 15 = 0 y = 157

6 a A (3, −2) B (4, −5) , C (1, 4)OA = 3 𝚤 − 2 𝚥 OB = 4 𝚤 − 5 𝚥 OC = 𝚤 + 4 𝚥AB = OB − OA BC = −3 𝚤 + 9 𝚥= 𝚤 − 3 𝚥BC = −3AB⇒ A,B,Ccollinear.

b A (5, −3) B (2, 1) C (8, −7)OA = 5 𝚤 − 3 𝚥 OB = 2 𝚤 + 𝚥 OC = 8 𝚤 − 7 𝚥AB = OB − OA

= −3 𝚤 + 4 𝚥BC = OC − OB

= 6 𝚤 − 8 𝚥BC = −2AB⇒A, B, C collinear.

7 a A (5, −2) B (−1, 3)

i OA = 5 𝚤 − 2 𝚥 OB = − 𝚤 + 13 𝚥AB = OB − OA

= (− 𝚤 + 3 𝚥) − (5 𝚤 − 2 𝚥)= −6 𝚤 + 5 𝚥

ii ||AB|| =√36 + 25 =√61

iii AB = 1

√61(−6 𝚤 + 5 𝚥)

b C (−4, 3) D (2, −5)

i OC = −4 𝚤 + 3 𝚥 OD = 2 𝚤 − 5 𝚥CD = OD − OC

= (2 𝚤 − 5 𝚥) − (−4 𝚤 + 3 𝚥)= 6 𝚤 − 8 𝚥

ii ||CD|| =√36 + 64 =√100 = 10

iii DC = −110

(6 𝚤 − 8 𝚥)

= 15

(−3 𝚤 + 4 𝚥)

8 a~ = 5 𝚤 + 2 𝚥 b~ = −3 𝚤 − 4 j⌢

c~ = 𝚤 − 8 𝚥c~ = ma~ + nb~𝚤 − 8 𝚥 = m (5 𝚤 + 2 𝚥) + n (−3 𝚤 − 4 𝚥)

= (5m − 3n) 𝚤 + (2m − 4n) 𝚥5m − 3n = 1 [1]2m − 4n = −8 [2][1] × 4 − 20m + 12n = 4

[2] × −3 − 6m + 12n = 2414m = 28m = 2 ⇒ 3n = 5m − 1

3n = 10 − 1 = 9

n = 39 r~ = x 𝚤 + y 𝚥 s~ = 4 𝚤 − 3 𝚥

2r~ + 3s~ = 2 (x 𝚤 + y 𝚥) + 3 (4 𝚤 − 3 𝚥) = 8 𝚤 + 𝚥= (2x + 12) 𝚤 + (2y − 9) 𝚥 = 8 𝚤 + 𝚥


ii


ii

ii


𝚤∶ 2x + 12 = 8

2x = −4x = −2

𝚥∶ 2y − 9 = 1

2y = 10

y = 5

10 r~ = x 𝚤 − 4 𝚥 s~ = 3 𝚤 + 5 𝚥

i || r~ || =√x2 + 16 = 6

x2 + 16 = 36

x2 = 20 = 4 × 5

x = ±2√5ii r~ = 𝜆s~ −5r~ = 4s~

⇒ −5x = 12

x = −125

11 a~ = 4 𝚤 + y 𝚥 b~ = 2 𝚤 − 5 𝚥

a || a~ || =√16 + y2

16 + y2 = 29

y2 = 13

y = ±√13

|| b~ || =√4 + 25 =√29

b 2b~ = a~−5 = 2y

y = −52

12 a a~ = x 𝚤 + 3 𝚥 b~ = −2 𝚤 + y 𝚥3a~ + b~ = 3 (x 𝚤 + 3 𝚥) + 4 (−2 𝚤 + y 𝚥)

= (3x − 8) 𝚤 + (9 + 4y) 𝚥 = 4 𝚤 + 𝚥⇒ 3x − 8 = 4

3x = 12

x = 4

9 + 4y = 1

4y = −8y = −2

b a~ = 4 𝚤 − 5 𝚥 b~ = −7 𝚤 + 3 𝚥c~ = ma~ + nb~

= m (4 𝚤 − 5 𝚥) + n (−7 𝚤 + 3 𝚥)= (4m − 7n) 𝚤 + (−5m + 3n) 𝚥= 8 𝚤 + 13 𝚥4m − 7n = 8 [1]

−5m + 3n = 13 [2][1] × 5 ⇒ 20m − 35n = 40[2] × 4 − 20m + 12n = 52add −23n = 92n = −4 ⇒ 4m = 7n + 8

= −28 + 8

= −20m = −5

c a~ = 5 𝚤 − 6 𝚥 b~ = −2 𝚤 + 4 𝚥c~ = ma~ + nb~

= m (5 𝚤 − 6 𝚥) + n (−2 𝚤 + 4 𝚥)= (5m − 2n) 𝚤 + (4n − 6m) 𝚥= 2 𝚥

5m − 2n = 0 [1]4n − 6m = 2 [2]

[1] × 2 10m − 4n = 0

4n − 6m = 2

4m = 2

m = 12

n = 5m2

n = 54

Exercise 2.5 — The scalar dot product

1 If a~ = 3 𝚤 + 3 𝚥|| a~ || =√32 + 32

=√18

= 3√2

and b~ = 6 𝚤 + 2 𝚥|| b~ || =√62 + 22

=√40

= 2√10

x

y

0 3

1

2

3

6

a~

b~α

θβ

tan 𝛽 = 26

= 13

⇒ 𝛽 = 18.4°𝜃 = 𝛼 − 𝛽

= 45° − 18.4°= 26.6°

cos 𝜃 = 0.8942∴ a~ ⋅ b~ = || a~ || || b~ || cos 𝜃

= 3√2 × 2√10 × 0.8942= 23.99

tan 𝛼 = 33

= 1

⇒ 𝛼 = 45°

2 a~ = 3 𝚤 + 3 𝚥 and b~ = 6 𝚤 + 2 𝚥a~ ⋅ b~ = 3 × 6 + 3 × 2

= 18 + 6

= 24This is more accurate as no angle is required.

