ALL CHAPTERS COVERED
with Sample Tests
, Table of Contents
Preface ........................................................................iii
Answers to Exercises………………………………….1
Chapter 1 Answer Key ……………………………….1
Chapter 2 Answer Key ……………………………...19
Chapter 3 Answer Key ……………………………...33
Chapter 4 Answer Key ……………………………...52
Chapter 5 Answer Key ……………………………...87
Chapter 6 Answer Key …………………………….118
Chapter 7 Answer Key …………………………….128
Chapter 8 Answer Key …………………………….145
Chapter 9 Answer Key …………………………….170
Appendix B Answer Key ………………………….185
Appendix C Answer Key ………………………….186
Appendix D Answer Key ………………………….187
Chapter Tests ……..……………………………….190
Chapter 1 – Chapter Tests …………………………191
Chapter 2 – Chapter Tests …………………………194
Chapter 3 – Chapter Tests …………………………197
Chapter 4 – Chapter Tests …………………………201
Chapter 5 – Chapter Tests …………………………204
Chapter 6 – Chapter Tests …………………………210
Chapter 7 – Chapter Tests …………………………213
Answers to Chapter Tests ………………………….219
Chapter 1 – Answers for Chapter Tests …………...219
Chapter 2 – Answers for Chapter Tests …………...222
Chapter 3 – Answers for Chapter Tests …………...225
Chapter 4 – Answers for Chapter Tests …………...228
Chapter 5 – Answers for Chapter Tests …………...234
Chapter 6 – Answers for Chapter Tests …………...240
Chapter 7 – Answers for Chapter Tests …………...245
ii
,Answers to Exercises Section 1.1
Answers to Exercises
Chapter 1
Section 1.1
√ √
(1) (a) [9 −4], distance = 97 (c) [−1 −1 2 −3 −4], distance = 31
√
(b) [−6 1 1], distance = 38
(2) (a) (3 4 2) (see Figure 1) (c) (1 −2 0) (see Figure 3)
(b) (0 5 3) (see Figure 2) (d) (3 0 0) (see Figure 4)
(3) (a) (7 −13) (b) (6 4 −8) (c) (−1 3 −1 4 6)
¡ 16 ¢
(4) (a) 3 − 13
3 8 (b) (− 20
3 −1 −6 −1)
h i
(5) (a) √3 − √5 √6 ; shorter, since length of original vector is 1
70 70 70
£ ¤
(b) − 67 27 0 − 7 ; shorter, since length of original vector is 1
3
(c) [06 −08]; neither, since given vector is a unit vector
h i √
(d) √111 − √211 − √111 √111 √211 ; longer, since length of original vector = 511 1
(6) (a) Parallel (b) Parallel (c) Not parallel (d) Not parallel
(7) (a) [−6 12 15] (c) [−7 1 11] (e) [−10 −32 −1]
(b) [10 6 −12] (d) [−9 −2 4] (f) [−35 3 20]
(8) (a) x + y = [1 1], x − y = [−3 9], y − x = [3 −9] (see Figure 5)
(b) x + y = [3 −5], x − y = [17 1], y − x = [−17 −1] (see Figure 6)
(c) x + y = [1 8 −5], x − y = [3 2 −1], y − x = [−3 −2 1] (see Figure 7)
(d) x + y = [−2 −4 4], x − y = [4 0 6], y − x = [−4 0 −6] (see Figure 8)
√ = (7 −3 6), = (11 −5 3), and = (10 −7 8), the length of side = √
(9) With length of side
= 29. The triangle is isosceles, but not equilateral, since the length of side is 30.
√
(10) (a) [10 −10] (b) [−5 3 −15] (c) [0 0] = 0
(11) See Figures 1.7 and 1.8 in Section 1.1 of the textbook.
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,Answers to Exercises Section 1.1
c 2016 Elsevier Ltd. All rights reserved.
Copyright ° 2
,Answers to Exercises Section 1.1
c 2016 Elsevier Ltd. All rights reserved.
Copyright ° 3
,Answers to Exercises Section 1.1
(12) See Figure 9. Both represent the same diagonal vector by the associative law of addition for vectors.
Figure 9: x + (y + z)
√ √
(13) [05 − 06 2 −04 2] ≈ [−03485 −05657]
√ √
(14) Net velocity = [−2 2 −3 + 2 2], resultant speed ≈ 283 km/hr
h √ √ i
(15) Net velocity = − 3 2 2 8−32 2 ; speed ≈ 283 km/hr
√ √
(16) [−8 − 2 − 2]
1 12 344 392
(17) Acceleration = 20 [ 13 − 65 65 ] ≈ [00462 −02646 03015]
£ 13 4
¤
(18) Acceleration = 2 0 3
180
(19) √
14
[−2 3 1] ≈ [−9622 14432 4811]
√
−
√
√ ]; b = [
√ √ 3 ]
(20) a = [ 1+ 3 1+ 3 1+ 3 1+ 3
(21) Let a = [1 ].
