Solutions Manual for Linear Algebra A Modern Introduction 4th Edition by David Poole

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Solutions Manual for Linear Algebra A Modern Introduction 4th Edition by David Poole

1. Complete Solutions Manual

Prepared by

Roger Lipsett





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Linear Algebra
A Modern Introduction

FOURTH EDITION




David Poole
Trent University












Solutions Manual for Linear Algebra A Modern Introduction 4th Edition by David Poole
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3. Contents
1 Vectors 3
1.1 The Geometry and Algebra of Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Length and Angle: The Dot Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exploration: Vectors and Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 Lines and Planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exploration: The Cross Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.4 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Chapter Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2 Systems of Linear Equations 53
2.1 Introduction to Systems of Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Direct Methods for Solving Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exploration: Lies My Computer Told Me . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exploration: Partial Pivoting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exploration: An Introduction to the Analysis of Algorithms . . . . . . . . . . . . . . . . . . . . . .
2.3 Spanning Sets and Linear Independence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5 Iterative Methods for Solving Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . .
Chapter Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3 Matrices 129
3.1 Matrix Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 Matrix Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3 The Inverse of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.4 The LU Factorization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.5 Subspaces, Basis, Dimension, and Rank . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.6 Introduction to Linear Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.7 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Chapter Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4 Eigenvalues and Eigenvectors 235
4.1 Introduction to Eigenvalues and Eigenvectors . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2 Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exploration: Geometric Applications of Determinants . . . . . . . . . . . . . . . . . . . . . . . . .
4.3 Eigenvalues and Eigenvectors ofnnMatrices . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4 Similarity and Diagonalization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.5 Iterative Methods for Computing Eigenvalues . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.6 Applications and the Perron-Frobenius Theorem . . . . . . . . . . . . . . . . . . . . . . . . .
Chapter Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1

4. 2 CONTENTS
5 Orthogonality 371
5.1 Orthogonality inR
n
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2 Orthogonal Complements and Orthogonal Projections . . . . . . . . . . . . . . . . . . . . . .
5.3 The Gram-Schmidt Process and theQRFactorization . . . . . . . . . . . . . . . . . . . . . .
Exploration: The ModiedQRProcess . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exploration: Approximating Eigenvalues with theQRAlgorithm . . . . . . . . . . . . . . . . . . .
5.4 Orthogonal Diagonalization of Symmetric Matrices . . . . . . . . . . . . . . . . . . . . . . . .
5.5 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Chapter Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6 Vector Spaces 451
6.1 Vector Spaces and Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.2 Linear Independence, Basis, and Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exploration: Magic Squares . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.3 Change of Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.4 Linear Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.5 The Kernel and Range of a Linear Transformation . . . . . . . . . . . . . . . . . . . . . . . .
6.6 The Matrix of a Linear Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exploration: Tiles, Lattices, and the Crystallographic Restriction . . . . . . . . . . . . . . . . . . .
6.7 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Chapter Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7 Distance and Approximation 537
7.1 Inner Product Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exploration: Vectors and Matrices with Complex Entries . . . . . . . . . . . . . . . . . . . . . . .
Exploration: Geometric Inequalities and Optimization Problems . . . . . . . . . . . . . . . . . . .
7.2 Norms and Distance Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.3 Least Squares Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.4 The Singular Value Decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.5 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Chapter Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8 Codes 633
8.1 Code Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.2 Error-Correcting Codes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.3 Dual Codes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.4 Linear Codes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.5 The Minimum Distance of a Code . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

5. Chapter 1
Vectors
1.1 The Geometry and Algebra of Vectors
1.H3, 0L
H2, 3L
H3,-2L
H-2, 3L
-2 -1 1 2 3
-2
-1
1
2
3
2.Since

2
3

+

3
0

=

5
3

;

2
3

+

2
3

=

4
0

;

2
3

+

2
3

=

0
0

;

2
3

+

3
2

=

5
5

;
plotting those vectors givesa
bc
d
1 2 3 4 5
-5
-4
-3
-2
-1
3

6. 4 CHAPTER 1. VECTORS
3.a
b
c
d
x
y
z
-1
0
1
2
3
-2 -1 0 1
2
-2
-1
0
1
2
4.Since the heads are all at (3;2;1), the tails are at
2
4
3
2
1
3
5
2
4
0
2
0
3
5=
2
4
3
0
1
3
5;
2
4
3
2
1
3
5
2
4
3
2
1
3
5=
2
4
0
0
0
3
5;
2
4
3
2
1
3
5
2
4
1
2
1
3
5=
2
4
2
4
0
3
5;
2
4
3
2
1
3
5
2
4
1
1
2
3
5=
2
4
4
3
3
3
5:
5.The four vectors
#
ABarea
b
c
d
1 2 3 4
-2
-1
1
2
3
In standard position, the vectors are
(a)
#
AB= [41;2(1)] = [3;3].
(b)
#
AB= [20;1(2)] = [2;1]
(c)
#
AB=

1
2
2;3
3
2

=


3
2
;
3
2

(d)
#
AB=

1
6

1
3
;
1
2

1
3

=


1
6
;
1
6

.a
b
c
d
-1 1 2 3
1
2
3

7. 1.1. THE GEOMETRY AND ALGEBRA OF VECTORS 5
6.Recall the notation that [a; b] denotes a move ofaunits horizontally andbunits vertically. Then during
the rst part of the walk, the hiker walks 4 km north, soa= [0;4]. During the second part of the
walk, the hiker walks a distance of 5 km northeast. From the components, we get
b= [5 cos 45

;5 sin 45

] =
"
5
p
2
2
;
5
p
2
2
#
:
Thus the net displacement vector is
c=a+b=
"
5
p
2
2
;4 +
5
p
2
2
#
:
7. +b=

3
0

+

2
3

=

3 + 2
0 + 3

=

5
3

.a
b
a+b
1 2 3 4 5
1
2
3
8. c=

2
3



2
3

=

2(2)
33

=

4
0

.b -c
b-c
1 2 3 4
1
2
3 9. c=

3
2



2
3

=

5
5

.d
-cd-c
1 2 3 4 5
-5
-4
-3
-2
-1 10. +d=

3
0

+

3
2

=

3 + 3
0 + (2)

=

6
2

.a
d
a+d
1 2 3 4 5 6
-2
-1
11.2a+ 3c= 2[0;2;0] + 3[1;2;1] = [20;22;20] + [31;3(2);31] = [3;2;3].
12.
3b2c+d= 3[3;2;1]2[1;2;1] + [1;1;2]
= [33;32;31] + [21;2(2);21] + [1;1;2]
= [6;9;1]:

8. 6 CHAPTER 1. VECTORS
13. = [cos 60

;sin 60

] =
h
1
2
;
p
3
2
i
, andv= [cos 210

;sin 210

] =
h

p
3
2
;
1
2
i
, so that
u+v=
"
1
2

p
3
2
;
p
3
2

1
2
#
;uv=
"
1
2
+
p
3
2
;
p
3
2
+
1
2
#
:
14.
#
AB=ba.
(b)Since
#
OC=
#
AB, we have
#
BC=
#
OCb= (ba)b=a.
(c)
#
AD=2a.
(d)
#
CF=2
#
OC=2
#
AB=2(ba) = 2(ab).
(e)
#
AC=
#
AB+
#
BC= (ba) + (a) =b2a.
(f)Note that
#
F Aand
#
OBare equal, and that
#
DE=
#
AB. Then
#
BC+
#
DE+
#
F A=a
#
AB+
#
OB=a(ba) +b=0:
15.2(a3b) + 3(2b+a)
property e.
distributivity
= (2a6b) + (6b+ 3a)
property b.
associativity
= (2a+ 3a) + (6b+ 6b) = 5a.
16.
3(ac) + 2(a+ 2b) + 3(cb)
property e.
distributivity
= (3a+ 3c) + (2a+ 4b) + (3c3b)
property b.
associativity
= (3a+ 2a) + (4b3b) + (3c+ 3c)
=a+b+ 6c:
17. a= 2(x2a) = 2x4a)x2x=a4a) x=3a)x= 3a.
18.
x+ 2ab= 3(x+a)2(2ab) = 3x+ 3a4a+ 2b)
x3x=a2a+ 2b+b)
2x=3a+ 3b)
x=
3
2
a
3
2
b:
19.We have 2u+ 3v= 2[1;1] + 3[1;1] = [21 + 31;2(1) + 31] = [5;1]. Plots of all three vectors areu
v
w
u
v
v
v
-1 1 2 3 4 5 6
-2
-1
1
2

9. 1.1. THE GEOMETRY AND ALGEBRA OF VECTORS 7
20.We haveu2v=[2;1]2[2;2] = [(2)22;12(2)] = [2;3]. Plots of all three
vectors areu
v
w
-v
-v
-u
-u
-2 -1 1 2
-2
-1
1
2
3
21.From the diagram, we see thatw=
2u+ 4v.u
v
w
-u
-u
v
v
v
-1 1 2 3 4 5 6
-1
1
2
3
4
5
6
22.From the diagram, we see thatw= 2u+
3v.u
u
u
v
v
v
w
-2 -1 1 2 3 4 5 6 7
1
2
3
4
5
6
7
8
9
23.Property (d) states thatu+ (u) =0. The rst diagram below showsualong withu. Then, as the
diagonal of the parallelogram, the resultant vector is0.
Property (e) states thatc(u+v) =cu+cv. The second gure illustrates this.

10. 8 CHAPTER 1. VECTORSu
-u u
v
u
v
u+v
cu
cu
cv
cv
cHu+vL
24.Letu= [u1; u2; : : : ; un] andv= [v1; v2; : : : ; vn], and letcanddbe scalars inR.
Property (d):
u+ (u) = [u1; u2; : : : ; un] + (1[u1; u2; : : : ; un])
= [u1; u2; : : : ; un] + [u1;u2; : : : ;un]
= [u1u1; u2u2; : : : ; unun]
= [0;0; : : : ;0] =0:
Property (e):
c(u+v) =c([u1; u2; : : : ; un] + [v1; v2; : : : ; vn])
=c([u1+v1; u2+v2; : : : ; un+vn])
= [c(u1+v1); c(u2+v2); : : : ; c(un+vn)]
= [cu1+cv1; cu2+cv2; : : : ; cun+cvn]
= [cu1; cu2; : : : ; cun] + [cv1; cv2; : : : ; cvn]
=c[u1; u2; : : : ; un] +c[v1; v2; : : : ; vn]
=cu+cv:
Property (f):
(c+d)u= (c+d)[u1; u2; : : : ; un]
= [(c+d)u1;(c+d)u2; : : : ;(c+d)un]
= [cu1+du1; cu2+du2; : : : ; cun+dun]
= [cu1; cu2; : : : ; cun] + [du1; du2; : : : ; dun]
=c[u1; u2; : : : ; un] +d[u1; u2; : : : ; un]
=cu+du:
Property (g):
c(du) =c(d[u1; u2; : : : ; un])
=c[du1; du2; : : : ; dun]
= [cdu1; cdu2; : : : ; cdun]
= [(cd)u1;(cd)u2; : : : ;(cd)un]
= (cd)[u1; u2; : : : ; un]
= (cd)u:
25. +v= [0;1] + [1;1] = [1;0].
26. +v= [1;1;0] + [1;1;1] = [0;0;1].

