Chapter 7 Cartesian Vectors Chapter 7 Prerequisite...

MHR • Calculus and Vectors 12 Solutions 679

Chapter 7 Cartesian Vectors Chapter 7 Prerequisite Skills Chapter 7 Prerequisite Skills Question 1 Page 358

Chapter 7 Prerequisite Skills Question 2 Page 358

Chapter 7 Prerequisite Skills Question 3 Page 358 a) 18°


b) 282°

c) 0°

d) 270°

e) 240°

f) 150°


g) 90°

h) 70°

i) 285°

j) 220°

k) 170°


l) 260°

Chapter 7 Prerequisite Skills Question 4 Page 358 a)

(3! 5)2+ (1! 6)2

= 4 + 25

= 29

The distance is 29 units. b)

(6 ! (!4))2+ (7 ! 3)2

= 102+ 42

= 116

= 2 29

The distance is 2 29 units. c)

(!5! (!1))2+ (8! 0)2

= (!4)2+ 82

= 80

= 4 5

The distance is 4 5 units. d)

(!3! 5)2+ (!9 ! (!2))2

= (!8)2+ (!7)2

= 113

The distance is 113 units.


Chapter 7 Prerequisite Skills Question 5 Page 358 Answers may vary. For example: Let a = 4, x = 3, y = 7, and z = 5. a) Verify: x + y = y + x.

L.S. = x + y = 3 + 7 = 10 R.S. = y + x = 7 + 3 = 10 Therefore, L.S. = R.S. In words, when adding two numbers, the order of the operation does not matter.

b) Verify: x × y = y × x.

L.S. = x × y = 3 × 7 = 21 R.S. = y × x = 7 × 3 = 21 Therefore, L.S. = R.S. In words, when multiplying two numbers, the order of the operation does not matter.

c) Verify: (x + y) + z = x + (y + z).

L.S. = (x + y) + z = (3 + 7) + 5 = 10 + 5 = 15 R.S. = x + (y + z) = 7 + (3 + 5) = 7 + 8 = 15 Therefore, L.S. = R.S. In words, when adding three numbers at a time, the grouping of the operations does not matter.


d) Verify: (x × y) × z = x × (y × z). L.S. = (x × y) × z = (3 × 7) × 5 = 21 × 5 = 105 R.S. = x × (y × z) = 7 × (3 × 5) = 7 × 15 = 105 Therefore, L.S. = R.S. In words, when multiplying three numbers at a time, the grouping of the operations does not matter.

e) Verify: a(x + y) = ax + ay.

L.S. = a(x + y) = 4(3 + 7) = 4(10) = 40 R.S. = ax + ay = 4(3) + 4(7) = 12 + 28 = 40 Therefore, L.S. = R.S. In words, when multiplying a binomial by a factor, you can multiply each of the terms in the binomial separately and then add the partial products together.

f) Verify: x – y ≠ y – x.

L.S. = x – y = 3 – 7 = –4 R.S. = y – x = 7 – 3 = 4 Therefore, L.S. ≠ R.S.

In words, when subtracting two numbers, the order of the operation does matter. When the order of operation is reversed the answer is of the opposite sign.

Chapter 7 Prerequisite Skills Question 6 Page 358 a) 5x + 3y = 11 2x + y = 4 –5x – 3y = –11 –1 6x + 3y = 12 3 x = 1 –1 + 3


Substitute x = 1 into equation . 2(1) + y = 4 y = 2 Therefore, (x, y) = (1, 2).

b) 2x + 6y = 14 x – 4y = –14 2x + 6y = 14 –2x + 8y = 28 –2 14y = 42 – 2 y = 3

Substitute y = 3 into equation .

x ! 4(3) = !14x = !2

Therefore, (x, y) = (–2, 3).

c) 3x – 5y = –5 –6x + 2y = 2 6x – 10y = –10 2 –6x + 2y = 2 –8y = –8 2 + y = 1


!6x + 2(1) = 2!6x = 0

x = 0

Therefore, (x, y) = (0, 1). d) –1.5x + 3.2y = 10 0.5x + 0.4y = 4 –1.5x + 3.2y = 10 1.5x + 1.2y = 12 3 4.4y = 22 + 3 y = 5


0.5x + 0.4(5) = 40.5x = 2

x = 4

Therefore, (x, y) = (4, 5).


Chapter 7 Prerequisite Skills Question 7 Page 358 B is the correct response. It is not equivalent.

A The equation can be multiplied by 13

to get this result.

B If you rearrange the equation into slope y-intercept form, the result is 9 6 18

6 9 18

9 18

6 6

33

2

x yy x

y x

y x

! =

! = ! +

!= +! !

= !

This is not the same as B.

C The equation can be multiplied by 13

! and rearranged to get this result.

D The equation can be multiplied by 19

to get this result.

9 6 18

9 6 18

9 9 9

22

3

x y

x y

x y

! =

! =

! =

Chapter 7 Prerequisite Skills Question 8 Page 359 Use the special triangles.

a) 21

b) 21

c) 23

d) 2

1

e) 2

1 f) 0 g) 1 h) 0

i) cos 120º = –cos 60º j) sin 300º = –sin 60º k) 1 l) –1

= 21

! = 23

!


Chapter 7 Prerequisite Skills Question 9 Page 359 a) 0.3 b) 0.7 c) −0.6 d) −0.9 e) 0.8 f) 0.8 Chapter 7 Prerequisite Skills Question 10 Page 359 a) Use the Pythagorean theorem

x2+ 52

= 132

x2= 169 ! 25

x2= 144

x = 12

Therefore, x = 12 cm. b) Use the Pythagorean theorem.

y2= 4.52

+ 8.92

y2= 20.25+ 79.21

y2= 99.46

y ! 9.97y ! 10.0

Therefore, y ! 10.0 cm.


a) osin3855sin38

h

h

=

= °

b) 2 2 23 4 5+ =

The missing side of the base triangle has a length of 4 units.

sin30! =

4h

h =

4sin30!

Chapter 7 Prerequisite Skills Question 12 Page 359 a)

(a1 + b1)2= (a1 + b1)(a1 + b1)

= a12

+ a1b1 + a1b1 + b12

= a12

+ 2a1b1 + b12


b)

(a1 + b1)(a1 ! b1) = a12! a1b1 + a1b1 ! b1

2

= a12! b1

2

c) (a1

2+ a2

2 )(b12

+ b22 ) = a1

2b12

+ a12b2

2+ a2

2b12

+ a22b2

2 d) (a1b1 + a2b2 + a3b3)3

= [(a1b1 + a2b2 + a3b3)(a1b1 + a2b2 + a3b3)](a1b1 + a2b2 + a3b3)

= [a12b1

2+ a1b1a2b2 + a1b1a3b3 + a2b2a1b1 + a2

2b22

+ a2b2a3b3 + a3b3a1b1 + a3b3a2b2 + a32b3

2 ](a1b1 + a2b2 + a3b3)

= a13b1

3+ a1

2a2b12b2 + a1

2a3b12b3 + a1

2a2b12b2 + a1a2

2b1b22

+ a1a2a3b1b2b3 + a12a3b1

2b3 + a1a2a3b1b2b3 + a1a32b1b3

2

+ a12a2b1

2b2 + a1a22b1b2

2+ a1a2a3b1b2b3 + a1a2

2b1b22

+ a23b2

3+ a2

2a3b22b3 + a1a2a3b1b2b3 + a2

2a3b22b3 + a2a3

2b2b32

+ a12a3b1

2b3 + a1a2a3b1b2b3 + a1a32b1b3

2+ a1a2a3b1b2b3 + a2

2a3b22b3 + a2a3

2b2b32

+ a1a32b1b3

2+ a2a3

2b2b32

+ a33b3

3

= a13b1

3+ a2

3b23+ a3

3b33+ 3a1

2a2b12b2 + 3a1a2

2b1b22

+ 3a12a3b1

2b3 + 3a1a32b1b3

2+ 3a2

2a3b22b3 + 3a2a3

2b2b32

+ 6a1a2a3b1b2b3


a)

A =12

0(2)! 0(0) + 0(2)! 2(2) + 2(1)! 3(2) + 3(!1)! 3(1) + 3(0)! 0(0)

=12!14

= 7

The area is 7 square units.

b)

A =12

1(!2)! 3(1) + 3(!4)! 0(!2) + 0(0)! (!5)(!4) + (!5)(4)! (!3)(0) + (!3)(1)!1(4)

=12!64

= 32


c)

A =12

4(!7)! 5(!10) + 5(2)! 3(!7) + 3(5)! 0(2) + 0(3)! (!2)(5) + (!2)(!10)! 4(3)

=12

86

= 43



Chapter 7 Prerequisite Skills Question 13 Page 359 Yes. There are similar formulas for triangles and hexagons. In fact, there are similar formulas for any polygon that does not intersect itself. For a triangle with vertices (x1,y1), (x2,y2), and (x3,y3), the area is

1 2 2 1 2 3 3 2 3 1 1 312

A x y x y x y x y x y x y= ! + ! + !

Answers may vary. For example, Consider a triangle with vertices (1, 1), (2, 7), and (–3, –4), the area is

A =12

1(7)! 2(1) + 2(!4)! (!3)(7) + (!3)(1)!1(!4)

=12

19

= 9.5

The area is 9.5 square units. For a hexagon with vertices (x1,y1), (x2,y2), (x3,y3), (x4,y4), (x5,y5), and (x6,y6), the area is

1 2 2 1 2 3 3 2 3 4 4 3 4 5 5 4 5 6 6 5 6 1 1 6

1

2A x y x y x y x y x y x y x y x y x y x y x y x y= ! + ! + ! + ! + ! + !

For a hexagon with vertices (0, 0), (3, 2), (2, 4), (0, 5), (–2, 4), and (–3, 2), the area is

A =12

0(2)! 3(0) + 3(4)! 2(2) + 2(5)! 0(4) + 0(4)! (–2)(5) + (–2)(2)! (!3)(4) + (!3)(0)! 0(2)

=12

36

= 18

The area is 18 square units. Chapter 7 Section 1 Cartesian Vectors Chapter 7 Section 1 Question 1 Page 367 a) ji +2 b) ji 53 ! c) ji 63 !! d) i5 e) ji 79 ! f) j8! g) 6i!

! i) 5.2 6.1i j! !

! !


Chapter 7 Section 1 Question 2 Page 367 a) [1, 1] b) [–4, 0] c) [0, 2] d) [3, 8] e) [–5, –2] f) [7, –4] g) [0, –8.2] h) [–2.5, 3.3] Chapter 7 Section 1 Question 3 Page 367 a) AB! "!!

= [2, –5] b) CD! "!!

= [5, 1] c) EF! "!

= [0, 7] d) GH! "!!

= [–2, 9] Chapter 7 Section 1 Question 4 Page 367 a)

AB! "!!

= 22+ (!5)2

= 29

The magnitude of the vector is 29 units.

b)

CD! "!!

= 52+12

= 26


c)

EF! "!

= 02+ 72

= 7


d)

GH! "!!

= !2( )2

+ 92

= 85


Chapter 7 Section 1 Question 5 Page 368 a) [ ]0,5=hv

[ ]1,0 !=vv


b) The magnitude of ,v v! !

, is 26 units.

Unit vectors collinear with v!

have the form 1 1 and v vv v

!

! !

! ! .

Therefore, the required unit vectors are !"

#$%

&'

261,

265

and !"

#$%

&'

261,

265

.

c) Let Q(x, y) be the point.

PQ! "!!!!!

= OQ! "!!

!OP! "!!

5,!1"# $% = [x, y]! !2,!7"# $%

5,!1"# $% = x + 2, y + 7"# $%

& x + 2 = 5 and y + 7 = !1x = 3 y = !8

The point is Q(3, –8).

d) Let L(x, y) be the point.

LM! "!!!!!

= OM! "!!!

!OL! "!!

5,!1"# $% = [5,8]! x, y"# $%

5,!1"# $% = 5! x,8! y"# $%

&5! x = 5 and 8! y = !1x = 0 y = 9

The point is L(0, 9).

Chapter 7 Section 1 Question 6 Page 368 a)

QP! "!!

= OP! "!!

!OQ! "!!

= !6,1"# $% ! !2,!1"# $%

= !4,2"# $%


b)

RP! "!!

= !6,1"# $% ! !3,4"# $%

= !3,!3"# $%

RP! "!!

= (!3)2+ (!3)2

= 18

= 3 2

The vector is 3 2 units.

c)

QP! "!!

= (!4)2+ 22

= 20

QR! "!!

= ((!3)! (!2))2+ (4 ! (!1))2

= 26

18RP =

!!!"

The perimeter is 18 units + 26 units + 20 units ! 13.8 units.


8u!

= 8 4,!1"# $%

= 32,!8"# $%

b)

!8u!

= ! 32,!8"# $%

= !32,8"# $%

c)

u!

+ v!

= 4,!1"# $% + 2,7"# $%

= 6,6"# $%

d)

v!

! u!

= v!

+ !u!

( )

= 2,7"# $% + !4,1"# $%

= !2,8"# $%

e)

5u!

! 3v!

= 5 4,!1"# $% ! 3 2,7"# $%

= 20,!5"# $% + !6,!21"# $%

= 14,!26"# $%


f)

!4u!

+ 7v!

= !4 4,!1"# $% + 7 2,7"# $%

= !16,4"# $% + 14,49"# $%

= !2,53"# $%

Chapter 7 Section 1 Question 8 Page 368 B is the correct response.

a!

= 2b!

= !d"!

= !23

e!

Therefore,

a!

, b!

, d"!

, and e!

are collinear vectors. Chapter 7 Section 1 Question 9 Page 368 a)

v!

= 500cos30o , 500sin30o!"

#$

= 433.0, 250!" #$

b)

v!

= 1000cos18o , 1000sin18o!"

#$

= 951.1, 309.0!" #$

c) [ ]0, 125v =

!

d)

v!

= 230, 0!" #$ e) [ ]0, 25v = !

!

f) [ ]650, 0v = !

!

Chapter 7 Section 1 Question 10 Page 368 The ship’s vector is

30cos(!102o ), 30sin(!102o )"#

$% = !6.237, ! 29.344"# $% .

The current’s vector is [14cos 158o , 14sin 158o ] = !12.981, 5.244"# $% . The resultant vector is [–19.219, –24.100]. The magnitude of the resultant is

(!19.219)2+ (!24.100)2

! 30.825 .

The angle of the resultant with the west direction is ! tan!1 24.100

19.219"

#$%

&'! !51.429o . (The resultant is pointing

SW.)


The bearing is 270º – 51.4º = 218.6º. The resultant is about 30.8 km/h on a bearing of 218.6°. Chapter 7 Section 1 Question 11 Page 368 11. a)

b)

a!

+ b!

= 42+ 72

+ 22+ (!9)2

= 57 + 85"= 17.3

a!

+ b!

= 6,!2"# $%

= 62+ 22

= 40"= 6.3

ba + is greater.

c) Answers may vary. For example: Yes, this will be true for all pairs of vectors, unless the two vectors are

collinear. Then a b a b+ = +

! ! ! !.



(7, 2) a

b

b) 10=+ !ba

a!

! b!

= !8, 4"# $%

= (!8)2+ 42

= 80"= 8.9

ba + is greater.

c) Answers may vary. For example: Yes this will be true for all pairs of vectors separated by an acute angle.

If the angle is 90º, a b a b+ = !

! ! ! !.

If the angle is obtuse, then a b a b+ < !

! ! ! !.



L.S. = u!

+ v!

( ) + w"!

= u1,u2!" #$ + v1,v2!" #$( ) + w1, w2!" #$

= u1 + v1, u2 + v2!" #$( ) + w1, w2!" #$

= u1 + v1 + w1, u2 + v2 + w2!" #$

R.S. = u!

+ v!

+ w"!

( )

= u1,u2!" #$ + v1,v2!" #$ + w1, w2!" #$( )

= u1,u2!" #$ + v1 + w1, v2 + w2!" #$

= u1 + v1 + w1, u2 + v2 + w2!" #$

L.S. = R.S.

Therefore,

u!

+ v!

( ) + w"!

= u!

+ v!

+ w"!

( ) .

b)

L.S. = k u!

+ v!

( )

= k u1, u2!" #$ + v1, v2!" #$( )

= k u1 + v1, u2 + v2!" #$

= k u1 + v1( ) , k u2 + v2( )!"

#$

= ku1 + kv1, ku2 + kv2!" #$

R.S. = ku!

+ kv!

= k u1, u2!" #$ + k v1, v2!" #$

= ku1, ku2!" #$ + kv1, kv2!" #$

= ku1 + kv1, ku2 + kv2!" #$

L.S. = R.S.

Therefore,

k u!

+ v!

( ) = ku!

+ kv!

.


c)

L.S. = u!

+ v!

= u1, u2!" #$ + v1, v2!" #$

= u1 + v1, u2 + v2!" #$

R.S. = v!

+ u!

= v1, v2!" #$ + u1, u2!" #$

= v1 + u1, v2 + u2!" #$

= u1 + v1, u2 + v2!" #$

Therefore, u

!

+ v!

= v!

+ u!

.

d)

L.S.= k + m( )u!

= k + m( ) u1, u2!" #$

= k + m( )u1, k + m( )u2!"

#$

= ku1 + mu1, ku2 + mu2!" #$

R.S. = ku!

+ mu!

= k u1, u2!" #$ + m u1, u2!" #$

= ku1, ku2!" #$ + mu1, mu2!" #$

= ku1 + mu1, ku2 + mu2!" #$

Therefore,

k + m( )u!

= ku!

+ mu!

. Chapter 7 Section 1 Question 14 Page 368

F!"

= 180cos 30o , 180sin 30o!"

#$

# 155.9, 90!" #$

Chapter 7 Section 1 Question 15 Page 368

F!"

= 250cos(!35o ), 250sin(!35o )"#

$%

# 204.8, !143.4"# $%


Chapter 7 Section 1 Question 16 Page 368 The airplane’s vector is

550cos 10o , 550sin 10o!"

#$ = 541.644, 95.506!" #$ .

The wind’s vector is [60cos 150o , 60sin 150o ] = !51.962, 30"# $% . The resultant vector is [489.682, 125.506].

The magnitude of the resultant is (489.682)2

+ (125.506)2! 505.51.

The angle of the resultant with the east direction is tan!1 125.506

489.682"

#$%

&'! 14.376o . (The resultant is

pointing NE.) The bearing is 90º – 14.4º = 75.6º. The resultant is about 505.5 km/h on a bearing of 075.6°. Chapter 7 Section 1 Question 17 Page 369

The vector for Emily’s kick is [120cos 60o , 120sin 60o ] = 60, 103.92!" #$ .

The vector for Claire’s kick is [200cos 120o , 200sin 120o ] = !100, 173.21"# $% . The resultant vector is [–40, 277.13].

The magnitude of the resultant is

!40( )2

+ 277.13( )2! 280.0 .

The angle of the resultant is tan!1 277.13

!40"

#$%

&'! !81.79o .

The correct angle is (the opposite direction) 180º – 81.8º = 98.2º. The resultant is about 280.0 N at an angle of 98.2° to the line between the centres of the goals. This is the same answer as for question 11 in section 6.4.

200 N

120 N

60º 60º

120º


Chapter 7 Section 1 Question 18 Page 369 Draw a diagram.

Use east as the reference direction. The three force vectors are [608, 0], [550cos 120º, 550sin 120º], and [0, –700]. The resultant vector is [333, –223.7]. The magnitude of the resultant is

3332+ (!223.7)2

! 401.2 .

The resultant angle is tan!1 !223.7

333"

#$%

&'! !33.9o .

The resultant force on the basketball is about 401.2 N in a direction 33.9º below Sam’s force (towards Nick’s force).

608 N Sam

700 N Nick

550 N Jason

120º 150º

90º



Place the 15 000 kg·m/s2 vector along the positive x–axis. The two vectors are [15 000, 0] and [12 000cos 15º, 12 000sin 15º]. The sum of the vectors is [26 591.1, 3105.8]. The magnitude of the resultant is

(26 591.1)2+ (3105.8)2

! 26 771.9 .

The resultant angle is tan!1 3105.8

26 591.1"

#$%

&'! 6.66o .

The resultant momentum is about 26 771.9 kg·m/s2 at an angle of 6.7º to the 15 000 kg·m/s2 vector. Chapter 7 Section 1 Question 20 Page 369

The three vectors are [4, 0], [0, 2], and [0, –5]. The vector sum is [4, –3].

a) The magnitude of the resultant is 42+ (!3)2

= 5" .

b) The angle of the resultant is tan!1 !3

4"

#$%

&'! !36.9o .

The resultant makes a 36.9º below the positive x–axis.

12 000 kg·m/s2

15º 15 000 kg·m/s2

4Ω

2Ω

5Ω


Chapter 7 Section 1 Question 21 Page 369 Draw a diagram.

The resultant vector must be

50012

cos 42o ,50012

sin 42o!

"#

$

%& .

The current vector is [15cos 58º, 15sin 58º]. The heading vector is the difference between the resultant vector and the current vector.

h!

=50012

cos 42o ,50012

sin 42o!

"#

$

%& ' [15cos 58º, 15sin 58º]

" [23.016, 15.160]

h!

= (23.016)2+ (15.160)2

" 27.560

! = tan"1 15.16023.016

#

$%&

'(

! 33.37

The bearing is 90º – 33.37º = 56.63º. The captain should set a speed of 27.6 km/h at a bearing of 056.6º.

50012

km/h

42º

32º

15 km/h

h!