3 a a~ ⋅ b~ = (2 𝚤 + 3 𝚥) ⋅ (3 𝚤 + 3 𝚥)= 2 × 3 + 3 × 3

= 6 + 9

= 15

b a~ ⋅ b~ = (4 𝚤 − 2 𝚥) ⋅ (5 𝚤 + 𝚥)= 4 × 5 + −2 × 1

= 20 − 2

= 18

c a~ ⋅ b~ = (− 𝚤 + 4 𝚥) ⋅ (3 𝚤 − 7 𝚥)= −1 × 3 + 4 × −7= −3 − 28

= −31


ii


ii

ii


d a~ ⋅ b~ = (5 𝚤 + 9 𝚥) ⋅ (2 𝚤 − 4 𝚥)= 5 × 2 + 9 × −4= 10 − 36

= −26e a~ ⋅ b~ = (−3 𝚤 + 𝚥) ⋅ ( 𝚤 + 4k~)

= −3 × 1 + 1 × 4

= 1

f a~ ⋅ b~ = (10 𝚤) ⋅ (−2 𝚤)= 10 × −2= −20

g a~ ⋅ b~ = (3 𝚤 + 5 𝚥) ⋅ ( 𝚤)= 3 × 1 + 5 × 0

= 3

h a~ ⋅ b~ = (6 𝚤 − 2 𝚥) ⋅ (− 𝚤 − 4 𝚥)= 6 × −1 + −2 × −4= −6 + 8

= 2

4 a~ ⋅ b~ = (3 𝚤 − 3 𝚥) ⋅ ( 𝚤 − 2 𝚥)= 3 × 1 + −3 × −2= 3 + 6

= 9The answer is D.

5 a~ ⋅ a~ = (x 𝚤 + y 𝚥) ⋅ (x 𝚤 + y 𝚥)= x × x + y × y

= x2 + y2

6 a~ ⋅ b~ = (2 𝚤 − 5 𝚥 + k~) ⋅ (− 𝚤 − 2 𝚥 + 4k~)= 2 × −1 + −5 × −2 + 1 × 4

= −2 + 10 + 4

= 12

7 c~ ⋅ (a~ − b~) = (5 𝚤 − 2 𝚥) ⋅ [3 𝚤 + 2 𝚥 − ( 𝚤 − 2 𝚥)]= (5 𝚤 − 2 𝚥) ⋅ (2 𝚤 + 4 𝚥)= 5 × 2 + −2 × 4

= 10 − 8

= 2

c~ ⋅ a~ − c~ ⋅ b~ = (5 𝚤 − 2 𝚥) ⋅ (3 𝚤 + 2 𝚥) − (5 𝚤 − 2 𝚥) ⋅ ( 𝚤 − 2 𝚥)= 5 × 3 + −2 × 2 − (5 × 1 + −2 × −2)= 15 − 4 − (5 + 4)= 15 − 4 − 9

= 2

⇒ c~ ⋅ (a~ − b~) = c~ ⋅ a~ − c~ ⋅ b~8 c~ ⋅ (a~ + b~) = (5 𝚤 − 2 𝚥) ⋅ (3 𝚤 + 2 𝚥 + 𝚤 − 2 𝚥)

= (5 𝚤 − 2 𝚥) ⋅ (4 𝚤)= 5 × 4 + −2 × 0

= 20

c~ ⋅ a~ + c~ ⋅ b~ = (5 𝚤 − 2 𝚥) ⋅ (3 𝚤 + 2 𝚥) + (5 𝚤 − 2 𝚥) ⋅ ( 𝚤 − 2 𝚥)= 5 × 3 + −2 × 2 + 5 × 1 + −2 × −2= 15 − 4 + 5 + 4

= 20

⇒ c~ ⋅ (a~ + b~) = c~ ⋅ a~ + c~ ⋅ b~9 Two vectors are perpendicular if their dot product is 0.

A: (5 𝚤 + 4 𝚥) ⋅ (−5 𝚤 − 4 𝚥)= 5 × −5 + 4 × −4= −41

B: (5 𝚤 + 4 𝚥) ⋅ (3 𝚤 + 4 𝚥)= 15 + 16

= 31

C: (5 𝚤 + 4 𝚥) ⋅ (−5 𝚤)= −25

D: (5 𝚤 + 4 𝚥) ⋅ (−4 𝚤 + 5 𝚥)= −20 + 20

= 0The answer is D.

10 (a~ − b~) ⋅ (a~ + b~) = 0

⇒ a~ ⋅ a~ − b~ ⋅ b~ = 0

u2 − v2 = 0

u2 = v2

The answer is B.11 (a~ − b~) ⋅ (a~ + b~) = || b~ ||

2

⇒ || a~ ||2 − || b~ ||2 = || b~ ||2

|| a~ ||2 = 2|| b~ ||

2

or || a~ || =√2|| b~ ||The answer is D.

12

6

545°

135°

a~

b~

a~ ⋅ b~ = || a~ || || b~ || cos 𝜃= 6 × 5 cos 135°

= −21.2The answer is C.

13 a~ ⋅ b~ = || a~ || || b~ || cos 𝜃|| a~ || = 7, || b~ || = 8 and 𝜃 = 180° − 50°

= 130°

a~ ⋅ b~ = 7 × 8 cos 130°

= −35.996≈ −36

14 a (4 𝚤 − 3 𝚥) ⋅ (7 𝚤 + 4 𝚥) = 4 × 7 − 3 × 4

= 28 − 12

= 16

b ( 𝚤 + 2 𝚥) ⋅ (−9 𝚤 + 4 𝚥) = −9 + 8

= −1c (8 𝚤 + 3 𝚥) ⋅ (2 𝚤 − 3 𝚥) = 16 − 9

= 7d (5 𝚤 − 5 𝚥) ⋅ (5 𝚤 + 5 𝚥)

= 25 − 25

= 0

15 a || a~ || =√42 + (−3)2

= 5

,|| b~ || =√72 + 42

=√65

cos 𝜃 = u ⋅ vuv

= 16

5 ×√65= 0.3969....

𝜃 = 67°


ii


ii

ii


b || a~ || =√12 + 22

=√5

|| b~ || =√(−9)2 + 42

=√97

cos 𝜃 = −1

√5√97= −0.045....

𝜃 = 93°

c || a~ || =√82 + 32,=√73

|| b~ || =√22 + (−3)2

=√13

cos 𝜃 =√7

√73√13= 0.227...

𝜃 = 77°

d || a~ || =√52 + (−5)2,=√50

|| b~ || =√52 + 52

=√50

cos 𝜃 = 0

√50√50= 0

𝜃 = 90°

16 a~ ⋅ b~ = (2 𝚤 + 3 𝚥) ⋅ (2 𝚤 − 3 𝚥)= 4 − 9

= −5|| a~ || =√22 + 32,

=√13

|| b~ || =√22 + (−3)2

=√13

cos 𝜃 = −5

√13√13

= − 513

= −0.384 62𝜃 = 112.6°

≈ 113°The answer is D.