11. 1.1. THE GEOMETRY AND ALGEBRA OF VECTORS 9
27. +v= [1;0;1;1] + [1;1;1;1] = [0;1;0;0].
28. +v= [1;1;0;1;0] + [0;1;1;1;0] = [1;0;1;0;0].
29.
+0 1 2 3
00 1 2 3
11 2 3 0
22 3 0 1
33 0 1 2
0 1 2 3
00 0 0 0
10 1 2 3
20 2 0 2
30 3 2 1
30.
+0 1 2 3 4
00 1 2 3 4
11 2 3 4 0
22 3 4 0 1
33 4 0 1 2
44 0 1 2 3
0 1 2 3 4
00 0 0 0 0
10 1 2 3 4
20 2 4 1 3
30 3 1 4 2
40 4 3 2 1
31.2 + 2 + 2 = 6 = 0 inZ3.
32.222 = 32 = 0 inZ3.
33.2(2 + 1 + 2) = 22 = 31 + 1 = 1 inZ3.
34.3 + 1 + 2 + 3 = 42 + 1 = 1 inZ4.
35.232 = 43 + 0 = 0 inZ4.
36.3(3 + 3 + 2) = 46 + 0 = 0 inZ4.
37.2 + 1 + 2 + 2 + 1 = 2 inZ3, 2 + 1 + 2 + 2 + 1 = 0 inZ4, 2 + 1 + 2 + 2 + 1 = 3 inZ5.
38.(3 + 4)(3 + 2 + 4 + 2) = 21 = 2 inZ5.
39.8(6 + 4 + 3) = 84 = 5 inZ9.
40.2
100
=

2
10

10
= (1024)
10
= 1
10
= 1 inZ11.
41.[2;1;2] + [2;0;1] = [1;1;0] inZ
3
3.
42.2[2;2;1] = [22;22;21] = [1;1;2] inZ
3
3.
43.2([3;1;1;2] + [3;3;2;1]) = 2[2;0;3;3] = [22;20;23;23] = [0;0;2;2] inZ
4
4.
2([3;1;1;2] + [3;3;2;1]) = 2[1;4;3;3] = [21;24;23;23] = [2;3;1;1] inZ
4
5.
44.x= 2 + (3) = 2 + 2 = 4 inZ5.
45.x= 1 + (5) = 1 + 1 = 2 inZ6
46.x= 2
1
= 2 inZ3.
47.No solution. 2 times anything is always even, so cannot leave a remainder of 1 when divided by 4.
48.x= 2
1
= 3 inZ5.
49.x= 3
1
4 = 24 = 3 inZ5.
50.No solution. 3 times anything is always a multiple of 3, so it cannot leave a remainder of 4 when
divided by 6 (which is also a multiple of 3).
51.No solution. 6 times anything is always even, so it cannot leave an odd number as a remainder when
divided by 8.

12. 10 CHAPTER 1. VECTORS
52.x= 8
1
9 = 79 = 8 inZ11
53.x= 2
1
(2 + (3)) = 3(2 + 2) = 2 inZ5.
54.No solution. This equation is the same as 4x= 25 =3 = 3 inZ6. But 4 times anything is even,
so it cannot leave a remainder of 3 when divided by 6 (which is also even).
55.Add 5 to both sides to get 6x= 6, so thatx= 1 orx= 5 (since 61 = 6 and 65 = 30 = 6 inZ8).
56.(a) All values. (b) All values. (c) All values.
57. Alla6= 0 inZ5have a solution because 5 is a prime number.
(b)a= 1 anda= 5 because they have no common factors with 6 other than 1.
(c)aandmcan have no common factors other than 1; that is, thegreatest common divisor, gcd, of
aandmis 1.
1.2 Length and Angle: The Dot Product
1.Following Example 1.15,uv=

1
2



3
1

= (1)3 + 21 =3 + 2 =1.
2.Following Example 1.15,uv=

3
2



4
6

= 34 + (2)6 = 1212 = 0.
3. v=
2
4
1
2
3
3
5
2
4
2
3
1
3
5= 12 + 23 + 31 = 2 + 6 + 3 = 11.
4. v= 3:21:5 + (0:6)4:1 + (1:4)(0:2) = 4:82:46 + 0:28 = 2:62.
5. v=
2
6
6
4
1
p
2
p
3
0
3
7
7
5

2
6
6
4
4

p
2
0
5
3
7
7
5
= 14 +
p
2(
p
2) +
p
30 + 0(5) = 42 = 2.
6. v=
2
6
6
4
1:12
3:25
2:07
1:83
3
7
7
5

2
6
6
4
2:29
1:72
4:33
1:54
3
7
7
5
=1:122:293:251:72 + 2:074:331:83(1:54) = 3:6265.
7.Finding a unit vectorvin the same direction as a given vectoruis callednormalizingthe vectoru.
Proceed as in Example 1.19:
kuk=
p
(1)
2
+ 2
2
=
p
5;
so a unit vectorvin the same direction asuis
v=
1
kuk
u=
1
p
5
"
1
2
#
=
"

1
p
5
2
p
5
#
:
8.Proceed as in Example 1.19:
kuk=
p
3
2
+ (2)
2
=
p
9 + 4 =
p
13;
so a unit vectorvin the direction ofuis
v=
1
kuk
u=
1
p
13
"
3
2
#
=
"
3
p
13

2
p
13
#
:

13. 1.2. LENGTH AND ANGLE: THE DOT PRODUCT 11
9.Proceed as in Example 1.19:
kuk=
p
1
2
+ 2
2
+ 3
2
=
p
14;
so a unit vectorvin the direction ofuis
v=
1
kuk
u=
1
p
14
2
6
6
4
1
2
3
3
7
7
5
=
2
6
6
4
1
p
14
2
p
14
3
p
14
3
7
7
5
:
10.Proceed as in Example 1.19:
kuk=
p
3:2
2
+ (0:6)
2
+ (1:4)
2
=
p
10:24 + 0:36 + 1:96 =
p
12:563:544;
so a unit vectorvin the direction ofuis
v=
1
kuk
u=
1
3:544
2
4
1:5
0:4
2:1
3
5
2
4
0:903
0:169
0:395
3
5:
11.Proceed as in Example 1.19:
kuk=
r
1
2
+
p
2

2
+
p
3

2
+ 0
2
=
p
6;
so a unit vectorvin the direction ofuis
v=
1
kuk
u=
1
p
6
2
6
6
6
6
4
1
p
2
p
3
0
3
7
7
7
7
5
=
2
6
6
6
6
4
1
p
6
p
2
p
6
p
3
p
6
0
3
7
7
7
7
5
=
2
6
6
6
6
4
1
p
6
1
p
3
1
p
2
0
3
7
7
7
7
5
=
2
6
6
6
6
4
p
6
6
p
3
3
p
2
2
0
3
7
7
7
7
5
12.Proceed as in Example 1.19:
kuk=
p
1:12
2
+ (3:25)
2
+ 2:07
2
+ (1:83)
2
=
p
1:2544 + 10:5625 + 4:2849 + 3:3489
=
p
19:45074:410;
so a unit vectorvin the direction ofuis
v
1
kuk
u=
1
4:410

1:123:25 2:071:83



0:2540:737 0:4690:415

:
13.Following Example 1.20, we compute:uv=

1
2



3
1

=

4
1

, so
d(u;v) =kuvk=
q
(4)
2
+ 1
2
=
p
17:
14.Following Example 1.20, we compute:uv=

3
2



4
6

=

1
8

, so
d(u;v) =kuvk=
q
(1)
2
+ (8)
2
=
p
65:
15.Following Example 1.20, we compute:uv=
2
4
1
2
3
3
5
2
4
2
3
1
3
5=
2
4
1
1
2
3
5, so
d(u;v) =kuvk=
q
(1)
2
+ (1)
2
+ 2
2
=
p
6:

14. 12 CHAPTER 1. VECTORS
16.Following Example 1.20, we compute:uv=
2
4
3:2
0:6
1:4
3
5
2
4
1:5
4:1
0:2
3
5=
2
4
1:7
4:7
1:2
3
5, so
d(u;v) =kuvk=
p
1:7
2
+ (4:7)
2
+ (1:2)
2
=
p
26:425:14:
17. vis a real number, sokuvkis the norm of a number, which is not dened.
(b) vis a scalar, whilewis a vector. Thusuv+wadds a scalar to a vector, which is not a
dened operation.
(c) is a vector, whilevwis a scalar. Thusu(vw) is the dot product of a vector and a scalar,
which is not dened.
(d)c(u+v) is the dot product of a scalar and a vector, which is not dened.
18.Letbe the angle betweenuandv. Then
cos=
uv
kuk kvk
=
3(1) + 01
p
3
2
+ 0
2
p
(1)
2
+ 1
2
=
3
3
p
2
=
p
2
2
:
Thus cos <0 (in fact,=
3
4
), so the angle betweenuandvis obtuse.
19.Letbe the angle betweenuandv. Then
cos=
uv
kuk kvk
=
21 + (1)(2) + 1(1)
p
2
2
+ (1)
2
+ 1
2
p
1
2
+ (2)
2
+ 1
2
=
1
2
:
Thus cos >0 (in fact,=

3
), so the angle betweenuandvis acute.
20.Letbe the angle betweenuandv. Then
cos=
uv
kuk kvk
=
41 + 3(1) + (1)1
p
4
2
+ 3
2
+ (1)
2
p
1
2
+ (1)
2
+ 1
2
=
0
p
26
p
3
= 0:
Thus the angle betweenuandvis a right angle.
21.Letbe the angle betweenuandv. Note that we can determine whetheris acute, right, or obtuse
by examining the sign of
uv
kukkvk
, which is determined by the sign ofuv. Since
uv= 0:9(4:5) + 2:12:6 + 1:2(0:8) = 0:45>0;
we have cos >0 so thatis acute.
22.Letbe the angle betweenuandv. Note that we can determine whetheris acute, right, or obtuse
by examining the sign of
uv
kukkvk
, which is determined by the sign ofuv. Since
uv= 1(3) + 21 + 32 + 4(2) =3;
we have cos <0 so thatis obtuse.
23.Since the components of bothuandvare positive, it is clear thatuv>0, so the angle between
them is acute since it has a positive cosine.
24.From Exercise 18, cos=
p
2
2
, so that= cos
1


p
2
2

=
3
4
= 135

.
25.From Exercise 19, cos=
1
2
, so that= cos
11
2
=

3
= 60

.
26.From Exercise 20, cos= 0, so that=

2
= 90

is a right angle.

15. 1.2. LENGTH AND ANGLE: THE DOT PRODUCT 13
27.As in Example 1.21, we begin by calculatinguvand the norms of the two vectors:
uv= 0:9(4:5) + 2:12:6 + 1:2(0:8) = 0:45;
kuk=
p
0:9
2
+ 2:1
2
+ 1:2
2
=
p
6:66;
kvk=
p
(4:5)
2
+ 2:6
2
+ (0:8)
2
=
p
27:65:
So ifis the angle betweenuandv, then
cos=
uv
kuk kvk
=
0:45
p
6:66
p
27:65

0:45
p
182:817
;
so that
= cos
1

0:45
p
182:817

1:537588:09

:
Note that it is important to maintain as much precision as possible until the last step, or roundo
errors may build up.
28.As in Example 1.21, we begin by calculatinguvand the norms of the two vectors:
uv= 1(3) + 21 + 32 + 4(2) =3;
kuk=
p
1
2
+ 2
2
+ 3
2
+ 4
2
=
p
30;
kvk=
p
(3)
2
+ 1
2
+ 2
2
+ (2)
2
=
p
18:
So ifis the angle betweenuandv, then
cos=
uv
kuk kvk
=
3
p
30
p
18
=
1
2
p
15
so that= cos
1


1
2
p
15

1:797:42

:
29.As in Example 1.21, we begin by calculatinguvand the norms of the two vectors:
uv= 15 + 26 + 37 + 48 = 70;
kuk=
p
1
2
+ 2
2
+ 3
2
+ 4
2
=
p
30;
kvk=
p
5
2
+ 6
2
+ 7
2
+ 8
2
=
p
174:
So ifis the angle betweenuandv, then
cos=
uv
kuk kvk
=
70
p
30
p
174
=
35
3
p
145
so that= cos
1

35
3
p
145

0:250214:34

:
30.To show that4ABCis right, we need only show that one pair of its sides meets at a right angle. So
letu=
#
AB,v=
#
BC, andw=
#
AC. Then we must show that one ofuv,uworvwis zero in
order to show that one of these pairs is orthogonal. Then
u=
#
AB= [1(3);02] = [4;2];v=
#
BC= [41;60] = [3;6];
w=
#
AC= [4(3);62] = [7;4];
and
uv= 43 + (2)6 = 1212 = 0:
Since this dot product is zero, these two vectors are orthogonal, so that
#
AB?
#
BCand thus4ABCis
a right triangle. It is unnecessary to test the remaining pairs of sides.

