1. No category

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1.1 Exercises, Sample Solutions
5. Equal vectors have the same magnitude and direction.
a) Opposite sides of a parallelogram are parallel and equal in length.
AD = BC, DC = AB
b) Opposite sides of a rectangle are parallel and equal in length. The diagonals of a rectangle
bisect each other.
QT = TS, PT = TR , SR = PQ, SP = RQ
c) J, L, and K are midpoints of sides AB, AC, and BC, respectively.
KJ = CL, KJ = LA, CL = LA, JL = BK, JL = KC, BK = BC, LK = AJ, LK = JB, AJ = JB
d) Opposite sides of a regular hexagon are parallel and equal in length.
ED = AB, CD = AF, CB = EF, DG = GA, BG = GE, FG = GC, AF = GE, BG = CD,
ED = GC, FG = AB, EF = GA, DG = CB, AF = BG, GE = CD, ED = FG, GC = AB,
EF = DG, GA = CB
6. Answers may vary.
Since equal vectors have the same magnitude and direction, DE = EF and AB = BC .
Since opposite vectors have the same magnitude and are opposite in direction, AB = −EF
and AB = −DE .
7. X is the midpoint of YZ, so XY = XZ. Also, XY and XZ are opposite in direction. Therefore,
XY = − XZ .
8. Answers may vary.
a) 50 km/h [north]
Scale 1 cm : 25 km/h.
b) 12 m/s [095°]
Scale: 1 cm : 4 m/s
1.1 Exercises, Sample Solutions
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c) 500 N [southeast]
Scale: 1 cm : 250 N
d) 2.5 m/s2 [335°]
Scale: 1 cm : 1.25 m/s2
e) 7 m [270°]
Scale: 1 cm : 3.5 m
9. a) 50 km/h [south]
Scale: 1 cm : 25 km/h
c) 500 N [northwest]
Scale : 1 cm : 250 N
b) 12 m/s [275°]
Scale: 1 cm : 4 cm/s
d) 2.5 m/s2 [155°]
Scale : 1 cm : 1.25 m/s2
1.1 Exercises, Sample Solutions
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e) 7 m [90°]
Scale: 1 cm : 3.5 cm
10. a) i) False, since AB and BC are not in the same direction.
ii) True, since the sides of a square are equal in length.
iii) False, since the magnitude of a vector is non-negative.
b) Use the Pythagorean theorem.
AC = 32 + 32
= 18
=3 2
11. a) Yes, equal vectors have equal magnitudes.
b) No, the vectors may not be in the same
direction.
2
4 6
12. The fractions 6 , 9 , … are equivalent representations of 3 so we can substitute any of these
2
fractions for 3 . Similarly, vectors with the same magnitude and direction are equivalent
vectors and any one of these vectors can be substituted for another.
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1.2 Exercises, Sample Solutions
4. a) AF + DB = AF + FE
= AE
c) FA + EB = CF + FD
= CD
e) AF + DE = AF + FC
= AC
b) DE + DB = AD + DE
= AE
d) DA + EC = BD + DF
= BF
f) EC + FD = EC + CE
= EE
G
=0
5.
a)
b)
c)
d)
e)
f)
6. a) KN + NR = KR
c) MN + MS = MR
b) RS + KR = KR + RS
= KS
d) KM + NK = NK + KM
= NM
1.2 Exercises, Sample Solutions
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e) KN + RS = KN + NM
= KM
7. a) DA = DB + BA or DA = DC + CA
c) CB = CD + DB
e) DB = DA + AB or DB = DC + CB
f) KR + NM + SK = (KR + RS) + SK
= KS + SK
= KK
G
=0
b) CD = CB + BD or CD = CA + AD
d) AB = AD + DB
f) BC = BD + DC
G G G
8. a) x + 0 = x
G
b) If x is a vector with tail at A and head at B, then:
G G
x + 0 = AB + BB
= AB
(the vector from the tail of AB to the head of BB )
G
= x
G G
G G G
G
G
9. Arrange u , v , and w head-to-tail. The resultant u + v + w is the vector from the tail of u to
G
the head of w .
G G G
G G G
10. Add the vectors in the order (a + b ) + c , then in the order a + (b + c ) .
G G G G G
G G G
G G
(a + b ) + c = m + c
a + (b + c ) = a + p
G
G
= n
= n
The same resultant is obtained regardless of the order in which we add the vectors. Hence,
vector addition is associative.
11. Diagrams may vary. Where necessary, arrange the vectors so that they are head-to-tail. The
resultant is the vector with the same tail as the first vector and the same head as the last
vector.
a) WX + XY + YZ = WZ
b) PQ + RP = RP + PQ
= RQ
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c) AB + CA = CA + AB
= CB
d) ST + US + VU = VU + US + ST
= VT
12. The vectors AB, BC, and CA are arranged head-to-tail so
AB + BC + CA is the vector from the tail of AB to the head
G
of CA . This is the vector AA , or 0 .
13. a) AC + CE + EB + BD + DA
The vectors are arranged sequentially from head-to-tail
so the sum is the vector from the tail of AC to the head
G
of DA . This is the vector AA , or 0 .
b) The vectors in part a have the same magnitude since the
diagonals of a regular pentagon are equal in length. When the heads
of the vectors are joined, 5 congruent isosceles triangles are obtained.
Thus, the heads of the vectors are the vertices of a regular pentagon.
G
14. a) Let r be the resultant velocity of the boat. Draw a vector diagram.
By the Pythagorean theorem:
G
r = 10 2 + 6 2
= 136
Ñ11.7
1.2 Exercises, Sample Solutions
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The resultant speed of the boat is approximately 11.7 km/h.
b) Let θ be the angle in the figure above.
6
tan θ = 10
θ Ñ31°
The resultant path of the boat makes an angle of
90° − 31°, or 59° with the shoreline.
c) Let x be how far downstream Pierre lands. The triangle in the
diagram at the right is similar to the one in part a, so:
x
6
=
120 10
6
x = 120 ×
10
= 72
Pierre lands 72 m downstream.
15. In order to travel directly across the river, Pierre must steer
slightly upstream, as shown in the diagram at the right.
6
a) sin θ = 10
θ Ñ36.9°
Pierre should head at an angle of 90° − 36.9° , or 53.1° to the shore.
G
b) Let r be the resultant velocity of the boat. By the Pythagorean
theorem:
G
r = 10 2 − 6 2
= 64
=8
The resultant speed of the boat is 8 km/h.
From part a, the width of the river is 120 m, or 0.12 km.
The time it takes to cross the river is:
width of river
Time = resultant speed
0.12
=
8
= 0.015
It will take Pierre 0.015 h, or approximately 0.9 minutes to cross the river.
16. To add the vectors, arrange them sequentially from head-to-tail. The resultant is the vector
from the tail of the first vector to the head of the last vector.
a) KR + NM + MK = NM + MK + KR
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= NR
b) KS + RN + RK = (RK + KS) + RN
= RS + SM
= RM
17. a) Let OA represent the 8-N force and OB represent the 11-N force.
Complete parallelogram OACB. Then OC represents the resultant
force.
b) ∠AOB = 30°
AOBC is a parallelogram, so:
∠OBC = 180° − 30°
= 150°
G
In ∆OBC, use the Cosine Law to calculate r .
G
r 2 = 82 + 112 − 2(8)(11)cos 150°
G
r Ñ 18.4
The magnitude of the resultant is approximately 18.4 N.
G
G G
G G
G
18. a) Yes. When a and b are in the same direction, a , b , and a + b
G G G G
lie on a straight line, and a + b = a + b . This is illustrated in
the diagram at the right.
G G
G G
G G G G
b) From part a, a + b = a + b when a , b , and a + b are in the same
direction.
G G
G G
G G
G G
When a , b , and a + b are in opposite directions, a , b , and a + b lie
G G G G
on a straight line, and a + b < a + b . This is illustrated in the diagram
at the right.
G G
G G
In all other cases, a , b , and a + b form a triangle whose side
lengths correspond to the magnitudes of the vectors. The sum
of any two sides of a triangle must be greater than the length
of the third side, or a triangle cannot be formed. Therefore,
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G G G G
G G G G
a + b > a + b , or a + b < a + b .
G G G G
Hence, in general, a + b ≤ a + b .
G
G
G G G G
c) No. From part b, a + b ≤ a + b for any vectors a and b .
G G
G G
G G G G
If a + b > a + b , then a , b , and a + b cannot form a triangle.
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1.3 Exercises, Sample Solutions
4.
a)
b)
c)
d)
e)
f)
G G
5. a) i) u + v = AB + BC
= AC
G G
iii) − u − v = − AB − BC
= BA + CB
= CD + DA
= CA
G G
ii) u − v = AB − BC
= AB + CB
= DC + CB
= DB
G G
iv) v − u = BC − AB
= BC + BA
= BC + CD
= BD
b) AC = AB + BC
G G
=u +v
Also, AC = AD + DC
G G
= v +u
G G G G
Thus u + v = v + u . This illustrates the commutative property.
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6. Answers may vary.
a) TQ = TR + RQ
b) RT = RS + ST
= TR − QR
= RS − TS
c) PS = TS + PT
= TS − TP
d) PR = TR + PT
= TR − TP
7. Draw the diagonals of the hexagon to locate the center O.
Since ABCDEF is a regular hexagon, the triangles in the
diagram at the right are equilateral triangles.
In ∆ABO, AB = AO + OB
= BC − CD
In ∆EFO, EF = EO + OF
= FA + DE
Therefore: AB − BC + CD − DE + EF − FA
= BC − CD − BC + CD − DE + FA + DE − FA
G
=0
G G G
8. a) u + v + w = AB + BC + CG
= AG
G G G
c) u − v + w = DC + CB + BF
= DF
G G G
b) u + v − w = EF + FG + GC
= EC
G G G
b) u − v − w = HG + GF + FB
= HB
G G
G G
9. The vectors a + b and a − b are diagonals of the parallelogram
G
K
OACB formed by a and b . This is illustrated in the diagram
at the right.
a) If OACB is a rectangle, then the diagonals will be congruent.
G G
G G
K
Therefore a + b and a − b have the same length when a and
G
b are perpendicular.
b) Diagonal OC is longer than diagonal AB when ∠BOA is acute.
G G
G G
Therefore, a + b is longer than a − b when the angle between
G
K
a and b is acute.
10. Refer to the diagram and explanation in exercise 9 above.
G G
G
G
G G
• u + v = u − v when u and v are perpendicular.
G
G
G G
G G
• u + v > u − v when the angle between u and v is acute
G
G
G G
G G
• u + v < u − v when the angle between u and v is obtuse
1.3 Exercises, Sample Solutions
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G
G
G G G G
11. a) If a and b are in opposite directions, then a − b = a + b .
G
G
G G G G
If a and b are in the same direction, then a − b < a + b .
G
G
G G
G G
If a and b are not collinear, then a , b , and a − b form a triangle. Since the sum of any
G G G G
two sides of a triangle must be greater than the length of the third side, a + b > a − b , or
G G G G
a −b < a + b.
G G G G G
G
Therefore, for any vectors a and b , a − b ≤ a + b .
b) i)
ii)
G
G
G G
G G
The only case that a − b ≤ a − b is when a and b are
G G
in the same direction and a ≥ b .
G G
G G
a − b ≥ a − b is always true.
G
G
G G G G
iii) The only case that a − b ≤ b − a is when a and b are
G G
in the same direction and a ≤ b .
iv)
G G G G
a − b ≥ b − a is always true.
1.3 Exercises, Sample Solutions
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1.4 Exercises, Sample Solutions
5. a) AB = 2AE
G
= 2u
c) CE = CB + BE
G G
= −v − u
b) AC = AB + BC
G G
= 2u + v
6. a) OR = OP + PR
= OP + 0.5OQ
b) OU = OP + PU
= OP + 2OQ
c) OW = OQ + QX + XW
= OQ + 2OP + 0.5OQ
d) OS = OQ + QS
= OQ + OP
= 2OP + 1.5OQ
e) OA =
1
OQ + 2OP
2
f) OY = OQ + QY
= OQ + 3OP
G
G
G
G
G
G
7. a) From page 27, DC = 4a − 2b and OF = 5a − 3b . Since 5a − 3b is not a scalar multiple of
G
G
4a − 2b , DC and OF are not parallel.
b) i) DE = OE − OD
ii) EF = OF − OE
G
G
G
G
G
G
G
G
= −1.5a − 2b − ( − a + 4b )
= 5a − 3b − ( − 1.5a − 2b )
G
G
G G
= −0.5a − 6b
= 6.5a − b
iii) DF = OF − OD
G
G
G
G
= 5a − 3b − ( − a + 4b )
G
G
= 6 a − 7b
G
G
8. a) i) OC = 2a + 4b
G
G
iii) OE = −2a − 2b
b) i) CD = OD − OC
G
G
G
G
= −3a + 3b − (2a + 4b )
G G
= −5a − b
iii) EF = OF − OE
G
G
G
G
= 4a − 2b − ( − 2a − 2b )
G
= 6a
G
G
ii) OD = −3a + 3b
G
G
iv) OF = 4a − 2b
ii) DE = OE − OD
G
G
G
G
= −2a − 2b − ( − 3a + 3b )
G
G
= a − 5b
iv) FC = OC − OF
G
G
G
G
= 2a + 4b − (4a − 2b )
G
G
= −2a + 6b
1.4 Exercises, Sample Solutions
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v)
DF = OF − OD
G
G
G
G
= 4a − 2b − ( − 3a + 3b )
G
G
= 7 a − 5b
vi) EC = OC − OE
G
G
G
G
= 2a + 4b − ( − 2a − 2b )
G
G
= 4a + 6b
9.
10. Answers may vary.
1.4 Exercises, Sample Solutions
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11. a)
b) The heads of the vectors lie on a straight line. This is the line l in the diagram in part a.
12. a) AM = AB + BM
1
= AB + AD
2
b) AN = AD + DN
1
= AD + AB
2
1
= AB + AD
2
b) From part a, we have:
1
AM = AB + AD c
2
1
AN = AB + AD d
2
To write AB in terms of AM and AN , eliminate AD from equations c and d.
−2 × c: − 2AM = −2AB − AD
1
d:
AN = AB + AD
2
3
Add: − 2AM + AN = − AB
2
4
2
Therefore, AB = AM − AN
3
3
To write AD in terms of AM and AN , eliminate AB from equations c and d.