17 a~ ⋅ b~ = (2 𝚤 − 3 𝚥) ⋅ (−4 𝚤 + 6 𝚥)= −8 − 18

= −26|| a~ || =√22 + (−3)2,

=√13

|| b~ || =√(−4)2 + 62

=√52

cos 𝜃 = −26

√13√52

= −2626

= −1𝜃 = 180°

The answer is D.

18 b~ ⋅ a~ = (m 𝚤 + 3 𝚥) ⋅ (6 𝚤 − 2 𝚥) = 06m − 6 = 0

6m = 6

m = 1

19 b~ ⋅ a~ = (m 𝚤 − 2) ⋅ (4 𝚤 − 3 𝚥) = 04m + 6 = 0

4m = −6

m = −64

m = −32

20 a~ = 2 𝚤 + 4 𝚥Let b~ = k (2 𝚤 + 4 𝚥)

= 2k 𝚤 + 4k 𝚥a~ ⋅ b~ = (2 𝚤 + 4 𝚥) ⋅ (2k 𝚤 + 4k 𝚥) = 404k + 16k = 40

20k = 40

k = 2b~ = 4 𝚤 + 8 𝚥

21 a~ = 4 𝚤 − 3 𝚥Let b~ = k (4 𝚤 − 3 𝚥)

= 4k 𝚤 − 3k 𝚥a~ ⋅ b~ = (4 𝚤 − 3 𝚥) ⋅ (4k 𝚤 − 3k 𝚥) = 8016k + 9k = 80

25k = 80

k = 8025

= 165

b~ = 645

𝚤 − 485

𝚥

Exercise 2.6 — Projection of vectors — scalar andvector resolutes

1 a u~ = 2 𝚤 + 3 𝚥 and a~ = 4 𝚤 + 5 𝚥

i || u~ || =√22 + 32

=√13

u~ = 1

√13(2 𝚤 + 3 𝚥)

u~ ⋅ a~ = 1

√13(2 𝚤 + 3 𝚥) ⋅ (4 𝚤 + 5 𝚥)

= 8

√13+ 15

√13

= 23

√13

=23√1313

ii || a~ || =√42 + 52

=√41

a~ = 1

√41(4 𝚤 + 5 𝚥)

a~ ⋅ u~ = 1

√41(4 𝚤 + 5 𝚥) ⋅ (2 𝚤 + 3 𝚥)

= 8

√41+ 15

√41

= 23

√41

=23√4141


ii


ii

ii


b u~ = 5 𝚤 − 2 𝚥 and a~ = 3 𝚤 − 𝚥

i || u~ || =√52 + (−2)2

=√29

u~ = 1

√29(5 𝚤 − 2 𝚥)

u~ ⋅ a~ = 1

√29(5 𝚤 − 2 𝚥) ⋅ (3 𝚤 − 𝚥)

= 15

√29+ 2

√29

= 17

√29

=17√2929

ii || a~ || =√32 + (−1)2

=√10

a~ = 1

√10(3 𝚤 − 𝚥)

a~ ⋅ u~ = 1

√10(3 𝚤 − 𝚥) ⋅ (5 𝚤 − 2 𝚥)

= 15

√10+ 2

√10

= 17

√10

=17√1010

c u~ = −2 𝚤 + 6 𝚥 and a~ = 𝚤 − 4 𝚥

i || u~ || =√(−2)2 + 62

=√40

= 2√10

u~ = 1

2√10(−2 𝚤 + 6 𝚥)

u~ ⋅ a~ = 1

2√10(−2 𝚤 + 6 𝚥) ⋅ ( 𝚤 − 4 𝚥)

= − 2

2√10− 24

2√10

= − 26

2√10

= −13√1010

ii || a~ || =√12 + (−4)2

=√17

a~ = 1

√17( 𝚤 − 4 𝚥)

a~ ⋅ u~ = 1

√17( 𝚤 − 4 𝚥) ⋅ (−2 𝚤 + 6 𝚥)

= −2

√17− 24

√17

= − 26

√17

= −26√1717

d u~ = 3 𝚤 − 2 𝚥 and a~ = −4 𝚤 − 3 𝚥

i || u~ || =√32 + (−2)2

=√13

u~ = 1

√13(3 𝚤 − 2 𝚥)

u~ ⋅ a~ = 1

√13(3 𝚤 − 2 𝚥) ⋅ (−4 𝚤 − 3 𝚥)

= − 12

√13+ 6

√13

= − 6

√13

= −6√1313

ii || a~ || =√(−4)2 + (−3)2

= 5

a~ = 15

(−4 𝚤 − 3 𝚥)

a~ ⋅ u~ = 15

(−4 𝚤 − 3 𝚥) ⋅ (3 𝚤 − 2 𝚥)

= −125

+ 65

= −65

e u~ = 8 𝚤 − 6 𝚥 and a~ = −5 𝚤 + 𝚥

i || u~ || =√82 + (−6)2

= 10

u~ = 110

(8 𝚤 − 6 𝚥)

u~ ⋅ a~ = 110

(8 𝚤 − 6 𝚥) ⋅ (−5 𝚤 + 𝚥)

= −4010

− 610

= −4610

= −235

ii || a~ || =√(−5)2 + 12

=√26

a~ = 1

√26(−5 𝚤 + 𝚥)

a~ ⋅ u~ = 1

√26(−5 𝚤 + 𝚥) ⋅ (8 𝚤 − 6 𝚥)

= − 40

√26− 6

√26

= − 46

√26

= −46√2626

= −23√2613


ii


ii

ii


2 a a~ = 3 𝚤 − 𝚥 and b~ = 2 𝚤 + 5 𝚥

i || a~ || =√32 + (−1)2

=√10

a~ = 1

√10(3 𝚤 − 𝚥)

a~ ⋅ b~ = 1

√10(3 𝚤 − 𝚥) ⋅ (2 𝚤 + 5 𝚥)

= 6

√10− 5

√10

= 1

√10 (√1010 )

ii b~ || = (a~ ⋅ b~) a~

= 1

√10× 1

√10(3 𝚤 − 𝚥)

= 110

(3 𝚤 − 𝚥)

= 310

𝚤 − 110

𝚥

iii b~ ⊥ = a~ − b~ ||

= 2 𝚤 + 5 𝚥 − (310

𝚤 − 110

𝚥)

= 1710

𝚤 + 5110

𝚥

b a~ = 4 𝚤 + 5 𝚥 and b~ = 8 𝚤 + 10 𝚥

i || a~ || =√42 + 52

=√41

a~ = 1

√41(4 𝚤 + 5 𝚥)

a~ ⋅ b~ = 1

√41(4 𝚤 + 5 𝚥) ⋅ (8 𝚤 + 10 𝚥)