16. 14 CHAPTER 1. VECTORS
31.To show that4ABCis right, we need only show that one pair of its sides meets at a right angle. So
letu=
#
AB,v=
#
BC, andw=
#
AC. Then we must show that one ofuv,uworvwis zero in
order to show that one of these pairs is orthogonal. Then
u=
#
AB= [31;21;(2)(1)] = [4;1;1];
v=
#
BC= [2(3);22;4(2)] = [5;0;2];
w=
#
AC= [21;21;4(1)] = [1;1;3];
and
uv=45 + 101(2) =18
uw=41 + 111(3) = 0:
Since this dot product is zero, these two vectors are orthogonal, so that
#
AB?
#
ACand thus4ABCis
a right triangle. It is unnecessary to test the remaining pair of sides.
32.As in Example 1.22, the dimensions of the cube do not matter, so we work with a cube with side length
1. Since the cube is symmetric, we need only consider one diagonal and adjacent edge. Orient the cube
as shown in Figure 1.34; take the diagonal to be [1;1;1] and the adjacent edge to be [1;0;0]. Then the
anglebetween these two vectors satises
cos=
11 + 10 + 10
p
3
p
1
=
1
p
3
;so= cos
1

1
p
3

54:74

:
Thus the diagonal and an adjacent edge meet at an angle of 54:74

.
33.As in Example 1.22, the dimensions of the cube do not matter, so we work with a cube with side length
1. Since the cube is symmetric, we need only consider one pair of diagonals. Orient the cube as shown
in Figure 1.34; take the diagonals to beu= [1;1;1] andv= [1;1;0][0;0;1] = [1;1;1]. Then the
dot product is
uv= 11 + 11 + 1(1) = 1 + 11 = 16= 0:
Since the dot product is nonzero, the diagonals are not orthogonal.
34.To show a parallelogram is a rhombus, it suces to show that its diagonals are perpendicular (Euclid).
But
d1d2=
2
4
2
2
0
3
5
2
4
1
1
3
3
5= 21 + 2(1) + 03 = 0:
To determine its side length, note that since the diagonals are perpendicular, one half of each diagonal
are the legs of a right triangle whose hypotenuse is one side of the rhombus. So we can use the
Pythagorean Theorem. Since
kd1k
2
= 2
2
+ 2
2
+ 0
2
= 8;kd2k
2
= 1
2
+ (1)
2
+ 3
2
= 11;
we have for the side length
s
2
=

kd1k
2

2
+

kd2k
2

2
=
8
4
+
11
4
=
19
4
;
so thats=
p
19
2
2:18.
35.SinceABCDis a rectangle, opposite sidesBAandCDare parallel and congruent. So we can use
the method of Example 1.1 in Section 1.1 to nd the coordinates of vertexD: we compute
#
BA=
[13;26;3(2)] = [2;4;5]. If
#
BAis then translated to
#
CD, whereC= (0;5;4), then
D= (0 + (2);5 + (4);4 + 5) = (2;1;1):

17. 1.2. LENGTH AND ANGLE: THE DOT PRODUCT 15
36.The resultant velocity of the airplane is the sum of the velocity of the airplane and the velocity of the
wind:
r=p+w=

200
0

+

0
40

=

200
40

:
37.Let thexdirection be east, in the direction of the current, and theydirection be north, across the
river. The speed of the boat is 4 mph north, and the current is 3 mph east, so the velocity of the boat
is
v=

0
4

+

3
0

=

3
4

:
38.Let thexdirection be the direction across the river, and theydirection be downstream. Sincevt=d,
use the given information to ndv, then solve fortand computed. Since the speed of the boat is 20
km/h and the speed of the current is 5 km/h, we havev=

20
5

. The width of the river is 2 km, and
the distance downstream is unknown; call ity. Thend=

2
y

. Thus
vt=

20
5

t=

2
y

:
Thus 20t= 2 so thatt= 0:1, and theny= 50:1 = 0:5. Therefore
(a)Ann lands 0:5 km, or half a kilometer, downstream;
(b)It takes Ann 0:1 hours, or six minutes, to cross the river.
Note that the river ow does not increase the time required to cross the river, since its velocity is
perpendicular to the direction of travel.
39.We want to nd the angle between Bert's resultant vector,r, and his velocity vector upstream,v. Let
the rst coordinate of the vector be the direction across the river, and the second be the direction
upstream. Bert's velocity vector directly across the river is unknown, sayu=

x
0

. His velocity vector
upstream compensates for the downstream ow, sov=

0
1

. So the resultant vector isr=u+v=

x
0

+

0
1

=

x
1

. Since Bert's speed is 2 mph, we havekrk= 2. Thus
x
2
+ 1 =krk
2
= 4;so thatx=
p
3:
Ifis the angle betweenrandv, then
cos=
rv
krk kvk
=
p
3
2
;so that= cos
1
p
3
2
!
= 60

:
40.We have
proj
uv=
uv
uu
u=
(1)(2) + 14
(1)(1) + 11

1
1

= 3

1
1

=

3
3

:
A graph of the situation is (with proj
uvin gray, and the perpendicular fromvto the projection also
drawn)

18. 16 CHAPTER 1. VECTORSu
v
proj
u
HvL
-3 -2 -1
1
2
3
4
41.We have
proj
uv=
uv
uu
u=
3
5
1 +


4
5
2

3
5

3
5
+


4
5




4
5

"
3
5

4
5
#
=
1
1
u=u:
A graph of the situation is (with proj
uvin gray, and the perpendicular fromvto the projection also
drawn)u
v
proj
u
HvL
1
1
2
42.We have
proj
uv=
uv
uu
u=
1
2
2
1
4
2
1
2
(2)
1
2

1
2
+


1
4


1
4

+


1
2


1
2

2
6
4
1
2

1
4

1
2
3
7
5=
8
3
2
6
4
1
2

1
4

1
2
3
7
5=
2
6
4
4
3

2
3

4
3
3
7
5:
43.We have
proj
uv=
uv
uu
u=
12 + (1)(3) + 1(1) + (1)(2)
11 + (1)(1) + 11 + (1)(1)
2
6
6
6
4
1
1
1
1
3
7
7
7
5
=
6
4
2
6
6
6
4
1
1
1
1
3
7
7
7
5
=
2
6
6
6
4
3
2

3
2
3
2

3
2
3
7
7
7
5
=
3
2
u:
44.We have
proj
uv=
uv
uu
u=
0:52:1 + 1:51:2
0:50:5 + 1:51:5

0:5
1:5

=
2:85
2:5

0:5
1:5

=

0:57
1:71

= 1:14u:

19. 1.2. LENGTH AND ANGLE: THE DOT PRODUCT 17
45.We have
proj
uv=
uv
uu
u=
3:011:340:334:25 + 2:52(1:66)
3:013:010:33(0:33) + 2:522:52
2
4
3:01
0:33
2:52
3
5
=
1:5523
15:5194
2
4
3:01
0:33
2:52
3
5
2
4
0:301
0:033
0:252
3
5
1
10
u:
46.Letu=
#
AB=

21
2(1)

=

1
3

andv=
#
AC=

41
0(1)

=

3
1

.
(a)To compute the projection, we need
uv=

1
3



3
1

= 6;uu=

1
3



1
3

= 10:
Thus
proj
uv=
uv
uu
u=
3
5
"
1
3
#
=
"
3
5
9
5
#
;
so that
vproj
uv=
"
3
1
#

"
3
5
9
5
#
=
"
12
5

4
5
#
:
Then
kuk=
p
1
2
+ 3
2
=
p
10;kvproj
uvk=
s

12
5

2
+

4
5

2
=
4
p
10
5
;
so that nally
A=
1
2
kuk kvproj
uvk=
1
2
p
10
4
p
10
5
= 4:
(b)We already knowuv= 6 andkuk=
p
10 from part (a). Also,kvk=
p
3
2
+ 1
2
=
p
10. So
cos=
uv
kuk kvk
=
6
p
10
p
10
=
3
5
;
so that
sin=
p
1cos
2
=
s
1

3
5

2
=
4
5
:
Thus
A=
1
2
kuk kvksin=
1
2
p
10
p
10
4
5
= 4:
47.Letu=
#
AB=
2
4
43
2(1)
64
3
5=
2
4
1
1
2
3
5andv=
#
AC=
2
4
53
0(1)
24
3
5=
2
4
2
1
2
3
5.
(a)To compute the projection, we need
uv=
2
4
1
1
2
3
5
2
4
2
1
2
3
5=3;uu=
2
4
1
1
2
3
5
2
4
1
1
2
3
5= 6:
Thus
proj
uv=
uv
uu
u=
3
6
2
6
4
1
1
2
3
7
5=
2
6
4

1
2
1
2
1
3
7
5;

20. 18 CHAPTER 1. VECTORS
so that
vproj
uv=
2
6
4
2
1
2
3
7
5
2
6
4

1
2
1
2
1
3
7
5=
2
6
4
5
2
1
2
1
3
7
5
Then
kuk=
p
1
2
+ (1)
2
+ 2
2
=
p
6;kvproj
uvk=
s

5
2

2
+

1
2

2
+ (1)
2
=
p
30
2
;
so that nally
A=
1
2
kuk kvproj
uvk=
1
2
p
6
p
30
2
=
3
p
5
2
:
(b)We already knowuv=3 andkuk=
p
6 from part (a). Also,kvk=
p
2
2
+ 1
2
+ (2)
2
= 3.
So
cos=
uv
kuk kvk
=
3
3
p
6
=
p
6
6
;
so that
sin=
p
1cos
2
=
v
u
u
t
1
p
6
6
!
2
=
p
30
6
:
Thus
A=
1
2
kuk kvksin=
1
2
p
63
p
30
6
=
3
p
5
2
:
48.Two vectorsuandvare orthogonal if and only if their dot productuv= 0. So we setuv= 0 and
solve fork:
uv=

2
3



k+ 1
k1

= 0)2(k+ 1) + 3(k1) = 0)5k1 = 0)k=
1
5
:
Substituting into the formula forvgives
v=
"
1
5
+ 1
1
5
1
#
=
"
6
5

4
5
#
:
As a check, we compute
uv=
"
2
3
#

"
6
5

4
5
#
=
12
5

12
5
= 0;
and the vectors are indeed orthogonal.
49.Two vectorsuandvare orthogonal if and only if their dot productuv= 0. So we setuv= 0 and
solve fork:
uv=
2
4
1
1
2
3
5
2
4
k
2
k
3
3
5= 0)k
2
k6 = 0)(k+ 2)(k3) = 0)k= 2;3:
Substituting into the formula forvgives
k= 2 :v1=
2
4
(2)
2
2
3
3
5=
2
4
4
2
3
3
5; k=3 :v2=
2
4
3
2
3
3
3
5=
2
4
9
3
3
3
5:
As a check, we compute
uv1=
2
4
1
1
2
3
5
2
4
4
2
3
3
5= 141(2) + 2(3) = 0;uv2=
2
4
1
1
2
3
5
2
4
9
3
3
3
5= 1913 + 2(3) = 0
and the vectors are indeed orthogonal.

21. 1.2. LENGTH AND ANGLE: THE DOT PRODUCT 19
50.Two vectorsuandvare orthogonal if and only if their dot productuv= 0. So we setuv= 0 and
solve foryin terms ofx:
uv=

3
1



x
y

= 0)3x+y= 0)y=3x:
Substitutingy=3xback into the formula forvgives
v=

x
3x

=x

1
3

:
Thus any vector orthogonal to

3
1

is a multiple of

1
3

. As a check,
uv=

3
1



x
3x

= 3x3x= 0 for any value ofx;
so that the vectors are indeed orthogonal.
51.As noted in the remarks just prior to Example 1.16, the zero vector0is orthogonal to all vectors in
R
2
. So if

a
b

=0, any vector

x
y

will do. Now assume that

a
b

6=0; that is, that eitheraorbis
nonzero. Two vectorsuandvare orthogonal if and only if their dot productuv= 0. So we set
uv= 0 and solve foryin terms ofx:
uv=

a
b



x
y

= 0)ax+by= 0:
First assumeb6= 0. Theny=
a
b
x, so substituting back into the expression forvwe get
v=
"
x

a
b
x
#
=x
"
1

a
b
#
=
x
b
"
b
a
#
:
Next, ifb= 0, thena6= 0, so thatx=
b
a
y, and substituting back into the expression forvgives
v=
"

b
a
y
y
#
=y
"

b
a
1
#
=
y
a
"
b
a
#
:
So in either case, a vector orthogonal to

a
b

, if it is not the zero vector, is a multiple of

b
a

. As a
check, note that

a
b



rb
ra

=rabrab= 0 for all values ofr:
52. The geometry of the vectors in Figure 1.26 suggests that ifku+vk=kuk+kvk, thenuandv
point in the same direction. This means that the angle between them must be 0. So we rst prove
Lemma 1.For all vectorsuandvinR
2
orR
3
,uv=kuk kvkif and only if the vectors point
in the same direction.
Proof.Letbe the angle betweenuandv. Then
cos=
uv
kuk kvk
;
so that cos= 1 if and only ifuv=kuk kvk. But cos= 1 if and only if= 0, which means
thatuandvpoint in the same direction.