θ


Chapter 7 Section 1 Question 22 Page 369 a) Solve for x.

x2+ (3x)2

= 9

10x2= 81 Square both sides.

x2=

8110

x = ±910

x = ±9 10

10

b)

2x, x!" #$ + x, 2x!" #$ = 6

3x, 3x!" #$ = 6

(3x)2+ (3x)2

= 6

18x2= 6

18x2= 36

x2= 2

x = ± 2

c)

3x, 7!" #$ + 5x, x!" #$ = 10x

8x, x + 7!" #$ = 10x

(8x)2+ (x + 7)2

= 10x

64x2+ x2

+14x + 49 = 10x

65x2+14x + 49 = 10x

65x2+14x + 49 = 100x2

35x2 %14x % 49 = 05x2 % 2x % 7 = 0

(5x % 7)(x +1) = 0

x =

75

or x = !1


Chapter 7 Section 1 Question 23 Page 369 Yes. Consider one unit vector starting at the origin. Its tip will lie somewhere on the unit circle (circle with centre the origin and radius one unit). Draw a second unit vector starting at the tip of the first vector and having its tip also on the unit circle. The resulting sum vector will start at the origin and have its tip on the unit circle. Clearly this is a unit vector.

Chapter 7 Section 1 Question 24 Page 369 Find the x-intercept of the circle.

(x ! 7)2+ (0 ! 4)2

= 25(x ! 7)2

= 9x ! 7 = ±3

x = 10 or x = 4 The centre of the circle is (7, 4). Solve for k. Set y = 0.

0 = 3(10)2! 42(10) + k

k = 120

Find the vertex of the parabola by completing the square.

y = 3x2! 42x +120

y = 3(x2!14x) +120

y = 3(x2!14x + 49 ! 49) +120

y = 3(x ! 7)2! 3(49) +120

y = 3(x ! 7)2! 27

The vertex of the parabola is (7, –27). The distance from the centre of the circle to the vertex of the parabola is 4 – (–27) or 31 units.

1v!

2v!

1 2v v+

! ! O


Chapter 7 Section 1 Question 25 Page 369 Method 1: Use a trigonometric identity. Let !A = 40o and !B = 20o .

sin A + sin B = 2sinA + B

2!

"#$

%&cos

A ' B2

!

"#$

%&

sin 40° + sin 20° = 2sin30°cos10°

= 212

!

"#$

%&sin 90°'10°( )

= sin80°

Method 2: Start with an equilateral triangle.

Using the sine law in the two smaller triangles, you get

ysin60o

=1

sin80o

=1! x

sin 20o

=x

sin 40o

From the second and last fractions,

o

o

sin 40sin80

x =

From the third and last actions,

o

o

sin 201sin 40

x! =

Adding these two equation together,

o o

o o

o o o

sin 40 sin 201sin80 sin80

sin80 sin 40 sin 20

= +

= +

60º

60º

20º

40º

80º

100º

1 1 y

1 – x

x


Chapter 7 Section 2 Dot Product Chapter 7 Section 2 Question 1 Page 375 a)

a!

!b!

= 70(115)cos70o

" 2753.3

b)

c!

!d"!

= 8(12)cos150o

# "83.1

c)

e!

! f"!

= 200(150)cos90o

= 0

d)

g!"

!h"

= 5000(4500)cos180o

# "22 500 000


u!

!v!

= 6(10)cos30o

" 52.0

b)

s!

! t!

= 30(15)cos120o

= "225

c)

f!"

! g!"

= 5.8(13.4)cos180o

# "77.7

d)

q!

! r!

= 4.0(6.1)cos90o

= 0

e)

a!

!b!

= 850(400)cos58o

" 180 172.5

f)

m!"

! p!"

= 16(2)cos153o

# "28.5


Chapter 7 Section 2 Question 3 Page 376 Answers may vary. For example:

a)

u!

! ku!

+ v!

( ) = k u!

!u!

( ) + u!

!v!

= k u! 2

+ u!

!v!

b)

ku!

! v!

( ) " lv!

( ) = kl u!

"v!

( )! l v!

"v!

( )

= kl u!

"v!

( )! l v! 2

c)

u!

! v!

( ) " u!

! v!

( ) = u!

"u!

! u!

"v!

! v!

"u!

+ v!

"v!

= u! 2

! 2u!

"v!

+ v! 2

d)

u!

+ v!

( ) ! w"!

+ x!

( ) = u!

!w"!

+ u!

! x!

+ v!

!w"!

+ v!

! x!


u!

!v!

= 2(3) + 4("1)= 2

b)

m!"

!n"

= "5(0) + ("7)(7)= "49

c)

s!

! t!

= 9(3) + (–3)(–3)= 36

d)

p!"

!q"

= "6(9) + 2(1)= "52

e)

a!

!b!

= 2(9) + 3("7)= "3

f)

s!

! t!

= 4("1) +1("1)= "5

Chapter 7 Section 2 Question 5 Page 376 a) wv ! is a scalar; a vector dotted with a scalar is not defined. b) This expression has meaning. It is the absolute value of a scalar (the dot product).


c) This expression has meaning. wv ! is a scalar. This is a vector multiplied by a scalar. d) This expression has meaning. This is the magnitude of a vector, squared. e) This expression does not have meaning. Multiplying two vectors has not been defined. f) This expression has meaning. u v!

! ! is a scalar which can be squared.

Chapter 7 Section 2 Question 6 Page 376 a) Using geometric vectors:

i!

! j!

= i!

j!

cos"

= 1(1)cos90°

= 1(0)= 0

Using Cartesian vectors:

i!

! j!

= 1, 0"# $% ! 0, 1"# $%

= 1(0) + 0(1)= 0 + 0= 0

b) The dot product of the two unit vectors i and j is 0 using either method. This will be true for any two

vectors. Chapter 7 Section 2 Question 7 Page 376 a)

u!

! v!

+ w"!

( ) = 3, " 5#$ %& ! "6, 1#$ %& + [4, 7]( )

= 3, " 5#$ %& ! "2, 8#$ %&

= 3("2) + ("5)(8)= "46

b)

u!

!v!

+ v!

!w"!

= 3, " 5#$ %& ! "6, 1#$ %& + "6, 1#$ %& ! 4, 7#$ %&

= 3("6) + ("5)(1) + ("6)(4) +1(7)= "40

c)

u!

+ v!

( ) ! u!

" v!

( ) = 3, " 5#$ %& + "6, 1#$ %&( ) ! 3, " 5#$ %& " ["6, 1]( )

= "3, " 4#$ %& ! 9, " 6#$ %&

= ("3)(9) + ("4)("6)= "3


d) This is not possible. It is the sum of a vector and a scalar, which is not defined. e)

!3v!

"2w"!

= !3 !6, 1#$ %&( ) " 2[4, 7]( )

= 18, ! 3#$ %& " 8, 14#$ %&

= 18(8) + (!3)(14)= 102

f)

5u!

! 2v!

" w"!

( ) = 5 3, " 5#$ %& ! 2 "6, 1#$ %& " [4, 7]( )

= 15, " 25#$ %& ! "16, " 5#$ %&

= 15("16) + ("25)("5)= "115

g) This is not possible. The dot product of three vectors has not been defined, nor has the dot product of a

scalar and a vector.

h)

u!

+ 2v!

( ) ! 3w"!

" u!

( ) = 3, " 5#$ %& + 2 "6, 1#$ %&( ) ! 3 4, 7#$ %& " [3, " 5]( )

= "9, " 3#$ %& ! 9, 26#$ %&

= ("9)(9) + ("3)(26)= "159

i)

u!

!u!

= 3, " 5#$ %& ! 3, " 5#$ %&

= 3(3) + ("5)("5)= 34

j)

v!

!v!

+ w"!

!w"!

= "6, 1#$ %& ! "6, 1#$ %& + 4, 7#$ %& ! 4, 7#$ %&

= ("6)("6) +1(1) + 4(4) + 7(7)= 102

Chapter 7 Section 2 Question 8 Page 376 a) The vectors for the three sides of !ABC are

AB! "!!

= !5, 2"# $% , BC! "!!

= 7, 3"# $% , and AC! "!!

= 2, 5"# $% .

AB! "!!

!AC! "!!

= ("5)(2) + 2(5)= 0

Since the dot product is zero, !ABC is a right-angle triangle where ! BAC is the right angle. The vectors for the three sides of ! STU are

ST! "!

= !7, 1"# $% , TU! "!!

= !2, !11"# $% , and SU! "!!

= !9, !10"# $% .

Since none of the pairs of vectors have a dot product of zero, ! STU is not a right-angled triangle.


b) The problem could be solved by plotting the points on a sheet of graph paper, then calculating the slopes of each of the sides. If the slopes of any of the two sides in the triangle are negative reciprocals, then the two sides would meet at a 90° angle and therefore the triangle would be a right-angled triangle.

Another solution would involve calculating the lengths of the three side vectors and determining if these three numbers satisfy the Pythagorean theorem relationship.

Chapter 7 Section 2 Question 9 Page 376 a) The two vectors have the same components in reverse order and the value of x in the second vector is

opposite to the value of y in the first vector. b) The two vectors are perpendicular. c) Using algebra,

7, ! 3"# $% & 3, 7"# $% = 7(3) + (!3)(7)

= 0

Therefore, the vectors are perpendicular. Chapter 7 Section 2 Question 10 Page 376 Many vectors are possible. One possibility is [4, –18].

9, 2!" #$ % 4, &18!" #$ = 9(4) + 2(&18)

= 0

Since the dot product is zero, [4, –18] is perpendicular to [9, 2]. Chapter 7 Section 2 Question 11 Page 376 If and u v! !

are perpendicular, then 0u v! =

! !.

and u v! !

are perpendicular if k = –10.

Chapter 7 Section 2 Question 12 Page 376 If and u v! !

are perpendicular, then 0u v! =

! !.

k, 3!" #$ % k, 2k!" #$ = 0

k 2+ 6k = 0

k(k + 6) = 0

k = 0 or k = 6

and u v! !

are perpendicular if k = –6. If k = 0, the vectors are [0, 3] and [0, 0]. These two vectors are not usually considered to be perpendicular, even though their dot product is zero.

2, 5!" #$ % k, 4!" #$ = 0

2k + 20 = 0k = &10


Chapter 7 Section 2 Question 13 Page 376 a) Use vectors [2, 3] and [3, –2].

If these non-zero vectors are plotted on grid paper, tail to tail, they will form a 90º angle. Also,

2, 3!" #$ % 3, & 2!" #$ = 2(3) + 3(–2)

= 0

b) Use vectors [ ] [ ]3, 5 and 2, 1a b= ! =

! !.

L.S. = a!

!b!

= 3, " 5#$ %& ! 2, 1#$ %&

= 3(2) + (–5)(1)= 1

R.S. = b!

!a!

= 2, 1#$ %& ! 3, " 5#$ %&

= 2(3) +1(–5)= 1

Therefore, L.S. = R.S. and a b b a! = !

! ! ! !.

c) Let 1 3, and 1, 32 2

u a! "

! "= =# $ % &% &

! !.

Since 2a u=

! !, the vectors are collinear.

u!

=12

!

"#$

%&

2

+3

2

!

"#

$

%&

2

=14

+34

= 1

Since !u = 1, u

! is a unit vector.


L.S. = a!

!u!

= 1, 3"#

$% !

12

,3

2

"

#&&

$

%''

= 112

(

)*+

,-+ 3

32

(

)*

+

,-

=12

+32

= 2

R.S. = a!

= 12+ 3( )

2

= 1+ 32

Therefore, L.S. = R.S. and a u a! =

! ! !.

Chapter 7 Section 2 Question 14 Page 376 a) Since this is an “if and only if” statement, two proofs are required.

Proof 1 (! ) Let [ ] [ ]1 2 1 2, and , so that 0a a a b b b a b= = ! =

! ! ! !.

Show a b!

! !.

Since 0a b! =

! !,

a1b1 + a2b2 = 0a1b1 = !a2b2

b2

b1

= !a1

a2

This statement says that the slope of b

! is the negative reciprocal of the slope of a

!.

Therefore, a b!

! !.

Proof 2 (! ) Let [ ] [ ]1 2 1 2, and , so that a a a b b b a b= = !

! ! ! !.

Show that 0a b! =

! !.

Since a b!

! !, consider the slopes of these two vectors; the slopes must be negative reciprocals of each

other.


b2

b1

= !a1

a2

a1b1 = !a2b2

a1b1 + a2b2 = 0

Therefore, 0a b! =

! !.

b) Let [ ] [ ]1 2 1 2, and ,a a a b b b= =

! ! be any two vectors.

L.S. = a!

!b!

= a1, a2"# $% ! b1, b2"# $%

= a1b1 + a2b2

R.S. = b!

!a!

= b1, b2"# $% ! a1, a2"# $%

= b1a1 + b2a2

= a1b1 + a2b2 commutuative property for multiplication in "

Therefore, L.S. = R.S. and a b b a! = !

! ! ! !.

c) Let [ ]

2 21 2 1 2, where 1u u u u u= + =

!.

Let

a!

= ku!

= ku1,ku2!" #$

L.S. = a!

!u!

= ku1, ku2"# $% ! u1, u2"# $%

= ku12

+ ku22

R.S. = a!

= ku1, ku2"# $%

= ku1( )2

+ ku2( )2

= k 2u12

+ k 2u22

= k 2 u12

+ u22

( )

= k 2 (1)= k

Therefore, L.S. = R.S. and a u a! =

! ! !.


Chapter 7 Section 2 Question 15 Page 377 Let three vectors be [ ]3, 0a =

!, [ ]0, 5b =

!, and [ ]0, 7c =

!.

In this case, 0a b! =

! ! and 0a c! =

! !, but b c!

! !.

Therefore, if a b a c! = !

! ! ! !, it is not always true that b c=

! !.

Chapter 7 Section 2 Question 16 Page 377 Use Cartesian vectors. Consider a circle with centre (0, 0) and radius a. Its equation will be 2 2 2x y a+ = .

AC! "!!

= [x + a, a2 ! x2 ] and BC! "!!

= x ! a, a2 ! x2"#$

%&'

.

AC! "!!

!BC! "!!

= (x + a)(x " a) + a2" x2 a2

" x2

= x2" a2

+ a2" x2

= 0

Therefore, AC

! "!!

! BC! "!!

and !ACB is a right angle. Alternative proof using geometric vectors.

CA! "!!

!CB! "!!

= CO! "!!

+ OA! "!!

( ) ! CO! "!!

+ OB! "!!

( )

= CO! "!!

+ OA! "!!

( ) ! CO! "!!

– OA! "!!

( )

= CO! "!!

!CO! "!!

– CO! "!!

!OA! "!!

+ CO! "!!

!OA! "!!

– OA! "!!

!OA! "!!

= CO! "!! 2

– OA! "!! 2

= 0 Since CO! "!!

= OA! "!!

because they are radii.

Since CA

! "!!

!CB! "!!

= 0 , therefore the angle between CA! "!!

and CB! "!!

is 90°. Therefore, ! ACB is a right angle.

A(–a, 0) B(a, 0)

C(x, 2 2a x! )


Chapter 7 Section 2 Question 17 Page 377 Let [ ] [ ] [ ]1, 2 , 3, 4 , 1, 4 , and 2u v w k= = = ! ! =

! ! "!.

a)

L.S. = ku!

( ) !v!

= 2 1, 2"# $%( ) ! 3, 4"# $%

= 2, 4"# $% ! 3, 4"# $%

= 2(3) + 4(4)= 22

M.S. = k u!

!v!

( )

= 2 1, 2"# $% ! 3, 4"# $%( )

= 2(1(3) + 2(4))= 2(11)= 22

R.S. = u!

! kv!

( )

= 1, 2"# $% ! 2 3, 4"# $%( )

= 1, 2"# $% ! 6, 8"# $%

= 1(6) + 2(8)= 22

Clearly all three sides are equal.

ku!

( ) !v!

= k u!

!v!

( )

= u!

! kv!

( )

b)

L.S. = u!

! v!

+ w"!

( )

= 1, 2"# $% ! 3, 4"# $% + &1, & 4"# $%( )

= 1, 2"# $% ! 2, 0"# $%

= 1(2) + 2(0)= 2

R.S. = u!

!v!

+ u!

!w"!

= 1, 2"# $% ! 3, 4"# $% + 1, 2"# $% ! &1, & 4"# $%

= 1(3) + 2(4) +1(–1) + 2(–4)= 2

Therefore, u!

! v!

+ w"!

( ) = u!

!v!

+ u!

!w"!

.


c)

L.S. = u!

+ v!

( ) !w"!

= 1, 2"# $% + 3, 4"# $%( ) ! &1, & 4"# $%

= 4, 6"# $% ! &1, & 4"# $%

= 4(&1) + 6(&4)= &28

R.S. = u!

!w"!

+ v!

!w"!

= 1, 2"# $% ! &1, & 4"# $% + 3, 4"# $% ! &1, & 4"# $%

= 1(&1) + 2(&4) + 3(&1) + 4(&4)= &28

Therefore,

u!

+ v!

( ) !w"!

= u!

!w"!

+ v!

!w"!

.

Chapter 7 Section 2 Question 18 7 Let [ ] [ ] [ ]1 2 1 2 1 2, , , , and ,u u u v v v w w w= = =

! ! "!.

a)

L.S. = ku!

( ) !v!

= k u1, u2"# $%( ) ! v1, v2"# $%

= ku1, ku2"# $% ! v1, v2"# $%

= ku1v1 + ku2v2

M.S. = k u!

!v!

( )

= k u1, u2"# $% ! v1, v2"# $%( )

= k u1v1 + u2v2( )

= ku1v1 + ku2v2

R.S. = u!

! kv!

( )

= u1, u2"# $% ! k v1, v2"# $%( )

= u1, u2"# $% ! kv1, kv2"# $%

= ku1v1 + ku2v2

Clearly all three sides are equal and

ku!

( ) !v!

= k u!

!v!

( ) = u!

! kv!

( ) .


b)

L.S. = u!

! v!

+ w"!

( )

= u1, u2"#

$% ! v1, v2"# $% + w1, w2"# $%( )

= u1, u2"#

$% ! v1 + w1, v2 + w2"# $%

= u1(v1 + w1) + u2 (v2 + w2 )= u1v1 + u1w1 + u2v2 + u2w2

R.S. = u!

!v!

+ u!

!w"!

= u1, u2"#

$% ! v1, v2"# $% + u1, u2

"#

$% ! w1, w2"# $%

= u1v1 + u2v2 + u1w1 + u2w2

= u1v1 + u1w1 + u2v2 + u2w2

Therefore,

u!

! v!

+ w"!

( ) = u!

!v!

+ u!

!w"!

.

c)

L.S. = u!

+ v!

( ) !w"!

= u1, u2"# $% + v1, v2"# $%( ) ! w1, w2"# $%

= u1 + v1, u2 + v2"# $% ! w1, w2"# $%

= (u1 + v1)w1 + (u2 + v2 )w2

= u1w1 + v1w1 + u2w2 + v2w2

R.S. = u!

!w"!

+ v!

!w"!

= u1, u2"# $% ! w1, w2"# $% + v1, v2"# $% ! w1, w2"# $%

= u1w1 + u2w2 + v1w1 + v2w2

= u1w1 + v1w1 + u2w2 + v2w2

Therefore,

u!

+ v!

( ) !w"!

= u!

!w"!

+ v!

!w"!

.

Chapter 7 Section 2 Question 19 Page 377 Solutions for Achievement Checks are shown in the Teacher’s Resource. Chapter 7 Section 2 Question 20 Page 377 a) P = V

!"

! I"

b)

P = V!"

! I"

= 120(5)cos15o

# 579.6

The power is approximately 579.6 W.


Chapter 7 Section 2 Question 21 Page 377 a) Let Q(x, y) be any point on the locus.

OQ! "!!

= x, y!" #$ and OP! "!!

= 5, 5!" #$ .

OP! "!!

!OQ! "!!

= 05x + 5y = 0

y = "x

The set of points Q will form the line with equation y x= ! .

b) Let Q(x, y) be any point on the locus.

OQ! "!!

= x, y!" #$ and OP! "!!

= 5, 5!" #$ .

OP! "!!

!OQ! "!!

= "2

2

5x + 5y = "2

2

y = "x " 210

The set of points Q form a line perpendicular to OP and have a y-intercept of 210

! .

c) Let Q(x, y) be any point on the locus.

OQ! "!!

= x, y!" #$ and OP! "!!

= 5, 5!" #$ .

OP! "!!

!OQ! "!!

= "3

2

5x + 5y = "3

2

y = "x " 310

The set of points Q form a line perpendicular to OP and have a y-intercept of 310

! .


Chapter 7 Section 2 Question 22 Page 377 There are 57 impossible totals less than $5.00. This problem is best done with a spreadsheet, taking care to round off answers correctly. The impossible totals are: 0.04 1.00 2.04 3.00 4.04 0.13 1.09 2.13 3.09 4.13 0.22 1.17 2.22 3.17 4.22 0.30 1.26 2.30 3.26 4.30 0.39 1.35 2.39 3.35 4.39 0.48 1.43 2.48 3.43 4.48 0.56 1.52 2.56 3.52 4.56 0.65 1.61 2.65 3.61 4.65 0.74 1.69 2.74 3.69 4.74 0.83 1.78 2.82 3.78 4.82 0.91 1.87 2.91 3.87 4.91 1.96 3.95 (5.00) Note that this is not what really happens in practice. The PST and GST are calculated and rounded off separately. Then they are added together. In the “real” situation 0.07 and 0.11 are impossible totals, while 0.04 and 0.13 are possible. The “real” answer to this problem is left as a challenge. Chapter 7 Section 2 Question 23 Page 377 No. The truck cannot make it under the bridge. Set up axes so that the vertex of the parabola is at (0, 5) and the x-intercepts are 3 and –3. The equation of the parabola is of the form 2 5; where 0y ax a= + < . Substitute (3, 0) to determine a.

