Ch05

Chapter 5 • Algebraic Vectors and Applications Solutions for Selected Problems

15. If two unit vectors are the sides of an equilateral

triangle then their sum as well as their difference could be a unit vector.

Exercise 5.1 13. c.



(9, 12) OG

2
(9)2  (12)2
OG

→

→

b

c


81  144
225

OG

5 12 tan 
. 9 Since G is in the second quadrant  ≅ 180  53  ≅ 127.

e. OJ

→

a

If a
 
 b
c
 where
a




 b

1 and the angle

 between a and b is 120 then
c


1. Also c
 
 b
a
 where as above
a




 b


c


1.

16. a. a

(2, 3, 2)

5 ,  5  2

60º

6


a

2
4  9  4



2
4  6
OJ 5


a


17 .

5


2

b.




2
OJ



Now

6 tan 
 . 2 Since J is in the fourth quadrant,  ≅ 360  51  ≅ 309. 14. a.

c.

If a

(12, 4, 6) then
a

2
144  16  36
196
a


14. 14 22 7 If c

, ,  27 27 27




1
c


1.

      17  17  17  2

2

3

2

2

2


1

1 therefore  a
 is a unit vector.

a
17. a.

v

2iˆ 3jˆ 6kˆ
(2, 3, 6)
v

2
4  9  36
49 therefore
v


7.

b. A unit vector in the direction of v
 is



196  484  49 then
c


 729 2



1 a

2, 3,  2 17 17   17 
a



1 v


 v

v
2 3 6 v

iˆ  jˆ  kˆ 7 7 7





2 3 6 v

, ,  . 7 7 7
 18. v
(3, 4, 12)
v

2
9  16  144
169
v


13. A unit vector in a direction opposite to 12 3 4 v
 is , ,  . 13 13 13





Chapter 5: Algebraic Vectors and Applications

53

19. Let a
 be a vector in two dimensions making an angle

23.

C c

 with the x-axis. Now a
 in component form is



,b,c)



P (a

a


a

cos ,
a

sin  . A unit vector in the 1 direction of a
 is  a

(cos , sin ) therefore any

a
unit vector in two dimensions can be written as (cos , sin ). 21. a. u

(a, b, c)

O b a A

B

In ∆OAB, ∠OAB
90 therefore OB2
OA2  AB2. In ∆OBP, ∠OBP
90 therefore OP2
OB2  BP2. Hence OP2
OA2  OB2  BP2 but OA
a, OB
b, BP
OC
c.

z

2


a Therefore
OP  b 2  c 2.

24. A vector in R4, u

(4, 2, 5, 2) might have a

α

y

→

u

a x

a cos α
, since α is acute, a is positive.
u



Exercise 5.2

b. u

(a, b, c)

8. a. Given the points P(15, 10), Q(6, 4), R(12, 8)

z

→

b

magnitude of
u

2
42  22  (5)2  22
u

2
16  4  25  4
u


 49,
u


7. It is very tempting to think that a geometric interpretation can be given. Mathematically we wish to consider vectors with n elements, n any integer, and a geometric interpretation is not possible for n ≥ 4.

u



(9, 6), PR

(27, 18) PQ
3( 9, 6).



3PQ
, P, Q, and R are Since PR collinear.

ß y

b. D(33, 5, 20), E(6, 4, 16), F(9, 3, 12)

x

b cos β
 since β is obtuse, b is negative.
u

22. Since the direction angles α, β, and γ are all equal, say

 then cos2   cos2   cos2 
1

1 cos2 
 3 11 cos 
 or 3

1 cos 
 3

 ≅ 55  ≅ 125 The direction angles are 55 or 125.

54

Chapter 5: Algebraic Vectors and Applications



(27, 9, 36), EF

(e, 1, 4) DE
9(3, 1, 4)
.
9EF

9EF
, D, E, and F are collinear. Since DE

9. b. A(0, 1, 0), B(4, 0, 1), C(5, 1, 2), D(2, 3, 5)

Let P2 (a2, b2) be a vertex of parallelogram AP2CB.



(4, 1, 1), CD

(3, 2, 3) AB
  k  CD
, k  R, AB is not parallel Since AB



BC
 and (a  5, b 3)
(2, 10) Now AP 2 2 2 therefore a2
3, b2
7.

to CD.

P3(a3, b3) is a vertex of parallelogram ABP3C.





AB 16  1 1
18 




AC
 and (a 5, b 2)
(12, 11) Now BP 3 3 3 therefore a3
17, b3
9.

32

The possible coordinates of the fourth vertex are P1(7, 9), P2(3, 7) and P3(17, 9).





CD 94 9
22 


CD

.
AB

P

12. C(4, 0, –1)

10. PQRS is a parallelogram. The coordinates are in cyclic



SR
. If R has coordinates (a, b) order therefore PQ

(10, 1) and SR

(a  3, b  4). PQ

SR
, a  3
10, a
13 Since PQ b  4
1, b
5 The coordinates of S are (13, 5). y

x

R(a, b)

S B(3, 6, 1)

O(0, 0, 0)

R

A(2, 4, –2)

Opposite faces of a parallelepiped are congruent parallelograms.

OC
  OB
 Now OP
(4, 0, 1)  (3, 6, 1)
 OP
(7, 6, 0).

  AR
 
 OQ
OA RQ

P(4, 2)

Q(–6, 1)

Q

S(–3, –4)

11. Let the three vertices be A(5, 3), B(5, 2), C(7, 8).

There will be 3 possible positions for the fourth vertex. Let one position be P1(a1, b1) for parallelogram ACBP1.

BP
 and (12, 11)
(a  5, b 2). CA 1 1 1 Therefore a1
7, b1
9.