1
c: AM = AB + AD
2
−2 × d: − 2AN = − AB − 2AD
3
Add: AM − 2AN = − AD
2
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4
2
Therefore, AD = − AM + AN
3
3
13. a) OD = OA + OE
G G
=u +v
c) AC = AB + BC
G G
=u +v
b) OC = OB + OE
G G
= 2u + v
d) EB = EO + OB
G
G
= −v + 2u
G G
= 2u − v
14. a) OD = OE + ED
G G
= v +u
G G
=u +v
c) OB = OC + CB
G G G
= 2u + v + u
G G
= 3u + v
b) OC = OD + DC
G G G
=u +v +u
G G
= 2u + v
d) AD = OE
G
=v
15. Use the results of part a to complete part b.
G
OA = u
b)
a)
AB = AF + FB
G G
= v +u
BC = OE
G
=v
CD = −OA
G
= −u
DE = −AB
G G
= −v − u
G
EO = −v
OB = OA + AB
G G
= 2u + v
AC = AB + BC
G
G
= u + 2v
BD = BC + CD
G G
= v −u
CE = CD + DE
G
G
= −v − 2u
DO = DE + EO
G G
= −2v − u
EA = EO + OA
G G
= v +u
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16. a) The diagram below left shows the vectors in exercise 15a arranged tail-to-tail.
The diagram below right shows the vectors in exercise 15b.
b) In each list, the heads of the vectors are equally spaced on a circle and form the vertices
of a regular hexagon. This is because in each list, the vectors have the same magnitude
and the angle between any pair of vectors is 60°.
G
G
17. a) AR = −u + 7v
G
G
BQ = −2u + 6v
G
G
CP = −3u + 5v
G
G
DO = −4u + 4v
G
G
EN = −5u + 3v
G
G
FM = −6u + 2v
G G
GL = −7u + v
b) The heads of the vectors lie on a straight line.
G
G
c) If we visualize u and v forming a parallelogram grid, the heads of the vectors in part a
correspond to the points (−1, 7), (−2, 6), (−3, 5), (−4, 4), (−5, 3), (−6, 2), (−7, 1) on the
grid. In each ordered pair, the second coordinates in 8 more than the first coordinate, so
the points lie on a straight line.
G
18. a) Multiplying by 0: 0 x = 0 .
Multiplying by 1: 1x = x .
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G
b) By the definition of scalar multiplication, 0 x = 0 .
The vector 1x has the same magnitude and direction as x . Therefore, 1x = x .
G
19. a) In the diagram below, let OA = a .
Points B and C are constructed on OA extended such that OB is m times as long as OA
G
G
and BC is n times as long as OA. Therefore, OB = ma and BC = na .
Since OC is (m + n) times as long as OA:
G
OC = (m + n)a
Also:
OC = OB + BC
G
G
= ma + na
G
G
G
Therefore, (m + n)a = ma + na .
G
b) In the diagram below, let OA = a .
Points B and C are constructed on OA extended such that OB is n times as long as OA
G
G
and OC is m times as long as OB. Therefore, OB = na and OC = m(na ) .
Since OC is collinear with OA and OB, and OC is m times as long as OB and OB is n
times as long as OA, OC is mn times as long as OA:
G
OC = (mn)a
G
G
Therefore, m(na ) = (mn)a .
20. Consider the diagram below in which ∆DOC is similar to ∆BOA and has sides |m| times as
long (m # 0).
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Since OD is |m| times as long as OB
and OD is oppositely directed to OB :
OD = mOB
G G
= m( a + b )
G
G G
G
Therefore, m( a + b ) = m a + m b , when m # 0.
Using theTriangle Law:
OD = OC + CD
G
G
= ma + mb
G
G
21. a) 3u = 3 u
= 3(1)
=3
1G 1 G
b) − v = v
4
4
1
= (1)
4
1
=
4
G G
c) The diagram for 2u − 3v is below left.
G G
2u − 3v 2 = 22 + 32 − 2(2)(3)cos 120°
G G
2u − 3v = 19
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G
G
d) The diagram for − 2u + 3v is above right.
G
G2
− 2u + 3v = 22 + 32 − 2(2)(3)cos 120°
G G
2u − 3v = 19
G
G
e) The diagram for 1.5u + 0.5v is below left.
G
G
1.5u + 0.5v 2 = (1.5)2 + (0.5)2 − 2(1.5)(0.5)cos 60°
7
G
G
1.5u + 0.5v = 2
f)
G
G
The diagram for − u + 2v is above right.
G
G2
− u + 2v = 12 + 22 − 2(1)(2) cos 120°
G
G
− u + 2v = 7
22. Use the given information to construct the following diagram.
G
G
Since u and v are collinear, AC || CE. Therefore, corresponding angles are equal:
∠ACB = ∠AED
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Because it is a common angle:
∠BAC = ∠DAE
Since two corresponding angles of ∆ABC and ∆ADE are equal, the third pair of angles are
equal.
Hence ∆ABC is similar to ∆ADE. Since the corresponding sides of similar triangles are
G
G
tb
sa
s t
AB BC
=
. Hence, G = G , so
= .
proportional.
ma
AD DE
m n
nb
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1.5 Exercises, Sample Solutions
G
5. a) i) 2 u = 2[3, 2]
= [6, 4]
G
iii) 5 u = 5[3, 2]
= [15, 10]
G
ii) 3 u = 3[3, 2]
= [9, 6]
G
iv) −4 u = −4[3, 2]
= [−12, −8]
b)
G
c) u = 3 2 + 2 2
= 13
G
G
G
G
Therefore, 2u = 2 13 , 3u = 3 13 , 5u = 5 13 , and − 4u = 4 13 .
6.
G
v = [4, 3]
G
v = 4 2 + 32
=5
G
a) 3v = 3[4, 3]
1G 1
v = [4,3]
2
2
 3
= [12, 9]
= 2, 
 2
G
G
c) Since v = 5, the vector 2v or [8, 6] has magnitude 10.
b)
1 G
G
 4 3
d) Since v = 5 , the vector 5 v or  ,  has length 1.
5 5
7. a) AB = [10 − 4, 3 − 1]
= [6, 2]
BC = [6 − 10, 5 − 3]
= [−4 , 2]
CD = [0 − 6, 3 − 5]
= [−6, −2]
DA = [4 − 0, 1 − 3]
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= [4, −2]
b) AB = 6 2 + 2 2
= 40
= 2 10
∴ CD = 2 10
BC = (−4) 2 + 2 2
= 20
=2 5
∴ DA = 2 5
c) ABCD is a parallelogram since opposite sides are parallel and equal.
8. a) AB = [−1 − (−2), 7 − (−1)]
= [1, 8]
BC = [6 − (−1), 3 − 7]
= [7 , −4]
CD = [5 − 6, −5 − 3]
= [−1, −8]
DA = [−2 − 5, −1 − (−5)]
= [−7, 4]
b)
AB = 12 + 8 2
BC = 7 2 + (−4) 2
65
= 65
∴ CD = 65
∴ DA = 65
=
c) ABCD is a rhombus since opposite sides are parallel and all sides are equal in length.
9. a) Form any 2 vectors from the 3 given points. If the vectors are scalar multiples of each
other, then the points are collinear.
b) i) PQ = [2 − (−3), 4 − 1]
= [5, 3]
PR = [5 − (−3), 6 − 1]
= [8, 5]
By inspection, PQ and PR are not scalar multiples of each other. Therefore, P, Q, and
R are not collinear.
ii) DE = [1 − 5, −5 − 1]
= [−4, −6]
DF = [−3 − 5, −11 − 1]
= [−8, −12]
By inspection, DF = 2 DE . Therefore, D, E, and F are collinear.
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1G
1
10. a) − u = − [−2, 4]
2
2
= [1, − 2]
G G
c) u + v = [−2, 4] + [3, − 1]
= [1, 3]
G
G
e) − u + 2v = −[−2, 4] + 2[3, − 1]
= [2, −4] + [6, −2]
= [8, −6]
G
b) 4v = 4[3, − 1]
= [12, − 4]
G G
d) u − v = [−2, 4] − [3, − 1]
= [− 5, 5]
G G
f) 2u − 3v = 2[−2, 4] − 3[3, − 1]
= [−4, 8] − [9, −3]
= [−13, 11]
11.
G
G
12. a) a + 3b = [5, 3] + 3[2, − 4]
= [5, 3] + [6, −12]
= [11, −9]
G
G
b) 2a − 4b = 2[5, 3] − 4[2, − 4]
= [10, 6] − [8, −16]
= [2, 22]
G
G
c) − 3a + 5b = −3[5, 3] + 5[2, − 4]
= [−15, −9] + [10, −20]
= [−5, −29]
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13.
G
G
G
14. a) 2u = 2(3i − 2 j )
G
G
= 6i − 4 j
G
G
G G
G G
c) u + v = 3i − 2 j + 2i + j
G G
= 5i − j
G
G
G G
G
G
e) 4u − 2v = 4(3i − 2 j ) − 2(2i + j )
G
G
G
G
= 12i − 8 j − 4i − 2 j
G
G
= 8i − 10 j
G G
G
b) − 3v = −3(2i + j )
G
G
= −6i − 3 j
G G
G
G
G G
d) u − v = 3i − 2 j − (2i + j )
G
G
= i −3j
G
G
G G
G
G
f) − 2u + 3v = −2(3i − 2 j ) + 3(2i + j )
G
G
G
G
= −6i + 4 j + 6i + 3 j
G
=7j
G G
15. a) a + b = [4,1] + [2, 3]
= [6, 4]
G G
a − b = [4,1] − [2, 3]
= [2, −2]
b)
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G G
c) a + b = 6 2 + 4 2
= 52
= 2 13
G G
a − b = 2 2 + (−2) 2
= 8
=2 2
K
G G
G
G G
d) The vectors a + b and a − b are diagonals of the parallelogram formed by a and b . Since
K
G
G G
the angle between a and b is acute, the diagonal corresponding to a + b is longer than
G G
G G
G G
the diagonal corresponding to a − b . Thus a + b is greater than a − b .
K
G
G G
G G
G G
e) If a and b are perpendicular, then a + b would be equal to a − b since a + b and
G G
a − b would be diagonals of a rectangle. This is illustrated in the diagram below left. If
K
G
G G
G G
the angle between a and b is obtuse, then a + b would be less than a − b as illustrated
in the diagram below right.
G
G
K
16. a) Let w = s u + t v for some scalars s and t.
[2, 8] = s[3, 0] + t[−1, 2]
= [3s − t, 2t]
Since the vectors are equal, their components are equal.
3s − t = 2
c
2t = 8, or t = 4
d
Substitute t = 4 in equation c to determine s.
3s − 4 = 2
3s = 6
s=2
G
G
K
Therefore, w = 2 u + 4 v .
b)
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G
G
K
17. From exercise 16, w = 2 u + 4 v .
G
G
K
a) w = 2 u + 4 v
G
G
K
2 u = −4 v + w
G
G 1G
u = −2v + w
2
b)
G
G
K
w = 2u + 4v
G
G
K
4 v = −2 u + w
G
1G 1G
v =− u+ w
2
4
G
G
K
18. a) Let w = s u + t v for some scalars s and t.
[12, −1] = s[2, 1] + t[−1, 3]
= [2s − t, s + 3t]
Since the vectors are equal, their components are equal.
2s − t = 12
c
s + 3t = −1
d
Solve equation d for s.
s = −3t − 1
e
Substitute s = −3t − 1 in equation d and solve for t.
2(−3t − 1) − t = 12
−6t − 2 − t = 12
−7t = 14
t = −2
Substitute t = −2 in equation e and solve for s.
s = −3(−2) − 1
=5
G
G
K
Therefore, w = 5 u − 2 v .
b)
G
G
K
19. From exercise 18, w = 5 u − 2 v .
G
G
K
a) w = 5 u − 2 v
G
G
K
5u = 2v + w
G 2G 1 G
u= v+ w
5
5
b)
G
G
K
w = 5u − 2v
G
G
K
2v = 5u − w
G 5G 1G
v= u− w
2
2
G
G
20. a) b + 3m = [0, 5] + 3[2,− 1]
= [0, 5] + [6, −3]
= [6, 2]
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G
To obtain the other vectors in the list, subtract m = [2, −1] from the preceding vector on
the list.
G
G
b + 2m = [6, 2] − [2,− 1]
= [4, 3]
G G
b + m = [4, 3] − [2,− 1]
= [2, 4]
G
G
b + 0m = [2, 4] − [2,− 1]
= [0, 5]
G G
b − m = [0, 5] − [2,− 1]
= [−2, 6]
G
G
b − 2m = [−2, 6] − [2,− 1]
= [−4, 7]
G
G
b − 3m = [−4, 7] − [2,− 1]
= [−6, 8]
b)
G
G
c) The first vector in the list is b + 3m = [6, 2] . To obtain the other vectors in the list,
G
repeatedly subtract m = [2, −1]. Therefore, the heads of the vectors lie on a line that passes
1
through the point (6, 2) and has slope − .
2
G
G
G
d) If b = [2, 4], then b + 3m = [8, 1] . Thus the heads of the vectors would lie on a line that
1
passes through (8, 1) and has slope − .
2
G
G
G
If b = [−1, 2], then b + 3m = [5, 1] . Thus the heads of the vectors would lie on a line that
1
passes through (5, −1) and has slope − .
2
G
G
21. a) − 2u + 3v = −2[3, − 1] + 3[1, 2]
= [−6, 2] + [3, 6]
= [−3, 8]
G G
To obtain the other vectors in the list, add u − v = [2, −3] to the preceding vector on
the list.
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b)
G
G
− u + 2v = [−3, 8] + [2,− 3]
= [−1, 5]
G
v = [−1, 5] + [2,− 3]
= [1, 2]
G
u = [1, 2] + [2,− 3]
= [3, −1]
G G
2u − v = [3, − 1] + [2,− 3]
= [5, −4]
G
G
3u − 2v = [5, − 4] + [2,− 3]
= [7, −7]
G
G
c) The first vector in the list is − 2u + 3v = [−3, 8] . To obtain the other vectors in the list,
G G
repeatedly add u − v = [2, −3]. Therefore, the heads of the vectors lie on a line that passes
3
through the point (−3, 8) and has slope − .
G2 G
d) Yes, since we are still repeatedly adding u − v to obtain each vector on the list.
G G G G
22. a) a + b = b + a
G G
LS = a + b
= [a1, a2] + [b1, b2]
= [a1 + b1, a2 + b2]
= [b1 + a1, b2 + a2]
G G G G
LS = RS, so a + b = b + a
G G G G G G
b) (a + b ) + c = a + (b + c )
G G G
LS = (a + b ) + c
= ([a1, a2] + [b1, b2]) + [c1, c2]
= [a1 + b1, a2 + b2] + [c1, c2]
G G
RS = b + a
= [b1, b2] + [a1, a2]
= [b1 + a1, b2 + a2]
G G G
RS = a + (b + c )
= [a1, a2] + ([b1, b2] + [c1, c2])
= [a1, a2] + [b1 + c1, b2 + c2]
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= [a1 + b1 + c1, a2 + b2 + c2]
G G G G G G
LS = RS, so (a + b ) + c = a + (b + c ) .
= [a1 + b1 + c1, a2 + b2 + c2]
G
G G
G
c) s (a + b ) = sa + sb
G G
LS = s (a + b )
= s([a1, a2] + [b1, b2])
= s[a1 + b1, a2 + b2]
= [s(a1 + b1), s(a2 + b2)]
= [sa1 + sb1, sa2 + sb2]
G
G G
G
LS = RS, so s (a + b ) = sa + sb .
G
G
RS = sa + sb
= s[a1, a2] + s[b1, b2]
= [sa1, sa2] + [sb1, sb2]
= [sa1 + sb1, sa2 + sb2]
G
G G
d) ( s + t )a = sa + ta
G
LS = (s + t) a
= (s + t)[a1, a2]
= [(s + t)a1, (s + t)a2]
= [sa1 + ta1, sa2 + ta2]
G
G G
LS = RS, so ( s + t )a = sa + ta .
G G
RS = sa + ta
= s[a1, a2] + t[a1,a2]
= [sa1, sa2] + [ta1, ta2]
= [sa1 + ta1, sa2 + ta2]
G G G
e) Since a + v = 0 :
G G G
v =0−a
= [0, 0] − [a1, a2]
= [0 − a1, 0 − a2]
= [− a1, − a2]
= −[a1, a2]
K
= –a
23. Let [2a, a] be the required vector.
Since the magnitude of the vector is 4:
(2a ) 2 + a 2 = 4
Square both sides and simplify.
4a2 + a2 = 16
5a2 = 16
16
a2 =
5
4
a=±
5
 8
 8
4 
4 
,−
,
There are two such vectors, 
 and −
.
5
5
5
 5