0 = a(3)2+ 5

a = !59

The equation of the parabola is 25 59

y x= ! + .

For the truck to pass safely, (1.5, 4) must be below the curve. Let x = 1.5, then y = 3.75. The truck needs at least 0.25 m of extra clearance to make it under the bridge.


Chapter 7 Section 3 Applications of the Dot Product Chapter 7 Section 3 Question 1 Page 384 a)

F!"

! s"

= 5, 2"# $% ! 7, 4"# $%

= 5(7) + 2(4)= 43

The work done is 43 N i m or 43 J.

b)

F!"

! s"

= 100, 400"# $% ! 12, 27"# $%

= 100(12) + 400(27)= 12 000

The work done is 12 000 N i m or 12 000 J.

c)

F!"

! s"

= 67.8, 3.9"# $% ! 4.7, 3.2"# $%

= 67.8(4.7) + 3.9(3.2)# 331.1

The work done is approximately 331.1 N i m or 331.1 J.


F!"

! s"

= 50(12)cos 10o

# 590.9

The work done is approximately 590.0 N i m or 590.0 J. b)

F!"

! s"

= 350(42)cos 30o

# 12 730.6

The work done is approximately 12 730.6 N i m or 12 730.6 J. c)

F!"

! s"

= 241(45.2)cos 80o

# 1891.6

The work done is approximately 1891.6 N i m or 1891.6 J. d)

F!"

! s"

= 1000(7)cos 20o

# 6577.8

The work done is approximately 6577.8 N i m or 6577.8 J.



a)

cos! =p!"

"q"

p!"

q"

=

7, 8#$ %& " 4, 3#$ %&7, 8#$ %& 4, 3#$ %&

=7(4) + 8(3)

72+ 82 42

+ 32

=52

53.1507

! = cos'1 5253.1507

(

)*+

,-

# 11.9º

b)

cos! =r!

" s!

r!

s!

=

#2, # 8$% &' " 6, #1$% &'#2, # 8$% &' 6, #1$% &'

=#2(6) + (–8)(–1)

(–2)2+ (–8)2 62

+ (–1)2

=#4

50.1597

! = cos#1 #450.1597

(

)*+

,-

" 94.6º

c)

cos! =t!

"u!

t!

u!

=

#7, 2$% &' " 6, 11$% &'#7, 2$% &' 6, 11$% &'

=#7(6) + 2(11)

(#7)2+ 22 62

+112

=#20

91.2195

! = cos#1 #2091.2195

(

)*+

,-

" 102.7º


d)

cos! =e!

" f"!

e!

f"!

=

2, 3#$ %& " 9, ' 6#$ %&

2, 3#$ %& 9, ' 6#$ %&

=2(9) + 3('6)

22+ 32 92

+ ('6)2

=0

39! = cos'1(0)

= 90º


a)

projv!u!

= u!

cos!1

v! v!"

#

$$

%

&

''

= 10cos 25o 118

v!"

#$%

&'

" 0.50v!

or 9.11

v! v!"

#

$$

%

&

''

The projection has magnitude 9.1 and has the same direction as v

!.


b)

projv!u!

= u!

cos!

= 7cos 110o

" "2.4

The projection has magnitude 2.4 and has direction opposite to v .

c)

projv!u!

= u!

cos!

= 20cos 90o

= 0

The projection has zero magnitude.


a)

projb! a!

=a!

!b!

b!

!b!

"

#$

%

&' b!

=

6, (1)* +, ! 11, 5)* +,11, 5)* +, ! 11, 5)* +,

"

#$

%

&' 11, 5)* +,

=61

146"

#$%

&'11, 5)* +,

" 4.6, 2.1)* +,

b)

projd!" c"

=c"

!d!"

c"

!d!"

"

#$

%

&' d!"

=

2, 7() *+ ! ,4, 3() *+,4, 3() *+ ! ,4, 3() *+

"

#$

%

&' ,4, 3() *+

=1325

"

#$%

&',4, 3() *+

= ,2.08, 1.56() *+

c)

projf!" e"

=e"

! f!"

f!"

! f!"

"

#$

%

&' f!"

=

(2, ( 5)* +, ! (5, 1)* +,(5, 1)* +, ! (5, 1)* +,

"

#$

%

&' (5, 1)* +,

=526

"

#$%

&'(5, 1)* +,

# (1.0, 0.2)* +,


d)

projh! g"!

=g"!

!h!

h!

!h!

"

#$

%

&' h!

=

10, ( 3)* +, ! 4, ( 4)* +,4, ( 4)* +, ! 4, ( 4)* +,

"

#$

%

&' 4, ( 4)* +,

=5232

"

#$%

&'4, ( 4)* +,

= 6.5, ( 6.5)* +,


F!"

! s"

= F!"

s"

cos"

= 50(8)cos 30o

# 346.4

The work done is approximately 346.4 J. Chapter 7 Section 3 Question 7 Page 385

a)

u!

=1

6, 1!" #$

6, 1!" #$

=137

6, 1!" #$

=637

,137

!

"%

#

$&

b)

f!"

= 25637

,137

!

"#

$

%&

=150

37,

2537

!

"#

$

%&


c)

s!

= 11, 0!" #$

W = F"!

% s!

=150

37,

2537

!

"&

#

$' % 11, 0!" #$

=150

37

(

)*+

,-(11) +

2537

(

)*+

,-(0)

# 271.3 J


W = F!"

s"

cos!

150 = F!"

(8)cos 20o

F!"

=150

8cos 20o

# 20.0

The force is approximately 20.0 N. Chapter 7 Section 3 Question 9 Page 385 The vectors representing the sides of the triangle are

AB! "!!

= !1, ! 8"# $% , BC! "!!

= !5, !1"# $% , and AC! "!!

= !6, ! 9"# $% .

Choose vectors carefully to find the interior angles of the triangle.

A

C B


cos B =AB! "!!

!CB! "!!

AB! "!!

CB! "!!

=

"1, " 8#$ %& ! 5, 1#$ %&"1, " 8#$ %& 5, 1#$ %&

="1(5) + ("8)(1)

("1)2+ ("8)2 52

+12

#"13

41.1096

'B = cos"1 "1341.1096

(

)*+

,-

# 108.4

Angle B is approximately 108.4 º.

cosC =AC! "!!

!BC! "!!

AC! "!!

BC! "!!

=

"6, " 9#$ %& ! "5, "1#$ %&"6, " 9#$ %& "5, "1#$ %&

="6(–5) + ("9)("1)

("6)2+ ("9)2 (–5)2

+ ("1)2

#39

55.1543

'C = cos"1 3955.1543

(

)*+

,-

# 45.0

Angle C is approximately 45.0º. ! A = 180º – 108.4º – 45.0º = 26.6º Therefore, ! ABC = 108.4°, ! BCA = 45.0°, and ! CAB = 26.6°.


Chapter 7 Section 3 Question 10 Page 385 The vectors representing the diagonals are

u!

= 5! 0, 3! 0"# $% = 5, 3"# $% and v!

= 3! 2, 0 ! 3"# $% = 1, ! 3"# $% .

cos! =

5, 3"# $% & 1, ' 3"# $%5, 3"# $% 1, ' 3"# $%

=5(1) + 3('3)

52+ 32 12

+ ('3)2

!'4

18.4391

! = cos'1 '418.4391

(

)*+

,-

! 102.5

The measure of ! is approximately 102.5º. The angles of intersection are 102.5° and 180º – 102.5º = 77.5º. Chapter 7 Section 3 Question 11 Page 385

a) Method 1:

OS! "!!

= OP! "!!

+ PS! "!

= OP! "!!

+ QR! "!!

= !2, 1"# $% + 10,–1"# $%

= 8, 0"# $%

S has coordinates (8, 0).

P(–2, 1)

Q(–6, 4) R(4, 3)

S

(2, 3)

(3, 0) (0, 0)

(5, 3) u!

v!

!


Method 2:

Let S have coordinates (x, y).

Since RS and PQ are parallel, their slopes are the same.

y ! 3x – 4

=4 !1

–6 ! (!2)

= –34

3x + 4y = 24 1

Since PS and QR are parallel, their slopes are the same.

y !1x + 2

=3! 4

4 ! (!6)

= –1

10

x + 10y = 8 2

1 – 32 gives:

–26y = 0

y = 0 Substitute y = 0 into 2. x + 10(0) = 8 x = 8 S has coordinates (8, 0).

b)

cos!PQR =QP! "!!

"QR! "!!

QP! "!!

QR! "!!

=

4, # 3$% &' " 10, #1$% &'

4, # 3$% &' 10, #1$% &'

=4(10) + (#3)(–1)

42+ (#3)2 102

+ (–1)2

#44

50.2494


!PQR = cos"1 4450.2494

#

$%&

'(

! 28.9

The interior angles are 29° and 180º – 29º = 151°.

c) Let θ be the angle between the diagonals PR and QS.

cos! =PR! "!!

"QS! "!!

PR! "!!

QS! "!!

=

6, 2#$ %& " 14, ' 4#$ %&6, 2#$ %& 14, ' 4#$ %&

=6(14) + 2('4)

62+ 22 142

+ ('4)2

#76

92.0869

! = cos'1 7692.0869

(

)*+

,-

# 34.4

Angle θ is approximately 34º. The angles between the diagonals are 34° and 180º – 34º = 146°.


PQ! "!!

= 8! 4, 3! 7"# $%

= 4, ! 4"# $%

Use [ ]1, 0i =

! as a vector along the x-axis.

cos! =PQ! "!!

" i"

PQ! "!!

i"

=

4, # 4$% &' " 1, 0$% &'

4, # 4$% &' (1)

=4(1) + (–4)(0)

42+ (–4)2

#4

5.6569


! = cos"1 45.6569

#

$%&

'(

! 45

The angle between PQ

!!!" and the positive x-axis is 45º. (This angle could be considered –45º since the vector

is below the positive x-axis.) Chapter 7 Section 3 Question 13 Page 385 proj cos

cos

cos

u v v

uv

u

v u

u

v uu

!

!

!

=

= "

=

#=

!

! !

!

!

!

! !

!

! !

!


1proj cos

1cos

cos

u v v uu

uv u

u u

v uu

u u

v u uu u

!

!

!

" #$ %=$ %& '" #$ %= ($ %& '

=

)=

)

!

! ! !

!

!

! !

! !

! !

!

! !

! !!

! !


a!

!b!

= 42, 23"# $% ! 115, 95"# $%

= 42(115) + 23(95)= 7015

The dot product is $7015. It represents the total revenue from the sales of the digital music players and the DVD players.



Using the diagram it is clear that [cos 10º, sin 10º] represents a vector parallel to the road surface. The force due to gravity is 1000 × 9.8 = 9800 N down.

proju! v!

=v!

!u!

u!

=

0, " 9800#$ %& ! cos 10o , sin 10o#$

%&

cos 10o , sin 10o#$

%&

="9800sin 10o

cos2 (10o ) + sin2 (10o )

""1701.8

1= 1701.8

The component of the force of gravity along the road vector is 1701.8 N.

10º

u!

= [cos 10º, sin 10º]

v!

= [0, –9800]

10º sin 10º

cos 10º


Chapter 7 Section 3 Question 17 Page 385 a) The hill (displacement) vector is u

!= [2500, 84].

The towing (force) vector is v!

= [30 000, 18 000].

proju! v!

=v!

!u!

u!

=

30 000, 18 000"# $% ! 2500, 84"# $%

2500, 84"# $%

=76 512 000

25002+ 842

" 30 587.5

The magnitude of the force drawing the car up the hill is about 30 587.5 N Use the Pythagorean theorem to find the force perpendicular to the hill.

x2+ 30 587.5( )

2= 30 0002

+18 0002( )2

x2= 288 404 843.8

x ! 16 982.5

The magnitude of the force perpendicular to the hill, tending to lift the car, is about 16 982.5 N.

b)

W = F!"

! s"

= 30 000, 18 000"# $% ! 2500, 84"# $%

= 30 000(2500) +18 000(84)= 76 512 000

The work done is 76 512 000 N . m (or J).

c) If only considering the raisin the altitude of the car, the displacement vector is [0, 84]. In this case,

W = F!"

! s"

= 30 000, 18 000"# $% ! 0, 84"# $%

= 30 000(0) +18 000(84)= 1512 000

The work done is 1512 000 N . m (or J)

d) Answers will vary. Work is determined by a force vector and a displacement vector. A change in the

displacement vector will lead to a change in the work done. Note that the answer in part c) is not realistic since it is not possible to raise the car directly upward by 84 m.


Chapter 7 Section 3 Question 18 Page 385 The force in this case is 25(9.8) = 245 N.

W = F!"

! s"

= F!"

s"

cos"

= 245(6)cos 30o

# 1273.1

The work done is about 1273.1 J. Chapter 7 Section 3 Question 19 Page 385

W = F!"

! s"

= F!"

s"

cos"

= 234(15)cos 12o

# 3433.3

The work done is about 3433.3 J. Chapter 7 Section 3 Question 20 Page 385

W = F!"

! s"

= F!"

s"

cos"

= 3(30)cos 25o+ 4(30)cos 5o

+ 5(30)cos 25o

# 337.1

The work done is about 337.1 J. Chapter 7 Section 3 Question 21 Page 385 The diagonals of the square are [ ]1 2[1, 1] and 1, 1d d= = !

!" !".

projd1

!"! i"

=i"

!d1

!"!

d1

!"!

!d1

!"!

"

#$

%

&' d1

!"!

=

1, 0() *+ ! 1, 1() *+1, 1() *+ ! 1, 1() *+

"

#$

%

&' 1, 1() *+

=12

"

#$%

&'1, 1() *+

=12

,12

(

),

*

+-

projd1

!"! j"

=j"

!d1

!"!

d1

!"!

!d1

!"!

"

#$

%

&' d1

!"!

=

0, 1() *+ ! 1,1() *+1, 1() *+ ! 1, 1() *+

"

#$

%

&' 1, 1() *+

=12

"

#$%

&'1, 1() *+

=12

,12

(

),

*

+-


projd2

!"! i"

=i"

!d2

!"!

d2

!"!

!d2

!"!

"

#$

%

&' d2

!"!

=

1, 0() *+ ! 1, ,1() *+1, ,1() *+ ! 1, ,1() *+

"

#$

%

&' 1, ,1() *+

=12

"

#$%

&'1, ,1() *+

=12

, ,12

(

)-

*

+.

projd2

!"! j"

=j"

!d2

!"!

d2

!"!

!d2

!"!

"

#$

%

&' d2

!"!

=

0, 1() *+ ! 1, ,1() *+1, ,1() *+ ! 1, ,1() *+

"

#$

%

&' 1, ,1() *+

=,12

"

#$%

&'1, ,1() *+

= ,12

,12

(

)-

*

+.

Chapter 7 Section 3 Question 22 Page 385 a) The following information is needed: magnitude of the force; distance along the ramp; angle between the

force and the ramp. b)

W = F!"

! s"

= F!"

s"

cos"

= 5000(5)cos15o

# 24 148.1

The work done is about 24 148.1 J.

Chapter 7 Section 3 Question 23 Page 385 a) The red vector indicates the answer.


b) The red vector indicates the answer.

c) The red vector indicates the answer.

Chapter 7 Section 3 Question 24 Page 385 Use components to represent the vectors.

u!

= [cos! , " sin!]

v!

= cos! , sin!#$ %&

w"!

= 0, 1#$ %&

L.S. = v!

R.S. = u!

" 2 u!

'w"!

( ) w"!

= [cos! , " sin!]" 2 [cos! , " sin!] ' 0, 1#$ %&( ) 0, 1#$ %&

= [cos! , " sin!]" 2(" sin! ) 0, 1#$ %&

= [cos! , " sin!]+ 0, 2sin!#$ %&

= cos! , sin!#$ %&

= v!

Therefore, L.S. = R.S.



a)

projb! a!

=a!

!b!

b!

!b!

"

#$

%

&' b!

=

6, 5() *+ ! 1, 3() *+1, 3() *+ ! 1, 3() *+

"

#$

%

&' 1, 3() *+

=2110

"

#$%

&'1, 3() *+

= 2.1, 6.3() *+

b) The component vector for the b

! direction is [2.1, 6.3].

The component vector in the perpendicular direction can be found by vector subtraction. [6, 5] – [[2.1, 6.3] = [3.9, –1.3] Check perpendicularity using the dot product. [2.1, 6.3] · [3.9, –1.3] = 2.1(3.9) + 6.3(–1.3) = 0 Therefore, the two perpendicular components are [2.1, 6.3] and [3.9, –1.3].


proju! F"!

=F"!

!u!

u!

!u!

"

#$

%

&' u!

=

25, 18() *+ ! 2, 5() *+2, 5() *+ ! 2, 5() *+

"

#$

%

&' 2, 5() *+

=14029

"

#$%

&'2, 5() *+

=28029

,70029

(

),

*

+-

# 9.7, 24.1() *+

As in question 25, find the perpendicular component vector by subtraction.

[25,18] –28029

,70029

!

"#

$

%& =

44529

,'178

29!

"#

$

%&

! [15.3,'6.1]

Use the dot product to check perpendicularity.

280 700 445 178, , 029 29 29 29

! " ! "# $ =% & % &

' ( ' (

The perpendicular component vectors are 280 700 445 178, and ,29 29 29 29

! " ! "#$ % $ %

& ' & '.


Chapter 7 Section 3 Question 27 Page 385 a) These projection vectors are in different directions unless u kv=

! !.

The projections will be equal only when k =1; that is, when u v=

! !.

The projections will also be equal if 0u v! =

! !; the two projection vectors are 0

! if the vectors and u v

! ! are

perpendicular. These two situations are demonstrated in the diagrams below.

b) proj proj

cos cos

or cos 0

u vv u

v u

v u

! !

!

=

=

= =

! !

! !

! !

! !

This is true whenever and u v! !

have the same magnitude or if the vectors and u v! !

are perpendicular. This first situation is demonstrated in the diagram below.

Chapter 7 Section 3 Question 28 Page 385 There are three situations to consider. Case 1: Suppose the “3” side has length 60.

Then the “4” side has length 80 and the area is

12

(60 cm)(80 cm) = 2400 cm2.

u!

v!

u!

v!

u!

v!


Case 2: Suppose the “4” side has length 60. Then the “3” side has length 45 and the area is clearly less than 2400 cm2 as in Case 1. Case 3: Suppose the altitude to the hypotenuse has length 60. The sides of the large triangle can be labelled as 3x, 4x, and 5x. Using similar triangles,

4x60

=5x3x

12x2! 300x = 0

x x ! 25( ) = 0

x = 0, 25

Therefore, x = 25 and the area is

12

(125 cm)(60 cm) = 3750 cm2.

The maximum possible area is Case 3 which is 3750 cm2. Chapter 7 Section 3 Question 29 Page 385

f (un ) =1471

nlogu (un ) +

538(un )un

=1471

nn( ) + 538

= 1471+ 538= 2009

5x 4x

3x

60


Chapter 7 Section 4 Vectors in Three-Space Chapter 7 Section 4 Question 1 Page 399 a) Answers may vary. For example: i) ii) iii)

b) Since only the z-coordinate is negative, P(2, 3, −5) is in the octant at the front right bottom of the

3-D grid. Since only the x-coordinate is negative, Q(−4, 1, 3) is in the octant at the back right top of the 3-D grid. Since only the y-coordinate is negative, R(6, −2, 1) is in the octant at the front left top of the 3-D grid.

c)

PS! "!

= (1! 2)2+ (1! 3)2

+ (!1+ 5)2= 21

QS! "!!

= (1+ 4)2+ (1!1)2

+ (!1! 3)2= 41

RS! "!!

= (1! 6)2+ (1+ 2)2

+ (!1!1)2= 38

P(2, 3, –5) is the closest point to S.

d) The z-coordinate gives the distance to the xy-plane.

R(6, −2, 1) is the closest point to the xy-plane.

e) The distance to the z-axis is determined by 2 2x y+ .

For P, d = 22+ 32

= 13.

For Q, d = !4( )2

+12= 17.

For R, d = 62+ !2( )

2= 40.

P(2, 3, –5) is the closest point to the z-axis..

Q(–4, 1, 3)

P(2, 2, –5)


Chapter 7 Section 4 Question 2 Page 399 a) The set of points P(x, y, z) where the x and y coordinates are equal. b) The set of points P(x, y, z) where the absolute values of the x, y, and z values are equal. (e.g., (2, 2, 2) or

(5, –5, –5)) Chapter 7 Section 4 Question 3 Page 399 Answers may vary. For example: a)

b)

c)


u!

= (!1)2+ 52

+ (!2)2

= 30

(–2, 0, 4)


b)

u!

= 32+ 32

+ 32

= 27

= 3 3

c)

u!

= 02+ (!2)2

+ (!4)2

= 20

= 2 5

Chapter 7 Section 4 Question 5 Page 399 a) kji 253 +! b) kji 963 +!! c) ki 75 ! Chapter 7 Section 4 Question 6 Page 399 a) [3, 8, 0] b) [−5, 0, −2] c) [7, −4, 9] Chapter 7 Section 4 Question 7 Page 399 Yes, since 2v u= !