  OB
  OC

OA
(2, 4, 2)  (3, 6, 1)  (4, 0, 1)
(9, 10, 2).


 OR
OA  OB
(2, 4, 2)  (3, 6, 1)
 OR
(5, 10, 1)

OA
  OC
 OS
(2, 4, 2)  (4, 0, 1)
 OS
(6, 4, 3) The other 4 coordinates are (7, 6, 0), (9, 10, 2), (5, 10, 1), and (6, 4, 3).

y

A(–5, 3)

B(5, 2) x

C(7, –8)

Chapter 5: Algebraic Vectors and Applications

55

13. Let the midpoint in each case be M and the position


. vector OM

1 2 Expanding and equating components 2x  12 2
0 x
5, 2 3y  1
0

b. 2(x, 1, 4) 3(4, y, 6) (4, 2, z)
(0, 0, 0)

a. A(5, 2), B(13, 4)


  OB
 OA

 OM  2

1 y
, 3

(5, 2)  (13, 4)
 2

1 8 18 z
0 2



(4, 3). OM

z
20.

b. C(3, 0), D(0, 7)


  OD
 OC

 OM  2 1
(3, 7) 2







3, 7 . OM 2 2 c. E(6, 4, 2), F(2, 8, 2)


  OF
 OE

 OM  2 1
(4, 12, 0) 2
 OM
(2, 6, 0).

15. Given points X(7, 4, 2) and Y(1, 2, 1)



(6, 2, 3). XY
 is
XY


 The magnitude of XY 36  4 9
7. A unit vector in a direction opposite to XY has 6 2 3 ˆ
6, 2, 3 . components , , and , or YX 7 7 7 7 7 7



16. a. A point on the y-axis has coordinates P(0, a, 0).

Since it is equidistant from A(2, 1, 1) and B(0, 1, 3)



BP

or
AP

2

BP

2
AP therefore 4  (a  1)2  1
(a  1)2  9 a2  2a  6
a2  2a  10 4a
4 a
1. The point on the y-axis equidistant from A and B is (0, 1, 0).

d. G(0, 16, 5), H(9, 7, 1)



 OG  OH

 OM  2 1
(9, 9, 6) 2







9, 9, 3 . OM 2 2 14. a. 3(x, 1) 2(2, y)
(2, 1)

(3x, 3)  (4, 2y)
(2, 1) (3x 4, 3 2y)
(2, 1) Equating components 3x  4
2, 3  2y
1 x
2 y
1.



b. The midpoint of AB is the point Q(1, 0, 2) which is

not on the y-axis and is equidistant from A and B. 17. a.

A(2, –3, –4) 2

G

C(1, 3, –7)

M

B(3, –4, 2)

Since AM is a median of ∆ABC, M will be the midpoint of BC. Therefore the coordinates of M





1 5 are 2, ,  . 2 2

56

Chapter 5: Algebraic Vectors and Applications



d. O(0, 0, 0), I(1, 0, 0), J(0, 1, 0), K(0, 0, 1)





0, 2, 3 Now AM 2



1 (1, 1, 1) OG 4

9   4 4

5
. 2 5 The length of median AM is . 2


and
AM



 1 1 1 The centroid is , , . 4 4 4

1, 1, 1 . OG 4 4 4

b. Let G be the centroid of ∆ABC.

2 AG
 3

5 AM
. 3 5 The distance from A to the centroid is . 3

Since AG:GM
2:1

2(0, 0)  3(4, 1)  5(1, 7)  1(11, 9) 2351



 19. a. OG (12  5  11, 3  35  9)
 11

18. In each case let the centroid be G with position




  O

 OA B  OC

 OG  3 1
(1  4  2, 2  1  2) 3

 



1, 1 . OG 3





1(1, 4, 1)  3(2, 0, 1)  7(1, 3, 10)

 b. OG 137

1
 (1  6  7, 4  21, 1  3  70) 11








  O

 OI J  OK

 OG  3

 1 1 1 The centroid is , , . 3 3 3 c. A1(3, 1), A2(1, 1), A3(7, 0), A4(4, 4)



1 (OA
  OA
  OA
  OA
 ) OG 1 2 3 4 4 1
 (3  1  7  4, 1  1  4) 4





15
, 1 . 4





15 The centroid is , 1 . 4



17 72

2,  OG ,  . 11 11 11

b. I(1, 0, 0), J(0, 1, 0), K(0, 0, 1)





41 8 The centre of mass is ,  . 11 11

1 The centroid is 1,  . 3



1, 1, 1 . OG 3 3 3



41

18,  OG  . 11 11


. vector OG a. A(1, 2), B(4, 1), C(2, 2)



17 72 2 The centre of mass is , ,  . 11 11 11

Exercise 5.3 7. b.

c

(1, 2, 3), d

(4, 2, 1) c
 · d


c


d

cos  c
 · d

4  4  3
3


c


 14 9
14 . d

16   4 1
21 . 3 cos 
 14 21  cos  ≅ 0.1750.

Chapter 5: Algebraic Vectors and Applications

57

10. u

3jˆ  4kˆ
(0, 3, 4)

8. c. ˆi
(1, 0, 0);
iˆ

1



(1, 1, 1);
m


3 m

1 ˆi · m


ˆi

m

cos  ˆi · m 1 cos 
 3  ≅ 55
 is 55. The angle between vectors ˆi and m d.
 p
(2, 4, 5);

 p



 4  16  25
45 

 q
(0, 2, 3);
q

4  9
13 
 p · q

8  15
7
 p · q



 p

q

cos 



(a, b, c). Let w
 ⊥ u
, w
 · u

0, and 3b  4c
0 Since w
 ⊥ u
, w
 · v

0, and 2a
0. w Solving these equations, we have a
0 and if b
4, c
3.
 is (0, 4, 3). A possible vector w

(10, y, z) are 11. Since a

(2, 3, 4) and b perpendicular a
 · b

0. Therefore 20  3y  4z
0 4 20 and y
z  . 3 3
 · v
 a. LS
u

The angle between vectors p
 and q
 is 73.