22. 20 CHAPTER 1. VECTORS
We can now show
Theorem 2.For all vectorsuandvinR
2
orR
3
,ku+vk=kuk+kvkif and only ifuandv
point in the same direction.
Proof.First assume thatuandvpoint in the same direction. Thenuv=kuk kvk, and thus
ku+vk
2
=uu+ 2uv+vv By Example 1.9
=kuk
2
+ 2uv+kvk
2
Sinceww=kwk
2
for any vectorw
=kuk
2
+ 2kuk kvk+kvk
2
By the lemma
= (kuk+kvk)
2
:
Sinceku+vkandkuk+kvkare both nonnegative, taking square roots givesku+vk=kuk+kvk.
For the other direction, ifku+vk=kuk+kvk, then their squares are equal, so that
(kuk+kvk)
2
=kuk
2
+ 2kuk kvk+kvk
2
and
ku+vk
2
=uu+ 2uv+vv
are equal. Butkuk
2
=uuand similarly forv, so that canceling those terms gives 2uv=
2kuk kvkand thusuv=kuk kvk. Using the lemma again shows thatuandvpoint in the same
direction.
(b)The geometry of the vectors in Figure 1.26 suggests that ifku+vk=kuk kvk, thenuandv
point in opposite directions. In addition, sinceku+vk 0, we must also havekuk kvk. If
they point in opposite directions, the angle between them must be. This entire proof is exactly
analogous to the proof in part (a). We rst prove
Lemma 3.For all vectorsuandvinR
2
orR
3
,uv= kuk kvkif and only if the vectors point
in opposite directions.
Proof.Letbe the angle betweenuandv. Then
cos=
uv
kuk kvk
;
so that cos=1 if and only ifuv= kuk kvk. But cos=1 if and only if=, which
means thatuandvpoint in opposite directions.
We can now show
Theorem 4.For all vectorsuandvinR
2
orR
3
,ku+vk=kuk kvkif and only ifuandv
point in opposite directions andkuk kvk.
Proof.First assume thatuandvpoint in opposite directions andkuk kvk. Thenuv=
kuk kvk, and thus
ku+vk
2
=uu+ 2uv+vv By Example 1.9
=kuk
2
+ 2uv+kvk
2
Sinceww=kwk
2
for any vectorw
=kuk
2
2kuk kvk+kvk
2
By the lemma
= (kuk kvk)
2
:
Now, sincekuk kvkby assumption, we see that bothku+vkandkuk kvkare nonnegative,
so that taking square roots givesku+vk=kuk kvk. For the other direction, ifku+vk=

23. 1.2. LENGTH AND ANGLE: THE DOT PRODUCT 21
kuk kvk, then rst of all, since the left-hand side is nonnegative, the right-hand side must be
as well, so thatkuk kvk. Next, we can square both sides of the equality, so that
(kuk kvk)
2
=kuk
2
2kuk kvk+kvk
2
and
ku+vk
2
=uu+ 2uv+vv
are equal. Butkuk
2
=uuand similarly forv, so that canceling those terms gives 2uv=
2kuk kvkand thusuv= kuk kvk. Using the lemma again shows thatuandvpoint in
opposite directions.
53.Prove Theorem 1.2(b) by applying the denition of the dot product:
uv=u1(v1+w1) +u2(v2+w2) + +un(vn+wn)
=u1v1+u1w1+u2v2+u2w2+ +unvn+unwn
= (u1v1+u2v2+ +unvn) + (u1w1+u2w2+ +unwn)
=uv+uw:
54.Prove the three parts of Theorem 1.2(d) by applying the denition of the dot product and various
properties of real numbers:
Part 1:For any vectoru, we must showuu0. But
uu=u1u1+u2u2+ +unun=u
2
1+u
2
2+ +u
2
n:
Since for any real numberxwe know thatx
2
0, it follows that this sum is also nonnegative, so
thatuu0.
Part 2:We must show that ifu=0thenuu= 0. Butu= 0 means thatui= 0 for alli, so that
uu= 00 + 00 + + 00 = 0:
Part 3:We must show that ifuu= 0, thenu=0. From part 1, we know that
uu=u
2
1+u
2
2+ +u
2
n;
and thatu
2
i
0 for alli. So if the dot product is to be zero, eachu
2
i
must be zero, which means
thatui= 0 for alliand thusu=0.
55.We must show d(u;v) =kuvk=kvuk= d(v;u). By denition, d(u;v) =kuvk. Then by
Theorem 1.3(b) withc=1, we havekwk=kwkfor any vectorw; applying this to the vectoruv
gives
kuvk=k(uv)k=kvuk;
which is by denition equal to d(v;u).
56.We must show that for any vectorsu,vandwthat d(u;w)d(u;v) + d(v;w). This is equivalent to
showing thatkuwk kuvk+kvwk. Now substituteuvforxandvwforyin Theorem
1.5, giving
kuwk=k(uv) + (vw)k kuvk+kvwk:
57.We must show that d(u;v) =kuvk= 0 if and only ifu=v. This follows immediately from
Theorem 1.3(a),kwk= 0 if and only ifw=0, upon settingw=uv.
58.Apply the denitions:
ucv= [u1; u2; : : : ; un][cv1; cv2; : : : ; cvn]
=u1cv1+u2cv2+ +uncvn
=cu1v1+cu2v2+ +cunvn
=c(u1v1+u2v2+ +unvn)
=c(uv):

24. 22 CHAPTER 1. VECTORS
59.We want to show thatkuvk kuk kvk. This is equivalent to showing thatkuk kuvk+kvk.
This follows immediately upon settingx=uvandy=vin Theorem 1.5.
60.Ifuv=uw, it doesnotfollow thatv=w. For example, since0v= 0 for every vectorvinR
n
,
the zero vector is orthogonal to every vectorv. So ifu=0in the above equality, we know nothing
aboutvandw. (as an example,0[1;2] =0[17;12]). Note, however, thatuv=uwimplies
thatuvuw=u(vw) =0, so thatuis orthogonal tovw.
61.We must show that (u+v)(uv) =kuk
2
kvk
2
for all vectors inR
n
. Recall that for anywinR
n
thatww=kwk
2
, and also thatuv=vu. Then
(u+v)(uv) =uu+vuuvvv=kuk
2
+uvuv kvk
2
=kuk
2
kvk
2
:
62. Letu;v2R
n
. Then
ku+vk
2
+kuvk
2
= (u+v)(u+v) + (uv)(uv)
= (uu+vv+ 2uv) + (uu+vv2uv)
=

kuk
2
+kvk
2

+ 2uv+

kuk
2
+kvk
2

2uv
= 2kuk
2
+ 2kvk
2
:
(b)Part (a) tells us that the sum of the
squares of the lengths of the diagonals
of a parallelogram is equal to the sum
of the squares of the lengths of its four
sides.u
v
u-v
u+v
63.Letu;v2R
n
. Then
1
4
ku+vk
2

1
4
kuvk
2
=
1
4
[(u+v)(u+v)((uv)(uv)]
=
1
4
[(uu+vv+ 2uv)(uu+vv2uv)]
=
1
4
h
kuk
2
kuk
2

+

kvk
2
kvk
2

+ 4uv
i
=uv:
64. Letu;v2R
n
. Then using the previous exercise,
ku+vk=kuvk , ku+vk
2
=kuvk
2
, ku+vk
2
kuvk
2
= 0
,
1
4
ku+vk
2

1
4
kuvk
2
= 0
,uv= 0
,uandvare orthogonal:

25. 1.2. LENGTH AND ANGLE: THE DOT PRODUCT 23
(b)Part (a) tells us that a parallelogram is
a rectangle if and only if the lengths of
its diagonals are equal.u
v
u-v
u+v
65. By Exercise 55, (u+v)(uv) =kuk
2
kvk
2
. Thus (u+v)(uv) = 0 if and only ifkuk
2
=kvk
2
.
It follows immediately thatu+vanduvare orthogonal if and only ifkuk=kvk.
(b)Part (a) tells us that the diagonals of a
parallelogram are perpendicular if and
only if the lengths of its sides are equal,
i.e., if and only if it is a rhombus.u
v
u-v
u+v
66.From Example 1.9 and the fact thatww=kwk
2
, we haveku+vk
2
=kuk
2
+ 2uv+kvk
2
. Taking
the square root of both sides yieldsku+vk=
q
kuk
2
+ 2uv+kvk
2
. Now substitute in the given
valueskuk= 2,kvk=
p
3, anduv= 1, giving
ku+vk=
r
2
2
+ 21 +
p
3

2
=
p
4 + 2 + 3 =
p
9 = 3:
67.From Theorem 1.4 (the Cauchy-Schwarcz inequality), we havejuvj kuk kvk. Ifkuk= 1 and
kvk= 2, thenjuvj 2, so we cannot haveuv= 3.
68. Ifuis orthogonal to bothvandw, thenuv=uw= 0. Then
u(v+w) =uv+uw= 0 + 0 = 0;
so thatuis orthogonal tov+w.
(b)Ifuis orthogonal to bothvandw, thenuv=uw= 0. Then
u(sv+tw) =u(sv) +u(tw) =s(uv) +t(uw) =s0 +t0 = 0;
so thatuis orthogonal tosv+tw.
69.We have
u(vproj
uv) =u

v

uv
uu

u

=uvu

uv
uu

u
=uv

uv
uu

(uu)
=uvuv
= 0:

26. 24 CHAPTER 1. VECTORS
70. proj
u(proj
uv) = proj
u

uv
uu
u

=
uv
uu
proj
uu=
uv
uu
u= proj
uv.
(b)Using part (a),
proj
u(vproj
uv) = proj
u

v
uv
uu
u

=

uv
uu

u

uv
uu

proj
uu
=

uv
uu

u

uv
uu

u=0:
(c)From the diagram, we see that
proj
uvku, so that proj
u(proj
uv) =
proj
uv. Also, (vproj
uv)?u, so
that proj
u(vproj
uv) =0.u
v
proj
u
HvL
v-proj
u
HvL -proj
u
HvL
71. We have
(u
2
1+u
2
2)(v
2
1+v
2
2)(u1v1+u2v2)
2
=u
2
1v
2
1+u
2
1v
2
2+u
2
2v
2
1+u
2
2v
2
2u
2
1v
2
12u1v1u2v2u
2
2v
2
2
=u
2
1v
2
2+u
2
2v
2
12u1u2v1v2
= (u1v2u2v1)
2
:
But the nal expression is nonnegative since it is a square. Thus the original expression is as well,
showing that (u
2
1+u
2
2)(v
2
1+v
2
2)(u1v1+u2v2)
2
0.
(b)We have
(u
2
1+u
2
2+u
2
3)(v
2
1+v
2
2+v
2
3)(u1v1+u2v2+u3v3)
2
=u
2
1v
2
1+u
2
1v
2
2+u
2
1v
2
3+u
2
2v
2
1+u
2
2v
2
2+u
2
2v
2
3+u
2
3v
2
1+u
2
3v
2
2+u
2
3v
2
3
u
2
1v
2
12u1v1u2v2u
2
2v
2
22u1v1u3v3u
2
3v
2
32u2v2u3v3
=u
2
1v
2
2+u
2
1v
2
3+u
2
2v
2
1+u
2
2v
2
3+u
2
3v
2
1+u
2
3v
2
2
2u1u2v1v22u1v1u3v32u2v2u3v3
= (u1v2u2v1)
2
+ (u1v3u3v1)
2
+ (u3v2u2v3)
2
:
But the nal expression is nonnegative since it is the sum of three squares. Thus the original
expression is as well, showing that (u
2
1+u
2
2+u
2
3)(v
2
1+v
2
2+v
2
3)(u1v1+u2v2+u3v3)
2
0.
72. Since proj
uv=
uv
uu
u, we have
proj
uv(vproj
uv) =
uv
uu
u

v
uv
uu
u

=
uv
uu

uv
uv
uu
uu

=
uv
uu
(uvuv)
= 0;