G G
24. u + v = [2, −1] + [x, 3]
= [2 + x, 2]
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G G
G G
Since, u + v = 5, u + v 2 = 25.
(2 + x)2 + 22 = 25
4 + 4x + x2 + 4 = 25
x2 + 4x − 17 = 0
− 4 ± 16 − 4(−17)
x=
2
− 4 ± 84
=
2
− 4 ± 2 21
=
2
= − 2 ± 21
G
25. It is not possible to express a as a linear
G
G
G
combination of b and c if c is collinear
G
to b . For example, consider the diagram at
G
G
G
G
G
the right, where c = 3b , or c = 0a + 3b .
G
In this situation, is not possible to express a
G
G
as a linear combination of b and c .
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1.6 Exercises, Sample Solutions
2. Method 1: Using geometric vectors
In the diagram below left, OB represents the 50 N force and OC represents the 35 N force.
Complete rectangle OBAC as shown in the diagram below right. Let OA represent the
resultant force
G
In ∆OBC, use the Pythagorean Theorem to calculate r .
G
r 2 = 502 + 352
G
r Ñ 61.0 N
G
Let θ = ∠BOA. Then the bearing of r is 270° − θ.
35
tan θ = 50
θ Ñ35.0°
G
The bearing of r is 270° − 35° = 235°.
The resultant force is 61. 0 N [235°].
Method 2: Using Cartesian vectors
Place the vectors on a coordinate system and represent each
vector algebraically.
G
G
Let f1 and f 2 represent the two forces.
G
Let r be the resultant force.
G
f1 = [−50, 0]
G
f 2 = [0, −35]
G
G
G
r = f1 + f 2
= [−50, 0] + [0, −35]
= [−50, −35]
The magnitude of the resultant is:
G
r = (−50)2 + (−35)2
Ñ 61.0 N
The direction of the resultant is:
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35
tan θ = 50
θ Ñ35.0°
G
Therefore, the bearing of r is 270° − 35° = 235°.
The resultant force is 61. 0 N [235°].
3. Method 1: Using geometric vectors
In the diagram below left, OA represents the 220 N force and OB represents the 400 N
force. Complete parallelogram OACB as shown in the diagram below right. Let OC
represent the resultant force
∠BOA = 55°
OACB is a parallelogram, so:
∠OAC = 180° − 55°
= 125°
G
In ∆OAC, use the Cosine Law to calculate r .
G
r 2 = 2202 + 4002 − 2(220)(400)cos 125°
G
r Ñ 556.2
Let θ be the angle the resultant makes with the 220 N force.
Use the Sine Law to determine θ.
sin θ sin 125°
400 = 556.2
400 sin 125°
sin θ =
556.2
θ Ñ36.1°
Therefore, the resultant force has a magnitude of 556.2 N and acts at an angle of 36.1° to the
220 N force.
Method 2: Using Cartesian vectors
Place the vectors on a coordinate system and represent each
vector algebraically.
G
G
Let f1 and f 2 represent the two forces.
G
Let r be the resultant force.
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G
f1 = [220, 0]
G
f 2 = [400 cos 55°, 400 sin 55°]
Ñ [229.4, 327.7]
G
G
G
r = f1 + f 2
= [220, 0]+ [229.4, 327.7]
= [449.4, 327.7]
The magnitude of the resultant is:
G
r = (449.4)2 + (327.7)2
Ñ 556.2
Let θ be the angle the resultant makes with the 220 N force.
327.7
tan θ = 449.4
θ Ñ36.1°
The resultant force has a magnitude of 556.2 N and acts at an
angle of 36.1° to the 220 N force.
4. Method 1: Using geometric vectors
In the diagram below left, OA represents the 20 N force and OB represents the 30 N force.
Complete parallelogram OACB as shown in the diagram below right. Let OC represent the
G
resultant force and e represent the equilibriant.
∠BOA = 120°
OACB is a parallelogram, so:
∠OBC = 180° − 120°
= 60°
G
In ∆OCB, use the Cosine Law to calculate r .
G
r 2 = 202 + 302 − 2(20)(30)cos 60°
G
r Ñ 26.5
Let θ be the angle the resultant makes with the 30 N force. Use the Sine Law to determine θ.
sin θ sin 60°
20 = 26.5
20 sin 60°
sin θ = 26.5
θ Ñ40.8°
The equilibriant is equal in magnitude but opposite in direction to the resultant. Hence, the
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equilibriant has magnitude 26.5 N and acts at an angle of 180° − 40.8°, or 139.2° to the 30 N
force.
Method 2: Using Cartesian vectors
Place the vectors on a coordinate system and represent
each vector algebraically.
G
G
Let f1 and f 2 represent the two forces.
G
Let r be the resultant force.
G
f1 = [30, 0]
G
f 2 = [20 cos 120°, 20 sin 120°]
Ñ [−10, 17.3]
G
G
G
r = f1 + f 2
= [30, 0] + [−10, 17.3]
= [20, 17.3]
The magnitude of the resultant is:
G
r = (20)2 + (17.3)2
Ñ 26.5
Let θ be the angle the resultant makes with the
30 N force.
17.3
tan θ = 20
θ Ñ40.8°
The equilibriant is equal in magnitude but opposite in
direction to the resultant. Hence, the equilibriant has
magnitude 26.5 N and acts at an angle of 180° − 40.8°,
or 139.2° to the 30 N force.
5. Method 1: Using geometric vectors
In the diagram below left, OA represents the unknown force and OB represents the 195 N
force. Complete parallelogram OACB as shown in the diagram below right. Let OC
represent the resultant force
∠BOA = 75°
OACB is a parallelogram, so:
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∠OBC = 180° − 75°
= 105°
Let θ be the angle the unknown force makes with the resultant.
Use the Sine Law to calculate θ.
sin θ sin 105°
195 = 225
195 sin 105°
sin θ =
225
θ Ñ 56.8°
Therefore, α = 180° − 105° − 56.8°
= 18.2°
Use the Sine Law to determine x, the magnitude of the unknown force.
225
x
=
sin 18.2° sin 105°
225 sin 18.2°
x=
sin 105°
Ñ 72.6
Therefore, the unknown force has a magnitude of 72.6 N and makes an angle of 56.8° with
the resultant.
Method 2: Using Cartesian vectors
Place the vectors on a coordinate system and represent each
vector algebraically.
G
G
Let f1 and f 2 represent the two forces.
G
Let r be the resultant force.
G
f1 = [x, 0]
G
f 2 = [195 cos 75°, 195 sin 75°]
Ñ [50.5, 188.4]
G
r = [225 cos θ, 225 sin θ]
G
G
G
r = f1 + f 2
[225 cos θ, 225 sin θ] = [x, 0] + [50.5, 188.4]
[225 cos θ, 225 sin θ] = [x + 50.5, 188.4]
Since the vectors are equal, their components are equal.
225 cos θ = x + 50.5 c
d
225 sin θ = 188.4
Solve equation d for θ.
188.4
sinθ = 225
θ Ñ 56.8°
Substitute θ Ñ 56.8° in equation c and solve for x.
225 cos 56.8° = x + 50.5
x = 225 cos 56.8° − 50.5
Ñ 72.6
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Therefore, the unknown force has a magnitude of 72.6 N and makes an angle of 56.8° with
the resultant.
6. Method 1: Using geometric vectors
In the diagram below left, OC represents the 40 N force applied by Paco and OL represents
the 30 N force applied by Louis. Complete parallelogram OCDL as shown in the diagram
G
below right. Let r represent the resultant of forces applied by Paco and Louis, and
G
e represent the force applied by Pepe.
∠LOC = 135°
OCDL is a parallelogram, so:
∠OCD = 180° − 135°
= 45°
G
In ∆OCD, use the Cosine Law to calculate r .
G
r 2 = 40 + 302 − 2(40)(30)cos 45°
G
r Ñ 28.3
Let θ be the angle the resultant makes with the 40 N force. Use the Sine Law to determine θ.
sin θ sin 45°
30 = 28.3
30 sin 45°
sin θ = 28.3
θ Ñ48.6°
Therefore, to keep his brothers efforts in equilibrium, Pepe should exert a force of 28.3 N at
an angle of (180° − 48.6°), or 131.4° to the force exerted by Paco.
Method 2: Using Cartesian vectors
Place the vectors on a coordinate system and represent each
vector algebraically.
G
G
Let f1 and f 2 represent the two forces.
G
Let r be the resultant force.
G
f1 = [40, 0]
G
f 2 = [30 cos 225°, 30 sin 225°]
Ñ [−21.2, −21.2]
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G
G
G
r = f1 + f 2
= [40, 0] + [−21.2, −21.2]
= [18.8, −21.2]
The magnitude of the resultant is:
G
r = (18.8)2 + (21.2)2
Ñ 28.4 N
The direction of the resultant is:
21.2
tan θ = 18.8
θ Ñ 48.4°
Therefore, to keep his brothers efforts in equilibrium, Pepe should exert a force of 28.3 N at
an angle of (180° − 48.6°), or 131.4° to the force exerted by Paco.
7. This problem is similar to Example 4, so it is simpler to use geometric vectors.
G
G
Let T1 and T2 represent the forces in the two ropes respectively. Draw a vector diagram and
the corresponding triangle diagram. Since the forces are in equilibrium, the resultant of the
forces in the two ropes is equal and opposite to the force exerted by the 225 N weight.
Use the Sine Law to find | T1 | and | T2 |.
T1
sin 55
D
=
T2
sin 66
D
=
225
sin(35 D + 24 D )
225 sin 55 D
225 sin 66 D
T
=
2
sin 59 D
sin 59 D
Ñ 215.0
Ñ 239.8
The tensions in the two ropes are 215. 0 N and 239.8 N respectively.
T1 =
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8. The simpler approach is to use geometric vectors.
G
The diagram, below left, shows the forces acting on the child. Let m represent the force
G
exerted by the mother and T represent the force in the chain. The weight of the child is the
equilibriant of these two forces.
Use the triangle diagram, above right.
G
m
100
tan 30° =
cos 30° = G
100
T
G
100
G
T =
m = 100 tan 30°
cos 30°
Ñ 115.5
Ñ 57.7
Therefore, the tension in the chain is 115.5 N. The magnitude of the force exerted by the
mother is 57.7 N.
9. The simpler approach is to use geometric vectors.
Draw a diagram. The forces are in equilibrium. The forces in the wires are along the wires
directed downwards. To keep the system in equilibrium, the nail exerts a force of 10 N
G
G
directed upwards. Let T1 and T2 represent the forces in the two wires. Since the picture is
G
G
G
G
hung symmetrically on the wires, | T1 | = | T2 |. Let | T1 | = | T2 | = T.
Use the Sine Law to determine T.
10
T
=
D
sin 120
sin 30 D
10 sin 30 D
T=
sin 120 D
Ñ 5.8 N
The tension in the wires is 5.8 N.
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10. Method 1: Using geometric vectors
Let OW represent the velocity of the wind.
Let OH represent the velocity of the plane in still air.
Complete parallelogram OWRH, where OR represents the velocity of the plane relative to
the ground.
From the given bearing:
∠WOH = 135°
Since OHRW is a parallelogram:
∠OHR = 180° − 135°
= 45°
G
Use the Cosine Law to calculate r .
G2
r = 9002 + 1002 − 2(900)(100)cos 45°
Ñ 832.3
Let ∠ROH = θ. The bearing of the plane is 135° − θ.
sin θ sin 45°
100 = 832.3
100 sin 45°
sin θ = 832.3
θ Ñ4.9°
Hence the bearing is 135° − 4.9° = 130.1°.
The plane’s speed relative to the ground is 832.3 km/h on a bearing of 130.1°.
Method 2: Using Cartesian vectors
The diagram at the right represents the situation.
A bearing of 135° corresponds to a direction angle of 315°.
G
Let a represent the velocity of the plane in still air.
G
Let w represent the velocity of the wind.
G
Let r represent the velocity of the plane relative to
the ground.
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G
a = [900 cos 315°, 900 sin 315°]
Ñ[636.4, −636.4]
G
w = [0, 100]
G
G
G
r = a + w
= [636.4, −636.4] + [0, 100]
Ñ[636.4, −536.4]
The magnitude of the resultant is:
G
r = (636.4)2 + (−536.4)2
Ñ 832.3
The bearing of the plane is 90° + θ .
536.4
tan θ = 636.4
θ Ñ 40.1°
Hence the bearing is 90° + 40.1° = 130.1°
The plane’s speed relative to the ground is 832.3 km/h on a bearing of 130.1°.
11. Method 1: Using geometric vectors
Let OW represent the velocity of the wind.
Let OH represent the velocity of the plane in still air.
Complete parallelogram OWRH, where OR represents the velocity of the plane relative to
the ground.
From the given bearings:
∠WOH = 113° − 30°
= 83°
Since OHRW is a parallelogram:
∠OWR = 180° − 83°
= 97°
G
Use the Cosine Law to calculate r .
G2
r = 6002 + 802 − 2(600)(80) cos 97°
Ñ 614.9
Let ∠ROH = θ. The bearing of the plane is 30° + θ.
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sin θ sin 97°
80 = 614.9
80 sin 97°
sin θ = 614.9
θ Ñ7.4°
Hence the bearing is 30° + 7.4° = 37.4°.
The plane’s speed relative to the ground is 614.9 km/h on a bearing
of 37.4°.
Method 2: Using Cartesian vectors
The diagram at the right represents the situation.
Bearings of 030° and 113° correspond to direction angles of
60° and 337° respectively.
Let
Let
Let
G
a represent the velocity of the plane in still air.
G
w represent the velocity of the wind.
G
r represent the velocity of the plane relative to the ground.
G
a = [600 cos 60°, 600 sin 60°]
Ñ[300, 519.6]
G
w = [80 cos 337°, 80 sin 337°]
Ñ[73.6, −31.3]
G
G
G
r = a + w
= [300, 519.6] + [73.6, −31.3]
Ñ[373.6, 488.3]
The magnitude of the resultant is:
G
r = (373.6)2 + (488.3)2
Ñ 614.9
The bearing of the plane is 90° − θ .
488.3
tan θ = 373.6
θ Ñ 52.6°
Hence the bearing is 90° − 52.6° = 37.4°
The plane’s speed relative to the ground is 614.9 km/h on a bearing of 37.4°.
12. The simpler approach is to use geometric vectors.
Let OW represent the velocity of the wind.
Let OR represent the velocity of the plane relative to the ground.
Complete parallelogram OWRH, where OH represents the velocity of the plane in still air.
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From the given bearings:
∠ROW = 125° − 75°
= 50°
a) Let ∠HOR = θ.
Use the Sine Law on ∆ORW to calculate θ.
sin θ sin 50°
80 = 550
80 sin 50°
sin θ =
550
θ Ñ6.4°
The heading the pilot should take is 075° − 6.4° = 68.6°.
b)
∠HOW = 50° + 6.4° = 56.4°.
Since OWRH is a parallelogram:
∠OHR = 180° − 56.4°
=123.6°
Use the Sine Law to calculate the speed of the plane relative to the ground.
G
r
80
=
sin 6.4° sin 123.6°
G 80 sin 123.6°
r =
sin 6.4
Ñ 597.8
The speed of the plane relative to the ground is 597.8 km/h.
distance
c) Time = speed
508
= 597.8
Ñ 0.85
The trip will take approximately 0.85 h, or approximately 51 min.
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13. Since a distance of 80 km was flown in 20 minutes, the speed of the plane relative to the