(a) kak2 = 21 + · · · + 2 is a sum of squares, which must be nonnegative. But then ||a|| ≥ 0 because
the square root of a nonnegative real number is a nonnegative real number.
(b) If ||a|| = 0, then kak2 = 21 + · · · + 2 = 0, which is only possible if every = 0. Thus, a = 0.
(22) In each part, suppose that x = [1 ], y = [1 ], and z = [1 ].
(a) x + (y + z) = [1 ] + [(1 + 1 ) ( + )] = [(1 + (1 + 1 )) ( + ( + ))]
= [((1 + 1 ) + 1 ) (( + ) + )] = [(1 + 1 ) ( + )] + [1 ] = (x + y) + z
(b) x + (−x) = [1 ] + [−1 − ] = [(1 + (−1 )) ( + (− ))] = [0 0]. Also,
(−x) + x = x + (−x) (by part (1) of Theorem 1.3) = 0, by the above.
(c) (x+y) = ([(1 +1 ) ( + )]) = [(1 +1 ) ( + )] = [(1 +1 ) ( + )] =
[1 ] + [1 ] = x + y
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Copyright ° 4
,Answers to Exercises Section 1.2
(d) ()x = [(()1 ) (() )] = [((1 )) (( ))] = [(1 ) ( )]
= ([1 ]) = (x)
(23) If = 0, done. Otherwise, ( 1 )(x) = 1 (0) =⇒ ( 1 · )x = 0 (by part (7) of Theorem 1.3) =⇒ x = 0
Thus either = 0 or x = 0.
(24) 1 x = 2 x =⇒ 1 x − 2 x = 0 =⇒ (1 − 2 )x = 0 =⇒ (1 − 2 ) = 0 or x = 0 by Theorem 1.4. But
since 1 6= 2 (1 − 2 ) 6= 0 Hence, x = 0
(25) (a) F (b) T (c) T (d) F (e) T (f) F (g) F (h) F
Section 1.2
27
(1) (a) arccos(− 5√37
), or about 1526◦ , or 266 radians
(b) arccos( √7446√29 ), or about 68◦ , or 012 radians
(c) arccos(0), which is 90◦ , or 2 radians
435√
(d) arccos(− √2175 87
) = arccos(−1), which is 180◦ , or radians (since x = −5y)
(2) The vector from 1 to 2 is [2 −7 −3], and the vector from 1 to 3 is [5 4 −6]. These vectors are
orthogonal.
(3) (a) [ ] · [− ] = (−) + = 0 Similarly, [ −] · [ ] = 0
(b) A vector in the direction of the line + + = 0 is [ −], while a vector in the direction of
− + = 0 is [ ].
√ √
1040 5
(4) (a) 15 joules (b) 9 ≈ 2584 joules (c) − 1895 15
≈ −1464 joules
(5) Note that y·z is a scalar, so x·(y·z) is not defined.
(6) In all parts, let x = [1 2 ] y = [1 2 ] and z = [1 2 ]
(a) x · y = [1 2 ] · [1 2 ] = 1 1 + · · · + = 1 1 + · · · +
= [1 2 ] · [1 2 ] = y · x
(b) x · x = [1 2 ] · [1 2 ] = 1 1 + · · · + = 21 + · · · + 2 . Now 21 + · · · + 2
is a sum of squares, each of which must be nonnegative. Hence, the sum is also nonnegative, and
³p ´2
so its square root is defined. Thus, 0 ≤ x · x = 21 + · · · + 2 = 21 + · · · + 2 = kxk2 .
(c) Suppose x · x = 0. From part (b), 0 = x · x = 21 + · · · + 2 ≥ 2 , for each , since all terms in the
sum are nonnegative. Hence, 0 ≥ 2 for each . But 2 ≥ 0, because it is a square. Hence each
= 0. Therefore, x = 0.
(d) (x · y) = ([1 2 ] · [1 2 ]) = (1 1 + · · · + )
= 1 1 + · · · + = [1 2 ] · [1 2 ] = (x) · y.
Next, (x · y) = (y · x) (by part (a)) = (y) · x (from above) = x · (y), by part (a)
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,Answers to Exercises Section 1.2
(e) (x + y) · z = ([1 2 ] + [1 2 ]) · [1 2 ]
= [1 + 1 2 + 2 + ] · [1 2 ]
= (1 + 1 )1 + (2 + 2 )2 + · · · + ( + )
= (1 1 + 2 2 + · · · + ) + (1 1 + 2 2 + · · · + )
Also, (x · z) + (y · z) = ([1 2 ] · [1 2 ]) + ([1 2 ] · [1 2 ])
= (1 1 + 2 2 + · · · + ) + (1 1 + 2 2 + · · · + ).