27. Exploration: Vectors and Geometry 25
so that proj
uvis orthogonal tovproj
uv. Since their vector sum isv, those three vectors form
a right triangle with hypotenusev, so by Pythagoras' Theorem,
kproj
uvk
2
kproj
uvk
2
+kvproj
uvk
2
=kvk
2
:
Since norms are always nonnegative, taking square roots giveskproj
uvk kvk.
(b)
kproj
uvk kvk ()




uv
uu

u


kvk
()






uv
kuk
2
!
u





kvk
()





uv
kuk
2





kuk kvk
()
juvj
kuk
kvk
() juvj kuk kvk;
which is the Cauchy-Schwarcz inequality.
73.Suppose proj
uv=cu. From the gure, we see that cos=
ckuk
kvk
. But also cos=
uv
kukkvk
. Thus these
two expressions are equal, i.e.,
ckuk
kvk
=
uv
kuk kvk
)ckuk=
uv
kuk
)c=
uv
kuk kuk
=
uv
uu
:
74.The basis for induction is the casesn= 1 andn= 2. Then= 1 case is the assertion thatkv1k kv2k,
which is obviously true. Then= 2 case is the Triangle Inequality, which is also true.
Now assume the statement holds forn=k2; that is, for anyv1,v2; : : :,vk,
kv1+v2+ +vkk kv1k+kv2k+ +kvkk:
Letv1,v2; : : :,vk,vk+1be any vectors. Then
kv1+v2+ +vk+vk+1k=kv1+v2+ + (vk+vk+1)k
kv1k+kv2k+ +kvk+vk+1k
using the inductive hypothesis. But then using the Triangle Inequality (or the casen= 2 in this
theorem),kvk+vk+1k kvkk+kvk+1k. Substituting into the above gives
kv1+v2+ +vk+vk+1k kv1k+kv2k+ +kvk+vk+1k
kv1k+kv2k+ +kvkk+kvk+1k;
which is what we were trying to prove.
Exploration: Vectors and Geometry
1.As in Example 1.25, letp=
#
OP. Thenpa=
#
AP=
1
3
#
AB=
1
3
(ba), so thatp=a+
1
3
(ba) =
1
3
(2a+b). More generally, ifPis the point
1
n
of the way fromAtoBalong
#
AB, thenpa=
#
AP=
1
n
#
AB=
1
n
(ba), so thatp=a+
1
n
(ba) =
1
n
((n1)a+b).
2.Use the notation that the vector
#
OXis writtenx. Then from exercise 1, we havep=
1
2
(a+c) and
q=
1
2
(b+c), so that
#
P Q=qp=
1
2
(b+c)
1
2
(a+c) =
1
2
(ba) =
1
2
#
AB:

28. 26 CHAPTER 1. VECTORS
3.Draw
#
AC. Then from exercise 2, we have
#
P Q=
1
2
#
AB=
#
SR. Also draw
#
BD. Again from exercise 2,
we have
#
P S=
1
2
#
BD=
#
QR. Thus opposite sides of the quadrilateralP QRSare equal. They are also
parallel: indeed,4BP Qand4BACare similar, since they share an angle andBP:BA=BQ:BC.
Thus\BP Q=\BAC; since these angles are equal,P QkAC. Similarly,SRkACso thatP QkSR. In
a like manner, we see thatP SkRQ. ThusP QRSis a parallelogram.
4.Following the hint, we ndm, the point that is two-thirds of the distance fromAtoP. From exercise
1, we have
p=
1
2
(b+c);so thatm=
1
3
(2p+a) =
1
3

2
1
2
(b+c) +a

=
1
3
(a+b+c):
Next we ndm
0
, the point that is two-thirds of the distance fromBtoQ. Again from exercise 1, we
have
q=
1
2
(a+c);so thatm
0
=
1
3
(2q+b) =
1
3

2
1
2
(a+c) +b

=
1
3
(a+b+c):
Finally we ndm
00
, the point that is two-thirds of the distance fromCtoR. Again from exercise 1,
we have
r=
1
2
(a+b);so thatm
00
=
1
3
(2r+c) =
1
3

2
1
2
(a+b) +c

=
1
3
(a+b+c):
Sincem=m
0
=m
00
, all three medians intersect at the centroid,G.
5.With notation as in the gure, we know that
#
AHis orthogonal to
#
BC; that is,
#
AH
#
BC= 0. Also
#
BHis orthogonal to
#
AC; that is,
#
BH
#
AC= 0. We must show that
#
CH
#
AB= 0. But
#
AH
#
BC= 0)(ha)(bc) = 0)hbhcab+ac= 0
#
BH
#
AC= 0)(hb)(ca) = 0)hchabc+ba= 0:
Adding these two equations together and canceling like terms gives
0 =hbhacb+ac= (hc)(ba) =
#
CH
#
AB;
so that these two are orthogonal. Thus all the altitudes intersect at the orthocenterH.
6.We are given that
#
QKis orthogonal to
#
ACand that
#
P Kis orthogonal to
#
CB, and must show that
#
RKis orthogonal to
#
AB. By exercise 1, we haveq=
1
2
(a+c),p=
1
2
(b+c), andr=
1
2
(a+b). Thus
#
QK
#
AC= 0)(kq)(ca) = 0)

k
1
2
(a+c)

(ca) = 0
#
P K
#
CB= 0)(kp)(bc) = 0)

k
1
2
(b+c)

(bc) = 0:
Expanding the two dot products gives
kcka
1
2
ac+
1
2
aa
1
2
cc+
1
2
ac= 0
kbkc
1
2
bb+
1
2
bc
1
2
cb+
1
2
cc= 0:
Add these two together and cancel like terms to get
0 =kbka
1
2
bb+
1
2
aa=

k
1
2
(b+a)

(ba) = (kr)(ba) =
#
RK
#
AB:
Thus
#
RKand
#
ABare indeed orthogonal, so all the perpendicular bisectors intersect at the circumcen-
ter.

29. 1.3. LINES AND PLANES 27
7.LetO, the center of the circle, be the origin. Thenb=aandkak
2
=kck
2
=r
2
whereris the radius
of the circle. We want to show that
#
ACis orthogonal to
#
BC. But
#
AC
#
BC= (ca)(cb)
= (ca)(c+a)
=kck
2
+ca kak
2
ac
= (acca) + (r
2
r
2
) = 0:
Thus the two are orthogonal, so that\ACBis a right angle.
8.As in exercise 5, we rst ndm, the point that is halfway fromPtoR. We havep=
1
2
(a+b) and
r=
1
2
(c+d), so that
m=
1
2
(p+r) =
1
2

1
2
(a+b) +
1
2
(c+d)

=
1
4
(a+b+c+d):
Similarly, we ndm
0
, the point that is halfway fromQtoS. We haveq=
1
2
(b+c) ands=
1
2
(a+d),
so that
m
0
=
1
2
(q+s) =
1
2

1
2
(b+c) +
1
2
(a+d)

=
1
4
(a+b+c+d):
Thusm=m
0
, so that
#
P Rand
#
QSintersect at their mutual midpoints; thus, they bisect each other.
1.3 Lines and Planes
1. The normal form isn(xp) = 0, or

3
2



x

0
0

= 0.
(b)Lettingx=

x
y

, we get

3
2



x
y



0
0

=

3
2



x
y

= 3x+ 2y= 0:
The general form is 3x+ 2y= 0.
2. The normal form isn(xp) = 0, or

3
4



x

1
2

= 0.
(b)Lettingx=

x
y

, we get

3
4



x
y



1
2

=

3
4



x1
y2

= 3(x1)4(y2) = 0:
Expanding and simplifying gives the general form 3x4y=5.
3. In vector form, the equation of the line isx=p+td, orx=

1
0

+t

1
3

.
(b)Lettingx=

x
y

and expanding the vector form from part (a) gives

x
y

=

1t
3t

, which yields
the parametric formx= 1t,y= 3t.
4. In vector form, the equation of the line isx=p+td, orx=

4
4

+t

1
1

.
(b)Lettingx=

x
y

and expanding the vector form from part (a) gives the parametric formx=4+t,
y= 4 +t.

30. 28 CHAPTER 1. VECTORS
5. In vector form, the equation of the line isx=p+td, orx=
2
4
0
0
0
3
5+t
2
4
1
1
4
3
5.
(b)Lettingx=
2
4
x
y
z
3
5and expanding the vector form from part (a) gives the parametric formx=t,
y=t,z= 4t.
6. In vector form, the equation of the line isx=p+td, orx=
2
4
3
0
2
3
5+t
2
4
2
5
0
3
5.
(b)Lettingx=
2
4
x
y
z
3
5and expanding the vector form from part (a) gives
2
4
x
y
z
3
5=
2
4
3 + 2t
5t
2
3
5, which
yields the parametric formx= 3 + 2t,y= 5t,z=2.
7. The normal form isn(xp) = 0, or
2
4
3
2
1
3
5
0
@x
2
4
0
1
0
3
5
1
A= 0.
(b)Lettingx=
2
4
x
y
z
3
5, we get
2
4
3
2
1
3
5
0
@
2
4
x
y
z
3
5
2
4
0
1
0
3
5
1
A=
2
4
3
2
1
3
5
2
4
x
y1
z
3
5= 3x+ 2(y1) +z= 0:
Expanding and simplifying gives the general form 3x+ 2y+z= 2.
8. The normal form isn(xp) = 0, or
2
4
2
5
0
3
5
0
@x
2
4
3
0
2
3
5
1
A= 0.
(b)Lettingx=
2
4
x
y
z
3
5, we get
2
4
2
5
0
3
5
0
@
2
4
x
y
z
3
5
2
4
3
0
2
3
5
1
A=
2
4
2
5
0
3
5
2
4
x3
y
z+ 2
3
5= 2(x3) + 5y= 0:
Expanding and simplifying gives the general form 2x+ 5y= 6.
9. In vector form, the equation of the line isx=p+su+tv, or
x=
2
4
0
0
0
3
5+s
2
4
2
1
2
3
5+t
2
4
3
2
1
3
5:
(b)Lettingx=
2
4
x
y
z
3
5and expanding the vector form from part (a) gives
2
4
x
y
z
3
5=
2
4
2s3t
s+ 2t
2s+t
3
5
which yields the parametric form the parametric formx= 2s3t,y=s+ 2t,z= 2s+t.