(4, y, 14) 9. a

(2, 3, 7); b
 are collinear if a

kb
, k  R a. a
 and b therefore (2, 3, 7)
k (4, y, 14) 3
ky,


(2, 0, 0).



(1, 5, 8), v

(1, 3, 2). 12. u

7 cos 
  45 13   ≅ 73

2
4k, 1 k
 2

v

2iˆ

7
14k 1 k
 2

1 Since k
, y
6, a
 and b
 will be collinear. 2
 · b

0 b. If the vectors are perpendicular a 8  3y  98
0 3y
106 106 y
. 3 106 If y
, a
 and b
 will be perpendicular. 3


1  15  16
2 RS
v
 · u

1  15  16
2 therefore u
 · v

v
 · u
.


 · u
 b. LS
u


1  25  64
90 RS

u

2
1  25  64
90 therefore u
 · u


u

2. LS
v
 · v

194
14 RS

v

2
194
14 therefore v
 · v


v

2.


  v
) · (u
  v
) c. LS
(u


(0, 8, 6) · (2, 2, 10)
16  60
76 RS

u

2 

v
2
90  14
76 therefore (u
 
v ) · (u
 
v )

u

2 

v
2.

58

Chapter 5: Algebraic Vectors and Applications

d. LS
u
 
 v  · u
 
v 


(0, 8, 6) · (0, 8, 6)
64  36
100 RS

u

2  2 u
 ·
v 

v
2
90  2 (2)  14
90  4  14
100 therefore (u
 
v ) · (u
 
v )

u

2  2 u
 ·
v 

v
2.

e.

(2u
) ·
v
(2, 10, 16) · (1, 3, 2)
2  30  32
4 u
 · (2v
)
(1, 5, 8) · (2, 6, 4)
2  30  32
4 z(u
 ·
v )
2 (2)
4 therefore (u
) ·
v
u
 · (2v
)
2(u
 ·
v ).



(1, 7, 8) 13. u

(2, 2, 1),
 v
(3, 1, 0), w
) LS
u
 · (u
  w
(2, 2, 1) · (4, 6, 8)
8  12  8
12
 RS
u
 ·
v  u
 · w
6  2  2  14  8
12

u
 ·
v  u
 · w
. therefore u
 · v
  w 14. a. (4iˆ  ˆj ) · ˆj
4iˆ · ˆj  ˆj · ˆj


0 
ˆj
2
1. ˆ )
ˆk · ˆj  3kˆ · ˆk b. ˆk · (jˆ  3k
0  3(1)
3.

ˆ ) · (iˆ  4kˆ)
ˆi · ˆi  8iˆ · bˆ  16kˆ · ˆk c. (iˆ  4k
1  0  16
17.

15. a. (3a
  4b
) · (5a
  6b
)


15a
 · a
  38a
 · b
  24b
 · b

15
a

2  38a
 · b
  24
b

2.

b. (2a
  b
) · (2a
  b
)


4a
 · a
  b
 · b

4
a

2 
b

2.

16. a

î  3jˆ  ˆk,

b

2î  4jˆ  5kˆ
(1, 3, 1)
(2, 4, 5)

 3a  b
(3, 9, 3)  (2, 4, 5)
(1, 5, 2)

 2b  4a
(4, 8, 10)  (4, 12, 4)
(8, 20, 14)

 (3a  b) · (2b
  4a
)
(1, 5, 2) · (8, 20, 14)


8  100  28



 (3a  b) · (2b  4a )
80. 17. Since 2a
  b
 is perpendicular to a
  3b
,

(2a
  b
) · (a
  3b
)
0

therefore 2
a

2  5a
 · b
  3
b

2
0 2
a

2  5
a


b

cos   3
b

2
0. But
a


2
b

. Substituting gives 8
b

2  10
b

2 cos   3
b

2
0 10 cos 
5 1 cos 
 2 
60. The angle between a and b is 60. 18. Since aˆ and bˆ are unit vectors, aˆ

bˆ

1. a. (6aˆ  bˆ) · (aˆ  2bˆ)
6
aˆ
2  11
aˆ

bˆ
cos   2
bˆ
2


6  11 cos   2
4  11 cos . But 
60 therefore 4  11 cos 
4  11 cos 60 11
4   2 3
 2 3 (6aˆ  bˆ) · (aˆ 2bˆ)
. 2

Chapter 5: Algebraic Vectors and Applications

59

b.

20. a.

 aˆ

→

→

→

b

b

b

ˆb

a →+

ˆb ˆa +

ˆb

C

B

180º –  O

Let  be the angle between the unit vectors aˆ and b
. From the cosine law
aˆ  bˆ
2

aˆ
2 
bˆ
2  2
aˆ

bˆ
cos(180  ).



A
, OB

AC

b
, OC

a
  b
. OA
 Since a is perpendicular to b
 ∠OAC
90.