 80 km 
 , or 240 km/h.
ground was 
 1

h 

 3

Draw a diagram.
Let OW represent the velocity of the wind.
Let OH represent the velocity of the plane in still air.
Let OR represent the velocity of the plane relative to the ground
Complete parallelogram OWRH.
∆OHR is a right triangle, so:
G
w
tan 10° =
240
G
w = 240 tan 10°
Ñ 42.3
The speed of the wind is 42.3 km/h.
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1.7 Exercises, Sample Solutions
G G
u ⋅v
6. Use the formula cosθ = G G .
uv
[0, 4] ⋅ [5,1]
a) cosθ =
0 2 + 4 2 5 2 + 12
0+4
=
4 26
1
=
26
θ Ñ 78.7°
3 2 + (−2) 2 (−1) 2 + 2 2
−3 − 4
=
13 5
−7
=
65
θ Ñ 150.3°
[4, − 1] ⋅ [−2, − 5]
c) cosθ =
[3, − 2] ⋅ [−1, 2]
b) cosθ =
4 2 + (−1) 2 (−2) 2 + (−5) 2
−8 + 5
=
17 29
−3
=
493
θ Ñ 97.8°
[6, 3] ⋅ [2, − 4]
d) cosθ =
6 2 + 3 2 2 2 + (−4) 2
12 − 12
=
45 20
=0
θ = 90°
G G
7. u ⋅ v is positive so the angle between the vectors is acute.
8. a) AB = [−2 − (−1), 1 − 0]
= [−1, 1]
AC = [1 − (−1), 4 − 0]
= [2 , 4]
BC = [1 − (−2), 4 − 1]
= [3, 3]
cos ∠A =
=
AB ⋅ AC
cos ∠B =
AB AC
[−1,1] ⋅ [2, 4]
(−1) + (1)
−2 + 4
=
2 20
2
=
40
∠A Ñ 71.6°
2
2
2 +4
2
2
=
BA ⋅ BC
BA BC
[1, − 1] ⋅ [3, 3]
1 + (−1) 2 3 2 + 3 2
3−3
=
2 18
2
=0
∠B = 90°
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∠C = 180° − ∠A − ∠B
= 180° − 71.6° − 90°
= 18.4°
b) PQ = [8 − 2, 3 − 6]
= [6, −3]
PR = [−4 − 2, 0 − 6]
= [−6 , −6]
QR = [−4 − 8, 0 − 3]
= [−12, −3]
cos ∠P =
=
PQ ⋅ PR
PQ PR
cos ∠Q =
[6, − 3] ⋅ [−6, − 6]
6 2 + (−3) 2 (−6) 2 + (−6) 2
−36 + 18
=
45 72
−18
=
3240
∠P Ñ 108.4°
QP ⋅ QR
QP QR
=
[−6, 3] ⋅ [−12, − 3]
(−6) 2 + (−3) 2 (−12) 2 + (−3) 2
72 − 9
=
45 153
63
=
6885
∠Q Ñ 40.6°
∠R = 180° − ∠P − ∠Q
= 180° − 108.4° − 40.6°
= 31.0°
9. a) See diagram below.
b) Quadrilateral ABCD
Use the diagram in part a.
AB = [9, 3], DC = [9, 3], BC = [1, −3], AD = [1, −3]
AB ⋅ AD = [9, 3]⋅[1, −3]
=9−9
=0
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BA ⋅ BC = [–9, –3]⋅[1, −3]
= –9 + 9
=0
DA ⋅ DC = [–1, 3]⋅[9, 3]
= –9 + 9
=0
CD ⋅ CB = [–9, –3]⋅[–1, 3]
=9–9
=0
Therefore, ABCD is a rectangle.
Quadrilateral PQRS
Use the diagram in part a.
PQ = [12, −4], SR = [12, −4], QR = [−2, −7], PS = [−2, −7]
Since opposite sides are equal, PQRS is a parallelogram.
PQ ⋅ PS = [12, −4]⋅[−2, −7]
= −24 + 28
=4
Therefore, ∠P ≠ 90°. Hence PQRS is a not rectangle.
10. The vertices of ∆PQR are P(−3, −2), Q(−1, 4), and R(5, 2).
PQ = [−1 − (−3), 4 − (−2)]
= [2, 6]
PR = [5 − (−3), 2 − (−2)]
= [8 , 4]
QR = [5 − (−1), 2 − 4]
= [6, −2]
ii) QP ⋅ QR = [−2, −6]⋅ [6 , −2]
a) i) PQ ⋅ PR = [2, 6]⋅ [8, 4]
= 16 + 24
= −12 + 12
= 40
=0
iii) RQ ⋅ RP = [−6, 2]⋅ [−8, −4]
= 48 − 8
= 40
G G G G
b) The geometric definition of the dot product is a ⋅ b = a b cosθ , so:
PQ ⋅ PR = | PQ || PR | cos ∠P
RQ ⋅ RP = | RQ || RP | cos ∠R
From the diagram, ∆PQR is isosceles with PQ = RQ and ∠P = ∠R. Therefore, PQ ⋅ PR
and RQ ⋅ RP are equal.
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c) From the geometric definition of the dot product:
PQ ⋅ PR = | PQ || PR | cos ∠P
QP ⋅ QR = | QP || QR | cos ∠Q
RQ ⋅ RP = | RQ || RP | cos ∠R
If ∆PQR is equilateral, PQ = RQ = PR and ∠P = ∠Q = ∠R. Therefore, PQ ⋅ PR ,
QP ⋅ QR , and RQ ⋅ RP will be equal.
11. The vertices of parallelogram ABCD are A(2, 0), B(6, 1), C(8, 4), and D(4, 3).
AB = [6 − 2, 1 − 0]
= [4, 1]
Since ABCD is a parallelogram, DC = AB = [4, 1].
AD = [4 − 2, 3 − 0]
= [2, 3]
Since ABCD is a parallelogram, BC = AD = [2, 3].
a) i) AB ⋅ AD = [4, 1]⋅[2, 3]
=8+3
= 11
iii) BA ⋅ BC = [−4, −1]⋅[2, 3]
= −8 − 3
= −11
ii) CB ⋅ CD = [−2, −3]⋅[−4, −1]
=8+3
= 11
iv) DA ⋅ DC = [−2, −3]⋅[4, 1]
= −8 − 3
= −11
b) There are two relationships:
AB ⋅ AD = CB ⋅ CD and BA ⋅ BC = DA ⋅ DC .
AB ⋅ AD = − BA ⋅ BC and CB ⋅ CD = − DA ⋅ DC .
G G G G
c) The geometric definition of a dot product is a ⋅ b = a b cosθ .
From part b, we have AB ⋅ AD = CB ⋅ CD and BA ⋅ BC = DA ⋅ DC .
AB ⋅ AD = | AB || AD | cos ∠A
CB ⋅ CD = | CB || CD | cos ∠C
In any parallelogram ABCD, opposite sides are equal so AB = DC and AD = BC.
Opposite angles are also equal, so ∠A = ∠C. Hence AB ⋅ AD and CB ⋅ CD are equal.
Similarly, BA ⋅ BC and DA ⋅ DC are equal.
From part b, we also have AB ⋅ AD = − BA ⋅ BC
AB ⋅ AD = | AB || AD | cos ∠A
− BA ⋅ BC = −| BA || BC | cos ∠B c
Adjacent angles in a parallelogram are supplementary, so ∠B = 180° − ∠A. Also
opposite sides are equal, so AD = BC. Clearly AB = BA. Therefore equation c becomes:
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− BA ⋅ BC = −| AB || AD | cos (180° − ∠A)
= −| AB || BC |(−cos ∠A)
= | AB || AD | cos ∠A
Hence AB ⋅ AD = − BA ⋅ BC . Similarly, CB ⋅ CD = − DA ⋅ DC .
d) If the parallelogram is a rectangle, AB ⋅ AD = CB ⋅ CD = BA ⋅ BC = DA ⋅ DC = 0.
12. a) Three vectors can be formed from the 3 points. If the dot product of any pair of vectors is
0, then the points form a right angle.
b) i) AB = [−2 − (−5), 1 − 5]
= [3, −4]
AC = [7 − (−5), 8 − 5]
= [12, 3]
BC = [7 − (−2), 8 − 1]
= [9, 7]
AB ⋅ AC = [3, −4]⋅[12, 3]
= 36 − 12
≠0
AB ⋅ BC = [3, −4]⋅ [9, 7]
= 27 − 28
≠0
AC ⋅ BC = [12, 3]⋅[9, 7]
= 108 − 21
≠0
Therefore, A, B, and C do not form a right angle.
ii) JK = [5 − (−3), 0 − (−4)]
= [8, 4]
JL = [2 − (−3), 6 − (−4)]
= [5, 10]
KL = [2 − 5, 6 − 0]
= [−3, 6]
JK ⋅ JL = [8, 4]⋅[5, 10]
= 40 + 40
≠0
JL ⋅ KL = [5, 10]⋅[−3, 6]
= −15 + 60
≠0
JK ⋅ KL = [8, 4]⋅[−3, 6]
= −24 + 24
=0
Therefore, J, K, and L form a right angle.
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13. a) Form any 2 vectors from the 3 given points, and find the angle between the vectors. If
the angle is 0° or 180°, the points are collinear.
b) i) DE = [2 − (−4), 3 − 7]
= [6, −4]
DF = [8 − (−4), −1 − 7]
= [12, −8]
cos ∠D =
=
DE ⋅ DF
DE DF
[6, − 4] ⋅ [12,−8]
6 + (−4) 2 12 2 + (−8) 2
72 + 32
=
52 208
104
=
10816
=1
∠D = 0°
Therefore, D, E, and F are collinear.
ii) RS = [4 − 7, 1 − 2]
= [−3, −1]
RT = [−4 − 7, −2 − 2]
= [−11, −4]
cos ∠R =
=
2
RS ⋅ RT
RS RT
[−3, − 1] ⋅ [−11,−4]
(−3) + (−1) 2 (−11) 2 + (−4) 2
33 + 4
=
10 137
37
=
1370
∠D Ñ 1.5°
Therefore, R, S, and T are not collinear.
2
14. Suppose the square is ABCD and AB = [5, 2].
Since BC is perpendicular to AB and AB has
2
5
slope , BC has slope − . Furthermore
5
2
AB = BC, so BC = [−2, 5] or [2, −5]. Thus the
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other three sides of the square are either
[5, 2], [2, −5], [2, −5] or [5, 2], [−2, 5], [−2, 5].
15. There are 2 cases to consider.
Case 1. [4, 2] represents the width of the rectangle.
The slope of the line segment representing the width is
2
, so the slope of the line segment
4
4
. Since the length is double the width, the vector representing
2
the length is either [4, −8] or [−4, 8]. Therefore, vectors that could represent the other 3 sides
of the rectangle are [4, 2], [4, −8], [4, −8] or [4, 2], [−4, 8], [−4, 8].
Case 2. [4, 2] represents the length of the rectangle.
Since the length is double the width, the vector representing the width is either [−1, 2] or