31. 1.3. LINES AND PLANES 29
10. In vector form, the equation of the line isx=p+su+tv, or
x=
2
4
6
4
3
3
5+s
2
4
0
1
1
3
5+t
2
4
1
1
1
3
5:
(b)Lettingx=
2
4
x
y
z
3
5and expanding the vector form from part (a) gives the parametric formx= 6t,
y=4 +s+t,z=3 +s+t.
11.Any pair of points on`determine a direction vector, so we usePandQ. We choosePto represent
the point on the line. Then a direction vector for the line isd=
#
P Q= (3;0)(1;2) = (2;2). The
vector equation for the line isx=p+td, orx=

1
2

+t

2
2

.
12.Any pair of points on`determine a direction vector, so we usePandQ. We choosePto represent the
point on the line. Then a direction vector for the line isd=
#
P Q= (2;1;3)(0;1;1) = (2;0;4).
The vector equation for the line isx=p+td, orx=
2
4
0
1
1
3
5+t
2
4
2
0
4
3
5.
13.We must nd two direction vectors,uandv. SinceP,Q, andRlie in a plane, we compute We get
two direction vectors
u=
#
P Q=qp= (4;0;2)(1;1;1) = (3;1;1)
v=
#
P R=rp= (0;1;1)(1;1;1) = (1;0;2):
Sinceuandvare not scalar multiples of each other, they will serve as direction vectors (if they were
parallel to each other, we would have not a plane but a line). So the vector equation for the plane is
x=p+su+tv, orx=
2
4
1
1
1
3
5+s
2
4
3
1
1
3
5+t
2
4
1
0
2
3
5.
14.We must nd two direction vectors,uandv. SinceP,Q, andRlie in a plane, we compute We get
two direction vectors
u=
#
P Q=qp= (1;0;1)(1;1;0) = (0;1;1)
v=
#
P R=rp= (0;1;1)(1;1;0) = (1;0;1):
Sinceuandvare not scalar multiples of each other, they will serve as direction vectors (if they were
parallel to each other, we would have not a plane but a line). So the vector equation for the plane is
x=p+su+tv, orx=
2
4
1
1
0
3
5+s
2
4
0
1
1
3
5+t
2
4
1
0
1
3
5.
15.The parametric and associated vector formsx=p+tdfound below are not unique.
(a)As in the remarks prior to Example 1.20, we start by lettingx=t. Substitutingx=tinto
y= 3x1 givesy= 3t1. So we get parametric equationsx=t,y= 3t1, and corresponding
vector form

x
y

=

0
1

+t

1
3

.
(b)In this case since the coecient ofyis 2, we start by lettingx= 2t. Substitutingx= 2tinto
3x+ 2y= 5 gives 32t+ 2y= 5, which givesy=3t+
5
2
. So we get parametric equationsx= 2t,
y=
5
2
3t, with corresponding vector equation
"
x
y
#
=
"
0
5
2
#
+t
"
2
3
#
:

32. 30 CHAPTER 1. VECTORS
Note that the equation was of the formax+by=cwitha= 3,b= 2, and that a direction vector
was given by

b
a

. This is true in general.
16.Note thatx=p+t(qp) is the line that passes throughp(whent= 0) andq(whent= 1). We
writed=qp; this is a direction vector for the line throughpandq.
(a)As noted above, the linep+tdpasses throughPatt= 0 and throughQatt= 1. So astvaries
from 0 to 1, the line describes the line segmentP Q.
(b)As shown inExploration: Vectors and Geometry, to nd the midpoint ofP Q, we start atP
and travel half the length ofP Qin the direction of the vector
#
P Q=qp. That is, the midpoint
ofP Qis the head of the vectorp+
1
2
(qp). Sincex=p+t(qp), we see that this line passes
through the midpoint att=
1
2
, and that the midpoint is in factp+
1
2
(qp) =
1
2
(p+q).
(c)From part (b), the midpoint is
1
2
([2;3] + [0;1]) =
1
2
[2;2] = [1;1].
(d)From part (b), the midpoint is
1
2
([1;0;1] + [4;1;2]) =
1
2
[5;1;1] =

5
2
;
1
2
;
1
2

.
(e)Again fromExploration: Vectors and Geometry, the vector whose head is
1
3
of the way from
PtoQalongP Qisx1=
1
3
(2p+q). Similarly, the vector whose head is
2
3
of the way fromPto
QalongP Qis also the vector one third of the way fromQtoPalongQP; applying the same
formula gives for this pointx2=
1
3
(2q+p). Whenp= [2;3] andq= [0;1], we get
x1=
1
3
(2[2;3] + [0;1]) =
1
3
[4;5] =

4
3
;
5
3

x2=
1
3
(2[0;1] + [2;3]) =
1
3
[2;1] =

2
3
;
1
3

:
(f)Using the formulas from part (e) withp= [1;0;1] andq= [4;1;2] gives
x1=
1
3
(2[1;0;1] + [4;1;2]) =
1
3
[6;1;4] =

2;
1
3
;
4
3

x2=
1
3
(2[4;1;2] + [1;0;1]) =
1
3
[9;2;5] =

3;
2
3
;
5
3

:
17.A line`1with slopem1has equationy=m1x+b1, orm1x+y=b1. Similarly, a line`2with slope
m2has equationy=m2x+b2, orm2x+y=b2. Thus the normal vector for`1isn1=

m1
1

, and
the normal vector for`2isn2=

m2
1

. Now,`1and`2are perpendicular if and only if their normal
vectors are perpendicular, i.e., if and only ifn1n2= 0. But
n1n2=

m1
1



m2
1

=m1m2+ 1;
so that the normal vectors are perpendicular if and only ifm1m2+1 = 0, i.e., if and only ifm1m2=1.
18.Suppose the line`has direction vectord, and the planePhas normal vectorn. Then ifdn= 0 (d
andnare orthogonal), then the line`is parallel to the planeP. If on the other handdandnare
parallel, so thatd=n, then`is perpendicular toP.
(a)Since the general form ofPis 2x+ 3yz= 1, its normal vector isn=
2
4
2
3
1
3
5. Sinced= 1n, we
see that`is perpendicular toP.

33. 1.3. LINES AND PLANES 31
(b)Since the general form ofPis 4xy+ 5z= 0, its normal vector isn=
2
4
4
1
5
3
5. Since
dn=
2
4
2
3
1
3
5
2
4
4
1
5
3
5= 24 + 3(1)15 = 0;
`is parallel toP.
(c)Since the general form ofPisxyz= 3, its normal vector isn=
2
4
1
1
1
3
5. Since
dn=
2
4
2
3
1
3
5
2
4
1
1
1
3
5= 21 + 3(1)1(1) = 0;
`is parallel toP.
(d)Since the general form ofPis 4x+ 6y2z= 0, its normal vector isn=
2
4
4
6
2
3
5. Since
d=
2
4
2
3
1
3
5=
1
2
2
4
4
6
2
3
5=
1
2
n;
`is perpendicular toP.
19.Suppose the planeP1has normal vectorn1, and the planePhas normal vectorn. Then ifn1n= 0
(n1andnare orthogonal), thenP1is perpendicular toP. If on the other handn1andnare
parallel, so thatn1=cn, thenP1is parallel toP. Note that in this exercise,P1has the equation
4xy+ 5z= 2, so thatn1=
2
4
4
1
5
3
5.
(a)Since the general form ofPis 2x+ 3yz= 1, its normal vector isn=
2
4
2
3
1
3
5. Since
n1n=
2
4
4
1
5
3
5
2
4
2
3
1
3
5= 4213 + 5(1) = 0;
the normal vectors are perpendicular, and thusP1is perpendicular toP.
(b)Since the general form ofPis 4xy+ 5z= 0, its normal vector isn=
2
4
4
1
5
3
5. Sincen1=n,
P1is parallel toP.
(c)Since the general form ofPisxyz= 3, its normal vector isn=
2
4
1
1
1
3
5. Since
n1n=
2
4
4
1
5
3
5
2
4
1
1
1
3
5= 411(1) + 5(1) = 0;
the normal vectors are perpendicular, and thusP1is perpendicular toP.

34. 32 CHAPTER 1. VECTORS
(d)Since the general form ofPis 4x+ 6y2z= 0, its normal vector isn=
2
4
4
6
2
3
5. Since
n1n=
2
4
4
1
5
3
5
2
4
4
6
2
3
5= 4416 + 5(2) = 0;
the normal vectors are perpendicular, and thusP1is perpendicular toP.
20.Since the vector form isx=p+td, we use the given information to determinepandd. The general
equation of the given line is 2x3y= 1, so its normal vector isn=

2
3

. Our line is perpendicular
to that line, so it has direction vectord=n=

2
3

. Furthermore, since our line passes through the
pointP= (2;1), we havep=

2
1

. Thus the vector form of the line perpendicular to 2x3y= 1
through the pointP= (2;1) is

x
y

=

2
1

+t

2
3

:
21.Since the vector form isx=p+td, we use the given information to determinepandd. The general
equation of the given line is 2x3y= 1, so its normal vector isn=

2
3

. Our line is parallel to that
line, so it has direction vectord=

3
2

(note thatdn= 0). Since our line passes through the point
P= (2;1), we havep=

2
1

, so that the vector equation of the line parallel to 2x3y= 1 through
the pointP= (2;1) is

x
y

=

2
1

+t

3
2

:
22.Since the vector form isx=p+td, we use the given information to determinepandd. A line is
perpendicular to a plane if its direction vectordis the normal vectornof the plane. The general
equation of the given plane isx3y+ 2z= 5, so its normal vector isn=
2
4
1
3
2
3
5. Thus the direction
vector of our line isd=
2
4
1
3
2
3
5. Furthermore, since our line passes through the pointP= (1;0;3), we
havep=
2
4
1
0
3
3
5. So the vector form of the line perpendicular tox3y+ 2z= 5 throughP= (1;0;3)
is
2
4
x
y
z
3
5=
2
4
1
0
3
3
5+t
2
4
1
3
2
3
5:
23.Since the vector form isx=p+td, we use the given information to determinepandd. Since the
given line has parametric equations
x= 1t; y= 2 + 3t; z=2t;it has vector form
2
4
x
y
z
3
5=
2
4
1
2
2
3
5+t
2
4
1
3
1
3
5:

35. 1.3. LINES AND PLANES 33
So its direction vector is
2
4
1
3
1
3
5, and this must be the direction vectordof the line we want, which is
parallel to the given line. Since our line passes through the pointP= (1;0;3), we havep=
2
4
1
0
3
3
5.
So the vector form of the line parallel to the given line throughP= (1;0;3) is
2
4
x
y
z
3
5=
2
4
1
0
3
3
5+t
2
4
1
3
1
3
5:
24.Since the normal form isnx=np, we use the given information to determinenandp. Note that
a plane is parallel to a given plane if their normal vectors are equal. Since the general form of the
given plane is 6xy+ 2z= 3, its normal vector isn=
2
4
6
1
2
3
5, so this must be a normal vector of the
desired plane as well. Furthermore, since our plane passes through the pointP= (0;2;5), we have
p=
2
4
0
2
5
3
5. So the normal form of the plane parallel to 6xy+ 2z= 3 through (0;2;5) is
2
4
6
1
2
3
5
2
4
x
y
z
3
5=
2
4
6
1
2
3
5
2
4
0
2
5
3
5or
2
4
6
1
2
3
5
2
4
x
y
z
3
5= 12:
25.Using Figure 1.34 in Section 1.2 for reference, we will nd a normal vectornand a point vectorpfor
each of the sides, then substitute intonx=npto get an equation for each plane.
(a)Start withP1determined by the face of the cube in thexy-plane. Clearly a normal vector for
P1isn=
2
4
1
0
0
3
5, or any vector parallel to thex-axis. Also, the plane passes throughP= (0;0;0),
so we setp=
2
4
0
0
0
3
5. Then substituting gives
2
4
1
0
0
3
5
2
4
x
y
z
3
5=
2
4
1
0
0
3
5
2
4
0
0
0
3
5orx= 0:
So the general equation forP1isx= 0. Applying the same argument above to the planeP2
determined by the face in thexz-plane gives a general equation ofy= 0, and similarly the plane
P3determined by the face in thexy-plane gives a general equation ofz= 0.
Now considerP4, the plane containing the face parallel to the face in theyz-plane but passing
through (1;1;1). SinceP4is parallel toP1, its normal vector is also
2
4
1
0
0
3
5; since it passes through
(1;1;1), we setp=
2
4
1
1
1
3
5. Then substituting gives
2
4
1
0
0
3
5
2
4
x
y
z
3
5=
2
4
1
0
0
3
5
2
4
1
1
1
3
5orx= 1:
So the general equation forP4isx= 1. Similarly, the general equations forP5andP6are
y= 1 andz= 1.