Now
aˆ  bˆ

3 and cos(180  )
cos  therefore 3
1  1  2
aˆ

bˆ
cos  1
2aˆ · bˆ 1 aˆ · bˆ
. 2 Now (2aˆ  5bˆ) · (bˆ  3aˆ)
13aˆ · bˆ  6
aˆ
2  5
bˆ
2



2

OA

2 
AC

2 In ∆OAC,
OC i.e.,
a
  b

2

a

2 
b

2. The usual name of this result is the Pythagorean Theorem. b.

13
  6  5 2 11 (2aˆ  5bˆ ) · (bˆ  3aˆ)
. 2 19. a

3iˆ  4jˆ  ˆk
(3, 4, 1) b
  2iˆ  3jˆ  6kˆ
(2, 3, 6) a
 · b

6  12  6

O

→

→

 →

A

a



a
, OB

b
 From ∆OAB, OA



 BA
c
a  b and ∠BOA
.

2

OA

2 
OB

2  2
OA


OB

cos  Now
BA


c

2

a

2 
b

2  2
a


b

cos . The result here is called the Cosine Law.

Let u
 and v
 be the sides of the rhombus.

Add: 2v

(5, 1, 7)

→

c=a –b

b

a rhombus.

u
  v

b

(2, 3, 6)

B →


0. Since a
 · b

0, a
 ⊥ b
 and the parallelogram will be

u
  v

a

(3, 4, 1)

A

→

a

21.

3x – y = 5

a, b y →

b

Subtract: 2u

(1, 7, 5) The angle between u
 and v
 is the same as the angle

x →

between 2u
 and 2v
. Therefore (2u
) · (2v
)

2u


2v

cos  5  7  35
(1  49  25) cos  23 cos 
 75  ≅ 108. The angles between the sides of the rhombus are 108 and 72. 53 Now
2u


 75
53,
u


 2 53 the lengths of the sides of the rhombus are . 2

60

Chapter 5: Algebraic Vectors and Applications

a →

c

In questions such as this a specific example can illustrate the desired result. Suppose a

(3, 1) and b

(3, 4) then a
 · b

5 Now, for c

(p, q), if a
 · c

a
 · b
 3p  q
5. There is an infinite number of possibilities for c
, one of which is b
.

However, others such as (1, 2) have c
 ≠ b
. In fact c
 is any vector having its end point on the line 3x  y
5.


  b
) · (a
 b
)
0, (a
  b
) ⊥ (a
  b
). 25. a. Since (a


 a  b
 and a
  b
 represent the diagonals of a parallelogram having sides
 a and b
 . Since the diagonals are perpendicular to each other, the a


b

. parallelogram is a rhombus with



22. Given vector a

4iˆ  3jˆ  ˆk


(4, 3, 1) A vector parallel to the xy-plane has the form u

(p, q, 0). Since a
 ⊥ u
, a
 · u

0 and 4p  3q
0. Choosing p
3 and q
4 gives vector u

(3, 4, 0) which is perpendicular to a
.

b. Since
u
  v



u
  v

,


u
  v

2

u
  v

2. But (u
  v
) · (u
  v
)

u
  v

2 and (u
  v
) · (u
  v
)

u
  v

2 therefore (u
  v
) · (u
  v
)
(u
  v
) · (u
  v
)

Now
u


 9  16
5 therefore a unit vector in the 3 4 xy-plane perpendicular to a
 is u

, , 0 . 5 5 23. Given that x
  y
  z
0 and
x


2,
y


3,
z

4.



u
 · u
  2u
 · v
  v
 · v

u
 · v
  2u
 · v
  v
 · v




4u
 · v

0 u
 · v

0. Therefore u
 ⊥ v
.

Now (x
  y
  z) · (x
  y
  z)
o
 · o

0.

u
 and v
 represent the sides of a rectangle whereas a
 and b
 were the sides of a rhombus.

Therefore x
 · x
  y
 · y
  z · z  2x
 · y
  2x
 · z  2y
 · z
0


x

2 
y

2 
z
2  2(x
 · y
  x
 · z  y
 · z)
0 4  9  16  2(x
 · y
  x · z  y
 · z)
0 29 and x
 · y
  x
 · z  y
 · z
. 2

C(0, 0, 1)

Q(0, 1, 1)

x

Equality holds when cos 
1; i.e.,
 a and b
 are collinear. If a

a1, a2 and b

b1, b2



then
a1b1  a2b2
≤  a12   a22

 b12   b22.

If a

a1, a2, a3 and b

b1, b2, b3

P(1, 1, 1) O


a
 · b

≤
a


b

.



z

24.


 · b
 

 26. Since a a

b

cos  and
cos 
≤ 1

B(0, 1, 0) y

A(1, 0, 0)



(1, 1, 1) and Two body diagonals of the cube are OP
 AQ
(1, 1, 1).

then
a1b1  a2b2  a3b3
≤  a12   a22   a32  b12   b22   b32. For a general solution to the Cauchy-Schwarz inequality refer to Exercise 12.2 question 18.

Exercise 5.4


 · AQ


OP


AQ

cos  where  is an angle Now OP between the body diagonals.

7. ˆi
(1, 0, 0), ˆj
(0, 1, 0), kˆ
(0, 0, 1) a. ˆi ˆj
(0, 0, 1)
kˆ . b. kˆ ˆj
(1, 0, 0)
iˆ.


 · AQ

1  1  1
1 OP

8. a.




3,
OP




3.
AQ

1 Therefore cos 
 and  ≅ 71 3 The body diagonals of a cube make angles of 71 and 119 to each other.

Let u

(u1, u2, u3) v

(v1, v2, v3) v

(v1, v2, v3) u
 v

(u2v3  v2u3, v1u3 u1v3, u1v2  v1u2)  v
 u

(v2u3  u2v3,  u1v3  v1u3,  v1u2  u1v2)
u
 v
.

