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°

MHR • Calculus and Vectors 12 Solutions 679

b) 282°

c) 0°

d) 270°

e) 240°

f) 150°

MHR • Calculus and Vectors 12 Solutions 680

g) 90°

h) 70°

i) 285°

j) 220°

k) 170°

MHR • Calculus and Vectors 12 Solutions 681

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.

MHR • Calculus and Vectors 12 Solutions 682

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.

MHR • Calculus and Vectors 12 Solutions 683

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

MHR • Calculus and Vectors 12 Solutions 684

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) = !14

x = !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

Substitute y = 1 into equation . !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

Substitute y = 5 into equation . 0.5x + 0.4(5) = 4 0.5x = 2 x = 4

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

MHR • Calculus and Vectors 12 Solutions 685

Chapter 7 Prerequisite Skills Question 7 Page 358

B is the correct response. It is not equivalent. 1 A The equation can be multiplied by to get this result. 3 B If you rearrange the equation into slope y-intercept form, the result is 9x ! 6y =18 !6y = !9x +18 !9 18 y = x + !6 !6 3 y = x ! 3 2 This is not the same as B. 1 C The equation can be multiplied by ! and rearranged to get this result. 3 1 D The equation can be multiplied by to get this result. 9 9x ! 6y =18 9 6 18 x ! y = 9 9 9 2 x ! y = 2 3

Chapter 7 Prerequisite Skills Question 8 Page 359

Use the special triangles.

1 1 3 1 a) b) c) d) 2 2 2 2

1 e) f) 0 g) 1 h) 0 2 i) cos 120º = –cos 60º j) sin 300º = –sin 60º k) 1 l) –1 1 3 = ! = ! 2 2

MHR • Calculus and Vectors 12 Solutions 686

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 ! 9.97 y 10.0 !

Therefore, y ! 10.0 cm.

Chapter 7 Prerequisite Skills Question 11 Page 359

h a) sin38o = 5 h = 5sin38° b) 32 + 42 = 52 The missing side of the base triangle has a length of 4 units.

4 sin30! = h 4 h = ! sin30

Chapter 7 Prerequisite Skills Question 12 Page 359

2 a) (a1 + b1 ) = (a1 + b1 )(a1 + b1 ) 2 2 = a1 + a1b1 + a1b1 + b1 = a 2 + 2a b + b 2 1 1 1 1

MHR • Calculus and Vectors 12 Solutions 687

2 2 b) (a1 + b1 )(a1 ! b1 ) = a1 ! a1b1 + a1b1 ! b1 = a 2 ! b 2 1 1 c) (a 2 + a 2 )(b 2 + b 2 ) = a 2b 2 + a 2b 2 + a 2b 2 + a 2b 2 1 2 1 2 1 1 1 2 2 1 2 2 d) (a b + a b + a b )3 1 1 2 2 3 3

= [(a1b1 + a2b2 + a3b3 )(a1b1 + a2b2 + a3b3 )](a1b1 + a2b2 + a3b3 ) 2 2 2 2 2 2 = [a1 b1 + a1b1a2b2 + a1b1a3b3 + a2b2a1b1 + a2 b2 + a2b2a3b3 + a3b3a1b1 + a3b3a2b2 + a3 b3 ](a1b1 + a2b2 + a3b3 ) 3 3 2 2 2 2 2 2 2 2 2 2 2 2 = a1 b1 + a1 a2b1 b2 + a1 a3b1 b3 + a1 a2b1 b2 + a1a2 b1b2 + a1a2a3b1b2b3 + a1 a3b1 b3 + a1a2a3b1b2b3 + a1a3 b1b3 2 2 2 2 2 2 3 3 2 2 2 2 2 2 + a1 a2b1 b2 + a1a2 b1b2 + a1a2a3b1b2b3 + a1a2 b1b2 + a2 b2 + a2 a3b2 b3 + a1a2a3b1b2b3 + a2 a3b2 b3 + a2a3 b2b3 2 2 2 2 2 2 2 2 2 2 2 2 3 3 + a1 a3b1 b3 + a1a2a3b1b2b3 + a1a3 b1b3 + a1a2a3b1b2b3 + a2 a3b2 b3 + a2a3 b2b3 + a1a3 b1b3 + a2a3 b2b3 + a3 b3 = a 3b 3 + a 3b 3 + a 3b 3 + 3a 2a b 2b + 3a a 2b b 2 + 3a 2a b 2b + 3a a 2b b 2 + 3a 2a b 2b + 3a a 2b b 2 + 6a a a b b b 1 1 2 2 3 3 1 2 1 2 1 2 1 2 1 3 1 3 1 3 1 3 2 3 2 3 2 3 2 3 1 2 3 1 2 3