36. 34 CHAPTER 1. VECTORS
(b)Letn=
2
4
x
y
z
3
5be a normal vector for the desired planeP. SincePis perpendicular to the
xy-plane, their normal vectors must be orthogonal. Thus
2
4
x
y
z
3
5
2
4
0
0
1
3
5=x0 +y0 +z1 =z= 0:
Thusz= 0, so the normal vector is of the formn=
2
4
x
y
0
3
5. But the normal vector is also
perpendicular to the plane in question, by denition. Since that plane contains both the origin
and (1;1;1), the normal vector is orthogonal to (1;1;1)(0;0;0):
2
4
x
y
0
3
5
2
4
1
1
1
3
5=x1 +y1 +z0 =x+y= 0:
Thusx+y= 0, so thaty=x. So nally, a normal vector toPis given byn=
2
4
x
x
0
3
5for
any nonzerox. We may as well choosex= 1, givingn=
2
4
1
1
0
3
5. Since the plane passes through
(0;0;0), we letp=0. Then substituting innx=npgives
2
4
1
1
0
3
5
2
4
x
y
z
3
5=
2
4
1
1
0
3
5
2
4
0
0
0
3
5;orxy= 0:
Thus the general equation for the plane perpendicular to thexy-plane and containing the diagonal
from the origin to (1;1;1) isxy= 0.
(c)As in Example 1.22 (Figure 1.34) in Section 1.2, useu= [0;1;1] andv= [1;0;1] as two vectors
in the required plane. Ifn=
2
4
x
y
z
3
5is a normal vector to the plane, thennu= 0 =nv:
nu=
2
4
x
y
z
3
5
2
4
0
1
1
3
5=y+z= 0)y=z;nv=
2
4
x
y
z
3
5
2
4
1
0
1
3
5=x+z= 0)x=z:
Thus the normal vector is of the formn=
2
4
z
z
z
3
5for anyz. Takingz=1 givesn=
2
4
1
1
1
3
5.
Now, the side diagonals pass through (0;0;0), so setp=0. Thennx=npyields
2
4
1
1
1
3
5
2
4
x
y
z
3
5=
2
4
1
1
1
3
5
2
4
0
0
0
3
5;orx+yz= 0:
The general equation for the plane containing the side diagonals isx+yz= 0.
26.Finding the distance between pointsAandBis equivalent to nding d(a;b), whereais the vector
from the origin toA, and similarly forb. Givenx= [x; y; z],p= [1;0;2], andq= [5;2;4], we want
to solve d(x;p) = d(x;q); that is,
d(x;p) =
p
(x1)
2
+ (y0)
2
+ (z+ 2)
2
=
p
(x5)
2
+ (y2)
2
+ (z4)
2
= d(x;q):

37. 1.3. LINES AND PLANES 35
Squaring both sides gives
(x1)
2
+ (y0)
2
+ (z+ 2)
2
= (x5)
2
+ (y2)
2
+ (z4)
2
)
x
2
2x+ 1 +y
2
+z
2
+ 4z+ 4 =x
2
10x+ 25 +y
2
4y+ 4 +z
2
8z+ 16)
8x+ 4y+ 12z= 40)
2x+y+ 3z= 10:
Thus all such points (x; y; z) lie on the plane 2x+y+ 3z= 10.
27.To calculate d(Q; `) =
jax0+by0cj
p
a
2
+b
2
, we rst put`into general form. Withd=

1
1

, we getn=

1
1

since thennd= 0. Then we have
nx=np)

1
1



x
y

=

1
1

1
2

= 1:
Thusx+y= 1 and thusa=b=c= 1. SinceQ= (2;2) = (x0; y0), we have
d(Q; `) =
j12 + 121j
p
1
2
+ 1
2
=
3
p
2
=
3
p
2
2
:
28.Comparing the given equation tox=p+td, we getP= (1;1;1) andd=
2
4
2
0
3
3
5. As suggested by
Figure 1.63, we need to calculate the length of
#
RQ, whereRis the point on the line at the foot of the
perpendicular fromQ. So ifv=
#
P Q, then
#
P R= proj
dv;
#
RQ=vproj
dv:
Now,v=
#
P Q=qp=
2
4
0
1
0
3
5
2
4
1
1
1
3
5=
2
4
1
0
1
3
5, so that
proj
dv=

dv
dd

d=

2(1) + 3(1)
2(2) + 33

2
6
4
2
0
3
3
7
5=
2
6
4
2
13
0

3
13
3
7
5:
Thus
vproj
dv=
2
6
4
1
0
1
3
7
5
2
6
4
2
13
0

3
13
3
7
5=
2
6
4

15
13
0

10
13
3
7
5:
Then the distance d(Q; `) fromQto`is
kvproj
dvk=
5
13






2
4
3
0
2
3
5






=
5
13
p
3
2
+ 2
2
=
5
p
13
13
:
29.To calculate d(Q;P) =
jax0+by0+cz0dj
p
a
2
+b
2
+c
2
, we rst note that the plane has equationx+yz= 0, so
thata=b= 1,c=1, andd= 0. Also,Q= (2;2;2), so thatx0=y0=z0= 2. Hence
d(Q;P) =
j12 + 12120j
p
1
2
+ 1
2
+ (1)
2
=
2
p
3
=
2
p
3
3
:

38. 36 CHAPTER 1. VECTORS
30.To calculate d(Q;P) =
jax0+by0+cz0dj
p
a
2
+b
2
+c
2
, we rst note that the plane has equationx2y+ 2z= 1, so
thata= 1,b=2,c= 2, andd= 1. Also,Q= (0;0;0), so thatx0=y0=z0= 0. Hence
d(Q;P) =
j1020 + 201j
p
1
2
+ (2)
2
+ 2
2
=
1
3
:
31.Figure 1.66 suggests that we letv=
#
P Q; thenw=
#
P R= proj
dv. Comparing the given line`to
x=p+td, we getP= (1;2) andd=

1
1

. Thenv=
#
P Q=qp=

2
2



1
2

=

3
0

. Next,
w= proj
dv=

dv
dd

d=

13 + (1)0
11 + (1)(1)

1
1

=
3
2

1
1

:
So
r=p+
#
P R=p+ proj
dv=p+w=
"
1
2
#
+
"
3
2

3
2
#
=
"
1
2
1
2
#
:
So the pointRon`that is closest toQis

1
2
;
1
2

.
32.Figure 1.66 suggests that we letv=
#
P Q; then
#
P R= proj
dv. Comparing the given line`tox=p+td,
we getP= (1;1;1) andd=
2
4
2
0
3
3
5. Thenv=
#
P Q=qp=
2
4
0
1
0
3
5
2
4
1
1
1
3
5=
2
4
1
0
1
3
5. Next,
proj
dv=

dv
dd

d=

2(1) + 3(1)
(2)
2
+ 3
2

2
6
4
2
0
3
3
7
5=
2
6
4
2
13
0

3
13
3
7
5:
So
r=p+
#
P R=p+ proj
dv=
2
6
4
1
1
1
3
7
5+
2
6
4
2
13
0

3
13
3
7
5=
2
6
4
15
13
1
10
13
3
7
5:
So the pointRon`that is closest toQis

15
13
;1;
10
13

.
33.Figure 1.67 suggests we letv=
#
P Q, wherePis some point on the plane; then
#
QR= proj
nv. The
equation of the plane isx+yz= 0, son=
2
4
1
1
1
3
5. Settingy= 0 shows thatP= (1;0;1) is a point
on the plane. Then
v=
#
P Q=qp=
2
4
2
2
2
3
5
2
4
1
0
1
3
5=
2
4
1
2
1
3
5;
so that
proj
nv=

nv
nn

n=

11 + 1111
1
2
+ 1
2
+ (1)
2

2
6
4
1
1
1
3
7
5=
2
6
4
2
3
2
3

2
3
3
7
5:
Finally,
r=p+
#
P Q+
#
P R=p+vproj
nv=
2
6
4
1
0
1
3
7
5+
2
6
4
1
2
1
3
7
5
2
6
4
2
3
2
3

2
3
3
7
5=
2
6
4
4
3
4
3
8
3
3
7
5:
Therefore, the pointRinPthat is closest toQis

4
3
;
4
3
;
8
3

.

39. 1.3. LINES AND PLANES 37
34.Figure 1.67 suggests we letv=
#
P Q, wherePis some point on the plane; then
#
QR= proj
nv. The
equation of the plane isx2y+ 2z= 1, son=
2
4
1
2
2
3
5. Settingy=z= 0 shows thatP= (1;0;0) is a
point on the plane. Then
v=
#
P Q=qp=
2
4
0
0
0
3
5
2
4
1
0
0
3
5=
2
4
1
0
0
3
5;
so that
proj
nv=

nv
nn

n=

1(1)
1
2
+ (2)
2
+ 2
2

2
6
4
1
2
2
3
7
5=
2
6
4

1
9
2
9

2
9
3
7
5:
Finally,
r=p+
#
P Q+
#
P R=p+vproj
nv=
2
6
4
1
0
0
3
7
5+
2
6
4
1
0
0
3
7
5
2
6
4

1
9
2
9

2
9
3
7
5=
2
6
4

1
9
2
9

2
9
3
7
5:
Therefore, the pointRinPthat is closest toQis


1
9
;
2
9
;
2
9

.
35.Since the given lines`1and`2are parallel, choose arbitrary pointsQon`1andPon`2, sayQ= (1;1)
andP= (5;4). The direction vector of`2isd= [2;3]. Then
v=
#
P Q=qp=

1
1



5
4

=

4
3

;
so that
proj
dv=

dv
dd

d=

2(4) + 3(3)
(2)
2
+ 3
2

2
3

=
1
13

2
3

:
Then the distance between the lines is given by
kvproj
dvk=





4
3

+
1
13

2
3




=




1
13

54
36




=
18
13





3
2




=
18
13
p
13:
36.Since the given lines`1and`2are parallel, choose arbitrary pointsQon`1andPon`2, sayQ=
(1;0;1) andP= (0;1;1). The direction vector of`2isd= [1;1;1]. Then
v=
#
P Q=qp=
2
4
1
0
1
3
5
2
4
0
1
1
3
5=
2
4
1
1
2
3
5;
so that
proj
dv=

dv
dd

d=

11 + 1(1) + 1(2)
1
2
+ 1
2
+ 1
2

2
4
1
1
1
3
5=
2
3
2
4
1
1
1
3
5:
Then the distance between the lines is given by
kvproj
dvk=







2
6
4
1
1
2
3
7
5+
2
3
2
6
4
1
1
1
3
7
5







=







2
6
4
5
3

1
3

4
3
3
7
5







=
1
3
p
5
2
+ (1)
2
+ (4)
2
=
p
42
3
:
37.SinceP1andP2are parallel, we choose an arbitrary point onP1, sayQ= (0;0;0), and compute
d(Q;P2). Since the equation ofP2is 2x+y2z= 5, we havea= 2,b= 1,c=2, andd= 5; since
Q= (0;0;0), we havex0=y0=z0= 0. Thus the distance is
d(P1;P2) = d(Q;P2) =
jax0+by0+cz0dj
p
a
2
+b
2
+c
2
=
j20 + 10205j
p
2
2
+ 1
2
+ 2
2
=
5
3
:

40. 38 CHAPTER 1. VECTORS
38.SinceP1andP2are parallel, we choose an arbitrary point onP1, sayQ= (1;0;0), and compute
d(Q;P2). Since the equation ofP2isx+y+z= 3, we havea=b=c= 1 andd= 3; since
Q= (1;0;0), we havex0= 1,y0= 0, andz0= 0. Thus the distance is
d(P1;P2) = d(Q;P2) =
jax0+by0+cz0dj
p
a
2
+b
2
+c
2
=
j11 + 10 + 103j
p
1
2
+ 1
2
+ 1
2
=
2
p
3
=
2
p
3
3
:
39.We wish to show that d(B; `) =
jax0+by0cj
p
a
2
+b
2
, wheren=

a
b

,na=c, andB= (x0; y0). Ifv=
#
AB=
ba, then
nv=n(ba) =nbna=

a
b



x0
y0

c=ax0+by0c:
Then from Figure 1.65, we see that
d(B; `) =kproj
nvk=




nv
nn

n


=
jnvj
knk
=
jax0+by0cj
p
a
2
+b
2
:
40.We wish to show that d(B; `) =
jax0+by0+cz0dj
p
a
2
+b
2
+c
2
, wheren=
2
4
a
b
c
3
5,na=d, andB= (x0; y0; z0). If
v=
#
AB=ba, then
nv=n(ba) =nbna=
2
4
a
b
c
3
5
2
4
x0
y0
z0
3
5d=ax0+by0+cz0d:
Then from Figure 1.65, we see that
d(B; `) =kproj
nvk=




nv
nn

n


=
jnvj
knk
=
jax0+by0+cz0dj
p
a
2
+b
2
+c
2
:
41.ChooseB= (x0; y0) on`1; since`1and`2are parallel, the distance between them is d(B; `2). Then
sinceBlies on`1, we havenb=

a
b



x0
y0

=ax0+by0=c1. ChooseAon`2, so thatna=c2. Set
v=ba. Then using the formula in Exercise 39, the distance is
d(`1; `2) = d(B; `2) =
jnvj
knk
=
jn(ba)j
knk
=
jnbnaj
knk
=
jc1c2j
knk
:
42.ChooseB= (x0; y0; z0) onP1; sinceP1andP2are parallel, the distance between them is d(B;P2).
Then sinceBlies onP1, we havenb=
2
4
a
b
c
3
5
2
4
x0
y0
z0
3
5=ax0+by0+cz0=d1. ChooseAonP2, so
thatna=d2. Setv=ba. Then using the formula in Exercise 40, the distance is
d(P1;P2) = d(B;P2) =
jnvj
knk
=
jn(ba)j
knk
=
jnbnaj
knk
=
jd1d2j
knk
:
43.SinceP1has normal vectorn1=
2
4
1
1
1
3
5andP2has normal vectorn2=
2
4
2
1
2
3
5, the anglebetween
the normal vectors satises
cos=
n1n2
kn1k kn2k
=
12 + 11 + 1(2)
p
1
2
+ 1
2
+ 1
2
p
2
2
+ 1
2
+ (2)
2
=
1
3
p
3
:
Thus
= cos
1