Chapter 5: Algebraic Vectors and Applications

61



kv
. b. If u
 and v
 are collinear, u

Let v

(a, b, c) then u

(ka, kb, kc) u
 v

(kbc  bkc, akc  kac, kbc  bkc)
(0, 0, 0) u
 v

0.

12.

→

→

→

b

→

n=a b

→

a


 ·
 9. (a a)(b
 · b
)  (a · b
)2


 a
2
b

2  (

 a

b

cos )2


 a
2
b

2 (1  cos2 )


 a
2
b

2 sin2  therefore RS


a
2
b

2  sin2 


 a

b


sin 
. But 0 ≤  ≤ 180 therefore sin  ≥ 0 and RS

a


b

sin 

a
 b


 ·

 · b
)2. therefore

 a

b


 (a b · b)  (a a)( 10.
 a
(2, 1, 0), b

(1, 0, 3), c

(4, 1, 1)


 b
 · c

(3, 6, 1) · (4, 1, 1) a. a
12  6  1
19. b. b
 c
 ·
 a
(3, 13, 1) · (2, 1, 0)


19. c. c

 a · b

(1, 2, 6) · (1, 0, 3)


19.
 b
) c

(3, 6, 1) (4, 1, 1) d. (a
(5, 1, 21).
 c
)
 e. (b a
(3, 13, 1) (2, 1, 0)
(1, 2, 23). f. (c

 a) b

(1, 2, 6) (1, 0, 3)


(6, 3, 2).

(p, q, r) 11. Let u

v

(a, b, c) and w



(br  qc, pc  ar, aq  bp) v
 w
)
[b(aq  bp)  c(pc  ar), u
 (v
 w c(br  qc)  a(aq bp), a(pc  ar)  b(br  qc)]

(0, 0, 0) u
 v

(0, 0, 0); (u
 v
) w
) ≠ (u
 v
) w
. hence u
 (v
 w

62

Chapter 5: Algebraic Vectors and Applications

→

→

→

(a b) a

a
 b
 is a vector that is perpendicular to both a
 and b
. Let n

a
 b
. Now n
 a
 is a vector that is perpendicular to both n
 and a
. Therefore n
 a
, n
 and a
 are perpendicular to each other; i.e., (a
 b
) a
, a
 b
 and a
 are mutually perpendicular. 13. Let u

(u1, u2, u3), v

(v1, v2, v3), and



(w , w , w ). w 1 2 3



(v w  w v , w v  v w , Now v
 w 2 3 2 3 1 3 1 3 v1w2 w1v2)
)
u v w  u w v  u w v  u v w u
 · (v
 w 1 2 3 1 2 3 2 1 3 2 1 3  u3v1w2  u3w1v2. Also u
 v

(u2v3  v2u3, v1u3  u1v3, u1v2  v1u2)

w u v  w v u  w v u and (u
 v
) · w 1 2 3 1 2 3 2 1 3  w2u1v3  w3u1v2  w3v1u2
).
u · (v
 w

(3, 1, 2), b
(1, 14. a. We show this by choosing a

1, 1), and c

(p, q, r). Now a
 b

(3, 5, 2) and a
 c

(r 2q, 2p  3r, 3q  p). If a
 b

a
 c
 then r  2q
3 ➀ 2p  3r
5 ➁ 3q  p
2 ➂ Choose q
k, from ➀ r
2k  3, from ➂ p
3k  2. These values for r and k satisfy ➁. This shows that there are an infinite number of possibilities for c
. For one such value choose k
2. r
1, p
4, q
2, and c

(4, 2, 1)
 a b

a
 c

(3, 5, 2) and b ≠ c.

→

b.

Exercise 5.5

b →

→

a b

z

5. a.

→

c

A(1, 1, 1)

→

a

O



(2, 1, 5), v

(3, y, z), 15. a. Given a

(1, 3, 1), b and a
 v

b
. a
 v

(3z  y, 3  z, y  a)
(2, 1, 5) Equating components gives y
4, z
2, and 3z  y
6  4
2. Therefore v

(3, 4, 2).

b. Let v

(a, b, c).

Now a
 v

(3c  b, a  c, b  3a)
(2, 1, 5). Equating components gives 3c  b
2 ➀ a  c
1 ➁ b  3a
5. ➂ From ➀ and ➁ we have b
2  3c and a
1  c. b  3a
2  3c  3 (1  c)
5 which satisfies ➂. This shows that choosing any value for c in ➀, substituting to find b and a from ➀ and ➁ will satisfy ➂ hence giving another vector v
. Let c
2, from ➀ b
8 and from ➁ a
1. Therefore v

(1, 8, 2) is another vector so that a
 v

b
. c. We see from part b that c is any real number, hence

there will be an infinite number of vectors v
.

x

y

B(1, 0, 0)



(1, 1, 1); an edge is OB
 A body diagonal is OA
 is
ˆi
(1, 0, 0). The projection of ˆi onto OA
A O
A ˆi · O (1, 0, 0) · (1, 1, 1) (1, 1, 1)   · 
 · 


OA

OA
3 3 1
(1, 1, 1) 3 1 1 1
(, , ). 3 3 3 b. The projection of a body diagonal onto an edge is


 onto ˆi which is ˆi
(1, 0, 0). the projection of OA

6. a. a

(1, 2, 2)

b

(1, 3, 0) The area of the parallelogram is
a
 b

. a
 b

(6, 2, 5) therefore
a
 b


 36  4  25
65 

b. c

(6, 4, 12)
2(3, 2, 6)

d

(9, 6, 18)
3(3, 2, 6). 2 Since c

d
, c
 and d
 are collinear therefore no 3 parallelogram is formed, hence its “area” is zero.

7. a. A triangle with vertices A(7, 3, 4), B(1, 0, 6) and

C(4, 5, 2).