Hence, (x + y) · z = (x · z) + (y · z).
(7) No; consider x = [1 0], y = [0 1], and z = [1 1].
(8) A method similar to the first part of the proof of Lemma 1.6 in the textbook yields:
2
ka − bk ≥ 0 =⇒ (a · a) − (b · a) − (a · b) + (b · b) ≥ 0 =⇒ 1 − 2(a · b) + 1 ≥ 0 =⇒ a · b ≤ 1.
(9) Note that (x + y)·(x − y) = (x·x) + (y·x) − (x·y) − (y·y) = kxk2 − kyk2 . Hence, (x + y)·(x − y) = 0
implies kxk2 = kyk2 , which means kxk = kyk (since both are nonnegative)
(10) Note that kx + yk2 = kxk2 + 2(x·y) + kyk2 , while kx − yk2 = kxk2 − 2(x·y) + kyk2 .
Hence, 21 (kx + yk2 + kx − yk2 ) = 12 (2kxk2 + 2kyk2 ) = kxk2 + kyk2
(11) (a) From the first equation in the solution to Exercise 10 above, kx + yk2 = kxk2 + kyk2 implies
2(x·y) = 0 which means x·y = 0.
(b) From the first equation in the solution to Exercise 10 above, x·y = 0 implies kx+yk2 = kxk2 +kyk2 .
(12) Note that kx + y + zk2 = k(x + y) + zk2
= kx + yk2 + 2((x + y)·z) + kzk2
= kxk2 + 2(x·y) + kyk2 + 2(x·z) + 2(y·z) + kzk2
= kxk2 + kyk2 + kzk2 since x, y, z are mutually orthogonal.
(13) From the first two equations in the solution for Exercise 10 above,
1
4 (kx + yk2 − kx − yk2 ) = 14 (4(x·y)) = x·y.
(14) Since x is orthogonal to both y and z, we have x·(1 y + 2 z) = 1 (x·y) + 2 (x·z) = 1 (0) + 2 (0) = 0
(15) Suppose y = x, for some 6= 0 Then, x · y = x·(x) = (x · x) = kxk2 = kxk ( kxk) = kxk (±|| kxk)
= ± kxk kxk = ± kxk kyk
√ √
3
(16) (a) Length = 3 (b) angle = arccos( 3 ) ≈ 547◦ , or 0955 radians
(17) (a) proja b = [− 35 − 10
3
− 32 ]; b − proja b = [ 85 43 3
10 − 2 ]; (b − proja b) · a = 0
(b) proja b = [− 65 1 25 ]; b − proja b = [− 14 8
5 −4 5 ]; (b − proja b) · a = 0
(c) proja b = [ 16 0 − 61 13 ]; b − proja b = [ 17 1 4
6 −1 6 − 3 ]; (b − proja b) · a = 0
(d) proja b = [−1 23 −2 − 23 ]; b − proja b = [6 − 12 −6 72 ]; (b − proja b) · a = 0
(18) (a) 0 (zero vector). The dropped perpendicular travels along b to the common initial point of a and
b.
(b) The vector b. The terminal point of b lies on the line through a, so the dropped perpendicular
has length zero.
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Copyright ° 6
,Answers to Exercises Section 1.2
(19) i, j, k
(20) (a) Parallel: [ 20 30 40 194 88 163
29 − 29 29 ], orthogonal: [− 29 29 29 ]
(b) Parallel: [− 12 1 − 21 ], orthogonal: [− 11 15
2 1 2 ]
(c) Parallel: [ 60 40 120 354 138 223
49 − 49 49 ], orthogonal: [− 49 49 49 ]
(21) From the lower triangle in the figure, we have (projr x) + (projr x − x) = reflection of x (see Figure
10).
Figure 10: Reflection of a vector x through a line.
(22) For the case kxk ≤ kyk: | kxk − kyk | = kyk − kxk = kx + y + (−x)k − kxk ≤ kx + yk + k − xk − kxk
(by the Triangle Inequality) = kx + yk + kxk − kxk = kx + yk.
The case kxk ≥ kyk is done similarly, with the roles of x and y reversed.
(23) (a) Note that projx y = [ 58 − 65 2] = 25 x and y − projx y = [ 75 − 24
5 −4] is orthogonal to x.
Let w = y − projx y Then since y = projx y + (y − projx y) we have y = 25 x + w where w is
orthogonal to x.