[1, −2]. Therefore, vectors that could represent the other 3 sides of the rectangle are
[4, 2], [−1, 2], [−1, 2] or [4, 2], [1, −2], [1, −2].
representing the length is −
16. a) In ∆OBC:
OC
cos θ = OB
OC = OB cos θ
G
= b cos θ
Area of rectangle OAED = (OA)(OD)
= (OA)(OC) since OD = OC
G G
= a b cosθ
G G
= a ⋅b
b) When θ = 90°, rectangle OAED cannot be constructed.
G
G
G G
When θ = 0°, a and b are parallel, and the area of the rectangle OAED is a b .
c) When θ = 45°, BC = OC = OD, and so D, B, and E are collinear.
17. a) See diagram for part c.
G G G
G G G
G
G
ii) (v ⋅ w)u = ([2, 1]⋅[1, 2]) u
b) i) (u ⋅ v ) w = ([2, 0]⋅[2, 1]) w
G
G
= (4 + 0) w
= (2 + 2) u
G
G
= 4 w or [4, 8]
= 4 u or [8, 0]
G G K
G
iii) ( w ⋅ u )v = ([1, 2]⋅[2, 0]) v
G
= (2 + 0) v
G
= 2 v or [4, 2]
G G
G
c) The expressions in part b are scalar multiples of u , v , and w .
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G
G
G G
18. If a and b are perpendicular, then a ⋅ b = 0.
a) [k, −2]⋅[−1, 2] = 0
−k − 4 = 0
b) [−3, 4]⋅[5, k] = 0
−15 + 4k = 0
15
k = −4
k= 4
G G G
G G G
19. a) a ⋅ (b + c ) = [2, −3]⋅([−1, 4] + [5, 2])
b) (a + b ) ⋅ c = ([2, −3] + [−1, 4])⋅[5, 2]
= [2, −3]⋅[4, 6]
= [1, 1]⋅[5, 2]
= 8 − 18
=5+2
= −10
=7
G G G G
c) (a + b ) ⋅ (a + c ) = ([2, −3] + [−1, 4])⋅([2, −3] + [5, 2])
= [1, 1]⋅[7, −1]
=7−1
=6
G
G
G
G
d) (2a + 3b ) ⋅ (5a − 2b ) = (2[2, −3] + 3[−1, 4])⋅(5[2, −3] − 2[−1, 4])
= [1, 6]⋅[12, −23]
= 12 − 138
= −126
20. a) The magnitude of the force in the direction of the displacement is the horizontal
G
component of the force, F cosθ .
b) The work done is the product of the magnitude of the displacement and the magnitude of
the force in the direction of the displacement. Thus:
G G
Work = d F cosθ
G G
= F d cosθ
G G
= F ⋅d
G G
21. Work = F ⋅ d
G G
= F d cosθ
= 30(100)cos 30°
Ñ 2598.1
The work done in moving the wagon 100 m is approximately 2598.1 joules.
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22.
23. a) To draw a different rectangle, construct C so that ∠OCA = 90°. Then construct
D so that OC = OD.
b) i) a) In ∆OAC:
OC
cos θ = OA
OC = OA cos θ
G
= a cosθ
Area of rectangle OBED = (OB)(OD)
= (OB)(OC) since OD = OC
G G
= b a cosθ
G G
= a ⋅b
ii) When θ = 90° or θ = 0°, ∆OAC and rectangle OBED cannot be constructed.
iii) When θ = 45°, DA = CA = BE, and so D, A, and E are collinear.
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1.8 Exercises, Sample Solutions
G G G
G G G G
1. a) a ⋅ (b + c ) = a ⋅ b + a ⋅ c
G G
G
G G
G G
c) u ⋅ (u + 2v ) = u ⋅ u + 2u ⋅ v
G2
G G
= u + 2u ⋅ v
G G G
G G G G
b) a ⋅ (a + b ) = a ⋅ a + a ⋅ b
G2 G G
= a + a ⋅b
G
G G
G G
G G
d) 3u ⋅ (2u − 3v ) = 6u ⋅ u − 9u ⋅ v
G2
G G
= 6 u − 9u ⋅ v
G G G G
G G G G G G
2. a) (a + b ) ⋅ (a − b ) = (a + b ) ⋅ a − (a + b ) ⋅ b
G G G G G G G G
= a ⋅a + b ⋅a − a ⋅b − b ⋅b
G G G G G G G G
= a ⋅a + a ⋅b − a ⋅b − b ⋅b
G2 G2
= a −b
G
G
G G G
G G G G G
b) (a − b ) ⋅ (a + 2b ) = (a − b ) ⋅ a + (a − b ) ⋅ 2b
G G
G G G G
G G
= a ⋅ a − b ⋅ a + 2a ⋅ b − 2b ⋅ b
G G
G G
G G G G
= a ⋅ a − a ⋅ b + 2a ⋅ b − 2b ⋅ b
G2
G2 G G
= a + a ⋅b − 2b
G
G
G G G
G G G
G G
c) (4a + b ) ⋅ (a + 2b ) = (4a + b ) ⋅ a + (4a + b ) ⋅ 2b
G G
G G
G G G G
= 4a ⋅ a + b ⋅ a + 8a ⋅ b + 2b ⋅ b
G G
G G G G
G G
= 4a ⋅ a + a ⋅ b + 8a ⋅ b + 2b ⋅ b
G2
G2
G G
= 4 a + 9a ⋅ b + 2 b
G
G
G G
G
G
G
G
G
G
d) (2a + 3b ) ⋅ (3a − 2b ) = (2a + 3b ) ⋅ 3a − (2a + 3b ) ⋅ 2b
G G
G G
G G
G G
= 6a ⋅ a + 9b ⋅ a − 4a ⋅ b − 6b ⋅ b
G G
G G
G G
G G
= 6a ⋅ a + 9a ⋅ b − 4a ⋅ b − 6b ⋅ b
G2
G2
G G
= 6 a + 5a ⋅ b − 6 b
G G G
3. a) The dot product does not satisfy the associative law because the expressions (a ⋅ b ) ⋅ c and
G G G
a ⋅ (b ⋅ c ) have no meaning.
G
G G G
• ( a ⋅ b ) ⋅ c has no meaning because a ⋅ b is a scalar and we cannot take its dot product
G
with vector c .
G G G
G
• a ⋅ (b ⋅ c ) also has no meaning because it too involves the dot product of vector ( a ) with
G G
a scalar ( b ⋅ c ).
1.8 Exercises, Sample Solutions
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G G G
G G G
G G G
b) No meaning can be given to the expression a ⋅ b ⋅ c since (a ⋅ b ) ⋅ c and a ⋅ (b ⋅ c )
have no meaning.
G G G
4. Proof that a ⋅ 0 = 0
Geometric proof
G G GG
a ⋅ 0 = a 0 cosθ
=0
Cartesian proof
G
Let a = [a1 , a 2 ]
G G
a ⋅ 0 = [a1 , a 2 ] ⋅ [0, 0]
= a1 (0) + a 2 (0)
=0
G G
G
G
G
Proof that a ⋅ u = | a | , where u is a unit vector in the direction of a
G
G
G
1 G
If u is a unit vector in the direction of a , then u = G a .
|a|
Geometric proof
Cartesian proof
G
G G G 1 G
a ⋅u = a ⋅ G a
Let a = [a1 , a 2 ]
|a|
G G G 1 G
1 G G
= G a⋅a
a ⋅u = a ⋅ G a
|a|
|a|
1 G2
1 G G
= G a
= G a⋅a
|a|
|a|
G
1
= |a|
=
[a1 , a 2 ] ⋅ [a1 , a 2 ]
2
2
a1 + a 2
2
=
a1 + a 2
2
2
2
2
2
a1 + a 2
= a1 + a 2
G
= |a|
G G G G
5. a) If a ⋅ c = b ⋅ c then:
G G G G
a ⋅c −b ⋅c = 0
G G G
c ⋅ (a − b ) = 0
G
Two vectors have a dot product of 0 if one of the vectors is 0 or if the vectors are
perpendicular. Therefore, there are three possibilities:
G
• c=0
G G
G G
• a − b = 0 or a = b
G
G G
• b ⊥ (a − c )
G G G G
G G
Hence if a ⋅ c = b ⋅ c , it does not necessarily follow that a = b .
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Less formally, we can also argue as follows:
G G
G
Suppose a , b , and c are the side vectors of an
equilateral triangle. Since the magnitudes of the
vectors are equal and the angle between the vectors
G G G G
G G
are equal, a ⋅ c = b ⋅ c . However, a ≠ b .
G G
G
ac
a ⋅c
a
for real numbers so it cannot be written as G . If
b) G G is not the same as
bc
b ⋅c
b
K
G
G
G
θ1 is the angle between a and c and θ2 is the angle between b and c , then
G G
G G
a c cosθ 1
a ⋅c
G G= G G
b ⋅ c b c cosθ 2
G
a cosθ 1
= G
b cosθ 2
AC = AB + BC
6. a) OB = OC + CB
= OC + OA
= OC − OA
G G
G G
=c+a
=c −a
G G G G
b) If (c + a ) ⋅ (c − a ) = 0 , then OB ⋅ AC = 0 . Therefore, diagonals OB and OC are
perpendicular, so parallelogram OABC is a rhombus.
7. These properties are common to dot products and the products of real numbers.
Real numbers
DotGproducts
G G
G
ab = ba
a ⋅b = b ⋅a
G G G
G G G G
a ⋅ (b + c ) = a ⋅ b + a ⋅ c
a(b + c) = ab + ac
G G G2
aa = a
a⋅a = a
G G G G
G G
(ka)b = a(kb) = k(ab)
(ka ) ⋅ b = a ⋅ kb = k (a ⋅ b )
K G
a(0) = 0
a ⋅0 = 0
These are properties of real numbers that do not have corresponding properties for dot
products: (ab)c = a(bc), a(1) = a.
G
G G
G G G G G G
8. a) (a + b) ⋅ (a + b ) = (a + b ) ⋅ a + (a + b ) ⋅ b
G G G G G G G G
= a ⋅ a + b ⋅a + a ⋅b + b ⋅b
G G G G G G G G
= a ⋅ a + a ⋅b + a ⋅b + b ⋅b
G G
G G G G
= a ⋅ a + 2a ⋅ b + b ⋅ b
(Distributive property)
(Commutative property)
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G
G G
G G
G G G G
b) (a + b) ⋅ (a + b ) = a ⋅ a + 2a ⋅ b + b ⋅ b
G2
G2
G G
= a + 2 a b cosθ + b
G
G2
G G G2
2
a + b = a + 2a ⋅ b + b
G2
G2
G G
= a + 2 a b cosθ + b
G2
G G2 G2
G G
c) a + b = a + 2 a b cosθ + b
G
G
G G
If a and b are perpendicular, then cos θ = 0 and 2 a b cosθ = 0.
2
G
G2 G2
Therefore, a + b = a + b .
This is the Pythagorean theorem.
G G
G G  a ⋅a G
9. a ↓ a =  G G a
 a⋅a 