(6, 3, 2) Two sides are defined by AB

(3, 2, 6) and AC
 AC

(14, 42, 21) AB
7(2, 6, 3)
 AC


7
AB 4  36 9
49 1

 the area of ∆ABC

AB AC
2 49
. 2 Chapter 5: Algebraic Vectors and Applications

63

b. A triangle with vertices P(1, 0, 0), Q(0, 1, 0),

R(0, 0, 1).

(1, 1, 0) Now two sides are PQ

(1, 0, 1) and PR
 PR

(1, 1, 1),
PQ
 PR


3 PQ .




110 N,
d


300 m, 
6, W
110 12. Since
F 300 cos 6 ≅ 32819. The work done is 32819 J. →

13.

|d| = 3

3 The area of ∆PQR is . 2

78.4 cos 70º 70º

8. Given a parallelepiped defined by a

(2, 5, 1)

20º

b

(4, 0, 1) c

(3, 1, 1). Now b
 c

(1, 7, 4).

8 kg ≅ 78.4 N

W
78.4 3 cos 70 ≅ 80. The work done against gravity is approximately 80 J.

The volume of the parallelepiped is
a
 · (b
 c
)


(2, 5, 1) · (1, 7, 4)


2  35  4


14.

F


29.

20º 12º


 · d


F


d

cos  9. Work W
F a.

W
220 15 cos 49 ≅ 2165 J.

b.

W
4.3 2.6 cos 85 ≅ 1.0 J.

c.

W
14 6 cos 110 ≅ 29 J.

d.


F


4000 kN
4 106 N


d


5 km
5 103 m


d

cos 90 W

F
0 J.
 must have 10. To overcome friction, the applied force F magnitude greater than 150 N. 
0, cos 
1.


d

cos  Therefore W >
F W > 150 1.5 1 W > 225. The work done is greater than 225 J.


30 9.8
294 N. 11.
F


d

cos  W

F
294 40 cos 52° ≅ 7240. The work done is 7240 J.

64

F 20º

Chapter 5: Algebraic Vectors and Applications

Consider the “same” force as a force acting at 20 to the direction of motion. The work done dragging the trunk up the ramp is 90(10) cos 20. The work done dragging the trunk horizontally is 90(15) cos 20. Total work done is 900 cos 20  1350 cos 20 ≅ 2114 J.

2iˆ 15. a. F


(2, 0)
 · d
 W
F W
10.

d

5iˆ  6jˆ
(5, 6)



4iˆ  ˆj b. F

d

3iˆ  10jˆ
(3, 10)



(800, 600) c. F

d

(20, 50)



12iˆ  5jˆ  6k ˆ d. F

ˆ d

2iˆ  8jˆ  4k
(2, 8, 4)


(4, 1)
 · d
 W
F
12  12 W
22.


 · d
 W
F
16000  30000 W
46000.


(12, 5, 6)
 · d
 W
F
24  40  24 W
88.

16. A 10 N force acts in the direction of a vector (1, 1).


 is A unit vector along F

10 F

2 , 2  therefore 1

1

2 , 2 
(52, 52) 1

1



(7, 5). the displacement vector is d

PQ

 W
F · d
352  252
602. The work done is 602 N.


F


50 N


r

20 cm


0.2 m


30


T


r F

T



r F



sin 

r

F
0.2 50 sin 30


T


5 The torque on the bolt is 5 N.

17.





r

F

sin , maximum torque can be b. Since
T

→

a

achieved when sin  is a maximum. This maximum


10 is 1 when 
90. Therefore
T



r

F

A(2, 1, 5)

B(3, –1, 2)

The 30 N force acts along a

(2, 1, 5)


a


 41  25
30  â


10 J.
 onto v
)
Proj(v
 onto u
) is a true statement 19. a. Proj(u when

2  , 1, 5 . 30 30   30 



and the maximum torque that can be achieved is



i) u

v
 or
. ii) u
 ⊥ v
, in which case the projection vector is O



30â The force vector F

(230 F , 30 , 530 )

(1, 2 3) The displacement d

AB
 · d
 Therefore W
F
230  230   1530 
1930 .


 onto v
)


Proj(v
 onto u
)
is a true b.
Proj(u statement when i) u
 ⊥ v
 in which case the projection has magnitude 0. ii) when
u



v

and the angle between u
 and v
 is 45 or 135 iii) when u

v
 or u

v
.

The work done in moving the object from A to B is 1930  J. 18. a. →

F

30º

Bolt

r→

Chapter 5: Algebraic Vectors and Applications

65

10. u
 has direction angles α1, β1, γ1. A unit vector along u


Review Exercise

is uˆ
(cos α1, cos β1, cos γ1). Similarly a unit vector along v
 is vˆ
(cos α2, cos β2, cos γ2). Since u
 ⊥ v
, uˆ ⊥ ˆv and u · v
0 therefore cos α1 cos α2  cos β1 cos β2  cos γ1 γ2
0.

ˆ, 7. Given a

6iˆ  3jˆ  2k a

(6, 3, 2) b

2iˆ  pjˆ  4kˆ, b

(2, p, 4) 4 and cos 
,  is the angle between a
 and b
. 21

11.


x
  y

2
(x
  y
) · (x
  y
)
x
 · y
  2 x
 · y
  y
 · y


x

2  2 x
 · y
 
y

2

Now a
 · b


a


b

cos .

2x
 · y


x
  y

2 
x

2 
y

2

Therefore 12  3p  8
4  36  9  4  4  p2  16  21



(2, 2, 0). BC
 · BC

0, AB
 ⊥ BC
, and ∠ABC
90 Since AB therefore ∆ABC is a right-angled triangle.