(b) Let = (x·y)(kxk2 ) (so that x = projx y), and let w = y − projx y which is orthogonal to x
by the argument before Theorem 1.11. Then y = x + w, where w is orthogonal to x.
(c) Suppose x+w = x+v. Then (−)x = w−v, and (−)x·(w−v) = (w−v)·(w−v) = kw−vk2 .
But ( − )x·(w − v) = ( − )(x·w) − ( − )(x·v) = 0, since v and w are orthogonal to x. Hence
kw − vk2 = 0 =⇒ w − v = 0 =⇒ w = v. Then, x = x =⇒ = , from Theorem 1.4, since x is
nonzero.
(24) If is the angle between x and y, and is the angle between projx y and projy x, then
³ ´ ³ ´ ³ ´³ ´
x·y y·x x·y y·x
projx y · projy x kxk 2 x · kyk 2 y kxk 2
kyk 2 (x · y)
cos = ° ° = °³ ´ ° °³ ´ °=³ ´ ³ ´
kprojx yk °projy x° ° x·y ° ° y·x
° kxk2 x° ° kyk 2 y°
° |x·y|
2 kxk |y·x|
2 kyk
kxk kyk
³ 3
´
(x·y)
kxk2 kyk2 (x · y)
= ³ ´ = = cos
(x·y)2 kxk kyk
kxkkyk
Hence = .
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Copyright ° 7
, Answers to Exercises Section 1.3
(25) (a) T (b) T (c) F (d) F (e) T (f) F
Section 1.3
(1) (a) We have k4x + 7yk ≤ k4xk + k7yk = 4kxk + 7kyk ≤ 7kxk + 7kyk = 7(kxk + kyk).
(b) Let = max{|| ||}. Then kx ± yk ≤ (kxk + kyk).
(2) (a) Note that 6 − 5 = 3(2( − 1)) + 1. Let = 2( − 1).
(b) Consider the number 4.
(3) Note that since x 6= 0 and y 6= 0, projx y = 0 iff (x·y)(kxk2 ) = 0 iff x·y = 0 iff y·x = 0 iff
(y·x)(kyk2 ) = 0 iff projy x = 0.
(4) If y = x (for 0), then kx + yk = kx + xk = (1 + )kxk = kxk + kxk = kxk + kxk = kxk + kyk.
On the other hand, if kx + yk = kxk + kyk, then kx + yk2 = (kxk + kyk)2 . Now kx + yk2 =
(x + y)·(x + y) = kxk2 + 2(x·y) + kyk2 , while (kxk + kyk)2 = kxk2 + 2kxkkyk + kyk2 . Hence
x·y = kxkkyk. By Result 4, y = x, for some 0.
(5) (a) Suppose y 6= 0. We must show that x is not orthogonal to y. Now kx + yk2 = kxk2 , so
kxk2 + 2(x·y) + kyk2 = kxk2 . Hence kyk2 = −2(x·y). Since y 6= 0, we have kyk2 6= 0, and so
x·y 6= 0.
(b) Suppose x is not a unit vector. We must show that x·y 6= 1.
Now projx y = x =⇒ ((x·y)(kxk2 ))x = 1x =⇒ (x·y)(kxk2 ) = 1 =⇒ x·y = kxk2 .
But then kxk 6= 1 =⇒ kxk2 6= 1 =⇒ x·y 6= 1.
(6) (a) Consider x = [1 0 0] and y = [1 1 0].
(b) If x 6= y, then x · y 6= kxk2 .
(c) Yes
(7) See the answer for Exercise 11(a) in Section 1.2.
(8) If kxk kyk, then kxk2 kyk2 , and so kxk2 − kyk2 0 But then (x + y)·(x−y) 0 and so the
cosine of the angle between (x + y) and (x−y) is positive. Thus the angle between (x + y) and (x−y)
is acute.
(9) (a) Contrapositive: If x = 0, then x is not a unit vector.
Converse: If x is nonzero, then x is a unit vector.
Inverse: If x is not a unit vector, then x = 0.
(b) (Let x and y be nonzero vectors.)
Contrapositive: If y 6= projx y, then x is not parallel to y.
Converse: If y = projx y, then x k y.
Inverse: If x is not parallel to y, then y 6= projx y.
(c) (Let x, y be nonzero vectors.)
Contrapositive: If projy x 6= 0, then projx y 6= 0.
Converse: If projy x = 0, then projx y = 0.
Inverse: If projx y 6= 0, then projy x 6= 0.
(10) (a) Converse: Let x and y be nonzero vectors in R . If kx + yk kyk, then x·y ≥ 0.
c 2016 Elsevier Ltd. All rights reserved.