=


G2
a
G2
a
G
a


G
=a
This is reasonable as illustrated in the diagram at the right.
G G
G G  a ⋅b G
10. a) a ↓ b =  G G b
 b ⋅b 
 [6, 4] ⋅ [8, − 4] 
[8, − 4]
= 
 [8,−4] ⋅ [8, − 4] 
G
G G  b ⋅ aG  G
b) b ↓ a =  G G a
 a⋅a 
 [8, − 4] ⋅ [6, 4] 
[6, 4]
= 
 [6, 4] ⋅ [6, 4] 
48 − 16
= 64 + 16 [8, −4]
48 − 16
= 36 + 16 [6, 4]
32
= 80 [8, −4]
32
= 52 [6, 4]
2
= 5 [8, −4]
8
= 13 [6, 4]
16 8 
=  ,− 
5
5
 48 32 
=  , 
 13 13 
G G 16 8 
c) a ↓ b =  , − 
5
5
= [3.2, −1.6]
G G  48 32 
b↓a= , 
 13 13 
Ñ [3.7, 2.5]
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G G G G
d) Since a ↓ b ≠ b ↓ a , vector projection is not
commutative.
G G
G G  a ⋅b G
11. Use the formula a ↓ b =  G G b .
 b ⋅b 
G G  [3,0] ⋅ [2, 3] 
[2, 3]
a) a ↓ b = 
 [2, 3] ⋅ [2, 3] 
6−0
= 4 + 9 [2,3]
6
= 13 [2, 3]
12 18 
= , 
13 13 
G G  [4, 5] ⋅ [−5, 4] 
[−5, 4]
b) a ↓ b = 
 [−5, 4] ⋅ [−5, 4] 
−20 + 20
= 25 + 16 [−5, 4]
= 0[8, −4]
G
= [0, 0] or 0
G G  [−4, − 2] ⋅ [3, 1] 
[3, 1]
c) a ↓ b = 
[
3
,
1
]
[
3
,
1
]
⋅