= 32

√41+ 50

√41

=82√4141

= 2√41

ii b~ || = (a~ ⋅ b~) a~

= 2√41 × 1

√41(4 𝚤 + 5 𝚥)

= 8 𝚤 + 10 𝚥iii b~ ⊥ = b~ − b~ ||

= 8 𝚤 + 10 𝚥 − (8 𝚤 + 10 𝚥)= 0~

c a~ = 4 𝚤 + 3 𝚥 and b~ = −3 𝚤 + 4 𝚥

i || a~ || =√42 + 32

= 5

a~ = 15

(4 𝚤 + 3 𝚥)

a~ ⋅ b~ = 15

(4 𝚤 + 3 𝚥) ⋅ (−3 𝚤 + 4 𝚥)

= −125

+ 125

= 0

ii b~ || = (a~ ⋅ b~) a~

= 0 × 15

(4 𝚤 + 3 𝚥)

= 0~iii b~ ⊥ = b~ − b~ ||

= −3 𝚤 + 4 𝚥 − 0~= −3 𝚤 + 4 𝚥

d a~ = 𝚤 + 𝚥 and b~ = 2 𝚤 + 𝚥

i || a~ || =√12 + 12

=√2

a~ = 1

√2( 𝚤 + 𝚥)

a~ ⋅ b~ = 1

√2( 𝚤 + 𝚥) (2 𝚤 + 𝚥)

= 2

√2+ 1

√2

= 3

√2

ii b~ || = (a~ ⋅ b~) a~

= 3

√2× 1

√2( 𝚤 + 𝚥)

= 32

𝚤 + 32

𝚥

iii b~ ⊥ = b~ − b~ ||

= 2 𝚤 + 𝚥 − (32

𝚤 + 32

𝚥)

= 12

𝚤 − 12

𝚥

e a~ = 2 𝚤 + 3 𝚥 and b~ = 2 𝚤 − 3 𝚥

i || a~ || =√22 + 32

=√13

a~ = 1

√13(2 𝚤 + 3 𝚥)

a~ ⋅ b~ = 1

√13(2 𝚤 + 3 𝚥) ⋅ (2 𝚤 − 3 𝚥)

= 4

√13− 9

√13

= − 5

√13

ii b~ || = (a~ ⋅ b~) a~

= − 5

√13× 1

√13(2 𝚤 + 3 𝚥)

= −1013

𝚤 − 1513

𝚥

iii b~ ⊥ = b~ − b~ ||

= 2 𝚤 − 3 𝚥 − (−1013

𝚤 − 1513

𝚥)

= 3613

𝚤 − 2413

𝚥


ii


ii

ii


f a~ = 3 𝚤 + 𝚥 and b~ = 2 𝚥

i || a~ || =√32 + 12

=√10

a~ = 1

√10(3 𝚤 + 𝚥)

a~ ⋅ b~ = 1

√10(3 𝚤 + 𝚥) ⋅ (2 𝚥)

= 0 + 2

√10

= 2

√10

ii b~ || = (a~ ⋅ b~) a~

= 2

√10× 1

√10(3 𝚤 + 𝚥)

= 610

𝚤 + 210

𝚥

= 35

𝚤 + 15

𝚥

iii b~ ⊥ = b~ − b~ ||

= 2 𝚥 − (35

𝚤 + 15

𝚥)

= −35

𝚤 + 115

𝚥

3

a Let the injured bushwalker’s position be denoted by b~ .Let the path of the searcher be denoted by a~ .b~ = 2 𝚤 + 3 𝚥a~ = k (3 𝚤 + 4 𝚥)

|| a~ || = k√32 + 42

= 5k

a~ = 15k

× k (3 𝚤 + 4 𝚥)

= 35

𝚤 + 45

𝚥

a~ ⋅ b = (35

𝚤 + 45

𝚥) ⋅ (2 𝚤 + 3 𝚥)

= 65

+ 125

= 185

b~ || = (a~ ⋅ b~) a~

= 185 (

35

𝚤 + 45

𝚥)

= 5425

𝚤 + 7225

𝚥

|| b~ || || =√(5425)

2

+ (7225)

2

= 3.6The searcher is 3.6 km from the camp when closest to thebushwalker.


b || b~ ⊥ represents the minimum distance between the searcherand the bushwalker.b~ ⊥ = b~ − b~ ||

= 2 𝚤 + 3 𝚥 − (5425

𝚤 + 7225

𝚥)

= − 425

𝚤 + 325

𝚥

= 125

(−4 𝚤 + 3 𝚥)

|| b~ ⊥ = 125

√(−4)2 + 32

= 125

× 5

= 0.2 km4

–3

–2

–1

321 4 5 6Cruiser

j

Yacht

i

b~

a~bll~

b⟂~

Let the yacht’s position be denoted by b~ .The path of the rescue boat be denoted by a~ .b~ = 5 𝚤 − 2 𝚥a~ = k (3 𝚤 − 𝚥)|| b~ ⊥ represents the closest distance the rescue boat gets to theyacht.|| a~ || = k√32 + (−1)2

= k√10

a~ = 1

k√10k (3 𝚤 − 𝚥)

= 1

√10(3 𝚤 − 𝚥)

a~ ⋅ b~ = 1

√10(3 𝚤 − 𝚥) ⋅ (5 𝚤 − 2 𝚥)

= 15

√10+ 2

√10

= 17

√10

b~ || = (a~ ⋅ b~) a~ = 17

√10× 1

√10(3 𝚤 − 𝚥)

= 1710

(3 𝚤 − 𝚥)

= 5110

𝚤 − 1710

𝚥

b~ ⊥ = b~ − b~ ||

= 5 𝚤 − 2 𝚥 − (5110

𝚤 − 1710

𝚥)

= − 110

𝚤 − 310

𝚥

= 110

(− 𝚤 − 3 𝚥)

|| b~ ⊥ = 110

√(−1)2 + (−3)2

= 110

√10

=√1010

km

or ≈ 0.316 km

||

||

||

||||||

||

1

2

3

4

321 4Camp

j

i

b~

a~

bll~

b┴~

ii


ii

ii


5 a~ = 4 𝚤 − 3 𝚥 b~ = 𝚤 − 2 𝚥

i || b~ || =√1 + 4 =√5

a~ ⋅ b~ = 10

√5a~ ⋅ b~ = 4 + 6 = 10

ii (a~ ⋅ b~) b~ = 10

√5× 1

√5( 𝚤 − 2 𝚥)

= 2 ( 𝚤 − 2 𝚥)iii a~ − (a~ ⋅ b~) b~ = (4 𝚤 − 3 𝚥) − 2 ( 𝚤 − 2 𝚥)