Copyright ° 8
with Sample Tests
, Table of Contents
Preface ........................................................................iii
Answers to Exercises………………………………….1
Chapter 1 Answer Key ……………………………….1
Chapter 2 Answer Key ……………………………...19
Chapter 3 Answer Key ……………………………...33
Chapter 4 Answer Key ……………………………...52
Chapter 5 Answer Key ……………………………...87
Chapter 6 Answer Key …………………………….118
Chapter 7 Answer Key …………………………….128
Chapter 8 Answer Key …………………………….145
Chapter 9 Answer Key …………………………….170
Appendix B Answer Key ………………………….185
Appendix C Answer Key ………………………….186
Appendix D Answer Key ………………………….187
Chapter Tests ……..……………………………….190
Chapter 1 – Chapter Tests …………………………191
Chapter 2 – Chapter Tests …………………………194
Chapter 3 – Chapter Tests …………………………197
Chapter 4 – Chapter Tests …………………………201
Chapter 5 – Chapter Tests …………………………204
Chapter 6 – Chapter Tests …………………………210
Chapter 7 – Chapter Tests …………………………213
Answers to Chapter Tests ………………………….219
Chapter 1 – Answers for Chapter Tests …………...219
Chapter 2 – Answers for Chapter Tests …………...222
Chapter 3 – Answers for Chapter Tests …………...225
Chapter 4 – Answers for Chapter Tests …………...228
Chapter 5 – Answers for Chapter Tests …………...234
Chapter 6 – Answers for Chapter Tests …………...240
Chapter 7 – Answers for Chapter Tests …………...245
ii
,Answers to Exercises Section 1.1
Answers to Exercises
Chapter 1
Section 1.1
√ √
(1) (a) [9 −4], distance = 97 (c) [−1 −1 2 −3 −4], distance = 31
√
(b) [−6 1 1], distance = 38
(2) (a) (3 4 2) (see Figure 1) (c) (1 −2 0) (see Figure 3)
(b) (0 5 3) (see Figure 2) (d) (3 0 0) (see Figure 4)
(3) (a) (7 −13) (b) (6 4 −8) (c) (−1 3 −1 4 6)
¡ 16 ¢
(4) (a) 3 − 13
3 8 (b) (− 20
3 −1 −6 −1)
h i
(5) (a) √3 − √5 √6 ; shorter, since length of original vector is 1
70 70 70
£ ¤
(b) − 67 27 0 − 7 ; shorter, since length of original vector is 1
3
(c) [06 −08]; neither, since given vector is a unit vector
h i √
(d) √111 − √211 − √111 √111 √211 ; longer, since length of original vector = 511 1
(6) (a) Parallel (b) Parallel (c) Not parallel (d) Not parallel
(7) (a) [−6 12 15] (c) [−7 1 11] (e) [−10 −32 −1]
(b) [10 6 −12] (d) [−9 −2 4] (f) [−35 3 20]
(8) (a) x + y = [1 1], x − y = [−3 9], y − x = [3 −9] (see Figure 5)
(b) x + y = [3 −5], x − y = [17 1], y − x = [−17 −1] (see Figure 6)
(c) x + y = [1 8 −5], x − y = [3 2 −1], y − x = [−3 −2 1] (see Figure 7)
(d) x + y = [−2 −4 4], x − y = [4 0 6], y − x = [−4 0 −6] (see Figure 8)
√ = (7 −3 6), = (11 −5 3), and = (10 −7 8), the length of side = √
(9) With length of side
= 29. The triangle is isosceles, but not equilateral, since the length of side is 30.
√
(10) (a) [10 −10] (b) [−5 3 −15] (c) [0 0] = 0
(11) See Figures 1.7 and 1.8 in Section 1.1 of the textbook.
c 2016 Elsevier Ltd. All rights reserved.
Copyright ° 1
,Answers to Exercises Section 1.1
c 2016 Elsevier Ltd. All rights reserved.
Copyright ° 2
,Answers to Exercises Section 1.1
c 2016 Elsevier Ltd. All rights reserved.
Copyright ° 3
,Answers to Exercises Section 1.1
(12) See Figure 9. Both represent the same diagonal vector by the associative law of addition for vectors.
Figure 9: x + (y + z)
√ √
(13) [05 − 06 2 −04 2] ≈ [−03485 −05657]
√ √
(14) Net velocity = [−2 2 −3 + 2 2], resultant speed ≈ 283 km/hr
h √ √ i
(15) Net velocity = − 3 2 2 8−32 2 ; speed ≈ 283 km/hr
√ √
(16) [−8 − 2 − 2]
1 12 344 392
(17) Acceleration = 20 [ 13 − 65 65 ] ≈ [00462 −02646 03015]
£ 13 4
¤
(18) Acceleration = 2 0 3
180
(19) √
14
[−2 3 1] ≈ [−9622 14432 4811]
√
−
√
√ ]; b = [
√ √ 3 ]
(20) a = [ 1+ 3 1+ 3 1+ 3 1+ 3
(21) Let a = [1 ].