−12 − 2
= 9 + 1 [3, 1]
14
= −10 [3, 1]
7
= −5 [3, 1]
 21 7 
= − , − 
5
 5
1.8 Exercises, Sample Solutions
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G G  [2,−3] ⋅ [6, 2] 
[6, 2]
d) a ↓ b = 
[
6
,
2
]
[
6
,
2
]
⋅


12 − 6
= 36 + 4 [6, 2]
6
= 40 [6, 2]
3
= 20 [6, 2]
9 G 3 G
9 3
i+
j
=  ,  or
10
10
10 10 
12. PQ = [−1 − (−4), 6 − 0]
= [3, 6]
PR = [3 − (−4), 4 − 0]
= [7, 4]
QR = [3 − (−1), 4 − 6]
= [4, −2]
a) See diagram in part b.
 PR ⋅ PQ 
PQ
b) i) PR ↓ PQ = 
 PQ ⋅ PQ 


 [7, 4] ⋅ [3, 6] 
[3, 6]
= 
 [3, 6] ⋅ [3, 6] 
21 + 24
= 9 + 36 [3, 6]
45
= 45 [3, 6]
= [3, 6]
= PQ
 PQ ⋅ PR 
PR
iii) PQ ↓ PR = 
 PR ⋅ PR 


 [3,6] ⋅ [7, 4] 
[7, 4]
= 
⋅
[
7
,
4
]
[
7
,
4
]


 RP ⋅ RQ 
RQ
ii) RP ↓ RQ = 
 RQ ⋅ RQ 


 [−7,−4] ⋅ [−4, 2] 
[−4, 2]
= 
 [−4,2] ⋅ [−4, 2] 
28 − 8
= 16 + 4 [−4, 2]
20
= 20 [−4, 2]
= [−4, 2]
= RQ
 QR ⋅ PR 
PR
ii) QR ↓ PR = 
 PR ⋅ PR 


 [4,−2] ⋅ [7, 4] 
[7, 4]
= 
⋅
[
7
,
4
]
[
7
,
4
]


21 + 24
= 49 + 16 [7, 4]
28 − 8
= 49 + 16 [7, 4]
45
= 65 [7, 4]
20
= 65 [7, 4]
1.8 Exercises, Sample Solutions
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4
= 13 [7, 4]
9
= 13 [7, 4]
=
9
PR
13
=
4
PR
13
9
4
PR + PR
13
13
= PR
c) PQ ↓ PR + QR ↓ PR =
G G
G G  a ⋅b G
13. Use the formula a ↓ b =  G G b
 b ⋅b 
 aG bG cosθ

=
G2

b

D G
G
 (4)(7) cos 60 
G
b
a ↓ b = 
72


G
14
= 49 b
2 G
=7 b
G
Ñ 0.3 b
G G
G G  u ⋅ v G
14. Use the formula u ↓ v =  G G v
 v ⋅v 
 uG vG cosθ
=
G2

v

G G  (8)(11) cos 135 D  G
v
u ↓ v = 
112


G

b


G
v


1.8 Exercises, Sample Solutions
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−44 2 G
= 121 v
4 2 G
= − 11 v
G
Ñ −0.5 v
G G
G
 a ⋅b 
G G
15. a ↓ b is a scalar multiple of b , where the scalar is  G G  .
 b ⋅b 
G
G G G
G G
G
a) Therefore, a ↓ b = 0 when a ⋅ b = 0 , that is, when a is perpendicular to b .This is
illustrated in the diagram, below left.
G G
G
G G
b) Therefore, a ↓ b is undefined when b ⋅ b = 0 . This occurs when b is the zero vector.
This is illustrated on the diagram, above right.
G
G G
16. a) (a ↓ b ) ↓ b
G
G
G G
Since a ↓ b is a scalar multiple of b , it is collinear to b .
G
G G
Therefore, projecting a ↓ b perpendicularly on b gives
G G
a↓b.
G
G G
b) b ↓ (a ↓ b )
Use the diagram in part a.
G
G
G G
Since a ↓ b is a scalar multiple of b , it is collinear to b .
G
G
G G
Therefore, projecting b perpendicularly on a ↓ b gives b .
G G
G
c) (a ↓ b ) ↓ a
G G
G
G G
G
(a ↓ b ) ↓ a is the projection of a ↓ b on a . It is
G G
a scalar multiple of a , k a , as shown in the diagram at
the right.
G
G G
d) a ↓ (a ↓ b )
Use the diagram in part a.
G
G G
G
G G
a ↓ (a ↓ b ) is the projection of a on a ↓ b . It is
G G
a ↓ b as illustrated in the diagram.
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G
G G G G
G
17. a) It is possible to have a ↓ b = b ↓ a when a and b are perpendicular. Then
G G G G
G G G G K
a ↓ b = b ↓ a = 0 . When a = b , clearly a ↓ b = b ↓ a .
G G
G G
G
G
G
b) Since b ↓ c is collinear to c , projecting a perpendicularly on c or on b ↓ c produces
G G
the same vector. This vector is a ↓ c .
G
G
G
G
b
18. Let u = [a, b]. Then u lies on a line with slope . Since v is perpendicular to u , it lies on a
a
G
G
G
G
a
line with slope − . If v is twice as long as u , then v = [−2b, 2a] or v = [2b, −2a]. Hence
b
there are 2 cases to consider.
G
Case 1. v = [−2b, 2a]
G
G
Since the sum of u and v is [6, 8]:
[a, b] + [−2b, 2a] = [6, 8]
[a − 2b, b + 2a] = [6, 8]
Since the vectors are equal, their components are equal.
a − 2b = 6 c
2a + b = 8 d
Eliminate b.
Eliminate a.
−2 × c: −2a + 4b = −12
c: a − 2b = 6
d:
2a + b = 8____
2 × d: 4a + 2b = 16
Add:
5b = −4
Add: 5a = 22
4
22
b= −
a=
5
5
G  22 4 
G  8 44 
Thus u =  ,−  and v =  ,  .
 5 5
5 5 
G
Case 2. v = [2b,− 2a]
G
G
Since the sum of u and v is [6, 8]:
[a, b] + [2b, −2a] = [6, 8]
[a + 2b, b − 2a] = [6, 8]
Since the vectors are equal, their components are equal.
a + 2b = 6 c
−2a + b = 8 d
Eliminate b.
Eliminate a.
2 × c: 2a + 4b = 12
c: a + 2b = 6
d: −2a + b = 8____
−2 × d: 4a − 2b = −16
Add:
5b = 20
Add:
5a = −10
b=4
a = −2
G
G
Thus u = [−2, 4] and v = [8, 4].
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G G2
G G2
G G G G
G G G G
19. a) 14 a + b − 14 a − b = 14 (a + b ) ⋅ (a + b ) − 14 (a − b ) ⋅ (a − b )
G G
G G G G
G G
G G G G
= 14 (a ⋅ a + 2a ⋅ b + b ⋅ b ) − 14 (a ⋅ a − 2a ⋅ b + b ⋅ b )
G2
G2
G G
G G
G2
G2
= 14 a + 12 a ⋅ b + 14 b − 14 a + 12 a ⋅ b − 14 b
G K
= a ⋅b
1
1
b) A similar equation for the product xy in algebra is xy = 4 (x + y)2 − 4 (x − y)2
1
1
1 2
1 2
2
2
2
2
4 (x + y) + 4 (x − y) = 4 (x + 2xy + y ) − 4 (x − 2xy + y )
1
1
1
1
1
1
= 4 x2 + 2 xy + 4 y2 − 4 x2 + 2 xy − 4 y2
= xy
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Chapter 1 Review Exercises, Sample Solutions
1. Three examples of scalar quantities are distance, speed, and mass. Three examples of vector
quantities are displacement, velocity, and weight.
2. Answers may vary.
a) 20 N [east]
Scale : 1 cm : 10 N
b) 24 m/s [135°]
Scale: 1 cm : 8 m/s
3. Answers may vary.
a) Equal vectors have the same magnitude and direction.
AO = OC, DO = OB, DC = AB, AD = BC
b) Opposite vectors have the same magnitude but are in opposite directions.
AO = −CO, DO = − BO, DC = − BA, AD = −CB
4. a) HE = HA + AE
c) DG = DH + HG
b) GF = GC + CF
d) DC = DG + GC
5. a) PG + PR = PQ
b) RA + RQ = RB
c) CD + RS − EF = CD + RS + FE
d) DR + QB − FS = DR + QB + SF
= (CD + DE) + FE
= DR + 2DR + DR
= CE + FE
= GC + CE
= GE
= 4 DR
6. AB + BC + CD + DA = AA
G
=0
Chapter 1 Review Exercises, Sample Solutions
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7.
8. Answers may vary.
a) FD = FA + AD
= −AF + AD
= AD − AF
b) EB = ED + DB
= −DE + DB
= DB − DE
c) CB = CA + AB
= −AC + AB
= AB − AC
b) AE = AB + BE
= −BA + BE
= BE − BA
9. Answers may vary.
10. a) AC = 2ED
G
= 2u
b) AD = AC + CD
= 2ED + CD
G G
= 2u + v
c) EA = EB + BA
= −CD − DE
G G
= −v − u
11. CG = CD + DG
= CD + 1 CB
2
G
12. a) i) 3u = 3[−1, 2]
= [−3, 6]
CM = CB + BM
= CB + 1 CD
2
G
ii) 2u = 2[−1, 2]
= [−2, 4]
Chapter 1 Review Exercises, Sample Solutions
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G
iii) − u = −[−1, 2]
= [1, −2]
G
b) u =
G
iv) − 4u = −4[−1, 2]
= [4, −8]
(−1)2 + 22
= 5
G
G
G
G
Therefore, 3u = 3 5 , 2u = 2 5 , − u = 5 , and 4u = 4 5 .
13. a) AB = [7 − 3, 4 − 2]
= [4, 2]
BC = [−1 − 7, 10 − 4]
= [−8, 6]
CA = [3 − (−1), 2 − 10]
= [4, −8]
b) | AB | = 42 + 22
| BC | = (−8)2 + 62
| CA | = 42 + (−8)2
= 20
= 100
= 80
2
2
2
| BC | = | AB | + | CA | since 100 = 20 + 80. Therefore, ∆ABC is a scalene, right triangle.
14. a)
G
G
K
Let w = s u + t v for some scalars s and t.
[4, 8] = s[0, 2] + t[1, −3]
= [t, 2s − 3t]
Since the vectors are equal, their components are equal.
t=4
c
2s − 3t = 8
d
Substitute t = 4 in equation d to determine s.
2s −3(4) = 8
2s = 20
s = 10
G
G
K
Therefore, w = 10 u + 4 v .
b)
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15. Method 1: Using geometric vectors
In the diagram below left, OA and OB represent the two 90 N forces.
Complete parallelogram OACB as shown in the diagram below right. Let OC represent the
resultant force.
∠BOA = 48°
OACB is a parallelogram, so:
∠OAC = 180° − 48°
= 132°
G
Use the Cosine Law to calculate r .
G
r 2 = 902 + 902 − 2(90)(90)cos 132°
G
r Ñ 164.4
Let θ = ∠COA. Use the Sine Law to determine θ.
sin θ sin 132°
90 = 164.4
90 sin 132°
sin θ = 164.4
θ Ñ24°
The resultant force has a magnitude of 164.4 N and acts at an angle of 24° with each force.
Therefore, a force of 164.4 N at an angle of 180° to the resultant must be applied to the
object to create equilibrium.
Method 2: Using Cartesian vectors
Place the vectors on a coordinate grid as shown in the
diagram at the right.
G
G
Let f1 and f 2 represent the two forces.
G
Let r be the resultant force.
G
f1 = [90, 0]
G
f 2 = [90 cos 48°, 90 sin 48°]
Ñ[60.2, 66.9]
G
G
G
r = f1 + f 2
= [90, 0] + [60.2, 66.9]
= [150.2, 66.9]
The magnitude of the resultant is:
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G
r = (150.2)2 + (66.9)2
Ñ 164.4
The direction of the resultant is:
66.9
tan θ = 150.2
The resultant force has a magnitude of 164.4 N and acts at an
angle of 24° with each force.
Therefore, a force of 164.4 N at an angle of 180° to the resultant
must be applied to the object to create equilibrium.
16. Method 1: Using geometric vectors
Let OW represent the velocity of the wind.
Let OH represent the velocity of the plane in still air.
Complete parallelogram OWRH, where OR represents the velocity of the plane relative to
the ground.
From the given bearings:
∠WOH = 220° − 120°
= 100°
Since OHRW is a parallelogram:
∠OHR = 180° − 100°
= 80°
G
Use the Cosine Law to calculate r .
G2
r = 550 + 502 − 2(550)(50)cos 80°
Ñ 543.6
Let ∠ROH = θ. The bearing of the plane is 120° + θ.
sin θ sin 80°
50 = 543.6
50 sin 80°
sin θ = 543.6
θ Ñ5.2°
Hence the bearing is 120° + 5.2° = 125.2°.
The plane’s speed relative to the ground is 543.6 km/h on a bearing of 125.2°.
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Method 2: Using Cartesian vectors
The diagram at the right represents the situation.
Bearings of 120° and 220° correspond to direction angles
of 330° and 230° respectively.
Let
Let
Let
G
a represent the velocity of the plane in still air.
G
w represent the velocity of the wind.
G
r represent the velocity of the plane relative to the ground.
G
a = [550 cos 330°, 550 sin 330°]
G
w = [50 cos 230°, 50 sin 230°]
G
G
G
r = a + w
= [550 cos 330° + 50 cos 230°, 550 sin 330° + 50 sin 230°]
Ñ[444.2, −313.3]
The magnitude of the resultant is:
G
r = (444.2)2 + (−313.3)2
Ñ 543.6
The bearing of the plane is 90° + θ .
313.3
tan θ = 444.2
θ Ñ 35.2°
Hence the bearing is 90° + 35.2° = 125.2°
The plane’s speed relative to the ground is 543.6 km/h on a bearing of 125.2°.
G
17. Let r be the resultant velocity of the boat.
G
r = 52 − 2 2
= 21
Ñ4.6
The resultant speed of the boat is approximately 4.6 m/s.
Since the river is 70 m wide, the time taken to cross the river,
70
s, or 15.3 s.
is
21
To head directly across the river, the boat must take the heading
shown in the diagram.
2
sin θ = 5
θ Ñ 23.6°
Therefore, the boat’s heading is 90° − 23.5°, or 66.4° to the shoreline.
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G G
a ⋅b
18. Use the formula cosθ = G G .
ab
a) cosθ =
[2, 0] ⋅ [4, 3]
2 4 2 + 32
8
= 10
θ Ñ 36.9°
c) cosθ =
[4, − 2] ⋅ [−1, − 3]
4 2 + (−2) 2 (−1) 2 + (−3) 2
−4 + 6
=
20 10
2
=
200
θ Ñ 81.9°
b) cosθ =
[−2,1] ⋅ [3, 5]
(−2) 2 + 12 3 2 + 5 2
−6 + 5
=
5 34
−1
=
5 34
θ Ñ 94.4°
d) cosθ =
[2, 6] ⋅ [−2, − 1]
2 2 + 6 2 (−2) 2 + (−1) 2
−4 − 6
=
40 5
−10
=
200
θ = 135°
19. AB = [8 − (−3), 1 − 5]
= [11, −4]
AC = [−2 − (−3), −1 − 5]
= [1 , −6]
BC = [−2 − 8, −1 − 1]
= [−10, −2]
cos ∠A =
=
=
=
AB ⋅ AC
AB AC
[11, − 4] ⋅ [1, − 6]
112 + (−4) 2 12 + (−6) 2
11 + 24
137 37
25
137 37
Ñ 60.6°
cos ∠B =
=
BA ⋅ BC
BA BC
[−11, 4] ⋅ [−10, − 2]
(−11) 2 + 4 2 (−10) 2 + (−2) 2
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=
=
110 − 8
137 104
102
137 37
Ñ 31.3°
∠C = 180° − ∠A − ∠B
= 180° − 60.6° − 31.3°
= 88.1°
G G G
20. a) a ⋅ (b + c ) = [−1, 3]⋅([4, 2] + [−2, −1])
= [−1, 3]⋅[2, 1]
= −2 + 3
=1
G G G
b) (a − b ) ⋅ c = ([−1, 3] −[4, 2])⋅[−2, −1]
= [−5, 1]⋅[−2, −1]
= 10 − 1
=9
G
G G G
c) (2a + c ) ⋅ (a − 3b ) = (2[−1, 3] +[−2, −1])⋅([−1, 3] − 3[4, 2])
= [−4, 5]⋅[−13, −3]
= 52 − 15
= 37
G G G
G G G G
21. a) a ⋅ (b + c ) = a ⋅ b + a ⋅ c
= [−1, 3]⋅[4, 2] + [−1, 3]⋅[−2, −1]
= −4 + 6 + 2 − 3
=1
G G G G G G G
b) (a − b ) ⋅ c = a ⋅ c − b ⋅ c
= [−1, 3]⋅[−2, −1] − [4, 2]⋅[−2, −1]
= 2 − 3 − (−8 −2)
=9
G
G G G
G G
G G G G G G
c) (2a + c ) ⋅ (a − 3b ) = 2a ⋅ a − 6a ⋅ b + c ⋅ a − 3c ⋅ b
= 2[−1, 3]⋅[−1, 3] − 6[−1, 3]⋅[4, 2] + [−2, −1]⋅[−1, 3] − 3[−2, −1]⋅[4, 2]
= 2(1 + 9) − 6(−4 + 6) + 2 − 3 − 3(−8 − 2)
= 37
G
G
G G
G G G G
G G
G G
22. a) (u + 3v ) ⋅ (2u + v ) = 2u ⋅ u + u ⋅ v + 6v ⋅ u + 3v ⋅ v
G2
G G
G2
= 2 u + 7u ⋅ v + 3 v
G
G
G G
G G
G
G
G G
G G
b) (3a + 4b ) ⋅ (3a − 4b ) = 9a ⋅ a − 12a ⋅ b + 12b ⋅ a − 16b ⋅ b
G2
G2
= 9 a − 16 b
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23. a)
G G
G G  a ⋅b G
b) a ↓ b =  G G b
 b ⋅b 
 aG bG cosθ  G