! !.


4,1,!7"# $% = 42+12

+ (!7)2

= 66

Unit vectors are of the form 1 vv

±

!

! .

Therefore, the unit vectors are 4 1 7, ,66 66 66

! "#$ %

& ' and 4 1 7, ,

66 66 66! "# #$ %& '

.



u!

and v!

are collinear if and only if u!

= kv!

for some k !" . a) [ ] [ ], 3, 6 8, 12,a k b= !

Comparing the y-coordinates gives k = 14

.

So,

a =14

!8( )

a = !2

6 =14

b

b = 24

b) [ ] [ ], 2, 0 3, 6,a k b= ! ! !

Comparing the y-coordinates gives k = 13

! .

So,

a = !13

(!3)

a = 1

0 = !13

(!b)

b = 0

Chapter 7 Section 4 Question 10 Page 399 Answers may vary. For example: a)

AB! "!!

= [3, 6, −2] b)

AB! "!!

= [−2, 5, 3]


c)

AB! "!!

= [0, −5, –6] d)

AB! "!!

= [3, 3, 4] Chapter 7 Section 4 Question 11 Page 399 a)

AB! "!!

= 32+ 62

+ (!2)2

= 49= 7

b)

AB! "!!

= (!2)2+ 52

+ 32

= 38

c)

AB! "!!

= 02+ (!5)2

+ (!6)2

= 61

d)

AB! "!!

= 32+ 32

+ 42

= 34



MN! "!!!

= ON! "!!

!OM! "!!!

2, 4, ! 7"# $% = ON! "!!

! !5, 0, 3"# $%

ON! "!!

= 2, 4, ! 7"# $% + !5, 0, 3"# $%

ON! "!!

= !3, 4, ! 4"# $%


DE! "!!

= OE! "!!

!OD! "!!

!4, 2, 6"# $% = 3, 3, 1"# $% !OD! "!!

OD! "!!

= 3, 3, 1"# $% ! !4, 2, 6"# $%

OD! "!!

= 7, 1, ! 5"# $%


AB! "!!

= 0 ! 0, 4 ! (!3), ! 4 ! 2"# $%

= 0, 7, ! 6"# $%

b)

CD! "!!

= !3! 4, ! 3! 5, 5! 0"# $%

= !7, ! 8, 5"# $%


7a!

= 7 !4, 1, 7"# $%

= !28, 7, 49"# $%

b)

a!

+ b!

+ c!

= !4, 1, 7"# $% + 2, 0, ! 3"# $% + 1, !1, 5"# $%

= !1, 0, 9"# $%

c)

b!

+ c!

+ a!

= 2, 0, ! 3"# $% + 1, !1, 5"# $% + !4, 1, 7"# $%

= !1, 0, 9"# $%

d)

b!

! c!

= 2, 0, ! 3"# $% ! 1, !1, 5"# $%

= 1, 1, ! 8"# $%


e)

3a!

! 2b!

+ 4c!

= 3 !4, 1, 7"# $% ! 2 2, 0, ! 3"# $% + 4 1, !1, 5"# $%

= !12, 3, 21"# $% + !4, 0, + 6"# $% + 4, ! 4, 20"# $%

= !12, !1, 47"# $%

f)

a!

!c!

= "4, 1, 7#$ %& ! 1, "1, 5#$ %&

= "4(1) +1("1) + 7(5)= 30

g)

b!

! a!

+ c!

( ) = 2, 0, " 3#$ %& ! "4, 1, 7#$ %& + 1, "1, 5#$ %&( )

= 2, 0, " 3#$ %& ! "3, 0, 12#$ %&

= 2("3) + 0(0) + "3(12)= "42

h)

b!

!c!

" a!

!c!

= 2, 0, " 3#$ %& ! 1, "1, 5#$ %& " "4, 1, 7#$ %& ! 1, "1, 5#$ %&

= 2(1) + 0("1) + "3(5)" ("4(1) +1(–1) + 7(5))= "43

i)

a!

+ b!

( ) ! a!

" b!

( ) = "4, 1, 7#$ %& + 2, 0, " 3#$ %&( ) ! "4, 1, 7#$ %& " 2, 0, " 3#$ %&( )

= "2, 1, 4#$ %& ! "6, 1, 10#$ %&

= "2("6) +1(1) + 4(10)= 53


cos! =g!"

"h"

g!"

h"

=

6, 1, 2#$ %& " '5, 3, 6#$ %&

62+12

+ 22 ('5)2+ 32

+ 62

='15

41 70# '0.2800

! = cos'1('0.2800)# 106.3

The angle has a measure of approximately 106.3º.


Chapter 7 Section 4 Question 17 Page 400 a) Let [x, y, z] be a vector orthogonal to [3, –1, 4].

[ ] [ ], , 3, 1, 4 03 4 0

x y zx y z! " =

" + =

Select any values for x and z and solve for y.

Let x = 2 and y = 1, then z = 5

4!

.

Let x = –3 and y = 5, then z = 72

.

Two orthogonal vectors are

2, 1, !54

"

#$

%

&' and

!3,5,

72

"

#$

%

&' .

b) Let [x, y, z] be a vector orthogonal to [–4, –9, 3].

[ ] [ ], , 4, 9, 3 04 9 3 0

x y zx y z! " " =

" " + =

Select any values for x and y and solve for z. Let x = 1 and y = 2.

!4(1)! 9(2) + 3z = 03z = 22

z =223

Let x = –3 and y = 2.

!4(!3)! 9(2) + 3z = 03z = 6z = 2

Two orthogonal vectors are

1,2,223

!

"#

$

%& and [–3, 2, 2].

Chapter 7 Section 4 Question 18 Page 400 a) Since the x- and y-coordinates are zero, each of the vectors is on the y-axis. b) Since the z-coordinates are zero, each of the vectors is in the x-y plane. c) Since the x-coordinates are zero, each of the vectors is in the y-z plane.



Unit vectors parallel to u!

have the form 1 uu

±

!

! .

a)

a!

= 52+ (!3)2

+ 22

= 38

The unit vectors parallel to a!

are 5 3 2, ,38 38 38

! "#$ %

& ' and 5 3 2, ,

38 38 38! "# #$ %& '

.

b)

PQ! "!!

= OQ! "!!

!OP! "!!

= 5+ 7, ! 2 ! 8, ! 2 ! 3"# $%

= 12, !10, ! 5"# $%

PQ! "!!

= 122+ (!10)2

+ (!5)2

= 269

The unit vectors parallel to PQ! "!!

are 12 10 5, ,269 269 269

! "# #$ %

& ' and

12 10 5, ,269 269 269

! "#$ %& '

.

c) [ ]

( )22 2

5, 6, 3

5 6 3

70

u

u

= !

= + + !

=

!

!

The unit vectors parallel to u!

are kjiu70

3

70

6

70

51 !+= and kjiu

70

3

70

6

70

52 +!!= .

d)

f!"

= !3, ! 2, ! 9"# $%

f!"

= (!3)2+ (!2)2

+ (!9)2

= 94

The unit vectors parallel to f!"

are kjif94

9

94

2

94

3

1!!!= and kjif

949

942

943

2 ++= .



L.S. = ku!

R.S. = ku!

1 + ku!

2 + ku!

3

= k u1, u2 , u3!" #$

= k u1, 0, 0!" #$ + 0, u2 , 0!" #$ + 0, 0, u3!" #$( )

= k u1i!

+ u2 j!

+ u 3k!

!"

#$

= ku1i!

+ ku2 j!

+ ku3 k!

= ku!

1 1, 0, 0!" #$ + ku!

2 0, 1, 0!" #$ + ku!

30, 0, 1!" #$

= ku!

1 + ku!

2 + ku!

3

Therefore, L.S. = R.S. Chapter 7 Section 4 Question 21 Page 400

L.S. = u!

+ v!

= u1, u2 , u3!" #$ + v1, v2 , v3!" #$

= u1, 0, 0!" #$ + 0, u2 , 0!" #$ + 0, 0, u3!" #$ + v1, 0, 0!" #$ + 0, v2 , 0!" #$ + 0, 0, v3!" #$

= u1i!

+ u2 j!

+ u3 k!

+ v1i!

+ v2 j!

+ v3 k!

= (u1 + v1)i!

+ (u2 + v2 ) j!

+ (u3 + v3)k!

= (u1 + v1) 1, 0, 0!" #$ + (u2 + v2 ) 0, 1, 0!" #$ + (u3 + v3) 0, 0, 1!" #$

= u1 + v1, 0, 0!" #$ + 0, u2 + v2 , 0!" #$ + 0, 0, u3 + v3!" #$

= u1 + v1, u2 + v2 , u3 + v3!" #$

R.S. = u1 + v1, u2 + v2 , u3 + v3!" #$

Therefore, L.S. = R.S. Chapter 7 Section 4 Question 22 Page 400

P1P2

! "!!!

= P1O! "!!

+ OP2

! "!!

= OP2

! "!!

+ P1O! "!!

= OP2

! "!!

!OP1

! "!!

= x2 , y2 , z2"# $% ! x1, y1, z1"# $%

= x2 ! x1, y2 ! y1, z2 ! z1"# $%



AB! "!!

= OB! "!!

!OA! "!!

= !4 ! 2, 8! 3, 1! (!5)"# $%

= !6, 5, 6"# $%

AB! "!!

= (!6)2+ 52

+ 62

= 97

BC! "!!

= 10, !12,!1"# $%

BC! "!!

= 245

AC! "!!

= 4, ! 7,5"# $%

AC! "!!

= 90

Since none of the sides have equal length, this is a scalene triangle. Chapter 7 Section 4 Question 24 Page 400 Proof 1 (! ) Suppose you are given 0

!.

0!

= 0,0,0!" #$

0!

= 02+ 02

+ 02

= 0 + 0 + 0= 0

Proof 2 (! ) Now suppose you are given a vector [x, y, z] whose length is zero.

2 2 2

2 2 2

0

0

x y z

x y z

+ + =

+ + =

The sum of three positive numbers is zero. Therefore each of the numbers must be zero.

x2= y2

= z2= 0

x = y = z = 0

Therefore, [x, y, z] is the zero vector. The magnitude of a vector is zero if and only if ( ! ) the vector is the zero vector.



AB! "!!

= !5, 5, 0"# $%

AB! "!!

= (–5)2+ 52

+ 02

= 50

AC! "!!

= !5, 0, 5"# $%

AC! "!!

= (–5)2+ 02

+ 52

= 50

BC! "!!

= 0, ! 5, 5"# $%

BC! "!!

= 02+ (–5)2

+ 52

= 50

Since

AB! "!!

= BC! "!!

= AC! "!!

, !ABC is an equilateral triangle.

b) Draw a diagram. It appears that the closest point to the origin is the centroid of the triangle

53

,53

,53

!

"#

$

%& .

In chapter 6, it was shown that the centroid of ΔABC is defined by the position vector

OX! "!!

=13

OA! "!!

+ OB! "!!

+ OC! "!!

( ) .

The point on the interior of the triangle closest to the origin is the point 5 5 5, ,3 3 3

! "# $% &

.

Chapter 7 Section 4 Question 26 Page 400 a) Assume the vectors are placed on a coordinate system as in the diagram in the question.

[ ] [ ] [ ]

[ ]

2 2 2

25, 0, 0 0, 35, 0 0, 0, 40

25, 35, 40

25 35 40

345058.7

R

R

= + +

=

= + +

=

!"

!"

#

The magnitude of the resultant is about 58.7 N.

x

y

z

C(0, 0, 5)

A(5, 0, 0)

B(0, 5, 0)


b)

cos! =u!

"v!

u!

v!

=

25, 35, 40#$ %& " 0, 35, 0#$ %&

252+ 352

+ 402 02+ 352

+ 02

=1225

3450(35)" 0.5959

! = cos"1(0.5959)= 53.4

The resultant makes a 53.4° angle with the 35-N force.


[ ]

[ ]

[ ]

2, 3, 5

6, 3, 1

5, 1, 7

a

b

c

= !

= !

= ! !

!

!

!

a)

a!

+ b!

( ) ! a!

" b!

( ) = 8, 0, 4#$ %& ! "4, " 6, 6#$ %&

= 8(–4) + 0(–6) + 4(6)= "8

b)

a!

! b!

+ c!

( ) = 2, " 3, 5#$ %& ! 1, 2, 6#$ %&

= 2(1) + ("3)(2) + 5(6)= 26

c)

b!

+ c!

( ) ! b!

+ a!

( ) = 1, 2, 6"# $% ! 8, 0, 4"# $%

= 1(8) + 2(0) + 6(4)= 32

d)

2c!

! 3a!

" 2b!

( ) = "10, " 2, 14#$ %& ! "6, "15, 17#$ %&

= –10 " 6 + (–2)("15) +14(17)= 328



L.S. = u!

!v!

= u1,u2 ,u3"# $% ! v1,v2 ,v3"# $%

= u1i!

+ u2 j!

+ u3 k!

( ) ! v1i!

+ v2 j!

+ v3 k!

( )

= u1v1i!

! i!

+ u1v2 i!

! j!

+ u1v3 i!

! k!

+ u2v1 j!

! i!

+ +u2v2 j!

! j!

+ +u2v3 j!

! k!

+ +u3v1k!

! i!

+ +u3v2 k!

! j!

+ u3v3 k!

! k!

Step 1

= u1v1i!

! i!

+ 0 + 0 + 0 + u2v2 j!

! j!

+ 0 + 0 + 0 + u3v3 k!

! k!

Step 2= u1v1 + u2v2 + u3v3

R.S. = u1v1 + u2v2 + u3v3

Therefore, L.S. = R.S. In the above proof, Step 1 is possible using the distributive property for geometric vectors. Step 2 is possible because 0 (perpendicular vectors)i j i k j i j k k i k j! = ! = ! = ! = ! = ! =

! ! ! ! ! ! ! ! ! ! ! !

and 1 (parallel unit vectors)i i j j k k! = ! = ! =

! ! ! ! ! !.


if and only if 0u v u v! " =

! ! ! !.

a) [ ] [ ]4, 1, 3 1, 5, 0

4 5 3 03 1

13

kkk

k

! " =

" + + =

= "

= "

b) [ ] [ ], 3, 6 5, , 8 0

5 3 48 08 48

6

k kk k

kk

! =

+ + =

= "

= "

c)

11, 3, 2k!" #$ % k, 4, k!" #$ = 0

11k +12 + 2k 2= 0

2k 2+11k +12 = 0

(2k + 3)(k + 4) = 0

k = !

32

or k = ! 4



i!

! j!

= 1, 0, 0"# $% ! 0, 1, 0"# $%

= 1(0) + 0(1) + 0(0)= 0

j!

! k!

= 0, 1, 0"# $% ! 0, 0, 1"# $%

= 0(0) +1(0) + 0(1)= 0

i!

! k!

= 1, 0, 0"# $% ! 0, 0, 1"# $%

= 1(0) + 0(0) + 0(1)= 0

Since 0=!=!=! kikjji , then i , j , and k are mutually orthogonal. Chapter 7 Section 4 Question 31 Page 401 a) i)

L.S. = u!

+ v!

= u1, u2 , u3!" #$ + v1, v2 , v3!" #$

= u1 + v1, u2 + v2 , u3 + v3!" #$

= v1 + u1, v2 + u2 , v3 + u3!" #$

= v1, v2 , v3!" #$ + u1, u2 , u3!" #$

R.S. = v!

+ u!

= v1, v2 , v3!" #$ + u1, u2 , u3!" #$

Therefore, L.S. = R.S. It is proven that uvvu +=+ .

ii)

L.S. = u!

!v!

= u1, u2 , u3"# $% ! v1, v2 , v3"# $%

= u1v1 + u2v2 + u3v3

= v1u1 + v2u2 + v3u3

R.S. = v!

!u!

= v1, v2 , v3"# $% ! u1, u2 , u3"# $%

= v1u1 + v2u2 + v3u3

Therefore, L.S. = R.S. It is proven that u v v u! = !

! ! ! !.

b) The proofs depend on the commutative properties of addition and multiplication for real numbers and

also on the definitions of vector addition and dot product for Cartesian vectors.



[ ] [ ] [ ]1 2 3 1 2 3 1 2 3Let , , , , , , and , , .a a a a b b b b c c c c= = =

! ! !

L.S. = a!

+ b!

( ) + c!

= a1, a2 , a3!" #$ + b1, b2 , b3!" #$( ) + c1, c2 , c3!" #$

= a1 + b1, a2 + b2 , a3 + b3!" #$ + c1, c2 , c3!" #$

= (a1 + b1) + c1, (a2 + b2 ) + c2 , (a3 + b3) + c3!" #$

= a1 + (b1 + c1), a2 + (b2 + c2 ), a3 + (b3 + c3)!" #$

R.S. = a!

+ b!

+ c!

( )

= a1, a2 , a3!" #$ + b1, b2 , b3!" #$ + c1, c2 , c3!" #$( )

= a1, a2 , a3!" #$ + b1 + c1, b2 + c2 , b3 + c3!" #$

= a1 + (b1 + c1), a2 + (b2 + c2 ), a3 + (b3 + c3)!" #$

Therefore, L.S. = R.S. It is proven that ( ) ( )a b c a b c+ + = + +

! ! ! ! ! !.

This proof depends on the associative property for addition of real numbers. Chapter 7 Section 4 Question 33 Page 401

[ ] [ ] [ ]1 2 3 1 2 3 1 2 3Let , , , , , , and , , .a a a a b b b b c c c c= = =

! ! !

a)

L.S. = k a!

+ b!

( )

= k a1, a2 , a3!" #$ + b1, b2 , b3!" #$( )

= k a1 + b1, a2 + b2 , a3 + b3!" #$

= k(a1 + b), k(a2 + b2 ), k(a3 + b3)!" #$

= ka1 + kb1, ka2 + kb2 , ka3 + kb3!" #$

R.S. = ka!

+ kb!

= k a1, a2 , a3!" #$ + k b1, b2 , b3!" #$

= ka1, ka2 , ka3!" #$ + kb1, kb2 , kb3!" #$

= ka1 + kb1, ka2 + kb2 , ka3 + kb3!" #$

Therefore, L.S. = R.S. It has been proven that ( )k a b ka kb+ = +

! ! ! !.


b)

L.S. = a!

! b!

+ c!

( )

= a1, a2 , a3"# $% ! b1, b2 , b3"# $% + c1, c2 , c3"# $%( )

= a1,a2,a3"#

$% ! b1 + c1, b2 + c2 , b3 + c3"# $%

= a1(b1 + c1) + a2 (b2 + c2 ) + a3(b3 + c3)= a1b1 + a1c1 + a2b2 + a2c2 + a3b3 + a3c3

R.S. = a!

!b!

+ a!

!c!

= a1, a2 , a3"# $% ! b1, b2 , b3"# $% + a1, a2 , a3"# $% ! c1, c2 , c3"# $%

= a1b1 + a2b2 + a3b3 + a1c1 + a2c2 + a3c3

= a1b1 + a1c1 + a2b2 + a2c2 + a3b3 + a3c3

Therefore, L.S. = R.S. It has been proven that ( )a b c a b a c! + = ! + !

! ! ! ! ! ! !.

c)

L.S. = k a!

!b!

( )

= k a1, a2 , a3"# $% ! b1, b2 , b3"# $%( )

= k a1b1 + a2b2 + a3b3( )

= k(a1b1) + k(a2b2 ) + k(a3b3)= ka1b1 + ka2b2 + ka3b3

M.S. = ka!

! b!

( )

= k a1, a2, , a3"#

$%( ) ! b1, b2 , b3"# $%

= ka1, ka2 , ka3"# $% ! b1, b2 , b3"# $%

= (ka1)b1 + (ka2 )b2 + (ka3)b3

= ka1b1 + ka2b2 + ka3b3

R.S. = a!

! kb!

( )

= a1, a2, , a3"#

$% ! k b1, b2 , b3"# $%( )

= a1, a2, , a3"#

$% ! kb1, kb2 , kb3"# $%

= a1(kb1) + a2 (kb2 ) + a3(kb3)= a1kb1 + a2kb2 + a3kb3

= ka1b1 + ka2b2 + ka3b3

Therefore, L.S. = M.S = R.S. It has been proven that ( ) ( ) ( )k a b ka b a kb! = ! = !

! ! ! ! ! !.



projv!u!

=u!

!v!

v!

!v!

"

#$

%

&' v!

=

3, 4, 7() *+ ! 1, 2, 3() *+1, 2, 3() *+ ! 1, 2, 3() *+

"

#$

%

&' 1, 2, 3() *+

=3214

"

#$%

&'1, 2, 3() *+

=167

,327

,487

(

),

*

+-

The component vector for the v!

direction is 16 32 48, ,7 7 7

! "# $% &

.

The component vector in the perpendicular direction can be found by vector subtraction.

[3, 4, 7] – 16 32 48, ,7 7 7

! "# $% &

= 5 4 1, ,7 7 7

!" #$ %& '

Check perpendicularity using the dot product.

[3, 4, 7] · 57

,!47

,17

"

#$

%

&'= 3

57

(

)*+

,-+ 4 !

47

(

)*+

,-+ 7

17

(

)*+

,-

=07

= 0

Therefore, the two perpendicular components are 16 32 48, ,7 7 7

! "# $% &

and 5 4 1, ,7 7 7

!" #$ %& '

.


Let east be the x-direction, north be the y-direction, and up be the z-direction. The original vector for the airplane is [200, 0, 200tan 14º]. The wind vector is [0, 20, 0].