(a) kak2 = 21 + · · · + 2 is a sum of squares, which must be nonnegative. But then ||a|| ≥ 0 because
the square root of a nonnegative real number is a nonnegative real number.
(b) If ||a|| = 0, then kak2 = 21 + · · · + 2 = 0, which is only possible if every = 0. Thus, a = 0.
(22) In each part, suppose that x = [1 ], y = [1 ], and z = [1 ].
(a) x + (y + z) = [1 ] + [(1 + 1 ) ( + )] = [(1 + (1 + 1 )) ( + ( + ))]
= [((1 + 1 ) + 1 ) (( + ) + )] = [(1 + 1 ) ( + )] + [1 ] = (x + y) + z
(b) x + (−x) = [1 ] + [−1 − ] = [(1 + (−1 )) ( + (− ))] = [0 0]. Also,
(−x) + x = x + (−x) (by part (1) of Theorem 1.3) = 0, by the above.
(c) (x+y) = ([(1 +1 ) ( + )]) = [(1 +1 ) ( + )] = [(1 +1 ) ( + )] =
[1 ] + [1 ] = x + y
c 2016 Elsevier Ltd. All rights reserved.
Copyright ° 4
,Answers to Exercises Section 1.2
(d) ()x = [(()1 ) (() )] = [((1 )) (( ))] = [(1 ) ( )]
= ([1 ]) = (x)
(23) If = 0, done. Otherwise, ( 1 )(x) = 1 (0) =⇒ ( 1 · )x = 0 (by part (7) of Theorem 1.3) =⇒ x = 0
Thus either = 0 or x = 0.
(24) 1 x = 2 x =⇒ 1 x − 2 x = 0 =⇒ (1 − 2 )x = 0 =⇒ (1 − 2 ) = 0 or x = 0 by Theorem 1.4. But
since 1 6= 2 (1 − 2 ) 6= 0 Hence, x = 0
(25) (a) F (b) T (c) T (d) F (e) T (f) F (g) F (h) F
Section 1.2
27
(1) (a) arccos(− 5√37
), or about 1526◦ , or 266 radians
(b) arccos( √7446√29 ), or about 68◦ , or 012 radians
(c) arccos(0), which is 90◦ , or 2 radians
435√
(d) arccos(− √2175 87
) = arccos(−1), which is 180◦ , or radians (since x = −5y)
(2) The vector from 1 to 2 is [2 −7 −3], and the vector from 1 to 3 is [5 4 −6]. These vectors are
orthogonal.
(3) (a) [ ] · [− ] = (−) + = 0 Similarly, [ −] · [ ] = 0
(b) A vector in the direction of the line + + = 0 is [ −], while a vector in the direction of
− + = 0 is [ ].
√ √
1040 5
(4) (a) 15 joules (b) 9 ≈ 2584 joules (c) − 1895 15
≈ −1464 joules
(5) Note that y·z is a scalar, so x·(y·z) is not defined.
(6) In all parts, let x = [1 2 ] y = [1 2 ] and z = [1 2 ]
(a) x · y = [1 2 ] · [1 2 ] = 1 1 + · · · + = 1 1 + · · · +
= [1 2 ] · [1 2 ] = y · x
(b) x · x = [1 2 ] · [1 2 ] = 1 1 + · · · + = 21 + · · · + 2 . Now 21 + · · · + 2
is a sum of squares, each of which must be nonnegative. Hence, the sum is also nonnegative, and
³p ´2
so its square root is defined. Thus, 0 ≤ x · x = 21 + · · · + 2 = 21 + · · · + 2 = kxk2 .
(c) Suppose x · x = 0. From part (b), 0 = x · x = 21 + · · · + 2 ≥ 2 , for each , since all terms in the
sum are nonnegative. Hence, 0 ≥ 2 for each . But 2 ≥ 0, because it is a square. Hence each
= 0. Therefore, x = 0.
(d) (x · y) = ([1 2 ] · [1 2 ]) = (1 1 + · · · + )
= 1 1 + · · · + = [1 2 ] · [1 2 ] = (x) · y.