➀

b. Since ∆ABC has a right angle at B, the area of

1

 ∆ABC

AB

BC
. 2

44 p
4 or p
 . 65 44 We see that p
  does not satisfy ➀ and 65 p
4 does; therefore the only value for p is 4. Let a

ˆi  ˆj  ˆk
(1, 1, 1) b

λ2ˆi  2λjˆ  ˆk
(λ2, 2λ, 1) Since a
 ⊥ b
, a
 · b

0 therefore λ2  2λ  1
0 (λ  1)2
0 λ
1 If a
 ⊥ b
 then λ
1.

9. If
x


3,
y


4 and the angle between x
 and y
 is

60 then (4x
  y
) · (2x
  3y
)
8
x

2  10 x
 · y
  3
y

2
72  10
x


y

cos 60  48 1
24  10(3)(4)  2 (4x
  y
) · (2x
  3y
)
84.



66

12. Given ∆ABC with vertices A(1, 3, 4), B(3, 1, 1),



(4, 4, 3), AC

(6, 2, 3), and a. AB

Squaring both sides: 81p2  216p  144
16p2  320 65p2  216p  176
0 (p  4)(65p  44)
0

8.



and C(5, 1, 1).

4 3p  4
 · 7  p2  20 21 9p  12
4  p2  20



1 therefore x
 · y


x
  x

2 
x

2 
y

2 . 2

Chapter 5: Algebraic Vectors and Applications




 Now
AB 16  1 69
41 



BC 44
22 1 ∆ABC
 41  22 2
 82 the area of ∆ABC is 82 .


c.
AC

 36  4 9


7 The perimeter of ∆ABC is AB  AC  BC
 41  7  22 ≅ 16.2. d. Let the fourth vertex to complete the rectangle be

D(a, b, c)

BA
 and (a  5, b  1, c  1)
(4, 4, 3) CD equating components, a
1, b
5, c
4, and the coordinates of the fourth vertex are (1, 5, 4).



(17, 3, 8). 13. Given the vector u a. The projection of u
 onto each of the coordinate

area will be 17iˆ, 3jˆ and 8kˆ. b. The projection of u

(17, 3, 8) onto the xy

plane is (17, 3, 0), onto the xz plane is (17, 0, 8), and onto the yz plane is (0, 3, 8).

14. Since the vertices lie in the xy plane, the coordinates

in R3 will be A(7, 3, 0), B(3, 1, 0), and C(2, 6, 0).

(10, 2, 0) Now AB

(9, 9, 0) AC
 AC

(0, 0, 72). and AB

D will be the foot of the perpendicular from the fourth vertex E to D. Let the coordinates of E be



63 , 12, a.


1 Now
OE 1 3 and     a 2
1 4 36 24 2 a2

 36 3

2 a
 3 the fourth vertex has coordinates

1

 Area of ∆ABC

AB AC
2 1
 72 2

or









63 , 12, 36 

63 , 12, 36 .

The coordinates of the fourth vertices are


36.

O(0, 0, 0), A(0, 1, 0), B

The area of ∆ABC is 36. 15. a. Consider the base of the tetrahedron as a triangle in

the xy-plane with O(0, 0, 0), A(0, 1, 0). A(0, 1, 0)

O(0, 0, 0)

y





63 12, 0 and



63 , 12, 36 .

E

b. The x-component of the centroid will be

3 3 3 1  0  0    
 4 2 6 6



D







1 1 1 1 the y-component is  1    
 4 2 2 2 B( 32 , 12 ,0)

C

6 6 1 the z-component is   
 . 4 3 12





The coordinates of the centroid are

x

Now ∆OBC is a 30, 60, 90 triangle with OB
1 3 1 therefore OC
 and CB
 2 2

3 1 hence the coordinates of B are ,, 0 .





6

2

63 ,12, 0.

The centroid of ∆ABC is D







63 , 12,  126 .

G

c. The distance from each vertex to the centroid will



. be the same, say
OG



OG



  

3  1  6 4 36 144 1 1    1 4 12 24 3  8

Chapter 5: Algebraic Vectors and Applications

67

6
. 4 6 The centroid is  units from each vertex. 4
 is a vector that is perpendicular to all 16. a. a
 b

vectors in the plane of a
 and b
. Let n

a
 b
. Now n
 c
 is a vector perpendicular to both n
 and c
.

Since n
 c
 is perpendicular to n
, as are a
 and b
, n
 c

(a
 b
) c
, a
 and b
 will be coplanar; i.e., (a
 b
) c
 lies in the plane of a
 and b
.

(b , b , b ), and b. Let a

(a1, a2, a3), b 1 2 3 c

(c1, c2, c3).


 AC
 is a vector perpendicular to the plane of Since AB ∆ABC, the height of the tetrahedron will be the magnitude
 on (AB
 AC
) of the projection of AD
 · (A

AD B A
 C )
therefore h
 .