14º


The air velocity is the sum of these 2 vectors, [200, 20, 200tan 14º]. The magnitude of the air velocity is 2002

+ 202+ (200 tan14o )2 or approximately 207.1 km/h.

The final ground velocity is the sum of the original ground velocity and the wind velocity, [200, 0, 0] + [0, 20, 0] = [200, 20, 0]. The magnitude of the final ground velocity is 2002

+ 202+ 02 or approximately 201.0 km/h.

Chapter 7 Section 4 Question 36 Page 401 This problem can be done with either geometric vectors or Cartesian vectors. Use Cartesian vectors. Set up a coordinate system so that south is the x-direction, east is the y-direction, and up is the z-direction.

The first person is located at

553tan 10o

, 0, 0!

"#$

%&.

The second person is located at

553tan 9o

cos 120o ,553

tan 9osin 120o , 0

!

"#$

%&.

The displacement vector is

553tan9o

cos 120o ,553

tan 9osin 120o , 0

!

"#$

%&'

553tan 10o

, 0, 0!

"#$

%&

=553

tan 9ocos 120o '

553tan 10o

,553

tan 9osin 120o , 0

(

)*

+

,-

! '4882.0, 3023.7. 0() +,

The distance between these points is

553tan 9o

cos 120o !553

tan 10o

"

#$%

&'

2

+553

tan 9osin 120o"

#$%

&'

2

+ 02! 5742.5 .

The distance between the two people is 5742.5 m.

10º

9º

60º 120º

553 m

x

z

y


Chapter 7 Section 4 Question 37 Page 401 a) Substitute in to the general formula for each case.

i) s!

(t) = 2t, 30t cos 20o , 30t sin 20o ! 9.8t2"#

$%

ii)

s!

(t) = 6t, 18t cos 35o , 18t sin 35o ! 9.8t2"#

$%

b) Let t = 1 in each case.

i) s!

(1) " 2, 28.2, 0.5!" #$ ii)

s!

(1) = 6, 14.7, 0.5!" #$ c) Let t = 3 in each case.

i) s!

(3) = 6, 84.6, ! 57.4"# $% ii)

s!

(3) = 18, 44.2, ! 57.2"# $% Note that the projectile is below the origin after 3 s.

d) The height is determined only by the z-coordinate.

i) h(t) = 30tsin 20º – 9.8t2 !h (t) = 30sin 20o

"19.6t

The maximum height occurs when !h (t) = 0.

30sin 20o!19.6t = 0

t =30sin 20o

19.6t ! 0.52

s!

(0.52) " 1.04, 14.66, 2.69!" #$ ii) h(t) = 18tsin 35º – 9.8t2 !h (t) = 18sin 35o

"19.6t

The maximum height occurs when !h (t) = 0.

18sin 35o!19.6t = 0

t =18sin 35o

19.6t ! 0.53

s!

(0.53) " 3.18, 7.81, 2.72!" #$



x = 5tan30º ! 2.9 fN

R1

!"!

= 52+ 2.92

# 5.8

The component of the force to the left is about 2.9 fN.

b)

z = 5tan 20º ! 1.8 fN The component of the force up is about 1.8 fN.

c) The magnitude of the resultant is found by combining the horizontal resultant from part a) and the

vertical component from part b).

Use the Pythagorean theorem.

R!"

= 5.82+1.82

# 6.1

The magnitude of the resultant force acting on the electron is about 6.1 fN.

20º 5.8 fN

1.8 fN R!"

5 fN 30º

x

1R!!"

20º 5 fN

z


Chapter 7 Section 4 Question 39 Page 402 Since GFDE is a parallelogram, then FG = DE and EG = FD. FG = DE (x – 6, y + 2, z – 4) = (–6, 1, –3) x – 6 = –6 y + 2 = 1 z – 4 = –3 x = 0 y = –1 z = 1

EG = FD (x + 4, y – 2, z – 0) = (–4, 3, –1) x + 4 = –4 y – 2 = –3 z – 0 = 1 x = –8 y = –1 z = 1

OG! "!!

= OD! "!!

+ OF! "!!

– OE! "!!

= [6 + 2 – (–4), –2 + 1 – 2, 4 + 3 – 0]

= [12, –3, 7]

The fourth point has coordinates (0, –1, 1), (–8, –1, 1), or (12, –3, 7).

F(6, –2, 4)

E(–4, 2, 0)

D(2, 1, 3)

G

x

y

z


Chapter 7 Section 4 Question 40 Page 402 The set of points that are 10 units from the x-axis form a cylinder around the x-axis line. The cylinder has a diameter of 20 units and infinite height. Similarly, the set of points 10 units from the y-axis also form a 20-unit-diameter cylinder but with the y-axis as its central core and having an infinite height. Points that are 10 units from both axes lie on the intersection of these two cylinders. This is not a simple curve. Some points that satisfy the conditions are (10, 10, 0), (10, –10, 0), (–10, –10, 0), –10, 10, 0), (6, 8, 6),

±5, ±5 3, ±5( ) , etc.

For a point (x, y, z) to satisfy the conditions, you need 2 2 2 2100 and 100x y y z+ = + = . The resulting surface is known as the Steinmetz solid. Apparently the surface area and volume of this solid were calculated by Chinese mathematicians in about 500 B.C.

Chapter 7 Section 4 Question 41 Page 402 Answers may vary. For example: The vector [ ]zyxwv ,,,= can take into consideration the position of the vector in 4-D space in terms of time as well as in terms of position. The vector [ ]zyxwv ,,,= can also represent a position vector in 4-D space, that is, a space that has four mutually perpendicular axes. Although it is not possible to visualize this, vectors in 4-D have the same operations and properties as those in 3-D such as addition, subtraction, scalar multiplication, dot product, commutative, associative, and distributive properties, etc. Chapter 7 Section 4 Question 42 Page 402 a) Let the required vector be [ ], ,v x y z=

!.

Since v!

! a!

, v!

"a!

= 0.

x, y, z!" #$ % 2, 3, 1!" #$ = 0

2x + 3y + z = 0

Since v!

! b!

, v!

"b!

= 0 . [x, y, z] ! [4, 5, –2] = 0 4x + 5y – 2z = 0 There are three variables but only two equations. Therefore, choose arbitrary values for z, and solve for x and y.


Let z = 0. 2x + 3y = 0 1 4x + 5y = 0 2 y = 0 21 – 2 x = 0

v!

= 0, 0, 0!" #$ Let z = 1. 2x + 3y = –1 1 4x + 5y = 2 2 y = –4 21 – 2

x =

112

v!

=112

, ! 4, 1"

#$

%

&'

Let z = 2. 2x + 3y = –2 1 4x + 5y = 4 2 y = –8 21 – 2 x = 11

v!

= 11, ! 8, 2"# $%

b) Let z = k. 2x + 3y = –k 1 4x + 5y = 2k 2 y = –4k 21 – 2

x =

112

k

v!

=11k

2, ! 4k, k

"

#$

%

&'

This last example shows that there is a vector v

! for every choice of parameter k. Therefore, there are

infinitely many solutions.



a)

cos! =a!

"b!

a!

b!

=

2, 1, 3#$ %& " '1, 2, 4#$ %&2, 1, 3#$ %& '1, 2, 4#$ %&

=2(–1) +1(2) + 3(4)

22+12

+ 32 (–1)2+ 22

+ 42

=12

17.1464

! = cos'1 1217.1464

(

)*+

,-

" 45.6º

cos! =b!

"c!

b!

c!

=

#1, 2, 4$% &' " 3, 4, 10$% &'#1, 2, 4$% &' 3, 4, 10$% &'

=#1(3) + 2(4) + 4(10)

(–1)2+ 22

+ 42 32+ 42

+102

=45

51.2348

! = cos#1 4551.2348

(

)*+

,-

" 28.6º

cos! =a!

"c!

a!

c!

=

2, 1, 3#$ %& " 3, 4, 10#$ %&2, 1, 3#$ %& 3, 4, 10#$ %&

=2(3) +1(4) + 3(10)

22+12

+ 32 32+ 42

+102

=40

41.8330

! = cos'1 4041.8330

(

)*+

,-

" 17.0º

The three angles are 45.6º, 28.6º, and 17.0º.


b)

cos! =d!"

"e"

d!"

e"

=

5, 1, 2#$ %& " '3, 1, 4#$ %&5, 1, 2#$ %& '3, 1, 4#$ %&

=5(–3) +1(1) + 2(4)

52+12

+ 22 ('3)2+12

+ 42

='6

27.9285

! = cos'1 '627.9285

(

)*+

,-

# 102.4º

cos! =e!

" f"!

e!

f"!

=

#3, 1, 4$% &' " 6, # 2, 3$% &'#3, 1, 4$% &' 6, # 2, 3$% &'

=#3(6) +1(#2) + 4(3)

(#3)2+12

+ 42 62+ (#2)2

+ 32

=#8

35.6931

! = cos#1 #835.6931

(

)*+

,-

# 103.0º

cos! =d!"

" f!"

d!"

f!"

=

5, 1, 2#$ %& " 6, ' 2, 3#$ %&5, 1, 2#$ %& 6, ' 2, 3#$ %&

=5(6) +1('2) + 2(3)

52+12

+ 22 62+ (–2)2

+ 32

=34

38.3406

! = cos'1 3438.3406

(

)*+

,-

# 27.5º

The three angles are 102.4º, 103.0º, and 27.5º.


Chapter 7 Section 4 Question 44 Page 402 Proof 1 (! ) Suppose u v!

! !.

Then 0 and 0u v v u! = ! =

! ! ! !.

L.S. = u!

+ v!

= u!

+ v!

( ) ! u!

+ v!

( )

= u!

!u!

+ u!

!v!

+ v!

!u!

+ v!

!v!

= u!

!u!

+ v!

!v!

R.S. = u!

" v!

= u!

" v!

( ) ! u!

" v!

( )

= u!

!u!

" u!

!v!

" v!

!u!

+ v!

!v!

= u!

!u!

+ v!

!v!

Therefore, L.S. = R.S. Proof 2 (! ) Suppose u v u v+ = !

! ! ! !.

Show that 0u v! =

! !.

( ) ( ) ( ) ( )

4 0

0

u v u v

u v u v u v u v

u u u v v u v v u u u v v u v v

u v u v u v u v

u v

u v

+ = !

+ " + = ! " !

" + " + " + " = " ! " ! " + "

" + " = ! " ! "

" =

" =

! ! ! !

! ! ! ! ! ! ! !

! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !

! ! ! ! ! ! ! !

! !

! !

Therefore, 0u v! =

! ! and u v!

! !.

Chapter 7 Section 4 Question 45 Page 402 a) Let a

!

= OA" !""

and b!

= OB" !""

be position vectors for two points in space and let r!

= OP" !""

be the position vector for any point satisfying the relationship.

r!

! a!

= AP" !""

and r!

! b!

= BR" !""

Since r

!

! a!

( ) " r!

! b!

( ) = 0, AP" !""

# BP" !""

.

Points P are all those points such that !APB is a right angle. This is a property of the sphere that has points A and B as endpoints for one of its diameters.


b) A and B are the two endpoints of a diameter of the sphere. ( and a b! !

are the position vectors for A and B.)

Chapter 7 Section 4 Question 46 Page 402 The last digit can only be an 8 or a 9 and the third digit can only be a 5 or a 6. The time must end in 58 or 59. The first two digits can only be 1 and 0. The time is either 10:58 or 10:59. In 12 h, there are 12 × 60 = 720 possible display times.

The probability is 2 1 or 720 360

or about 0.28%.

Chapter 7 Section 4 Question 47 Page 402 The equation of the line through Q and R is the line y = 1. Find the points P and S using algebra. Since Q and R are intersection points, (x – 1) and (x – 4) are factors. Using long division or synthetic division, x4 – 10x3 + 24x2 + 5x – 20 = (x – 1)(x – 4)(x2 – 5x – 5). The other intersection points, P and S, have

x =5 ± (–5)2

! 4(1)(–5)2

=5 ± 45

2

=5 ± 3 5

2

Then,

QRRS

=3

5+ 3 52

!

"#

$

%& ' 4

=6

5+ 3 5 ' 8

=6

'3+ 3 5(

3+ 3 53+ 3 5

=

6(3) 1+ 5( )'9 + 45

=1+ 5

2

RQQP

=3

1! 5! 3 52

"

#$

%

&'

=6

2 ! 5+ 3 5

=6

!3+ 3 5(

3+ 3 53+ 3 5

=

6(3) 1+ 5( )!9 + 45

=1+ 5

2

5 3 52

! 5 3 52

+ 1 4

R Q P S


Chapter 7 Section 5 The Cross Product and Its Properties Chapter 7 Section 5 Question 1 Page 410 a)

u!

! v!

= u!

v!

(sin" )n̂

= 60(80)sin55o n̂" 3931.9n̂

b)

u!

! v!

= u!

v!

(sin" )n̂

= 112(128)sin 164o n̂" 3951.5n̂


a!

! b!

= 3, " 2, 9#$ %& ! 1, 1, 6#$ %&

= "2(6)"1(9), 9(1)" 6(3), 3(1)"1("2)#$ %&

= "21, " 9, 5#$ %&

b)

a!

! b!

= 6, 3, 2"# $% ! &5, 5, 9"# $%

= 3(9)& 5(2), 2(&5)& 9(6), 6(5)& (&5)(3)"# $%

= 17, & 64, 45"# $%

c)

a!

! b!

= "8, 10, 3#$ %& ! 2, 0, 5#$ %&

= 10(5)" 0(3), 3(2)" 5("8), ("8)(0)" 2(10)#$ %&

= 50, 46, " 20#$ %&

d)

a!

! b!

= 4.3, 5.7, " 0.2#$ %& ! 12.3, " 4.9, 8.8#$ %&

= 5.7(8.8)" ("4.9)("0.2), ("0.2)(12.3)" 8.8(4.3), 4.3("4.9)"12.3(5.7)#$ %&

= 49.18, " 40.3, " 91.18#$ %&



u!

! v!

= 5, " 3, 7#$ %& ! "1, 6, 2#$ %&

= "3(2)" 6(7), 7("1)" 2(5), 5(6)" (–1)(–3)#$ %&

= "48, "17, 27#$ %&

Use the dot product to check if u v!

! ! is orthogonal to and u v

! !.

u!

! v!

"u!

= #48, #17, 27$% &' " 5, # 3, 7$% &'

= #48(5) + (#17)(–3) + 27(7)= 0

u!

! v!

"v!

= #48, #17, 27$% &' " #1, 6, 2$% &'

= #48(–1) + (#17)(6) + 27(2)= 0

Therefore, u v!


! !.

b)

u!

! v!

= "2, 1, 5#$ %& ! 3, 2, 0#$ %&

= 1(0)" 2(5), 5(3)" 0("2), " 2(2)" 3(1)#$ %&

= "10, 15, " 7#$ %&



! !.

u!

! v!

"u!

= #10, 15, # 7$% &' " #2, 1, 5$% &'

= #10(–2) +15(1) + (–7)5= 0

u!

! v!

"v!

= #10, 15, # 7$% &' " 3, 2, 0$% &'

= #10(3) +15(2) + (–7)(0)= 0

Therefore, u v!


! !.

c)

u!

! v!

= 4, " 6, 2#$ %& ! 6, 8, " 3#$ %&

= –6(–3)" 8(2), 2(6)" (–3)(4), 4(8)" 6(–6)#$ %&

= 2, 24, 68#$ %&



! !.


u!

! v!

"u!

= 2, 24, 68#$ %& " 4, ' 6, 2#$ %&

= 2(4) + 24(–6) + 68(2)= 0

u!

! v!

"v!

= 2, 24, 68#$ %& " 6, 8, ' 3#$ %&

= 2(6) + 24(8) + 68(–3)= 0

Therefore, u v!


! !.

Chapter 7 Section 5 Question 4 Page 411 The area of the parallelogram defined by and p q

!" "

is sinp q p q !" =

!" " !" ".

a)

p!"

! q"

= 6, 3, 8"# $% ! 3, 3, 5"# $%

= 3(5)& 3(8), 8(3)& 5(6), 6(3)& 3(3)"# $%

= &9, & 6, 9"# $%

p!"

! q"

= (&9)2+ (&6)2

+ 92

# 14.1

The area of the parallelogram is about 14.1 square units.

b)

u!

! v!

= u!

v!

sin"

= 43(27)sin32o

" 615.2


Chapter 7 Section 5 Question 5 Page 411 Let the cross product a b!

! ! represent the situation where the wrench undergoes a clockwise rotation and the

direction of the cross product is up. Then situation where the wrench undergoes a counterclockwise rotation and the direction of the cross product would be down is b a!

! !. Since the two cross products are vector

quantities that are opposite in direction, therefore ( )a b b a! = " !

! ! ! !.


Chapter 7 Section 5 Question 6 Page 411 a) Let [ ]3, 4, 7u = !

! and [ ]2, 8, 3v =

!.

u!

! v!

= "3, 4, 7#$ %& ! 2, 8, 3#$ %&

= 4(3)" 8(7), 7(2)" 3("3), ("3)(8)" 2(4)#$ %&

= "44, 23, " 32#$ %&

u!

! v!

= ("44)2+ 232

+ ("32)2

= 3489

u!

= (–3)2+ 42

+ 72

= 74

v!

= 22+ 82

+ 32

= 77

u!

!v!

= –3(2) + 4(8) + 7(3)= 47

u! 2

v! 2

! u!

"v!

( )2

= 74(77)! 472

= 3489

Therefore, it has been verified that ( )22 2

u v u v u v! = " #

! ! ! ! ! !.

b) Let and u v

! !be any two vectors.

L.S. = u!

! v!

= u!

v!

sin"

R.S. = u! 2

v! 2

# u!

$v!

( )2

= u! 2

v! 2

# u!

v!

cos"( )2

= u! 2

v! 2

# u! 2

v! 2

cos2"

= u! 2

v! 2

(1# cos2" )

= u! 2

v! 2

sin2" sin2

" + cos2" = 1

= u!

v!

sin"




a!

! b!

+ c!

( ) = 2, " 6, 3#$ %& ! "1, 5, 8#$ %& + "4, 5, 6#$ %&( )

= 2, " 6, 3#$ %& ! "5, 10, 14#$ %&

= "6(14)"10(3), 3("5)"14(2), 2(10)" (–5)(–6)#$ %&

= "114, " 43, "10#$ %&

b)

b!

+ c!

( )! a!

= "1, 5, 8#$ %& + "4, 5, 6#$ %&( )! 2, " 6, 3#$ %&

= "5, 10, 14#$ %& ! 2, " 6, 3#$ %&

= 10(3)" (–6)(14), 14(2)" 3(–5), " 5(–6)" 2(10)#$ %&

= 114, 43, 10#$ %&

c)

a!

! b!

" a!

! c!

( ) = 2, " 6, 3#$ %& ! "1, 5, 8#$ %& " 2, " 6, 3#$ %& ! "4, 5, 6#$ %&

= "6(8)" 5(3), 3(–1)" 8(2), 2(5)" (–1)(–6)#$ %& " "6(6)" 5(3), 3(–4)" 6(2), 2(5)" (–4)(–6)#$ %&

= "63, "19, 4#$ %& " "51, " 24, "14#$ %&

= "12, 5, 18#$ %&

d)

a!

! 5a!

= 2, " 6, 3#$ %& ! 10, " 30, 15#$ %&

= "6(15)" (–30)(3), 3(10)"15(2), 2("30)"10(–6)#$ %&

= 0, 0, 0#$ %&

e)

a!

! c!

= 2, " 6, 3#$ %& ! "4, 5, 6#$ %&

= "6(6)" 5(3), 3(–4)" 6(2), 2(5)" (–4)(–6)#$ %&

= "51, " 24, "14#$ %&

= (–51)2+ (–24)2

+ (–14)2

= 3373

f)

b!

! c!

" a!

( ) = "1, 5, 8#$ %& ! "4, 5, 6#$ %& " 2, " 6, 3#$ %&( )

= "1, 5, 8#$ %& ! "6, 11, 3#$ %&

= 5(3)"11(8), 8(–6)" 3(–1), –1(11)" (–6)(5)#$ %&

= "73, " 45, 19#$ %&

= (–73)2+ (–45)2

+192

= 7715


Chapter 7 Section 5 Question 8 Page 411 Use and c d d c! !

! "! "! !.

c!

! d"!

= 4, 6, "1#$ %& ! "2, 10, 11#$ %&

= 6(11)"10(–1), –1(–2)"11(4), 4(10)" (–2)(6)#$ %&

= 76, " 42, 52#$ %&

Two possible orthogonal vectors are [ ]76, 42, 52! and [ ]76, 42, 52! ! . Chapter 7 Section 5 Question 9 Page 411

u!

! v!

= 3, " 4, 1#$ %& ! 2, 3, " 4#$ %&

= –4(–4)" 3(1), 1(2)" (–4)(3), 3(3)" 2(–4)#$ %&

= 13, 14, 17#$ %&

u!

! v!

= 132+142

+172

= 654

p"!

=1654

u!

! v!

( )

=13654

,14654

,17654

#

$'

%

&(

A vector orthogonal to both and u v! !

is 13 14 17, ,654 654 654

! "# $% &

.


PQ! "!!

= OQ! "!!

!OP! "!!

= 1, 6, 2"# $% ! 0, 2, 5"# $%

= 1, 4, ! 3"# $%

and

SR! "!!

= OR! "!!

!OS! "!!

= 7, 4, 2"# $% ! 6, 0, 5"# $%

= 1,4,!3"# $%

Therefore,

PQ! "!!

= SR! "!!