Next, (x · y) = (y · x) (by part (a)) = (y) · x (from above) = x · (y), by part (a)
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,Answers to Exercises Section 1.2
(e) (x + y) · z = ([1 2 ] + [1 2 ]) · [1 2 ]
= [1 + 1 2 + 2 + ] · [1 2 ]
= (1 + 1 )1 + (2 + 2 )2 + · · · + ( + )
= (1 1 + 2 2 + · · · + ) + (1 1 + 2 2 + · · · + )
Also, (x · z) + (y · z) = ([1 2 ] · [1 2 ]) + ([1 2 ] · [1 2 ])
= (1 1 + 2 2 + · · · + ) + (1 1 + 2 2 + · · · + ).
Hence, (x + y) · z = (x · z) + (y · z).
(7) No; consider x = [1 0], y = [0 1], and z = [1 1].
(8) A method similar to the first part of the proof of Lemma 1.6 in the textbook yields:
2
ka − bk ≥ 0 =⇒ (a · a) − (b · a) − (a · b) + (b · b) ≥ 0 =⇒ 1 − 2(a · b) + 1 ≥ 0 =⇒ a · b ≤ 1.
(9) Note that (x + y)·(x − y) = (x·x) + (y·x) − (x·y) − (y·y) = kxk2 − kyk2 . Hence, (x + y)·(x − y) = 0
implies kxk2 = kyk2 , which means kxk = kyk (since both are nonnegative)
(10) Note that kx + yk2 = kxk2 + 2(x·y) + kyk2 , while kx − yk2 = kxk2 − 2(x·y) + kyk2 .
Hence, 21 (kx + yk2 + kx − yk2 ) = 12 (2kxk2 + 2kyk2 ) = kxk2 + kyk2
(11) (a) From the first equation in the solution to Exercise 10 above, kx + yk2 = kxk2 + kyk2 implies
2(x·y) = 0 which means x·y = 0.
(b) From the first equation in the solution to Exercise 10 above, x·y = 0 implies kx+yk2 = kxk2 +kyk2 .
(12) Note that kx + y + zk2 = k(x + y) + zk2
= kx + yk2 + 2((x + y)·z) + kzk2
= kxk2 + 2(x·y) + kyk2 + 2(x·z) + 2(y·z) + kzk2
= kxk2 + kyk2 + kzk2 since x, y, z are mutually orthogonal.
(13) From the first two equations in the solution for Exercise 10 above,
1
4 (kx + yk2 − kx − yk2 ) = 14 (4(x·y)) = x·y.
(14) Since x is orthogonal to both y and z, we have x·(1 y + 2 z) = 1 (x·y) + 2 (x·z) = 1 (0) + 2 (0) = 0
(15) Suppose y = x, for some 6= 0 Then, x · y = x·(x) = (x · x) = kxk2 = kxk ( kxk) = kxk (±|| kxk)
= ± kxk kxk = ± kxk kyk
√ √
3
(16) (a) Length = 3 (b) angle = arccos( 3 ) ≈ 547◦ , or 0955 radians
(17) (a) proja b = [− 35 − 10
3
− 32 ]; b − proja b = [ 85 43 3
10 − 2 ]; (b − proja b) · a = 0
(b) proja b = [− 65 1 25 ]; b − proja b = [− 14 8
5 −4 5 ]; (b − proja b) · a = 0
(c) proja b = [ 16 0 − 61 13 ]; b − proja b = [ 17 1 4
6 −1 6 − 3 ]; (b − proja b) · a = 0
(d) proja b = [−1 23 −2 − 23 ]; b − proja b = [6 − 12 −6 72 ]; (b − proja b) · a = 0
(18) (a) 0 (zero vector). The dropped perpendicular travels along b to the common initial point of a and
b.
(b) The vector b. The terminal point of b lies on the line through a, so the dropped perpendicular
has length zero.
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,Answers to Exercises Section 1.2
(19) i, j, k
(20) (a) Parallel: [ 20 30 40 194 88 163
29 − 29 29 ], orthogonal: [− 29 29 29 ]
(b) Parallel: [− 12 1 − 21 ], orthogonal: [− 11 15
2 1 2 ]
(c) Parallel: [ 60 40 120 354 138 223
49 − 49 49 ], orthogonal: [− 49 49 49 ]
(21) From the lower triangle in the figure, we have (projr x) + (projr x − x) = reflection of x (see Figure
10).
Figure 10: Reflection of a vector x through a line.
(22) For the case kxk ≤ kyk: | kxk − kyk | = kyk − kxk = kx + y + (−x)k − kxk ≤ kx + yk + k − xk − kxk
(by the Triangle Inequality) = kx + yk + kxk − kxk = kx + yk.
The case kxk ≥ kyk is done similarly, with the roles of x and y reversed.
(23) (a) Note that projx y = [ 58 − 65 2] = 25 x and y − projx y = [ 75 − 24
5 −4] is orthogonal to x.