A
B AC
 · (A

AD 1 1

 B A
 C )
The volume V
 
AB AC
 

  3 2
AB AC
1


 V

AD · (AB AC )
. 6
 AC

(19, 26, 42), AD

(2, 4, 6) Now AB where D is the fourth vertex, D (1, 5, 8)
 · (AB
 AC
)
38  104  252 AD
394 1 V
 294 6

Now a
 b

(a2b3  a3b2, b1a3  a1b3, a1b2  a2 b1) and LS
(a
 b
) c

(c3b1a3  c3a1b3  c2a1b2  c2a2b1, c1a1b2  c1b1a2  c3a2b3  c3b2a3, c2a2b3  c2a3b2  c1b1a3  c1a1b3) a
 · c

a1c1  a2c2  a3c3 (a
 · c
)b
[(a1c1  a2c2  a3c3)b1, (a1c1  a2c2  a3c3)b2, (a1c1  a2c2  a3c3)b3] and (b
 · c
)a

[(b1c1  b2c2  b3c3)a1, (b1c1  b2c2  b3c3)a2, (b1c1  b2c2  b3c3)a3] RS
(a
 · c
)b
  (b
 · c
)a

(a2c2b1  a3c3b1  b2c2a1  b3c3a1, a1c1b2  a3c3b2  b1c1a2  b3c3a2, a1c1b3  a2c2b3  b1c1a3  b2c2a3). Since LS
RS, (a
 b
) c

(a
 · c
)b
  (b
 · c
)a
. 17. The volume of a tetrahedron is given by the formula

1 1 v
(area of the base)(height)
 Ah. 3 3 Consider the base to be the triangle with vertices A(1, 1, 2), B(3, 4, 6), C(7, 0, 1)

(2, 5, 4) and AC

(8, 1, 3) now AB 1



AC
. the area of the base will be A

AB 2 68

Chapter 5: Algebraic Vectors and Applications

197
 3 197 The volume of the tetrahedron is . 3

Chapter 5 Test 1. a. If u
 · v

0 then u
 is perpendicular to v
. b. If u
 · v


u


v

then cos 
1, 
0 and u
 and v


will have the same direction, i.e., u

kv
, k > 0.

c. If u
 v


 0 then u
 and v
 are collinear,

i.e., u

kv
, k > R.

d. If
u
 v



u


v

then sin 
1, 
90 and

u
 ⊥ v
.

e. If (u
 v
) · u

0, no conclusion can be made about u


and v
 since u
 v
 is perpendicular to both u
 and v
 and the dot product of perpendicular vectors is zero.

f. If (u
 v
) u


 0 then u
 v
 and u
 are collinear. But

u
 v
 is perpendicular to both u
 and v
. This is true only if u
 v

0 in which case u
 and v


are collinear.

ˆ
(6, 3, 2) 2. Given u

6iˆ  3jˆ  2k

4. a.

v

3iˆ  4jˆ  ˆk
(3, 4, 1)

C B

a. 4u
  3 v

(24, 12, 8)  (9, 12, 3)
(33, 0, 5) ˆ.
33iˆ  5k b.

c.

d.

u
 · v

(6, 3, 2) · (3, 4, 1)
18  12  2
4.

D A

ABCD is a parallelogram with coordinates A(1, 2, 1), B(2, 1, 3), C(p, q, r), D(3, 1, 3).

DC
 Now AB therefore (3, 3, 4)
(p  3, q  1, r  3) and p
0, q
2, r
1 and the coordinates of C are (0, 2, 1).

u
 v

(5, 12, 33)
5iˆ  12jˆ  33kˆ.


u
 v


 25  1 44  1089 
1258 .

A unit vector perpendicular to both u
 and v
 is

b. To determine the angle at A we use the dot product

5 12  ,  , 33 .  1258  1258  1258





3. a.


 · AD


AB


AD

cos A AB
 AB
(3, 3, 4)



AB 99  16

z A


34 

P(3, –2, 5)



(2, 1, 2) AD



AD 41 4 O

B

y

3

x


3
 · AD

6  3  8 AB
11 11 cos A
 334 


 is the position vector of point P(3, 2, 5) i) OP


 onto the z-axis is ii) the projection of OP



(0, 0, 5) OA
 onto the xy plane is iii) the projection of OP

(3, 2, 0). OB


5 b.
OA



OB 9  4
13 .

A ≅ 129 The angle at A is approximately 129.
 AD

c. The area of parallelogram ABCD

AB
 AD

(10, 2, 9) AB
 AD



AB 100  4  81
 185 The area of parallelogram ABCD is 185. 

Chapter 5: Algebraic Vectors and Applications

69

→

5.

c.

F

→

F

30º 35º →

d

→

r


 acting at a direction of 35 to the horizontal A force F


75 N moves a box a distance has magnitude
F
d


16 m. The work done is


d

cos  W

F

1 Since sin 30
 a force applied at an angle of 2 30 will produce half the maximum torque


T



r

F

sin 30


75 16 cos 35.

1
0.18 50  2 →


T


4.5 J.

F 15º

→

20º

d

B

C

→

20º



b

The same force acting at 35 to the horizontal has a


75 N and acts at an angle of 15 magnitude of
F

to the line of motion where
d


8 m. The work done is W
75 8 cos 15. Total work done is 75 16 cos 35  75 8 cos 15 ≅ 1562.5 J. 6.

7.

→

F

→

T

A

→

D

a



a
  b
 Diagonal AC

a
  b
. and BD
 and BD
 be . Let the angle between AC
 BD


AC


BD

cos  Now AC
 · BD

(a
  b
) · (a
  b
) AC

a

2 
b

2






AC a

2 
 b

2




BD a

2 
 b

2

→

r

a. The force should act at right angles to the wrench

to produce maximum torque.

r F
 b. T


T



r F



r

F

sin .

If 
90, maximum torque is
T


0.18 50
9 J.
 is perpendicular to the plane The direction of T
 so that r, F
, and r F
 form a rightof r and F handed system.

70

Chapter 5: Algebraic Vectors and Applications

therefore
a

2 
b

2




2 cos  
 a

2 
 b

2 
a

2 
b


a

2 
b

2
(
a

2 
b

2) cos 
a

2 
b

2  and cos 

 2  .
a

b

2
a

2 
b

2  for
a

2 >
b

2. For 0 <  < 90, cos 

a

2 
b

2