=
b
G2


b


 (5)(3)(cos 150 D )  G
b
= 
2
3


G
Ñ−1.44 b
G
G G  b ⋅ aG  G
b ↓ a =  G G a
 a⋅a 
G
 b aG cosθ

=
G2

a


G
a

 (3)(5)(cos 150 D )  G
a
= 
2
5


G
Ñ −0.52 a
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Chapter 1 Self-Test, Sample Solutions
1. a) CD − DA = CD + AD
= BC + CD
= BD
c) DC − CB = DC + BC
= AD + DC
= AC
b) AD + DC + CP + PB = AB
2. AB − BC + CD − DA = AB + CB + CD + AD
= (AB + CD) + (CB + AD)
= (AB + BA) + (CB + BC)
G G
= 0+0
G
=0
3. Method 1: Using geometric vectors
Let OW represent the velocity of the wind.
Let OH represent the velocity of the plane in still air.
Complete parallelogram OWRH, where OR represents the velocity of the plane relative to
the ground.
From the given bearings:
∠WOH = 140° − 40°
= 100°
Since OHRW is a parallelogram:
∠OHR = 180° − 100°
= 80°
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G
Use the Cosine Law to calculate r .
G2
r = 4202 + 402 − 2(420)(40)cos 80°
Ñ 414.9
Let ∠ROH = θ. The bearing of the plane is 140° − θ.
sin θ sin 80°
40 = 414.9
40 sin 80°
sin θ = 414.9
θ Ñ5.5°
Hence the bearing is 140° − 5.5° = 134.5°
The plane’s speed relative to the ground is 414.9 km/h on a bearing of 134.5°.
Method 2: Using Cartesian vectors
The diagram at the right represents the situation.
Bearings of 140° and 40° correspond to direction angles of 310°
and 50° respectively.
Let
Let
Let
G
a represent the velocity of the plane in still air.
G
w represent the velocity of the wind.
G
r represent the velocity of the plane relative to the ground.
G
a = [420 cos 310°, 420 sin 310°]
G
w = [40 cos 50°, 40 sin 50°]
G
G
G
r = a + w
= [420 cos 310° + 40 cos 50°, 420 sin 310° + 40 sin 50°]
Ñ[295.7, −291.1]
The magnitude of the resultant is:
G
r = (295.7)2 + (−291.1)2
Ñ 414.9
The bearing of the plane is 90° + θ .
291.1
tan θ = 295.7
θ Ñ 44.6°
Hence the bearing is 90° + 44.6° = 134.6°
The plane’s speed relative to the ground is 414.9 km/h on a bearing of 134.6°.
G
1 G
4. a) The vector − G b has unit length and is opposite in direction to b .
b
G
b = 12 + 22
= 5
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Therefore:
1 G
1
− G b =−
[1, 2]
5
b
 1
2 
= −
,−

5
5

G
b) a = 42 + 62
= 52
= 2 13
Therefore, the required vector has magnitude 2 13 . Since this vector makes an angle of
60° with the positive x-axis, its components are [2 13 cos 60°, 2 13 sin 60°] Ñ
[3.61, 6.24].
G G
G G  a ⋅b G
c) a ↓ b =  G G b
 b ⋅b 
 [4,6] ⋅ [1, 2] 
[1,2]
= 
 [1,2] ⋅ [1, 2] 
4 + 12
= 1 + 4 [1,2]
16
= 5 [1, 2]
16 32 
=  , 
5 5
G G
5. a − b = [k, 2] − [7, 6]
= [k − 7, −4]
G G
Since a − b = 5:
(k − 7)2 + (−4)2 = 5
Square both sides and simplify.
k2 − 14k + 49 + 16 = 25
k2 − 14k + 40 = 0
(k − 10)(k − 4) = 0
k = 10 or k = 4
G
G
6. We are given that AB = a and BC = b . Since DE and EF are parallel to AB and BC
respectively, DE is a scalar multiple of AB and EF is a scalar multiple of BC . Thus, we
G
G
can write DE = k a and EF = l b .
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AC = AB + BC
G
G
= a +b
FD = FEG+ ED
G
= −l b − k a
G
G
= −k a − l b
c
d
Since AC is parallel to FD :
AC = s FD
e
Substitute c and d in e.
G
G
G
G
a + b = s(−k a − l b )
G
G
G
G
a + b = −sk a − sl b
G
G
Equate the coefficients of a and b to obtain:
1 = −sl
1 = −sk and
1
1
s=−k
s = −l
1
1
Therefore −k = − l , or k = l.
7. a) See diagram at right.
b) The heads of the vectors lie on a straight line.
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