= 2 𝚤 + 𝚥6 r~ = 2 𝚤 − 3 𝚥 s~ = 3 𝚤 − 4 𝚥r~ ⋅ s~ = 6 + 12 = 8 || s~ || =√9 + 16 = 5

i (r~ ⋅ s~) s~ = 185

× 15

(3 𝚤 − 4 𝚥)

= 1825

(3 𝚤 − 4 𝚥)

ii r~ − (r~ ⋅ s~) s~ = (2 𝚤 − 3 𝚥) − 1825

(3 𝚤 − 4 𝚥)

= 125 [25 (2 𝚤 − 3 𝚥) − 18 (3 𝚤 − 4 𝚥)]

= 125

(25 × 2 − 18 × 3) 𝚤 + (25 × −3 + 18 × 4) 𝚥

= 125

(−4 𝚤 − 3 𝚥)

7 A (5, 2) B (4, −3)OA = 5 𝚤 + 2 𝚥 OB = 4 𝚤 − 3 𝚥OA ⋅ OB = (5 𝚤 + 2 𝚥) ⋅ (4 𝚤 − 3 𝚥)

= 20 − 6

= 14||OB|| =√16 + 9 =√25 = 5

a ||OE|| = OA ⋅ OB||OB||

= 145

= 2.8

bOC = −3 𝚤 − 4 𝚥

||OC|| =√(−3)2 + (−4)2 =√9 + 16 =√25 = 5 = ||OB||OC ⋅ OB = (−3 𝚤 − 4 𝚥) ⋅ (4 𝚤 − 3 𝚥)

= −12 + 12 = 0

so, OC is ⊥ to OB shown

c ||OE|| × ||OC|| = 145

× 5 = 14 = OA ⋅ OB shown

23

1

–2–3

–1

–4–5–6

–2–4 2 31 4 5 6 7–5 –3 –1 0

y

x

A

BE

O

D

C

i~

j~

8 A (−1, 5) B (3, −2)OA = − 𝚤 + 5 𝚥 OB = 3 𝚤 − 2 𝚥OA ⋅ OB = −3 − 10

= −13||OB|| =√32 + (−2)2 =√13

a ||OE|| = OA ⋅ OB||OB||

= −13

√13×

√13

√13= −√13

b OC = −2 𝚤 − 3 𝚥||OC|| =√(−2)2 + (−3)2 =√13 = ||OB||

OC ⋅ OB = (−2 𝚤 − 3 𝚥) ⋅ (3 𝚤 − 2 𝚥)= −6 + 6 = 0

so OC is ⊥ OB shown.c ||OE|| × ||OC|| = −√13 ×√13 = −13 = OA ⋅ OB shown

4

21

3

5

–2–3–4

–1–2–4 21 3 54–6 –5 –3 –1 0

y

x

A

E

D

CB

i~

j~

9 a a~ = 3 𝚤 − 𝚥 b~ = 𝚤 + 𝚥 || b~ || =√1 + 1 =√2

i b~ = 1

√2( 𝚤 + 𝚥)

ii a~ ⋅ b~ = 3 − 1 = 2 a~ ⋅ b~ = 2

√2×

√2

√2=√2

iii (a~ ⋅ b~) b~ = 𝚤 + 𝚥iv a~ − (a~ ⋅ b~) b~ = (3 𝚤 − 𝚥) − ( 𝚤 + 𝚥)

= 2 𝚤 − 2 𝚥b u~ = 3 𝚤 + 4 𝚥 v~ = 2 𝚤 − 𝚥 || v~ || +√4 + 1 =√5

v~ = 1

√5(2 𝚤 − 𝚥)

u~ ⋅ v~ = 6 − 4 = 2

(u~ ⋅ v~) v~ = 25

(2 𝚤 − 𝚥)

(u~ ⋅ v~) = 2

√5

c r~ = 4 𝚤 + 𝚥 s~ = −3 𝚤 + 4 𝚥 || s~ || =√9 + 16 = 5

s~ = 15

(−3 𝚤 + 4 𝚥)

r~ ⋅ s~ = −12 + 4 = −8

r~ ⋅ s~ = −85

r~ − (r~ ⋅ s~) s~ = 4 𝚤 + 𝚥 + 825

(−3 𝚤 + 4 𝚥)

= 125

[25 (4 𝚤 + 𝚥) + 8 (−3 𝚤 + 4 𝚥)]

= 125

[(100 − 24) 𝚤 + (25 + 32) 𝚥]

= 125

(76 𝚤 + 57 𝚥)


ii


ii

ii

CHAPTER 2 Vectors in the plane • REVIEW 2.7 33

10 a || a~ || = 4, || b~ || = 3, a~ ⋅ b~ = 0

i || a~ + b~ ||2 = (a~ + b~) ⋅ (a~ + b~)= a~ ⋅ a~ + b~ ⋅ a~ + a~ ⋅ b~ + b~ ⋅ b~= || a~ ||

2 + || b~ ||2

= 16 + 9

= 25|| a~ + b~ || = 5

|| a~ − b~ ||2 = (a~ − b~) ⋅ (a~ − b~)= a~ ⋅ a~ − b~ ⋅ a~ − a~ ⋅ b~ + b~ ⋅ b~= || a~ ||

2 + || b~ ||2

= 16 + 9 = 25|| a~ + b~ || = 5

b || a~ || = 5 || b~ || = 12 a~ ⋅ b~ = 0⇒ || a~ + b~ || = || a~ − b~ || = 13

u~

v~

c || r~ || = 7 || s~ || = 24 r~ ⋅ s~ = 0⇒ || r~ + s~ || = || r~ − s~ || = 25

d || a~ || = a || b~ || = b a~ ⋅ b~ = 0

|| a~ + b~ || = || a~ − b~ ||=√a2 + b2

a

b

11 a || a~ || = 4 || b~ || = 3 a~ ⋅ b~ = 2

i (a~ + b~) ⋅ (a~ − b~) = a~ ⋅ a~ − a~ ⋅ b~ + b~ ⋅ a~ − b~ ⋅ b~= || a~ ||

2 − ||b⌢ ||2

= 42 − 32

= 7

ii (a~ + 2b~) ⋅ (a~ − b~) = a~ ⋅ a~ + 2b~ ⋅ a~ − a~ ⋅ b~ − 2.b~ ⋅ b~= || a~ ||

2 + b~ ⋅ a~ − 2|| b~ ||2

= 42 + 2 − 2 × 32

= 0

iii (a~ + b~) ⋅ (a~ − 2b~) = a~ ⋅ a~ + b~ ⋅ a~ − 2b~ ⋅ a~ − 2b~ ⋅ b~= || a~ ||

2 − b~ ⋅ a~ − 2|| b~ ||2

= 42 − 2 − 2 × 32

= −4

b || a~ || = 6 || b~ || = 7 a~ ⋅ b~ = −4

i || a~ + b~ ||2 = (a~ + b~) ⋅ (a~ + b~)= || a~ ||

2 + 2a~ ⋅ b~ + || b~ ||2

= 62 + 2 × −4 + 72 = 77

|| a~ + b~ || =√77

ii || a~ − b~ ||2 = (a~ − b~) ⋅ (a~ − b~)= || a~ ||

2 − 2a~ ⋅ b~ + || b~ ||2

= 62 − 2 × −4 + 72 = 93

|| a~ − b~ || =√93

iii ||3a~ − 2b~ ||2 = (3a~ − 2b~) (3a~ − 2b~)