. This gives one pair of side equal and parallel. This is a sufficient condition to prove that quadrilateral PQRS is a parallelogram.


b) The area is found by calculating PQ! "!!

! PS! "!

.

PQ! "!!

! PS! "!

= 1, 4, " 3#$ %& ! 6, " 2, 0#$ %&

= 4(0)" (–2)(–3), – 3(6)" 0(1), 1(–2)" 6(4)#$ %&

= 6, "18, " 26#$ %&

PQ! "!!

! PS! "!

= (–6)2+ (–18)2

+ (–26)2

= 2 259# 32.2


c) The easiest test is to check if two adjacent sides are perpendicular. Use the dot product.

PQ! "!!

!PS! "!

= 1, 4, " 3#$ %& ! 6, " 2, 0#$ %&

= 1(6) + 4(–2) + (–3)(0)= "2' 0

No. The parallelogram is not a rectangle since

PQ! "!!

is not perpendicular to PS! "!

. Chapter 7 Section 5 Question 11 Page 411

g!"

! h"

= 4, 5, 2"# $% ! &2, 6, 1"# $%

= 5(1)& 6(2), 2(–2)&1(4), 4(6)& (–2)(5)"# $%

= [&7, & 8, 34]

g!"

! h"

= (–7)2+ (–8)2

+ 342

= 1269

2 2 24 5 2

45

g = + +

=

!"

h!

= (–2)2+ 62

+12

= 41

g!"

!h"

= 4(–2) + 5(6) + 2(1)= 24


Using the cross product:

sin! =

!g "!h

!g!h

=1269

45 41

! = 56.0º

Using the dot product:

cos! =g!"

"h"

g!"

h"

=24

45 41

! # 56.0°

Chapter 7 Section 5 Question 12 Page 411 The area of the parallelogram is sina b a b !" =

! ! ! ! where and a b

! ! are two adjacent sides and θ is the angle

between the sides.

85 = 10(9) sin!

sin! =8590

! = sin"1±

8590

#

$%&

'(

! ! 70.8o and 109.2o

Since adjacent interior angles of a parallelogram add to 180º, the four interior angles have measures 70.8º, 109.2º, 70.8º, and 109.2º. Chapter 7 Section 5 Question 13 Page 411

L.S. = u!

! v!

( )! w"!

= 3, 2, 9"# $% ! 8, 0, 3"# $%( )! 6, 2, 6"# $%

= 2(3)& 0(9), 9(8)& 3(3), 3(0)& 8(2)"# $% ! 6, 2, 6"# $%

= 6, 63, &16"# $% ! 6, 2, 6"# $%

= 63(6)& 2(&16), (&16)(6)& 6(6), 6(2)& 6(63)"# $%

= 410,&132,&366"# $%


R.S. = u!

! v!

! w"!

( )

= 3, 2, 9"# $% ! 8, 0, 3"# $% ! 6, 2, 6"# $%( )

= 3, 2, 9"# $% ! 0(6)& 2(3), 3(6)& 6(8), 8(2)& 6(0)"# $%

= 3, 2, 9"# $% ! &6, & 30, 16"# $%

= 2(16)& (–30)(9), 9(–6)&16(3), 3(–30)& (–6)(2)"# $%

= 302, &102, & 78"# $%

Therefore, L.S. ≠ R.S. Therefore,( ) ( )u v w u v w! ! " ! !

! ! "! ! ! "!.

Chapter 7 Section 5 Question 14 Page 411 Solutions for Achievement Checks are shown in the Teacher Resource. Chapter 7 Section 5 Question 15 Page 411 The area of the parallelogram is sina b a b !" =

! ! ! ! where and a b

! ! are two adjacent sides and θ is the angle

between the sides. The area will be zero when either of the two sides have length zero or when sin 0! = which means that

o o0 or 180! = and the two sides are parallel. The latter case occurs when the vertices of the parallelogram are collinear. For example, if three of the points of a parallelogram are A = (3, 0, 0), B = (4, 0, 0), and C = (5, 0, 0), then

AB! "!!

= 1, 0, 0!" #$ ,

AC! "!!

= 1, 0, 0!" #$ , and AB! "!!

! AC! "!!

= 0"

, and the area is zero. Chapter 7 Section 5 Question 16 Page 411 a) Verify a specify case of the left distributive law for vector cross product over vector addition.

L.S. = a!

! b!

+ c!

( )

= "2, 4, 3#$ %& ! 6, 1, 2#$ %& + 5, " 3, " 2#$ %&( )

= "2, 4, 3#$ %& ! 11, " 2, 0#$ %&

= 4(0)" (–2)(3), 3(11)" 0(–2), – 2(–2)"11(4)#$ %&

= 6, 33, " 40#$ %&


R.S. = a!

! b!

+ a!

! c!

= "2, 4, 3#$ %& ! 6, 1, 2#$ %& + "2, 4, 3#$ %& ! 5, " 3, " 2#$ %&

= 4(2)"1(3), 3(6)" 2(–2), – 2(1)" 6(4)#$ %& + 4(–2)" (–3)(3), 3(5)" (–2)(–2), – 2(–3)" 5(4)#$ %&

= 5, 22, " 26#$ %& + 1, 11, "14#$ %&

= 6, 33, " 40#$ %&

Therefore, L.S. = R.S. Therefore,

a!

! b!

+ c!

( ) = a!

! b!

+ a!

! c!

.

b) Let [ ]1 2 3, ,a a a a=

!, [ ]1 2 3, ,b b b b=

!, and [ ]1 2 3, ,c c c c=

!.

L.S. = a!

! b!

+ c!

( )

= a1, a2 , a3"# $% ! b1, b2 , b3"# $% + c1, c2 , c3"# $%( )

= a1, a2 , a3"# $% ! b1 + c1, b2 + c2 , b3 + c3"# $%

= a2 (b3 + c3)& (b2 + c2 )a3, a3(b1 + c1)& (b3 + c3)a1, a1(b2 + c2 )& (b1 + c1)a2"# $%

= a2b3 + a2c3 & a3b2 & a3c2 , a3b1 + a3c1 & a1b3 & a1c3, a1b2 + a1c2 & a2b1 & a2c1"# $%

R.S. = a!

! b!

+ a!

! c!

= a1, a2 , a3"# $% ! b1, b2 , b3"# $% + a1, a2 , a3"# $% ! c1, c2 , c3"# $%

= a2b3 & a3b2 , a3b1 & a1b3, a1b2 & a2b1"# $% + a2c3 & a3c2 , a3c1 & a1c3, a1c2 & a2c1"# $%

= a2b3 + a2c3 & a3b2 & a3c2 , a3b1 + a3c1 & a1b3 & a1c3, a1b2 + a1c2 & a2b1 & a2c1"# $%

Therefore, L.S. = R.S. Therefore, ( )a b c a b a c! + = ! + !

! ! ! ! ! ! !.

c) Verify a specify case of the right distributive law for vector cross product over vector addition.

L.S. = a!

+ b!

( )! c!

= "2, 4, 3#$ %& + 6, 1, 2#$ %&( )! 5, " 3, " 2#$ %&

= 4, 5, 5#$ %& ! 5, " 3, " 2#$ %&

= 5(–2)" (–3)(5), 5(5)" (–2)(4), 4(–3)" 5(5)#$ %&

= 5, 33, " 37#$ %&


R.S. = a!

! c!

+ b!

! c!

= "2, 4, 3#$ %& ! 5, " 3, " 2#$ %& + 6, 1, 2#$ %& ! 5, " 3, " 2#$ %&

= 4(–2)" (–3)(3), 3(5)" (–2)(–2), – 2(–3)" 5(4)#$ %& + 1(–2)" (–3)(2), 2(5)" (–2)(6), 6(–3)" 5(1)#$ %&

= 1, 11, "14#$ %& + 4, 22, " 23#$ %&

= 5, 33, " 37#$ %&

Therefore, L.S. = R.S. Therefore, ( )a b c a c b c+ ! = ! + !

! ! ! ! ! ! !.

d) Let [ ]1 2 3, ,a a a a=

!, [ ]1 2 3, ,b b b b=

!, and [ ]1 2 3, ,c c c c=

!.

L.S. = a!

+ b!

( )! c!

= a1, a2 , a3"# $% + b1, b2 , b3"# $%( )! c1, c2 , c3"# $%

= a1 + b1, a2 + b2 , a3 + b3"# $% ! c1, c2 , c3"# $%

= (a2 + b2 )c3 & c2 (a3 + b3), (a3 + b3)c1 & c3(a1 + b1), (a1 + b1)c2 & c1(a2 + b)"# $%

= a2c3 + b2c3 & a3c2 & b3c2 , a3c1 + b3c1 & a1c3 & b1c3, a1c2 + b1c2 & a2c1 & b2c1"# $%

R.S. = a!

! c!

+ b!

! c!

= a1, a2 , a3"# $% ! c1, c2 , c3"# $% + b1, b2 , b3"# $% ! c1, c2 , c3"# $%

= a2c3 & a3c2 , a3c1 & a1c3, a1c2 & a2c1"# $% + b2c3 & b3c2 , b3c1 & b1c3, b1c2 & b2c1"# $%

= a2c3 + b2c3 & a3c2 & b3c2 , a3c1 + b3c1 & a1c3 & b1c3, a1c2 + b1c2 & a2c1 & b2c1"# $%

Therefore, the L.S. = R.S. Therefore, ( )a b c a c b c+ ! = ! + !

! ! ! ! ! ! !.

Chapter 7 Section 5 Question 17 Page 411 a) Verify a specify case.

L.S. = k u!

! v!

( )

= 2 "1, 4, 1#$ %& ! 3, " 2, 4#$ %&( )

= 2 4(4)" (–2)(1), 1(3)" 4(–1), "1(–2)" 3(4)#$ %&

= 2(18), 2(7), 2(–10)#$ %&

= 36, 14, " 20#$ %&


M.S. = ku!

( )! v!

= 2 "1, 4, 1#$ %&( )! 3, " 2, 4#$ %&

= "2, 8, 2#$ %& ! 3, " 2, 4#$ %&

= 8(4)" (–2)(2), 2(3)" 4(–2), – 2(–2)" 3(8)#$ %&

= 36, 14, " 20#$ %&

R.S. = u!

! kv!

( )

= "1, 4, 1#$ %& ! 2 3, " 2, 4#$ %&( )

= "1, 4, 1#$ %& ! 6, " 4, 8#$ %&

= 4(8)" (–4)(1), 1(6)" 8(–1), "1(–4)" 6(4)#$ %&

= 36, 14, " 20#$ %&

Therefore, L.S. = M.S. = R.S. Therefore, ( )k u v!

! != ( )ku v!

! ! = ( )u kv!

! !.

b) Let [ ]1 2 3, ,u u u u=

! and [ ]1 2 3, ,v v v v=

!.

L.S. = k u!

! v!

( )

= k u1, u2 , u3"# $% ! v1, v2 , v3"# $%( )

= k u2v3 & v2u3, u3v1 & v3u1, u1v2 & v1u2"# $%

= k(u2v3 & v2u3), k(u3v1 & v3u1), k(u1v2 & v1u2 )"# $%

= ku2v3 & kv2u3, ku3v1 & kv3u1, ku1v2 & kv1u2"# $%

M.S. = ku!

( )! v!

= k u1, u2 , u3"# $%( )! v1, v2 , v3"# $%

= ku1, ku2 , ku3"# $% ! v1, v2 , v3"# $%


R.S. = u!

! kv!

( )

= u1, u2 , u3"# $% ! k v1, v2 , v3"# $%( )

= u1, u2 , u3"# $% ! kv1, kv2 , kv3"# $%


Therefore, L.S. = M.S. = R.S. Therefore, ( ) ( ) ( )k u v ku v u kv! = ! = !

! ! ! ! ! !.


Chapter 7 Section 5 Question 18 Page 412 No. Choose three collinear vectors. Let [ ] [ ] [ ]1, 2, 3 , 2, 4, 6 , and 3, 6, 9a b c= = =

! ! !.

a!

! b!

= 1, 2, 3"# $% ! 2, 4, 6"# $%

= 2(6)& 4(3), 3(2)& 6(1), 1(4)& 2(2)"# $%

= 0, 0, 0"# $%

a!

! c!

= 1, 2, 3"# $% ! 3, 6, 9"# $%

= 2(9)& 6(3), 3(3)& 9(1), 1(6)& 3(2)"# $%

= 0, 0, 0"# $%

Clearly but a b a c b c! = ! "

! ! ! ! ! !. Any three collinear vectors will show this.


a!

! b!

= a!

b!

sin" and a!

#b!

= a!

b!

cos".

Therefore, a!

! b!

< a!

#b!

if sin" < cos".

Examining the related functions sin and cosy y! != = in the interval [0º, 180º], it can be seen

that 0º < θ < 45º when sin! < cos! .

a!

! b!

< a!

"b!

if # < 45o.

a!

! b!

= a!

"b!

if # = 45o.

a!

! b!

> a!

"b!

if # > 45o.

The easy test is to determine the measure of θ is cos a ba b

!"

=

! !

! ! .


b)

2, 1, !1"# $% & !1, ! 2, 1"# $% = !5

2, 1, !1"# $% = 6

!1, ! 2, 1"# $% = 6

! = cos"1 "56 6

#

$%&

'(

! 146.4o

In this case, a b a b! > "

! ! ! !.

c) [ ] [ ]

[ ]

[ ]

2, 1, 1 3, 1, 2 9

2, 1, 1 6

3, 1, 2 14

! =

=

=

1

o

96 14

10.9

cos! " # $= % &

' (!

In this case, a b a b! < "

! ! ! !.

d) The most likely random case is a b a b! > "

! ! ! ! since θ is more likely to be in the interval (45º, 180º] than

in the interval [0º, 45º). Chapter 7 Section 5 Question 20 Page 412

a) From question 6, it is known that ( )22 2

u v u v u v! = " #

! ! ! ! ! !.

The expression under the square root symbol must be greater than or equal to zero (since u v!! !

is always defined).

( )

( )

22 2

22 2

0u v u v

u v u v

u v u v

! " #

# "

# "

! ! ! !

! ! ! !

! ! ! !

[Note: Actually, the stronger result, u v u v! "

! ! ! !, is true.]


b) Proof 1 (! ) Let and u v! !

be collinear. Then u kv=

! !.

u!

! v!

= u! 2

v! 2

" u!

#v!

( )2

kv!

! v!

= u! 2

v! 2

" u!

#v!

( )2

0 = u! 2

v! 2

" u!

#v!

( )2

u!

! v!

= 0!

for collinear vectors.

0 = u! 2

v! 2

" u!

#v!

( )2

u!

#v!

( )2

= u! 2

v! 2

u!

#v!

= u!

v!

Take the positive square root of both sides.

Proof 2 (! ) Let u v u v! =

! ! ! !

u!

!v!

= u!

v!

u!

!v!

( )2

= u! 2

v! 2

Square both sides

0 = u! 2

v! 2

" u!

!v!

( )2

0 = u! 2

v! 2

" u!

!v!

( )2

0 = u!

# v!

(From question 6.)

0 = u!

v!

sin$

If sin 0! = , then o0! = and and u v

! !are collinear.

If 0 or 0u v= =

! !, then 0 or 0u v= =

! ! ! ! and the vectors are (trivially) collinear.

Therefore, and u v

! !be collinear if and only if ( ! ) u v u v! =

! ! ! !.

Chapter 7 Section 5 Question 21 Page 412 Vectors , , and ca b

! ! ! are position vectors for the three vertices of the triangle.

Two of the sides of the triangle can be represented by and b a c a! !

! ! ! !.

The area of the parallelogram defined by and b a c a! !

! ! ! ! is

b!

! a!

( )" c!

! a!

( ) .

The area of the triangle is one-half of this expression.


!Area =12

b!

" a!

( )# c!

" a!

( )

=12

b!

# c!

" b!

# a!

" a!

# c!

+ a!

# a!

Distributive property

=12

b!

# c!

" b!

# a!

" a!

# c!

a!

# a!

= 0!

=12

b!

# c!

+ a!

# b!

+ c!

# a!

a!

# b!

= "b!

# a!

=12

a!

# b!

+ b!

# c!

+ c!

# a!

Commutative property

Chapter 7 Section 5 Question 22 Page 412 a) Two examples of ( ) ( )a b c a b c! ! = ! !

! ! ! ! ! !:

Since i!

! j!

! k!

( ) = i!

! i!

= 0!

and i!

! j!

( )! k!

= k!

! k!

= 0,"!

i!

! j!

! k!

( ) = i!

! j!

( )! k!

and

Since j!

! i!

! k!

( ) = j!

! " j!

( ) = 0!

and j!

! i!

( )! k!

= "k!

( )! k!

= 0!

,

j!

! i!

! k!

( ) = j!

! i!

( )! k!

b) Let [ ]1 2 3, ,a a a a=

!, [ ]1 2 3, ,b b b b=

!, and [ ]1 2 3, ,c c c c=

!.

L.S. = a!

! b!

! c!

( )

= a1, a2 , a3"# $% ! b1, b2 , b3"# $% ! c1, c2 , c3"# $%( )

= a1, a2 , a3"# $% ! b2c3 & b3c2 , b3c1 & b1c3, b1c2 & b2c1"# $%

= a2 (b1c2 & b2c1)& a3 (b3c1 & b1c3) a3(b2c3 & b3c2 )& a1(b1c2 & b2c1), a1(b3c1 & b1c3)& a2 (b2c3 & b3c2 )"# $%

= a2b1c2 & a2b2c1 & a3b3c1 + a3b1c3, a3b2c3 & a3b3c2 & a1b1c2 + a1b2c1, a1b3c1 & a1b1c3 & a2b2c3 + a2b3c2"# $%

R.S. = a!

!c!

( )b!

" a!

!b!

( )c!

= a1, a2 , a3#$ %& ! c1, c2 , c3#$ %&( ) b1, b2 , b3#$ %& " a1, a2 , a3#$ %& ! b1, b2 , b3#$ %&( ) c1, c2 , c3#$ %&

= (a1c1 + a2c2 + a3c3) b1, b2 , b3#$ %& " (a1b1 + a2b2 + a3b3) c1, c2 , c3#$ %&

= a1b1c1 + a2b1c2 + a3b1c3, a1b2c1 + a2b2c2 + a3b2c3, a1b3c1 + a2b3c2 + a3b3c3#$ %&

" a1b1c1 + a2b2c1 + a3b3c1, a1b1c2 + a2b2c2 + a3b3c2 , a1b1c3 + a2b2c3 + a3b3c3#$ %&

= a2b1c2 " a2b2c1 " a3b3c1 + a3b1c3, a3b2c3 " a3b3c2 " a1b1c2 + a1b2c1, a1b3c1 " a1b1c3 " a2b2c3 + a2b3c2#$ %&

Therefore, L.S. = R.S. Therefore,

a!

! b!

! c!

( ) =

a!

!c!

( )b!

" a!

!b!

( )c!

.


Chapter 7 Section 5 Question 23 Page 412 See Question 12 on page 359. For a pentagon with vertices (x1,y1), (x2,y2), (x3,y3), (x4,y4), and (x5,y5), the area is

1 2 2 1 2 3 3 2 3 4 4 3 4 5 5 4 5 1 1 5

1

2A x y x y x y x y x y x y x y x y x y x y= ! + ! + ! + ! + !

Therefore,

A =12

0(7)! 2(5) + 2(6)! 5(7) + 5(4)! 6(6) + 6(2)!1(4) +1(5)! 0(2)

=12!36

= 18

The area is 18 square units. Chapter 7 Section 5 Question 24 Page 412 Assume the interior of each square is 20 mm by 20 mm. The total area of the grid is (120 + 7)(100 + 6). The centre of the marble can pass through a square that is 10 mm by 10 mm in each grid hole. Therefore, the area for “success” is 30(100).

The probability it will pass through (success) is

30(100)(127)(106)

! 0.223or 22.3%.


Chapter 7 Section 6 Applications of the Dot Product and Cross Product Chapter 7 Section 6 Question 1 Page 418 a)

!

!

= r!

F"!

sin"

= 0.15(90)sin 70o

# 12.7

The torque is approximately 12.7 Nim . b) The torque vector is upward from the material. Therefore, the head of the bolt is being loosened and it

will move upward out of the material. [Note: In a real situation, this is true only if the bolt has “regular” right-handed thread.]


a)

projv!u!

=u!

!v!

v!

!v!

"

#$

%

&' v!

=

3, 1, 4() *+ ! 6, 2, 7() *+6, 2, 7() *+ ! 6, 2, 7() *+

"

#$

%

&' 6, 2, 7() *+

=3(6) +1(2) + 4(7)6(6) + 2(2) + 7(7)

"

#$%

&'6, 2, 7() *+

=4889

6, 2, 7() *+

[ ]

2 2 2

48proj 6, 2, 78948 6 2 78948 89

895.1

v u =

= + +

=

!

!

"

b)

projv!u!

=u!

!v!

v!

!v!

"

#$

%

&' v!

=

5, ( 4, 8)* +, ! 3, 7, 6)* +,3, 7, 6)* +, ! 3, 7, 6)* +,

"

#$

%

&' 3, 7, 6)* +,

=5(3) + (–4)(7) + 8(6)

3(3) + 7(7) + 6(6)"

#$%

&'3, 7, 6)* +,

=3594

3, 7, 6)* +,


[ ]

2 2 2

35proj 3, 7, 69435 3 7 69435 94

943.6

v u =

= + +

=

!

!

"

c)

projv!u!

=u!

!v!

v!

!v!