1
3
p
3

78:9

:

41. 1.3. LINES AND PLANES 39
44.SinceP1has normal vectorn1=
2
4
3
1
2
3
5andP2has normal vectorn2=
2
4
1
4
1
3
5, the anglebetween
the normal vectors satises
cos=
n1n2
kn1k kn2k
=
3114 + 2(1)
p
3
2
+ (1)
2
+ 2
2
p
1
2
+ 4
2
+ (1)
2
=
3
p
14
p
18
=
1
p
28
:
This is an obtuse angle, so the acute angle is
=cos
1


1
p
28

79:1

:
45.First, to see thatPand`intersect, substitute the parametric equations for`into the equation forP,
giving
x+y+ 2z= (2 +t) + (12t) + 2(3 +t) = 9 +t= 0;
so thatt=9 represents the point of intersection, which is thus (2 + (9);12(9);3 + (9)) =
(7;19;6). Now, the normal toPisn=
2
4
1
1
2
3
5, and a direction vector for`isd=
2
4
1
2
1
3
5. So ifis
the angle betweennandd, thensatises
cos=
nd
knk kdk
=
11 + 1(2) + 21
p
1
2
+ 1
2
+ 1
2
p
1
2
+ (2)
2
+ 1
2
=
1
6
;
so that
= cos
1

1
6

80:4

:
Thus the angle between the line and the plane is 90

80:4

9:6

.
46.First, to see thatPand`intersect, substitute the parametric equations for`into the equation forP,
giving
4xyz= 4t(1 + 2t)(2 + 3t) =t3 = 6;
so thatt=9 represents the point of intersection, which is thus (9;1 + 2(9);2 + 3(9)) =
(9;17;25). Now, the normal toPisn=
2
4
4
1
1
3
5, and a direction vector for`isd=
2
4
1
2
3
3
5. So if
is the angle betweennandd, thensatises
cos=
nd
knk kdk
=
411213
p
4
2
+ 1
2
+ 1
2
p
1
2
+ 2
2
+ 3
2
=
1
p
18
p
14
:
This corresponds to an obtuse angle, so the acute angle between the two is
=cos
1


1
p
18
p
14

86:4

:
Thus the angle between the line and the plane is 90

86:4

3:6

.
47.We havep=vcn, so thatcn=vp. Take the dot product of both sides withn, giving
(cn)n= (vp)n)
c(nn) =vnpn)
c(nn) =vn(sincepandnare orthogonal))
c=
nv
nn
:

42. 40 CHAPTER 1. VECTORS
Note that another interpretation of the gure is thatcn= proj
nv=

nv
nn

n, which also implies that
c=
nv
nn
.
Now substitute this value ofcinto the original equation, giving
p=vcn=v

nv
nn

n:
48. A normal vector to the planex+y+z= 0 isn=
2
4
1
1
1
3
5. Then
nv=
2
4
1
1
1
3
5
2
4
1
0
2
3
5= 11 + 10 + 1(2) =1
nn=
2
4
1
1
1
3
5
2
4
1
1
1
3
5= 11 + 11 + 11 = 3;
so thatc=
1
3
. Then
p=v

nv
nn

n=
2
6
4
1
0
2
3
7
5+
1
3
2
6
4
1
1
1
3
7
5=
2
6
4
4
3
1
3

5
3
3
7
5:
(b)A normal vector to the plane 3xy+z= 0 isn=
2
4
3
1
1
3
5. Then
nv=
2
4
3
1
1
3
5
2
4
1
0
2
3
5= 3110 + 1(2) = 1
nn=
2
4
3
1
1
3
5
2
4
3
1
1
3
5= 331(1) + 11 = 11;
so thatc=
1
11
. Then
p=v

nv
nn

n=
2
6
4
1
0
2
3
7
5
1
11
2
6
4
3
1
1
3
7
5=
2
6
4
8
11
1
11

23
11
3
7
5:
(c)A normal vector to the planex2z= 0 isn=
2
4
1
0
2
3
5. Then
nv=
2
4
1
0
2
3
5
2
4
1
0
2
3
5= 11 + 002(2) = 5
nn=
2
4
1
0
2
3
5
2
4
1
0
2
3
5= 11 + 002(2) = 5;

43. Exploration: The Cross Product 41
so thatc= 1. Then
p=v

nv
nn

n=
2
6
4
1
0
2
3
7
5
2
6
4
1
0
2
3
7
5=0:
Note that the projection is0because the vector is normal to the plane, so its projection onto the
plane is a single point.
(d)A normal vector to the plane 2x3y+z= 0 isn=
2
4
2
3
1
3
5. Then
nv=
2
4
2
3
1
3
5
2
4
1
0
2
3
5= 2130 + 1(2) = 0
nn=
2
4
2
3
1
3
5
2
4
2
3
1
3
5= 223(3) + 11 = 14;
so thatc= 0. Thusp=v=
2
4
1
0
2
3
5. Note that the projection is the vector itself because the
vector is parallel to the plane, so it is orthogonal to the normal vector.
Exploration: The Cross Product
1. v=
2
4
u2v3u3v2
u3v1u1v3
u1v2u2v1
3
5=
2
4
130(1)
1302
0(1)13
3
5=
2
4
3
3
3
3
5.
(b) v=
2
4
u2v3u3v2
u3v1u1v3
u1v2u2v1
3
5=
2
4
1121
2031
31(1)0
3
5=
2
4
3
3
3
3
5.
(c) v=
2
4
u2v3u3v2
u3v1u1v3
u1v2u2v1
3
5=
2
4
2(6)3(4)
32(1)(6)
1(4)22
3
5=0.
(d) v=
2
4
u2v3u3v2
u3v1u1v3
u1v2u2v1
3
5=
2
4
1312
1113
1211
3
5=
2
4
1
2
1
3
5.
2.We have
e1e2=
2
4
1
0
0
3
5
2
4
0
1
0
3
5=
2
4
0001
0010
1100
3
5=
2
4
0
0
1
3
5=e3
e2e3=
2
4
0
1
0
3
5
2
4
0
0
1
3
5=
2
4
1100
0001
0010
3
5=
2
4
1
0
0
3
5=e1
e3e1=
2
4
0
0
1
3
5
2
4
1
0
0
3
5=
2
4
0010
1100
0001
3
5=
2
4
0
1
0
3
5=e2:

44. 42 CHAPTER 1. VECTORS
3.Two vectors are orthogonal if their dot product equals zero. But
(uv)u=
2
4
u2v3u3v2
u3v1u1v3
u1v2u2v1
3
5
2
4
u1
u2
u3
3
5
= (u2v3u3v2)u1+ (u3v1u1v3)u2+ (u1v2u2v1)u3
= (u2v3u1u1v3u2) + (u3v1u2u2v1u3) + (u1v2u3u3v2u1) = 0
(uv)v=
2
4
u2v3u3v2
u3v1u1v3
u1v2u2v1
3
5
2
4
v1
v2
v3
3
5
= (u2v3u3v2)v1+ (u3v1u1v3)v2+ (u1v2u2v1)v3
= (u2v3v1u2v1v3) + (u3v1v2u3v2v1) + (u1v2v3u1v3v2) = 0:
4. By Exercise 1, a vector normal to the plane is
n=uv=
2
4
0
1
1
3
5
2
4
3
1
2
3
5=
2
4
121(1)
1302
0(1)13
3
5=
2
4
3
3
3
3
5:
So the normal form for the equation of this plane isnx=np, or
2
4
3
3
3
3
5
2
4
x
y
z
3
5=
2
4
3
3
3
3
5
2
4
1
0
2
3
5= 9:
This simplies to 3x+ 3y3z= 9, orx+yz= 3.
(b)Two vectors in the plane areu=
#
P Q=
2
4
2
1
1
3
5andv=
#
P R=
2
4
1
3
2
3
5. So by Exercise 1, a vector
normal to the plane is
n=uv=
2
4
2
1
1
3
5
2
4
1
3
2
3
5=
2
4
1(2)13
112(2)
2311
3
5=
2
4
5
5
5
3
5:
So the normal form for the equation of this plane isnx=np, or
2
4
5
5
5
3
5
2
4
x
y
z
3
5=
2
4
5
5
5
3
5
2
4
0
1
1
3
5= 0:
This simplies to5x+ 5y+ 5z= 0, orxyz= 0.
5. u=
2
4
v2u3v3u2
v3u1u3v1
v1u2v2u1
3
5=
2
4
u2v3u3v2
u3v1u1v3
u1v2u2v1
3
5=(uv).
(b) 0=
2
4
u1
u2
u3
3
5
2
4
0
0
0
3
5=
2
4
u20u30
u30u10
u10u20
3
5=
2
4
0
0
0
3
5=0.
(c) u=
2
4
u2u3u3u2
u3u1u1u3
u1u2u2u1
3
5=
2
4
0
0
0
3
5=0.
(d) kv=
2
4
u2kv3u3kv2
u3kv1u1kv3
u1kv2u2kv1
3
5=k
2
4
u2v3u3v2
u3v1u1v3
u1v2u2v1
3
5=k(uv).

45. Exploration: The Cross Product 43
(e) ku=k(uu) =k(0) =0by parts (d) and (c).
(f)Compute the cross-product:
u(v+w) =
2
4
u2(v3+w3)u3(v2+w2)
u3(v1+w1)u1(v3+w3)
u1(v2+w2)u2(v1+w1)
3
5
=
2
4
(u2v3u3v2) + (u2w3u3w2)
(u3v1u1v3) + (u3w1u1w3)
(u1v2u2v1) + (u1w2u2w1)
3
5
=
2
4
u2v3u3v2
u3v1u1v3
u1v2u2v1
3
5+
2
4
u2w3u3w2
u3w1u1w3
u1w2u2w1
3
5
=uv+uw:
6.In each case, simply compute:
(a)
u(vw) =
2
4
u1
u2
u3
3
5
2
4
v2w3v3w2
v3w1v1w3
v1w2v2w1
3
5
=u1v2w3u1v3w2+u2v3w1u2v1w3+u3v1w2u3v2w1
= (u2v3u3v2)w1+ (u3v1u1v3)w2+ (u1v2u2v1)w3
= (uv)w:
(b)
u(vw) =
2
4
u1
u2
u3
3
5
2
4
v2w3v3w2
v3w1v1w3
v1w2v2w1
3
5
=
2
4
u2(v1w2v2w1)u3(v3w1v1w3)
u3(v2w3v3w2)u1(v1w2v2w1)
u1(v3w1v1w3)u2(v2w3v3w2)
3
5
=
2
4
(u1w1+u2w2+u3w3)v1(u1v1+u2v2+u3v3)w1
(u1w1+u2w2+u3w3)v2(u1v1+u2v2+u3v3)w2
(u1w1+u2w2+u3w3)v3(u1v1+u2v2+u3v3)w3
3
5
= (u1w1+u2w2+u3w3)
2
4
v1
v2
v3
3
5(u1v1+u2v2+u3v3)
2
4
w1
w2
w3
3
5
= (uw)v(uv)w:
(c)
kuvk
2
=






2
4
u2v3u3v2
u3v1u1v3
u1v2u2v1
3
5






2
= (u2v3u3v2)
2
+ (u3v1u1v3)
2
+ (u1v2u2v1)
2
= (u
2
1+u
2
2+u
2
3)
2
(v
2
1+v
2
2+v
2
3)
2
(u1v1+u2v2+u3v3)
2
=kuk
2
kvk
2
(uv)
2
:
Solutions Manual for Linear Algebra A Modern Introduction 4th Edition by David Poole
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