= 9|| a~ ||2 − 12a~b~ + 4|| b~ ||

2

= 9 × 36 − 12 × −4 + 4 × 49

||3a~ − 2b~ || =√568

= 2√142

12 a Let a~ = 2 𝚤 + 𝚥 b~ = 𝚤 + y 𝚥|| a~ || =√5

a~ ⋅ b~ = 2 + y

|| b~ || =√1 + y2

cos (45°) = a~ ⋅ b~|| a~ || || b~ ||

=2 + y

√5√1 + y2=

√22

⇒ 2 (2 + y) =√2 × 5 (1 + y2) (square both sides)

4 (4 + 4y + y2) = 10 (1 + y2)16 + 16y + 4y2 = 10 + 10y2

6y2 − 16y − 6 = 0

3y2 − 8y − 3 = 0

(3y + 1) (y = 3) = 0

⇒ y = 3 or y = −13

b Let a~ = x 𝚤 − 2√3 𝚥 b = −3 𝚤 +√3 𝚥|| a~ || =√x2 + 12 || b~ || =√9 + 3 =√12 = 2√3

a~ ⋅ b~ = −3x − 6

= −3 (x + 2)

cos (150°) = −√32

= a~ ⋅ b~|| a~ || || b~ ||

= −3 (x + 2)

√x2 + 12 × 2√3

−√32

× 2√3 = −3 (x + 2)

√x2 + 12

⇒√x2 + 12 = x + 2

x2 + 12 = x2 + 4x + 4

4x = 8

x = 213 a Let a~ = 3 𝚤 − 4 𝚥 b~ = x 𝚤 + y 𝚥

|b~ | = 1 ⇒ x2 + y2 = 1 [1]a~ ⋅ b~ = 0 ⇒ 3x − 4y = 0 [2]

Solving x = −45

y = −35

x = 45

y = 35

⇒ b~ = ±15

(4 𝚤 + 3 𝚥)b Let a~ = −5 𝚤 + 12 𝚥 b~ = x 𝚤 + y 𝚥

|b~ | = 1 ⇒ x2 + y2 = 1 [1]a~ ⋅ b~ = 0 ⇒ −5x + 12y = 0 [2]

Solving x = 1213

y = 513

, or x = −1213

, y = −513

b~ = ± 113

(12 𝚤 + 5 𝚥)c Let a~ = 7 𝚤 + 24 𝚥 b~ = x 𝚤 + y 𝚥

|b~ | = 1 ⇒ x2 + y2 = 0 [1]a~ ⋅ b~ = 0 ⇒ 7x + 24y = 0 [2]

⇒ x = −2425

y = 725

or x = 2425

y = −725

b~ = ± 125

(24 𝚤 − 7 𝚥)d Let a~ = a 𝚤 + b 𝚥

b~ = ±1

√a2 + b2(b 𝚤 − a 𝚥)


ii


ii

ii

34 CHAPTER 2 Vectors in the plane • REVIEW 2.7

2.7 Review: exam practice

1 a~ = 4 𝚤 − 3 𝚥 and b~ = 2 𝚤 + 4 𝚥

4a~ − 2.5b~ = 4 (4 𝚤 − 3 𝚥) − 2.5 (2 𝚤 + 4 𝚥)= 16 𝚤 − 12 𝚥 − 5 𝚤 − 10 𝚥= 11 𝚤 − 22 𝚥

2

x

Xy

5 km

10 km

5 kmH

N

S

EW

45°

a HX = 5 𝚤 + 10 cos 45° 𝚥 + 10 sin 45° 𝚥 + 5 𝚥= 5 𝚤 + 5√2 𝚤 + 5√2 𝚥 + 5 𝚥= (5 + 5√2) 𝚤 + (5 + 5√2) 𝚥

b || HX || = (5 + 5√2)√2

= 17.07 km

3 a~ = 3 𝚤 − 2 𝚥 and b~ = 16 𝚤 + 24 𝚥

cos 𝜃 = a~ ⋅ b~|| a~ || || b~ ||

= 48 − 48

√13√832= 0

𝜃 = cos−1 (0)90°

4 A Magnitude =√42 + 32

=√25

= 5Not a unit vector.

B Magnitude =√0

= 0Not a unit vector.

C Magnitude =√0.82 + 0.62

=√1

= 1A unit vector.and (3 𝚤 − 4 𝚥) ⋅ (0.8 𝚤 + 0.6 𝚥)

= 2.4 − 2.4= 0

So the vectors are perpendicular.The answer is C.

5 a~ = 3 𝚤 + a 𝚥 and b~ = 2a 𝚤 − a 𝚥(3 𝚤 + a 𝚥) ⋅ (2a 𝚤 − a 𝚥) = 0

⇒ 6a − a2 = 0

a (6 − a) = 0a = 0 or a = 6Reject a = 0 as v~ = 0 in this case.

6 a~ = 4 𝚤 + 3 𝚥 and b~ = − 𝚤 + 2 𝚥cos 𝜃 = a~ ⋅ b~

|| a~ || || b~ ||

=(4 𝚤 + 3 𝚥) ⋅ (− 𝚤 + 2 𝚥)

√42 + 32√(−1)2 + 22

= −4 + 6

√25√5

= 2

5√5

=2√525

𝜃 = cos−1

(2√525 )

= 1.39097 a~ = 3 𝚤 − 5 𝚥 and b~ = −4 𝚤 + 𝚥

a a~ + b~ = 3 𝚤 − 5 𝚥 + −4 𝚤 + 𝚥= − 𝚤 − 4 𝚥

b a~ − b~ = 3 𝚤 − 5 𝚥 − (−4 𝚤 + 𝚥)= 7 𝚤 − 6 𝚥

c a~ ⋅ b~ = (3 𝚤 − 5 𝚥) ⋅ (−4 𝚤 + 𝚥)= −12 − 5

= −17d a~ = a~

|| a~ ||=

3 𝚤 − 5 𝚥

√32 + (−5)2

= 1

√34(3 𝚤 − 5 𝚥)