"

#$

%

&' v!

=

(2, ( 7, 3)* +, ! 6, 1, ( 8)* +,6, 1, ( 8)* +, ! 6, 1, ( 8)* +,

"

#$

%

&' 6, 1, ( 8)* +,

=(2(6) + (–7)(1) + 3(–8)6(6) +1(1) + (–8)(–8)

"

#$%

&'6, 1, ( 8)* +,

=(43101

6, 1, ( 8)* +,

projv!u!

=!43101

6, 1, ! 8"# $%

=43

10162

+12+ (–8)2

=43 101

101" 4.3

d)

projv!u!

=u!

!v!

v!

!v!

"

#$

%

&' v!

=

1, 0, (1)* +, ! 9, 1, 0)* +,9, 1, 0)* +, ! 9, 1, 0)* +,

"

#$

%

&' 9, 1, 0)* +,

=1(9) + 0(1) + (–1)(0)

9(9) +1(1) + 0(0)"

#$%

&'9, 1, 0)* +,

=9

829, 1, 0)* +,


[ ]

2 2 2

9proj 9, 1, 0829 9 1 0829 82

821.0

v u =

= + +

=

!

!

"

Chapter 7 Section 6 Question 3 Page 418 a) To calculate the work done against gravity, use only the vertical components of and F s

!" ".

Wgravity = F!"

! s"

= 0, 0, 12"# $% ! 0, 0, 8"# $%

= 96

The work done against gravity is 96 J.

W = F!"

! s"

= 3, 5, 12"# $% ! 0, 0, 8"# $%

= 96

The work done in the direction of travel is 96 J.

b)

Wgravity = F!"

! s"

= 0, 0, 12"# $% ! 0, 0, 10"# $%

= 120


W = F!"

! s"

= 3, 5, 12"# $% ! 2, 0, 10"# $%

= 126


c)

Wgravity = F!"

! s"

= 0, 0, 12"# $% ! 0, 0, 6"# $%

= 72



W = F!"

! s"

= 3, 5, 12"# $% ! 2, 1, 6"# $%

= 83



a!

! b!

"c!

= a!

! b!

( ) "c!

= #2, 3, 5$% &' ! 4, 0, #1$% &' " 2, # 2, 3$% &'

= 3(–1)# 0(5), 5(4)# (–1)(–2), # 2(0)# 4(3)$% &' " 2, # 2, 3$% &'

= #3, 18, #12$% &' " 2, # 2, 3$% &'

= #3(2) +18(–2) + (–12)(3)= #78

b)

a!

!b!

" c!

= a!

! b!

" c!

( )

= #2, 3, 5$% &' ! 4, 0, #1$% &' " 2, # 2, 3$% &'

= #2, 3, 5$% &' ! 0(3)# (–2)(#1), (#1)(2)# 3(4), 4(–2)# 2(0)$% &'

= #2, 3, 5$% &' ! #2, #14, # 8$% &'

= #2(–2) + 3(–14) + 5(–8)= #78

c)

a!

! c!

"b!

= a!

! c!

( ) "b!

= #2, 3, 5$% &' ! 2, # 2, 3$% &' " 4, 0, #1$% &'

= 3(3)# (–2)(5), 5(2)# 3(–2), # 2(–2)# 2(3)$% &' " 4, 0, #1$% &'

= 19, 16, # 2$% &' " 4, 0, #1$% &'

= 19(4) +16(0) + (–2)(#1)= 78

d)

b!

!a!

" c!

= b!

! a!

" c!

( )

= 4, 0, #1$% &' ! #2, 3, 5$% &' " 2, # 2, 3$% &'

= 4, 0, #1$% &' ! 3(3)# (–2)(5), 5(2)# 3(–2), – 2(–2)# 2(3)$% &'

= 4, 0, #1$% &' ! 19, 16, # 2$% &'

= 4(19) + 0(16) + (–1)(–2)= 78



V = w!"

!u"

" v"

= 1, 2, 7#$ %& ! 1, 4, 3#$ %& " 2, 5, 6#$ %&

= 1, 2, 7#$ %& ! 4(6)' 5(3), 3(2)' 6(1), 1(5)' 2(4)#$ %&

= 1, 2, 7#$ %& ! 9, 0, ' 3#$ %&

= '12

= 12

The volume is 12 cubic units.

b)

V = w!"

!u"

" v"

= 1, 3, 5#$ %& ! '2, 5, 1#$ %& " 3, ' 4, 2#$ %&

= 1, 3, 5#$ %& ! 5(2)' (–4)(1), 1(3)' 2(–2), – 2(–4)' 3(5)#$ %&

= 1, 3, 5#$ %& ! 14, 7, ' 7#$ %&

= 0

= 0


c)

V = w!"

!u"

" v"

= #2, 0, 5$% &' ! 1, 1, 9$% &' " 0, 0, 4$% &'

= #2, 0, 5$% &' ! 1(4)# 0(9), 9(0)# 4(1), 1(0)# 0(1)$% &'

= #2, 0, 5$% &' ! 4, # 4, 0$% &'

= #8

= 8



Chapter 7 Section 6 Question 6 Page 418 The area of the parallelogram defined by and a b

! ! is a b!! !

where a!

= AB" !""

and b!

= AC" !""

.

For triangle !ABC , calculate 12

a b!! !

.

AB! "!!

= 9, 7, ! 7"# $%

AC! "!!

= 7, !1, !1"# $%

a"

& b"

= 9, 7, ! 7"# $% & 7, !1,!1"# $%

= 7(–1)! (–1)(–7), – 7(7)! (–1)(9), 9(–1)! 7(7)"# $%

= !14, ! 40, ! 58"# $%

a"

& b"

= (–14)2+ (–40)2

+ (–58)2

# 71.83

A =12

(71.83)

# 35.9

The area of the triangle is about 35.9 square units. Chapter 7 Section 6 Question 7 Page 418

!

!

= r!

F"!

sin"

= 0.15(75)sin 80o

# 11.1

The magnitude of the torque is about 11.1 N·m. Chapter 7 Section 6 Question 8 Page 418 The torques created by each chills should be equal and opposite. Let x be the distance of the girl from the balance point. The angles made by the children will be θ and –θ.

65(0.6)sin! = "40(x)sin("! )

x =65(0.6)

40x = 0.975

The girl should sit 0.975 m form the balance point.



u!

! v!

= 2, 2, 3"# $% ! 1, 3, 4"# $%

= 2(4)& 3(3), 3(1)& 4(2), 2(3)&1(2)"# $%

= &1, & 5, 4"# $%

v!

! w"!

= 1, 3, 4"# $% ! 6, 2, 1"# $%

= 3(1)& 2(4), 4(6)&1(1), 1(2)& 6(3)"# $%

= &5, 23, &16"# $%

u!

! w"!

= 2, 2, 3"# $% ! 6, 2, 1"# $%

= 2(1)& 2(3), 3(6)&1(2), 2(2)& 6(2)"# $%

= &4, 16, & 8"# $%

a)

u!

! v! 2

" w"!

#w"!

( )2

= ["1, " 5, 4]2" 6, 2, 1$% &' # 6, 2, 1$% &'( )

2

= ("1)2+ ("5)2

+ 42( )2

" (6(6) + 2(2) +1(1))2

= 1+ 25+16 "1681= "1639

b)

u!

! u!

+ u!

"u!

= 0!

+ 2, 2, 3#$ %& " 2, 2, 3#$ %&

= 0 + 22+ 22

+ 32

= 17

c)

u!

! v!

"v!

! w"!

= #1, # 5, 4$% &' " #5, 23, #16$% &'

= #1(#5) + (#5)(23) + 4(–16)= #174

d)

u!

! v!

"u!

! w"!

= #1, # 5, 4$% &' " #4, 16, # 8$% &'

= #1(–4) + (–5)(16) + 4(–8)= #108



Given vectors a

! and b!

. Area ABEC = a b!

! !.

Let AD! "!!

= a"

! b"

. Let AF! "!!

be the product of the projection of b!

on a!

and the magnitude of a!

.

AF! "!!

= b"

cos!( ) a"

= a"

"b"

b) By the Pythagorean theorem, AD2 + AF2 = DF2.

Therefore, 2 22DF a b a b= ! + "

! ! ! !.

c) The proof or verification of 2 2 2 2

a b a b a b! + " =

! ! ! ! ! ! can be done geometrically or using coordinates.

First show a proof using coordinates. Let [ ] [ ]1 2 3 1 2 3, , and , ,a a a a b b b b= =

! !.

L.S. = a!

! b! 2

+ a!

"b! 2

= a1, a2 , a3#$ %& ! b1, b2 , b3#$ %&2

+ a1, a2 , a3#$ %& " b1, b2 , b3#$ %&2

= a2b3 ' a3b2 , a3b1 ' a1b3, a1b2 ' a2b1#$ %&2

+ a1b1 + a2b2 + a3b3

2

= (a2b3 ' a3b2 )2+ (a3b1 ' a1b3)2

+ (a1b2 ' a2b1)2+ (a1b1 + a2b2 + a3b3)2

= a22b3

2 ' 2a2b3a3b2 + a32b2

2+ a3

2b12 ' 2a1b3a3b1 + a1

2b32

+ a12b2

2 ' 2a2b1a1b2 + a22b1

2

+ a12b1

2+ a2

2b22

+ a32b3

2+ 2a1b1a2b2 + 2a2b2a3b3 + 2a1b1a3b3

= a12b1

2+ a1

2b22

+ a12b3

2+ a2

2b12

+ a22b2

2+ a2

2b32

+ a32b1

2+ a3

2b22

+ a32b3

2

b!

a!

A

B

D

E

F

θ C


R.S. = a! 2

b! 2

= a1, a2 , a3!" #$2

b1, b2 , b3!" #$2

= (a12

+ a22

+ a32 )(b1

2+ b2

2+ b3

2 )

= a12b1

2+ a1

2b22

+ a12b3

2+ a2

2b12

+ a22b2

2+ a2

2b32

+ a32b1

2+ a3

2b22

+ a32b3

2

Therefore, L.S. = R.S. Now show a proof using geometric vectors.

L.S. = a!

! b! 2

+ a!

"b! 2

= a!

b!

sin#2

+ a!

b!

cos#2

= a! 2

b! 2

sin2# + a! 2

b! 2

cos2#

= a! 2

b! 2

(sin2# + cos2

# )

= a! 2

b! 2

(1)

= a! 2

b! 2

R.S. = a! 2

b! 2


Chapter 7 Section 6 Question 11 Page 418 Assume each force makes a 90º angle with the radius. However, one angle is clockwise (positive) and one is counterclockwise (negative).

!

!

= r!

F"!

sin"

!

!

total = (0.75)10sin90o+ 0.35(5)sin(#90o )

# 5.75

The magnitude of the total torque is about 5.75 N·m. The direction will be that of the 10-N torque (along the axle, away from you).



!

!

= r!

F"!

sin"

10 = 0.20(80)sin"

sin" =10

(0.20)80" # 38.7

The force is applied to the wrench at about a 38.7º angle. Chapter 7 Section 6 Question 13 Page 419 Answers will vary.

• Torque is the force responsible for getting a vehicle to start moving from a stop position and it pulls the vehicle up steep hills. This measurement is important to know when

Horsepower is responsible for moving a vehicle along and allows it to cruise on the highway and accelerate in normal conditions.

A higher torque rating is more important than a higher horsepower rating when pulling a trailer, hauling a heavy load, or driving on a long, steep road.

A higher horsepower is more important than a higher torque rating when determining how much time it

takes to go from stoplight to stoplight or if performing a quick acceleration. • Torque is measured in pound-feet.

The abbreviated measurement unit for horsepower is hp. 1 hp = 550 foot-pounds per second

Horsepower =

torque ! engine speed5252

• When plotting graphs involving torque, the independent variable or torque is measured in Newton•metres.

The dependent variable may be measured in revolutions per minute—a measure of engine speed or amperes—a measure of current.

Chapter 7 Section 6 Question 14 Page 419 The statement is false.

0e f! =

! "! if e and f are collinear and the angle between them is 0° or 180°.

0e f! =

! "! if e!

and f!"

are perpendicular and the angle between them is 90°.



!

!

= r!

F"!

sin"

100 = 0.30 F"!

sin 40o

F"!

=100

(0.30)sin 40o

# 518.6

The magnitude of the force is about 518.6 N. Chapter 7 Section 6 Question 16 Page 419 a)

b)

V =12!a!b !c

=12

14 20 180

= 7.5 14" 28.1

The volume of the prism is approximately 28.1 cubic units.

c) If the prism was not a right triangular prism, the volume of the prism can be determined by the same formula but the perpendicular distance between the faces must be found first before the volume formula can be applied.

The volumes of the right and non-right prism would not be the equivalent given equal bases and height of

the triangle.

!a

!c

!b


Chapter 7 Section 4 Question 17 Page 419 Let [ ] [ ]1 2 3 1 2 3, , and , ,u u u u v v v v= =

! !

L.S. = u!

! v!

"w"!

= u!

! v!

" ku!

+ mv!

( )

= u1, u2 , u3#$ %& ! v1, v2 , v3#$ %& " k u1, u2 , u3#$ %& + m v1, v2 , v3#$ %&#$

%&

= u2v3 ' u3v2 , u3v1 ' u1v3, u1v2 ' u2v1#$ %& " ku1 + mv1, ku2 + mv2 , ku3 + mv3#$ %&

= (u2v3 ' u3v2 )(ku1 + mv1) + (u3v1 ' u1v3)(ku2 + mv2 ) + (u1v2 ' u2v1)(ku3 + mv3)= ku1u2v3 + mu2v1v3 ' ku1u3v2 ' mu3v1v2 + ku2u3v1 + mu3v1v2 ' ku1u2v3 ' mu1v2v3

+ ku1u3v2 + mu1v2v3 ' ku2u3v1 ' mu2v1v3

= 0R.S. = 0

Therefore, L.S. = R.S. Chapter 7 Section 6 Question 18 Page 419 For an informal proof, draw a diagram.

Suppose and a b

! ! lie in a plane and c

! is a vector mot in this plane.

a b!! !

is perpendicular to both and a b! !

. ( )a b c! !

! ! ! is perpendicular a b!

! ! and c

!. But the vectors perpendicular to a b!

! ! can only be in the plane

containing both and a b! !

. Therefore ( )a b c! !

! ! ! lies in the plane containing and a b

! !.

For a more formal proof, show that the volume of the parallelepiped formed by ( )a b c! !

! ! !, , and a b! !

has

volume zero; i.e., ( )a b c! !

! ! !, , and a b! !

are coplanar vectors. Let [ ] [ ] [ ]1 2 3 1 2 3 1 2 3, , , , , , and , ,a a a a b b b b c c c c= = =

! ! !.

a!

b!

c!

a b!! !

( )a b c! !! ! !


V = a!

! b!

( )! c!

( ) "a!

! b!

= a1, a2 , a3#$ %& ! b1, b2 , b3#$ %&( )! c1, c2 , c3#$ %&( ) " a1, a2 , a3#$ %& ! b1, b2 , b3#$ %&

= a2b3 ' a3b2 , a3b1 ' a1b3, a1b2 ' a2b1#$ %& ! c1, c2 , c3#$ %& " a2b3 ' a3b2 , a3b1 ' a1b3, a1b2 ' a2b1#$ %&

=

(a3b1 ' a1b)c3 ' c2 (a1b2 ' a2b1), (a1b2 ' a2b)c1 ' c3(a2b3 ' a3b2 ), (a2b3 ' a3b2 )c2 ' c1(a3b1 ' a1b3)#$ %&

" a2b3 ' a3b2 , a3b1 ' a1b3, a1b2 ' a2b1#$ %&

=

a3b1c3 ' a1b3c3 ' a1b2c2 + a2b1c2 , a1b2c1 ' a2b1c1 ' a2b3c3 + a3b2c3, a2b3c2 ' a3b2c2 ' a3b1c1 + a1b3c1#$ %&

" a2b3 ' a3b2 , a3b1 ' a1b3, a1b2 ' a2b1#$ %&

=(a3b1c3 ' a1b3c3 ' a1b2c2 + a2b1c2 )(a2b3 ' a3b2 ) + (a1b2c1 ' a2b1c1 ' a2b3c3 + a3b2c3)(a3b1 ' a1b3)+(a2b3c2 ' a3b2c2 ' a3b1c1 + a1b3c1)(a1b2 ' a2b1)

=

a2a3b1b3c3 ' a1a2b3b3c3 ' a1a2b2b3c2 + a2a2b1b3c2 ' a3a3b1b2c3 + a1a3b2b3c3 + a1a3b2b2c2 ' a2a3b1b2c2

+a1a3b1b2c1 ' a2a3b1b1c1 ' a2a3b1b3c3 + a3a3b1b2c3 ' a1a1b2b3c1 + a1a2b1b3c1 + a1a2b3b3c3 ' a1a3b2b3c3

+a1a2b2b3c2 ' a1a3b2b2c2 ' a1a3b1b2c1 + a1a1b2b3c1 ' a2a2b1b3c2 + a2a3b1b2c2 + a2a3b1b1c1 ' a1a2b1b3c1

= 0

= 0

Chapter 7 Section 6 Question 19 Page 419 If , , and wu v! ! ""!

are mutually orthogonal, they could be considered to form axes for 3-space in the same way

as do , , and i j k! ! !

. They create three orthogonal planes: uv- plane, uw-plane, and vw-plane.

Then you can say that u v+

! !lies in the uv- plane, u w+

! "! lies in the uw-plane, and v w+

! "! lies in the vw-plane.

In general, these three new vectors are not themselves orthogonal. For example, if , , and i j k! ! !

are the

original vectors, then the new vectors are [ ] [ ] [ ]1, 1, 0 , 0, 1, 1 , and 1, 0, 1i j j k i k+ = + = + =

! ! ! ! ! !.

i!

+ j!

( ) ! j!

+ k!

( ) = 1, 1, 0"# $% ! 0, 1, 1"# $%

= 1& 0

Therefore, i j+

! ! is not perpendicular to j k+

! !.


Chapter 7 Section 6 Question 20 Page 419 Draw a regular pentagon ABCDE. The interior angles are 540º ÷ 5 = 108º.

I

J

G

K

H

B

A

E

DC

There are many isosceles triangles in the diagram. This leads to labelling all of the angles as either 36º, 72º, or 108º. ΔBKG is an isosceles triangle having the same shape as the triangle in the contest question. ΔACD is similar to ΔBKG. (equal angles) Let CD = 1 and AD = x. By similar triangles, x is the measure needed for the question. Consider the triangles ΔHAE ~ ΔEAD.

HAAE

=DEAD

HA1

=1x

HA =1x

ΔDHE is isosceles, so DE = DH = 1. Therefore,

1+1x

= x

x2! x !1= 0

x =1± 5

2(Quadratic formula)

Since x must be positive, 1 52

x += cm.


Chapter 7 Section 6 Question 21 Page 419 The first term is 333; the second term counts the symbols involved in the first term, namely, 3 threes. The second term is 33; the third term counts these symbols: 2 threes. The third term is 23; the fourth term counts these symbols: 1 two and 1 three. Additional terms are 112113, 11121113, 1111211113, 111112111113, …


Chapter 7 Review Chapter 7 Review Question 1 Page 420 a)

v!

= !6, 3"# $%

= !6i!

+ 3 j!

b)

v!

= (–6)2+ 33

= 45

= 3 5

Unit vectors collinear with v!

have the form 1 vv

±

!

!

The required unit vectors are 2 1,5 5

! "#$ %& '

and 2 1,5 5

! "#$ %

& '.

c) Let B(x, y) be the point.

AB! "!!

= OB! "!!

!OA! "!!

= x, y"# $% ! 2, 9"# $%

= x ! 2, y ! 9"# $%

= !6, 3"# $%

x ! 2 = !6 y ! 9 = 3x = !4 y = 12

Therefore, B(–4, 12) is the required point.

Chapter 7 Review Question 2 Page 420 a) [ ]5 5, 2! ! = [−25, 10] b)

u!

+ v!

= 5, ! 2"# $% + 8, 5"# $%

= [13,3]

c)

4u!

+ 2v"!"

= 4 5, ! 2"# $% + 2 8, 5"# $%

= 20, ! 8"# $% + 16, 10"# $%

= 36, 2"# $%


d)

3u!

! 7v!

= 3 5, ! 2"# $% ! 7 8, 5"# $%

= 15, ! 6"# $% ! 56, 35"# $%

= !41, ! 41"# $%

Chapter 7 Review Question 3 Page 420 The airplane’s vector is

345cos 50º , 345sin 50º!" #$ ! 221.8, 264.3!" #$ .

The wind’s vector is [18cos 183o , 18sin 183o ] ! !17.98, ! 0.94"# $% .

The resultant vector is [203.82, 263.36].

The magnitude of the resultant is

203.82( )2

+ 263.36( )2! 333.02 .

The angle of the resultant with the east direction is tan!1 263.36

203.82"

#$%

&'! 52.3o .

The bearing is 90º – 52.3º = 37.7º. The ground velocity of the airplane is about 333.0 km/h on a bearing of 37.7°.

Chapter 7 Review Question 4 Page 420 a)

u!

!v!

= 20(15)cos 70o

" 102.6

b)

p!"

!q"

= 425(300)cos 110o

# "43 607.6


u!

!v!

= 5(–6) + 2(7)= "16

b)

u!

!v!

= (–3)(3) + 2(7)= 5


c)

u!

!v!

= 3(4) + 2(–6)= 0

Chapter 7 Review Question 6 Page 420 The vectors in part c) are orthogonal since 0u v! =

! !.