Let w = y − projx y Then since y = projx y + (y − projx y) we have y = 25 x + w where w is
orthogonal to x.
(b) Let = (x·y)(kxk2 ) (so that x = projx y), and let w = y − projx y which is orthogonal to x
by the argument before Theorem 1.11. Then y = x + w, where w is orthogonal to x.
(c) Suppose x+w = x+v. Then (−)x = w−v, and (−)x·(w−v) = (w−v)·(w−v) = kw−vk2 .
But ( − )x·(w − v) = ( − )(x·w) − ( − )(x·v) = 0, since v and w are orthogonal to x. Hence
kw − vk2 = 0 =⇒ w − v = 0 =⇒ w = v. Then, x = x =⇒ = , from Theorem 1.4, since x is
nonzero.
(24) If is the angle between x and y, and is the angle between projx y and projy x, then
³ ´ ³ ´ ³ ´³ ´
x·y y·x x·y y·x
projx y · projy x kxk 2 x · kyk 2 y kxk 2
kyk 2 (x · y)
cos = ° ° = °³ ´ ° °³ ´ °=³ ´ ³ ´
kprojx yk °projy x° ° x·y ° ° y·x
° kxk2 x° ° kyk 2 y°
° |x·y|
2 kxk |y·x|
2 kyk
kxk kyk
³ 3
´
(x·y)
kxk2 kyk2 (x · y)
= ³ ´ = = cos
(x·y)2 kxk kyk
kxkkyk
Hence = .
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, Answers to Exercises Section 1.3
(25) (a) T (b) T (c) F (d) F (e) T (f) F
Section 1.3
(1) (a) We have k4x + 7yk ≤ k4xk + k7yk = 4kxk + 7kyk ≤ 7kxk + 7kyk = 7(kxk + kyk).
(b) Let = max{|| ||}. Then kx ± yk ≤ (kxk + kyk).
(2) (a) Note that 6 − 5 = 3(2( − 1)) + 1. Let = 2( − 1).
(b) Consider the number 4.
(3) Note that since x 6= 0 and y 6= 0, projx y = 0 iff (x·y)(kxk2 ) = 0 iff x·y = 0 iff y·x = 0 iff
(y·x)(kyk2 ) = 0 iff projy x = 0.
(4) If y = x (for 0), then kx + yk = kx + xk = (1 + )kxk = kxk + kxk = kxk + kxk = kxk + kyk.
On the other hand, if kx + yk = kxk + kyk, then kx + yk2 = (kxk + kyk)2 . Now kx + yk2 =
(x + y)·(x + y) = kxk2 + 2(x·y) + kyk2 , while (kxk + kyk)2 = kxk2 + 2kxkkyk + kyk2 . Hence
x·y = kxkkyk. By Result 4, y = x, for some 0.
(5) (a) Suppose y 6= 0. We must show that x is not orthogonal to y. Now kx + yk2 = kxk2 , so
kxk2 + 2(x·y) + kyk2 = kxk2 . Hence kyk2 = −2(x·y). Since y 6= 0, we have kyk2 6= 0, and so
x·y 6= 0.
(b) Suppose x is not a unit vector. We must show that x·y 6= 1.
Now projx y = x =⇒ ((x·y)(kxk2 ))x = 1x =⇒ (x·y)(kxk2 ) = 1 =⇒ x·y = kxk2 .
But then kxk 6= 1 =⇒ kxk2 6= 1 =⇒ x·y 6= 1.
(6) (a) Consider x = [1 0 0] and y = [1 1 0].
(b) If x 6= y, then x · y 6= kxk2 .
(c) Yes
(7) See the answer for Exercise 11(a) in Section 1.2.
(8) If kxk kyk, then kxk2 kyk2 , and so kxk2 − kyk2 0 But then (x + y)·(x−y) 0 and so the
cosine of the angle between (x + y) and (x−y) is positive. Thus the angle between (x + y) and (x−y)
is acute.
(9) (a) Contrapositive: If x = 0, then x is not a unit vector.
Converse: If x is nonzero, then x is a unit vector.
Inverse: If x is not a unit vector, then x = 0.
(b) (Let x and y be nonzero vectors.)
Contrapositive: If y 6= projx y, then x is not parallel to y.
Converse: If y = projx y, then x k y.
Inverse: If x is not parallel to y, then y 6= projx y.
(c) (Let x, y be nonzero vectors.)
Contrapositive: If projy x 6= 0, then projx y 6= 0.
Converse: If projy x = 0, then projx y = 0.
Inverse: If projx y 6= 0, then projy x 6= 0.
(10) (a) Converse: Let x and y be nonzero vectors in R . If kx + yk kyk, then x·y ≥ 0.
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