= 3

√34𝚤 − 5

√34𝚥

e cos 𝜃 = a~ ⋅ b~|| a~ || || b~ ||

= −17

√32 + (−5)2√(−4)2 + 12

= −17

√34√17

= −17

17√2

= −1

√2⇒ 𝜃 = 135°

8 [p 𝚤 + 2 (1 − 3p) 𝚥] ⋅ (2p 𝚤 + 3 𝚥) = 0

2p2 + 6 (1 − 3p) = 0

2p2 + 6 − 18p = 0

p2 − 9p + 3 = 0

p =9 ±√(−9)2 − 4 (1) (3)

2

=9 ±√69

2


ii


ii

ii


9 a~ = 𝚤 − 2 𝚥 and b~ = 2 𝚤 + 3 𝚥|| b~ || =√23 + 32

=√13

b~ = 1

√13(2 𝚤 + 3 𝚥)

b~ ⋅ a~ = 1

√13(2 𝚤 + 3 𝚥) ⋅ ( 𝚤 − 2 𝚥)

= 1

√13(2 − 6)

= − 4

√13

10 || a~ || =√12 + (−2)2

=√5

a~ = 1

√5( 𝚤 − 2 𝚥)

(a~ ⋅ b~) a~ = 1

√5( 𝚤 − 2 𝚥) ⋅ (2 𝚤 + 3 𝚥) × 1

√5( 𝚤 − 2 𝚥)

= 15

(2 − 6) ( 𝚤 − 2 𝚥)

= −45

( 𝚤 − 2 𝚥)

11 a A (2, 5) B (−4, 7)OA = 2 𝚤 + 5 𝚥OB = −4 𝚤 + 7 𝚥OM = OA + AM

= OA + 12AB

= OA + 12 (OB − OA)

= 12 (OA + OB)

= 12

((2 𝚤 + 5 𝚥) + (−4 𝚤 + 7 𝚥))

= − 𝚤 + 6 𝚥

B

M

A

O

b Let A (−2, −6) , B (1, −3) , C (5, 1)OA = −2 𝚤 − 6 𝚥 OB = 𝚤 − 3 𝚥 OC = 5 𝚤 − 1 𝚥AB = OB − OA

= 3 𝚤 + 3 𝚥

AB = 34BC

AC = OC − OB

= 4 𝚤 + 4 𝚥

𝜆 = 34

⇒ AB is // to BC, B point in common so A, B, Ccollinear.

c Let A (3, −5) B (−3, 4)

P

B

A

O

OP = OA + AP

= OA + 13AB

= OA + 13 (OB − OA)

= 13 (OB + 2OA)

= 13 [(−3i + 4i) + 2 (3 𝚤 − 5 𝚥)]

= 𝚤 − 2 𝚥12 a Let A (−6, 4) B (2, 0) C (−2, −4)

AB = OB − OA

= 8 𝚤 − 4 𝚥||AB|| =√82 + (−4)2

=√80 =√16×5

= 4√5

AC = OC − OA

= 4 𝚤 − 8 𝚥||AC|| =√(4)2 + (−8)2

= 4√5

Since ||AB|| = ||AC|| two sides are equal in length ⇒ABC isan isosceles triangle.

b A (2√3, 4√3) B (4, −2) C (−4, 2)

AB = OB − OA

= (4 − 2√3) 𝚤 + (−2 − 4√3) 𝚥

||AB||2

= (4 − 2√3)2

+ (−2 + 4√3)2

= 16 − HHH16√3 + 12 + 4 + HHH16√3 + 48

= 80 (= 16 × 5)||AB|| = 4√5

BC = OC − OB

= −8 𝚤 + 4 𝚥||BC|| =√64 + 16 =√80 = 4√5

AC = OC − OA

= (−4 − 2√3) 𝚤 + (2 − 4√3) 𝚥||AC|| =√80 = 4√5

⇒ ||AB|| = ||AC|| = ||BC||A, B, C is equilateral triangle.

c A (−7, 5) B (1, −3) C (6, 2)

AC = OC − OA

= 13 𝚤 − 3 𝚥BC = OC − OB

= 5 𝚤 − 5 𝚥AB = OB − OA

= 8 𝚤 − 8 𝚥Now BC ⋅ AB = 5 × 8 − 5 × 8 = 0⇒ BC⊥ to AB, triangle ABC is right-angle at B.

13 a A (3, −4) B (x, y) C (7, 8)AB = OB − OA = (x − 3) i + (y + 4) jBC = OC − OB = (7 − x) 𝚤 + (8 − y) 𝚥⇒ x − 3 = 7 − x

2x = 10

x = 5

y + 4 = 8 − y

2y = 4

y = 2


ii


ii

ii


b A (x, 3) B (2, −1) C (7, −2) D (2, −4)AB = OB − OA = (2 − x) 𝚤 − 4 𝚥DC = OC − OD = 5 𝚤 + 2 𝚥AB = 𝜆DC ⇒ 𝜆 = −2⇒ 2 − x = −10

x = 12c A (3, 5) B (−1, y) C (2, 7) D (−6, 3)

AB = OB − OA = −4 𝚤 + (y − 5) 𝚥DC = OC − OD = 8 𝚤 + 4 𝚥DC = −2AB4 = −2 (y − 5)

y − 5 = −2y = 3

14 a

x

y

O Z (5, 0)

X (2, 7) Y (7, 7)

5i~

2î + 7ĵ

b ZY = 7 𝚤 + 7 𝚥 − 5 𝚤= 2 𝚤 + 7 𝚥

YX = 2 𝚤 + 7 𝚥 − (7 𝚤 + 7 𝚥)= −5 𝚤

c OY = 7 𝚤 + 7 𝚥 − 0 𝚤 + 0 𝚥= 7 𝚤 + 7 𝚥

ZX = 2 𝚤 + 7 𝚥 − 5 𝚤= −3 𝚤 + 7 𝚥

d cos 𝜃 =(−3 𝚤 + 7 𝚥) ⋅ (7 𝚤 + 7 𝚥)

√(−3)2 + 72√72 + 72

= −21 + 49

√58√98

= 28

√5684

= 282√1421

= 28

2 × 7 ×√29

= 2

√29

=2√2929

e tan 𝜃 = 72

= 3.5𝜃 = 74.05

≈ 74.1°f The vector resolute of OX on the x-axis is the x-componentof X.= 2 𝚤

g Let P be the point such thatOP = x 𝚤 + 7 𝚥 (P lies onXY)