Chapter 7 Review Question 7 Page 420

cos! =a!

"b!

a!

b!

=20

5.2(7.3)

! = cos#1 205.2(7.3)

$

%&'

()

" 58.2o


a)

cos! =a!

"b!

a!

b!

=

6, # 5$% &' " 7, 2$% &'6, # 5$% &' 7, 2$% &'

=6(7) + (–5)(2)

62+ (–5)2 72

+ 22

=32

56.8595

! = cos#1 3256.8595

(

)*+

,-

" 55.8o


b)

cos! =p!"

"q"

p!"

q"

=

#9, # 4$% &' " 7, # 3$% &'#9, # 4$% &' 7, # 3$% &'

=–9(7) + (–4)(–3)

(–9)2+ (–4)2 72

+ (–3)2

=#51

75.0067

! = cos#1 #5175.0067

(

)*+

,-

# 132.8o


projv!u!

= u!

cos!

= 56cos 125o

" "32.1

The projection has magnitude 32.1 and has direction opposite to v .

b)

projv!u!

=u!

!v!

v!

!v!

"

#$

%

&' v!

=

7, 1() *+ ! 9, , 3() *+9, , 3() *+ ! 9, , 3() *+

"

#$

%

&' 9, , 3() *+

=6090

"

#$%

&'9, , 3() *+

" 6, , 2() *+



F!"

!d!"

= 16, 12"# $% ! 3, 9"# $%

= 16(3) +12(9)= 156

The work done is 156 N !m or 156 J.

b)

F!"

!d!"

= 200, 2000"# $% ! 3, 45"# $%

= 200(3) + 2000(45)= 90 600

The work done is 90 600 N !m or 90 6000 J.

Chapter 7 Review Question 11 Page 420 a) The total revenue can be represented by [125, 70]·[229, 329]. b)

R = 125(229) + 70(329)= 51 655

The total revenue from sales is $51 655.00


AB! "!!

= OB! "!!

!OA! "!!

= !5, 9, !1"# $% ! 2, 7, 8"# $%

= !7, 2, ! 9"# $%

AB! "!!

= (–7)2+ 22

+ (–9)2

= 134

b)

PQ! "!!

= OQ! "!!

!OP! "!!

= 4, ! 9, 7"# $% ! 0, 3, 6"# $%

= 4, !12, 1"# $%

PQ! "!!

= 42+ (!12)2

+12

= 161



5a!

! 4b!

+ 3c!

= 5 3, ! 7, 8"# $% ! 4 !6, 3, 4"# $% + 3 2, 5, 7"# $%

= 15, ! 35, 40"# $% ! !24, 12, 16"# $% + 6, 15, 21"# $%

= 45, ! 32, 45"# $%

b)

!5a!

"c!

= !5 3, ! 7, 8#$ %& " 2, 5, 7#$ %&

= !15, 35, ! 40#$ %& " 2, 5, 7#$ %&

= –15(2) + 35(5) + (–40)(7)= !135

c)

b!

! c!

" a!

( ) = "6, 3, 4#$ %& ! 2, 5, 7#$ %& " 3, " 7, 8#$ %&( )

= "6, 3, 4#$ %& ! "1, 12, "1#$ %&

= (–6)(–1) + 3(12) + 4(–1)= 38

Chapter 7 Review Question 14 Page 420 Orthogonal vectors have a dot product equal to zero. [ ] [ ]6, 1, 8 , 4, 5 0

6 4 40 06 36

6

kk

kk

! " =

" + =

= "

= "


u!

! v!

= u!

v!

(sin" )n̂

= 200(350)sin 110o n̂" 65 778.5n̂

b)

u!

! v!

= 4, 1, " 3#$ %& ! 3, 7, 8#$ %&

= 1(8)" 7(–3), – 3(3)" 8(4), 4(7)" 3(1)#$ %&

= 29, " 41, 25#$ %&



A = u!

! v!

= 6, 8, 9"# $% ! 3, &1, 2"# $%

= 8(2)& (–1)(9), 9(3)& 2(6), 6(–1)& 3(8)"# $%

= 25, 15, & 30"# $%

= 252+152

+ (–30)2

" 41.8

Chapter 7 Review Question 17 Page 420 Let [ ] [ ] [ ]1, 0, 0 , 1, 1, 1 , and 0, 0, 1a b c= = =

! ! !.

L.S. = a!

! b!

+ c!

( )

= 1, 0, 0"# $% ! 1, 1, 1"# $% + 0, 0, 1"# $%( )

= 1, 0, 0"# $% ! 1, 1, 2"# $%

= 0(2)&1(0), 0(1)& 2(1), 1(1)&1(0)"# $%

= 0, & 2, 1"# $%

R.S. = a!

! b!

+ a!

! c!

= 1, 0, 0"# $% ! 1, 1, 1"# $% + 1, 0, 0"# $% ! 0, 0, 1"# $%

= 0(1)&1(0), 0(1)&1(1), 1(1)&1(0)"# $% + 0(1)& 0(0), 0(0)&1(1), 1(0)& 0(0)"# $%

= 0, &1, 1"# $% + 0, &1, 0"# $%

= 0, & 2, 1"# $%

Therefore, L.S. = R.S. and ( )a b c a b a c! + = ! + !

! ! ! ! ! ! !.


!

!

= r!

F"!

sin"

= 0.10(200)sin 80o

# 19.7

The torque is approximately 19.7 N !m .

b) The torque vector points downward into the material, along the direction of the bolt. The bolt is being

tightened. (Note this practical example assumes that the bolt has normal right-handed thread.)



projv!u!

=u!

!v!

v!

!v!

"

#$

%

&' v!

=

(2, 5, 3)* +, ! 4, ( 8, 9)* +,4, ( 8, 9)* +, ! 4, ( 8, 9)* +,

"

#$

%

&' 4, ( 8, 9)* +,

=–2(4) + 5(–8) + 3(9)

4(4) + (–8)(–8) + 9(9)"

#$%

&'4, ( 8, 9)* +,

=(21161

4, ( 8, 9)* +,

projv!u!

=!21161

4, ! 8, 9"# $%

=21

16142

+ (!8)2+ 92

=21 161

161" 1.7


Chapter 7 Practice Test Chapter 7 Practice Test Question 1 Page 422 B is the best answer. [ ] [ ]3, 4 6, 0

18 4 04 18

4.5

kkkk

! " =

! =

=

=

Chapter 7 Practice Test Question 2 Page 422 D is incorrect. Chapter 7 Practice Test Question 3 Page 422 D is the best answer. A = 1(1)sin 90º = 1 B = 1(1)sin 60º = 0.9 C = 4(1)sin 180º = 0 D = 4(0.5)sin 90º = 2 Chapter 7 Practice Test Question 4 Page 422 D is the best answer. Position vectors for points in the yz-plane are of the form [0, y, z]. Some examples are [0, 1, 1], [0, 4, 3], and [0, –2, –5]. Chapter 7 Practice Test Question 5 Page 422 C is the best answer. A [–8, –6]·[–6, 8] = 0 B [8, 6]·[–6, 8] = 0 C [4, –3]·[–6, 8] = –48 D [–4, –3]·[–6, 8] = 0


Chapter 7 Practice Test Question 6 Page 422 C is the best answer.

AB! "!!

= OB! "!!

!OA! "!!

= 0, 2, ! 4"# $% ! 1, 3, ! 7"# $%

= !1, !1, 3"# $%

AB! "!!

= (–1)2+ (–1)2

+ 32

= 11

Chapter 7 Practice Test Question 7 Page 422 a)

u!

!v!

= 8, 3"# $% ! 2, 7"# $%

= 8(2) + 3(7)= 37

b)

cos! =u!

"v!

u!

v!

=37

8, 3#$ %& 2, 7#$ %&

=37

82+ 32 22

+ 72

=37

62.2013

! = cos'1 3762.2013

(

)*+

,-

" 53.5o

c)

proju! v!

=v!

!u!

u!

!u!

"

#$

%

&' u!

=37

8, 3() *+ ! 8, 3() *+

"

#$

%

&' 8, 3() *+

=37

8(8) + 3(3)"

#$%

&'8, 3() *+

=3773

8, 3() *+

" 4.1, 1.5() *+


d) No. The projections are not equal. Proj

v!u!

is in the direction of v!

while the direction of proju v!!

is in the

direction of u!

, and these are clearly different directions. Chapter 7 Practice Test Question 8 Page 422 Let the price vector be p

!"= [399, 129] and the sales vector be s

!= [100, 240].

p!"

! s"

= 399, 129"# $% ! 100, 240"# $%

= 399(100) +129(240)= 70 860

The total revenue is $70 860. Chapter 7 Practice Test Question 9 Page 422 a)

u!

! v!

= [5, 8, 2]! "7, 3, 6#$ %&

= 8(6)" 3(2), 2(–7)" 6(5), 5(3)" (–7)(8)#$ %&

= 42, " 44, 71#$ %&

u!

'v!

= [5, 8, 2] ' "7, 3, 6#$ %&

= 5(–7) + 8(3) + 2(6)= 1

b)

u!

! v!

= [1, 2, " 5]! 3, " 4, 0#$ %&

= 2(0)" (–4)(–5), – 5(3)" 0(1), 1(–4)" 3(2)#$ %&

= "20, "15, "10#$ %&

u!

'v!

= [1, 2, " 5] ' 3, " 4, 0#$ %&

= 1(3) + 2(–4) + (–5)(0)= "5

c)

u!

! v!

= ["3, 0, 0]! 7, 0, 0#$ %&

= 0(0)" 0(0), 0(7)" 0(–3), – 3(0)" 7(0)#$ %&

= 0, 0, 0#$ %&

u!

'v!

= ["3, 0, 0] ' 7, 0, 0#$ %&

= "3(7) + 0(0) + 0(0)= "21


d)

u!

! v!

= [1, " 9, 7]! 9, 1, 0#$ %&

= "9(0)"1(7), 7(9)" 0(1), 1(1)" 9(–9)#$ %&

= "7, 63, 82#$ %&

u!

'v!

= [1, " 9, 7] ' 9, 1, 0#$ %&

= 1(9) + (–9)(1) + 7(0)= 0

Chapter 7 Practice Test Question 10 Page 422 a) The vectors in part c are collinear since 0u v! =

! ! !.

b) The vectors in part d are orthogonal since 0u v! =

! !.

Chapter 7 Practice Test Question 11 Page 423 a) Position vectors for points in the yz-plane are of the form [0, y, z].

Some examples of points in the plane are (0, 1, 4), (0, 4, 3), and (0, –2, –5). The diagram shows some of the points with positive y- and z-coordinates.

b) These points form a cylinder in 3-space with the x-axis as the central axis of symmetry of the cylinder.

The cylinder is defined by the points (x, 10cos! , 10sin! ) . Each point on the cylinder is 10 units from the x-axis.

The equation of the cylinder is 2 2 100y z+ = .

x

z

y

(0, 4, 3)

(0, 1, 4)

(0, 8, 1)


Chapter 7 Practice Test Question 12 Page 423 a)

u!

= 22+ 42

+ (–9)2

= 101

Unit vectors collinear with u!

have the form 1 uu

±

!

! .

The two unit vectors collinear with u!

are 2 4 9, ,101 101 101

! "#$ %

& 'and

2 4 9, ,101 101 101

! "# #$ %& '

.

b) Choose any vector v

! that is perpendicular to u

!. You want a vector that makes the dot product 0u v! =

! !.

Choose a convenient vector such as [ ]2, 1, 0v = !

!.

v!

= 22+ (–1)2

+ 02

= 5

Therefore two possible unit vectors collinear with u!

are 2 1 2 1, , 0 and , , 05 5 5 5

! " ! "# #$ % $ %

& ' & '.

c) The choice of vector v

! was somewhat arbitrary. You just need the dot product 0u v! =

! !.

In general, if [ ]1 2 3, ,v v v v=

!, then

1 2 32 4 9 0v v v+ ! = . For every choice of 1 2 and v v , there will be a

(unique) value for 3v . Every v!

created in this way, will lead to two orthogonal (to u!

) unit vectors. Hence, there are an infinite number of solutions.

Chapter 7 Practice Test Question 13 Page 423 The area of the triangle is one-half of the area of the parallelogram defined by two of the sides of the triangle.

A!

=12

a!

" b!

=12

a!

b!

sin#

=12

(14)(17)sin 24o

" 48.4

The area of the triangle is approximately 48.4 square units.


Chapter 7 Practice Test Question 14 Page 423 One vector perpendicular to and is .c d c d!

! "! ! "!

c!

! d"!

= "5, 6, 2#$ %& ! 3, 3, " 8#$ %&

= 6(–8)" 3(2), 2(3)" (–8)(–5), – 5(3)" 3(6)#$ %&

= "54, " 34, " 33#$ %&

c!

! d"!

= ("54)2+ ("34)2

+ ("33)2

= 5161

The required unit vectors are 54 34 33 54 34 33, , and , ,5161 5161 5161 5161 5161 5161

! " ! "# # #$ % $ %& ' & '

.

Chapter 7 Practice Test Question 15 Page 423 a) The standard formula for computing the area of a parallelogram requires 3-D vectors. You can achieve

this by embedding the given vectors in 3-space by including a z-coordinate of zero for each vector.

A = a!

! b!

= 1, " 4, 0#$ %& ! 3, 5, 0#$ %&

= "4(0)" 5(0), 0(3)" 0(1), 1(5)" 3(–4)#$ %&

= 0, 0, 17#$ %&

= 17


b)

A = u!

! v!

= "3, 2, 0#$ %& ! 6, " 4, 2#$ %&

= 2(2)" (–4)(0), 0(6)" 2(–3), – 3(–4)" 6(2)#$ %&

= 4, 6, 0#$ %&

= 42+ 62

+ 02

" 7.2

The area is approximately 7.2 square units.


Chapter 7 Practice Test Question 16 Page 423

V = u!

!v!

" w"!

= 1, 0, # 4$% &' ! 0, # 3, 2$% &' " 2, # 2, 0$% &'

= 1, 0, # 4$% &' ! #3(0)# (–2)(2), 2(2)# 0(0), 0(–2)# 2(–3)$% &'

= 1, 0, # 4$% &' ! 4, 4, 6$% &'

= 1(4) + 0(4) + (–4)(6)

= #20

= 20

The volume is 20 cubic units. Chapter 7 Practice Test Question 17 Page 423 Many solutions are possible. Check for right angles using the dot product.

AB! "!!

= !8, 4"# $%

BC! "!!

= 4, ! 9"# $%

AC! "!!

= !4, ! 5"# $%

AB! "!!

!BC! "!!

= ["8, 4] ! 4, " 9#$ %&

= "68' 0

AB! "!!

!AC! "!!

= ["8, 4] ! "4, " 5#$ %&

= 12' 0

BC! "!!

!AC! "!!

= 4, " 9#$ %& ! "4, " 5#$ %&

= 29' 0

Therefore, !ABC is not a right triangle.


Chapter 7 Practice Test Question 18 Page 423 If you assume the force is parallel to the ramp, then the angle between o and is 0F s

!" ".

W = F!"

! s"

= F!"

s"

cos"

= 40(5)cos 0o

= 200

The work done in pulling the cart up the ramp is 200 J. Chapter 7 Practice Test Question 19 Page 423 Represent the velocities as vectors in 3-space. Let the y-axis point north, the x-axis point east, and the z-axis point vertical. The jet’s vector is [0, 450, 450tan 12º]. The wind vector is [–12, 0, 0]. The resultant air velocity is the vector sum [–12, 450, 95.65]. The magnitude of this vector (the air speed) is (!12)2

+ 4502+ 95.652 or approximately 460.2 km/h.

The ground velocity vector can be calculated by ignoring the vertical (z) components of the vectors.

The ground velocity vector is [–12, 450, 0]. Its magnitude is (!12)2+ 4502

+ 02 or approximately 450.2 kmh

x

y

z

12º [0, 450, 95.65]

[–12, 0, 0]


Chapter 7 Practice Test Question 20 Page 423 Draw a diagram of the situation.

The force can be resolved into rectangular components, 120cos 22º (vertical) and 20sin 22º (horizontal). To calculate the mechanical work done against gravity, consider only the vertical component of the force and the vertical distance moved.

Wg!" = (120sin 22o )(8sin 12o )

# 74.8

The work against gravity is approximately 74.8 J. The mechanical work done pulling the cart up the ramp is 120(8)cos 10o or approximately 945.4 J. Chapter 7 Practice Test Question 21 Page 423 Draw a diagram.

In component form, the vectors are

70cos 40o , 70sin 40o!"

#$ and 125cos 65o , 125sin 65o!

"#$ .

The resultant vector is the vector sum.

12º 10º

120 N

8 8sin 12º

40º

65º

125 N

70 N


R!"

= 70cos 40o+125cos 65o , 70sin 40o

+125sin 65o!" #$# 106.45, 158.28!" #$

R!"

= 106.52+158.32

# 190.7

% = tan&1 158.28106.45

'

()*

+,

# 56.1o

The resultant has a magnitude of 190.7 N and its direction is 56.1° (above the horizontal). The equilibrant vector is the opposite vector, having a magnitude of 190.7 N and a direction of −123.9° (below the horizontal). Chapter 7 Practice Test Question 22 Page 423 One component can be found using the projection formula.

projv!u!

=u!

!v!

v!

!v!

"

#$

%

&' v!

=

7, 9() *+ ! 6, 2() *+6, 2() *+ ! 6, 2() *+

"

#$

%

&' 6, 2() *+

=6040

6, 2() *+

= 9, 3() *+

The other component can be found by vector subtraction. [ ] [ ] [ ]7, 9 9, 3 2, 6! = ! The two rectangular components are [ ] [ ]9, 3 and 2,6! . Chapter 7 Practice Test Question 23 Page 423 Let the θ be the angle between the two sides and a b

! ! of the parallelogram.

A = a!

! b!

A = a!

b!

sin"

200 = 90(30)sin"

sin" =2002700

" = sin#1 227

$

%&'

()

" " 4.2o

The interior angles of the parallelogram are 4.2º and 180º – 4.2º = 175.8º.


Chapter 7 Practice Test Question 24 Page 423 Use the formula for torque.

!

!

= r!

F"!

sin"

175 = 0.25 F"!

sin 78o

F"!

=175

0.25sin78o

F"!

# 715.6

The magnitude of the force is about 715.6 N. Chapter 7 Practice Test Question 25 Page 423 The statement is false.

( )u v v u! = " !

! ! ! !

If u v v u! = !

! ! ! !, then 0u v! =

! ! !.

The cross product of two vectors is the zero vector if and only if the vectors are collinear; that is, the vectors have either an angle of 0º or 180º between them. Chapter 7 Practice Test Question 26 Page 423 Draw a diagram of the situation.

Resolve the forces along the surface of the ramp.

20sin! = 5+ 4cos20o

sin! =5+ 4cos20o

20

! = sin"1 5+ 4cos20o

20#

$%&

'(

! ! 26.0o

θ

20º

4 N 5 N

20 N θ


Chapter 7 Practice Test Question 27 Page 423 a) i)

u!

! v!

"w"!

= 4, # 8, 12$% &' ! #11, 17, 12$% &' " 15, 10, 8$% &'

= –8(12)#17(12), 12(–11)#12(4), 4(17)# (–11)(–8)$% &' " 15, 10, 8$% &'

= #300, #180, # 20$% &' " 15, 10, 8$% &'

= #300(15) + (–180)(10) + (–20)(8)= #6460

ii)

u!

!v!

" w"!

= 4, # 8, 12$% &' ! #11, 17, 12$% &' " 15, 10, 8$% &'

= 4, # 8, 12$% &' ! 17(8)#10(12), 12(15)# 8(–11), #11(10)#15(17)$% &'

= 4, # 8, 12$% &' ! 16, 268, # 365$% &'

= 4(16) + (–8)(269) +12(–365)= #6460

b) Yes, the two answers are the same. The absolute value of each expression represents the volume of the

parallelepiped defined by three vectors , , and wu v! ! ""!

. These expressions should be either equal or opposite to each other.

Chapter 7 Practice Test Question 28 Page 423 Let [ ] [ ] [ ]1 2 1 2 1 2, , , , and ,a a a b b b c c c= = =

! ! !.

a)

L.S. = a!

+ b!

( ) + c!

= a1,a2!" #$ + b1, b2!" #$( ) + c1,c2!" #$

= a1 + b1, a2 + b2!" #$ + c1,c2!" #$

= (a1 + b1) + c1, (a2 + b2 ) + c2!" #$

R.S. = a!

+ b!

+ c!

( )

= a1,a2!" #$ + b1, b2!" #$ + c1,c2!" #$( )

= a1,a2!" #$ + b1 + c1. b2 + c2!" #$

= a1 + (b1 + c1), a2 + (b2 + c2 )!" #$

= (a1 + b1) + c1, (a2 + b2 ) + c2!" #$ Associative property for real numbers



b)

L.S. = k a!

+ b!

( )

= k a1,a2!" #$ + b1, b2!" #$( )

= k a1 + b1, a2 + b2!" #$

= k(a1 + b1), k(a2 + b2 )!" #$

= ka1 + kb1, ka2 + kb2!" #$ Distributive property for real numbers

R.S. = ka!

+ kb!

= k a1,a2!" #$ + k b1, b2!" #$

= ka1,ka2!" #$ + kb1. kb2!" #$

= ka1 + kb1, ka2 + kb2!" #$


Chapter 7 Cartesian Vectors Chapter 7 Prerequisite...

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