⇒ PZ = 5 𝚤 − (x 𝚤 + 7 𝚥)= (5 − x) 𝚤 − 7 𝚥

OY = 7 𝚤 + 7 𝚥⇒ PZ ⋅ OY = [(5 − x) 𝚤 − 7 𝚥] ⋅ (7 𝚤 + 7 𝚥) = 0

⇒ 7 (5 − x) − 49 = 0

35 − 7x − 49 = 0

7x = −14x = −2

⇒ OP = −2 𝚤 + 7 𝚥Coordinates of P are (−2, 7) .

h Area = bh

where b = 5, h = 7

Area = 5 × 7

= 35 square units15 a~ = 5 𝚤 − 6 𝚥 b~ = x 𝚤 + 3 𝚥

i || a~ || = √52 + (−6)2 =√61

|| b~ || = √x2 + 9

|| a~ || = || b~ ||⇒√x2 + 9 =√61

x2 + 9 = 61

x2 = 52

x = ±√52 = ±√4 × 13

x = ±2√13

ii a~⊥b~ ⇒ a~ ⋅ b~ = 0

5x − 18 = 0

5x = 18

x = 185

iii a~ ∥ to b~ a~ = 𝜆b~5 𝚤 − 6 𝚥 = 𝜆 (x 𝚤 + 3 𝚥)

𝜆 = −2 ⇒ −2x = 5

x = −52

iv b~ ⋅ a = b ⋅ a~|| a~ ||

= 5x − 18

√61= 2

61

⇒ 5x − 18 = 2

5x = 20

x = 4

16 a~ = 4 𝚤 − 5 𝚥 b~ = 2 𝚤 − 𝚥

a || b~ || =√(2)2 + (−1)2 =√5

b~ = 1

√5(2 𝚤 − 𝚥)

b a~ ⋅ b~ = 8 + 5 = 13

scalar resolute a~ ⋅ b~ = 13

√5


ii


ii

ii


c || a~ || =√16 + 25 =√41

cos (𝜃) = a~ ⋅ b~|| a~ |||| b~ ||

𝜃 = cos−1

(13

√41√5)= cos−1 (0.9080) = 24.775°

𝜃 = 24°47′

d (a~ ⋅ b~) b~ = 13

√5× 1

√5(2 𝚤 − 𝚥)

= 135

(2 𝚤 − 𝚥)

e a~ − (a~ ⋅ b~) b~(4 𝚤 − 5 𝚥) − 13

5(2 𝚤 − 𝚥)

= 15 [5 (4 𝚤 − 5 𝚥) − 13 (2 𝚤 − 𝚥)]

= 15

(−6 𝚤 − 12 𝚥)

= −65

( 𝚤 + 2 𝚥)

17 a OA = a~OB = b~

M

B

A

O

b~

a~

M is mid-point of AB

AM = 12AB

= 12 (OB − OA)

= 12

(b~ − a~)

OM = OA + AM

= OA + 12AB

= OA + 12 (OB − OA) = a~ + 1

2(b~ − a~)

= 12

(a~ + b~)

LHS ||OA||2

+ ||OB||2

= || a~ ||2 + || b~ ||2

RHS 2 ||AM||2

+ 2 ||OM||2

= 2 [12

(b~ − a~) .12

(b~ − a~) + 12

(a~ + b~) .12

(a~ + b~)]

= 2 [14 (|| b~ ||

2 + || a~ ||2 − 2XXa~ ⋅ b~ ) + 14 (|| a~ || + || b~ ||2 + 2XXa~ ⋅ b~ )]

= || a~ ||2 + || b~ ||2

b a~ = cos (A) 𝚤 + sin (A) 𝚥 b~ = cos (B) 𝚤 + sin (B) 𝚥a~ ⋅ b~ = || a~ |||| b~ || cos (𝜃)|| a~ || =√cos2 (A) + sin2 (A) || b~ || =√cos2 (B) + sin2 (B)

= 1 = 1

a~ ⋅ b~ = cos (A) cos (B) + sin (A) sin (B)𝜃 = A − B

angle between a~ and b~

x

y

B

A

b~

a~

î

ĵ

cos (A − B) = cos (A) cos (B) + sin (A) sin (B)18 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 = OE − OC

= 5.5 𝚥 + 4k~ − (3 𝚤 + 5.5 𝚥)= −3 𝚤 + 4k~

c Other diagonal = 3 𝚤 + 5.5 𝚥 + 4k~ − 5.5 𝚥= 3 𝚤 + 4k~

cos 𝜃 = (−3 𝚤 + 4k~) ⋅ (3 𝚤 + 4k~)√(−3)2 + 42√32 + 42

= −9 + 165 × 5

= 725

𝜃 = 73.7°

d V = 3 × 5.5 × 4

= 66 cm3

in litresV = 66 ÷ 1000

= 0.066 litrese d~ = 3 𝚤 + 5.5 𝚥 + 4k~f || d~ || =√32 + 5.52 + 42

=√55.25= 7.43 cm

g Other longest diagonal (connecting D to the y-axis) isgiven by:

b~ = −3 𝚤 + 5.5 𝚥 − 4k~|| b~ || =√55.25

= 7.43

cos 𝜃 = d~ ⋅ b~|| d~ |||| b~ ||

= −9 + 30.25 − 16

(7.43)2

= 5.2555.25

= 0.095 02𝜃 = 84.5°


ii


ii

ii


19 a~ = x 𝚤 + y 𝚥 b~ = 2 𝚤 + 3 𝚥a~ ⋅ b~ = 9 ⇒ 2x + 3y = 9 [1]|| a~ || =√x2 + y2 =√34 ⇒ x2 + y2 = 34 [2]Solving x = −3 y = 5

or y = 7513

, y = −1113


20 a~ = x 𝚤 − 2 𝚥 b~ = −3 𝚤 + y 𝚥2a~ + 3b~ = 2 (x 𝚤 − 2 𝚥) + 3 (−3 𝚤 + y 𝚥)

= (2x − 9) 𝚤 + (3y − 4) 𝚥3a + 2b~ = 3 (x 𝚤 − 2 𝚥) + 2 (−3 𝚤 + y 𝚥)

= (3x − 6) 𝚤 + (2y − 6) 𝚥|2a~ + 3b~ || =√65⇒ (2x − 9)2 + (3y − 4)2 = 65 [1]

|3a~ + 2b~ || =√85⇒ (3x − 6)2 + (2y − 6)2 = 85 [2]

Use technology to graph [1] and [2]Points of intersection (5, 4), (3.24, −1.22)⇒ x = 5, y = 4

x = 3.24, y = −1.22

chapter2—vectorsintheplane - weebly · i i “c02vectorsintheplane_ws_print” — 2018/10/16 —...

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