Chapter 7 Prerequisite Skills Question 13 Page 359

1 a) A = 0(2) ! 0(0) + 0(2) ! 2(2) + 2(1) ! 3(2) + 3(!1) ! 3(1) + 3(0) ! 0(0) 2 1 = !14 2 = 7

The area is 7 square units.

1 b) A = 1(!2) ! 3(1) + 3(!4) ! 0(!2) + 0(0) ! (!5)(!4) + (!5)(4) ! (!3)(0) + (!3)(1) !1(4) 2 1 = !64 2 = 32

The area is 32 square units.

1 c) A = 4(!7) ! 5(!10) + 5(2) ! 3(!7) + 3(5) ! 0(2) + 0(3) ! (!2)(5) + (!2)(!10) ! 4(3) 2 1 = 86 2 = 43

The area is 43 square units.

MHR • Calculus and Vectors 12 Solutions 688

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 A = x y ! x y + x y ! x y + x y ! x y 2 1 2 2 1 2 3 3 2 3 1 1 3

Answers may vary. For example, Consider a triangle with vertices (1, 1), (2, 7), and (–3, –4), the area is 1 A = 1(7) ! 2(1) + 2(!4) ! (!3)(7) + (!3)(1) !1(!4) 2 1 = 19 2 = 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 A = x y ! x y + x y ! x y + x y ! x y + x y ! x y + x y ! x y + x y ! x y 2 1 2 2 1 2 3 3 2 3 4 4 3 4 5 5 4 5 6 6 5 6 1 1 6

For a hexagon with vertices (0, 0), (3, 2), (2, 4), (0, 5), (–2, 4), and (–3, 2), the area is 1 A = 0(2) ! 3(0) + 3(4) ! 2(2) + 2(5) ! 0(4) + 0(4) ! (–2)(5) + (–2)(2) ! (!3)(4) + (!3)(0) ! 0(2) 2 1 = 36 2 = 18

The area is 18 square units.

Chapter 7 Section 1 Cartesian Vectors

Chapter 7 Section 1 Question 1 Page 367 a) 2i + j b) 3i ! 5 j c) ! 3i ! 6 j d) 5i e) 9i ! 7 j f) !8 j

! ! ! g) !6i i) !5.2i ! 6.1j

MHR • Calculus and Vectors 12 Solutions 689

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

The magnitude of the vector is 26 units.

!!" c) EF = 02 + 72 = 7

The magnitude of the vector is 7 units.

!!!" 2 d) GH = (!2) + 92

= 85

The magnitude of the vector is 85 units.

Chapter 7 Section 1 Question 5 Page 368

a) v h = [5,0]

vv = [0,!1]

MHR • Calculus and Vectors 12 Solutions 690

! ! b) The magnitude of v, v , is 26 units. ! 1 ! 1 ! Unit vectors collinear with v have the form ! v and ! ! v . v v & 5 1 # & 5 1 # Therefore, the required unit vectors are $ ,' ! and $' , ! . % 26 26 " % 26 26 " c) Let Q(x, y) be the point. !!!!!!" !!!" !!!" PQ = OQ ! OP #"5,!1%$ = [x, y]! #"!2,!7%$ #"5,!1%$ = #"x + 2, y + 7%$

& x + 2 = 5 and y + 7 = !1 x 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 = !1 x 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$ # %

MHR • Calculus and Vectors 12 Solutions 691

!!!" 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 !!!" RP = 18

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

Chapter 7 Section 1 Question 7 Page 368

! a) 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$ # %

MHR • Calculus and Vectors 12 Solutions 692

! ! 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 2 ! = ! e 3

! ! "! ! Therefore, a, b , d, and e are collinear vectors.

Chapter 7 Section 1 Question 9 Page 368

! a) ! o , 500sin30o # v = "500cos30 $ = !433.0, 250# " $

! b) ! o , 1000sin18o # v = "1000cos18 $ = !951.1, 309.0# " $

! c) v = [0, 125]

! d) v = !230, 0# " $

! e) v = [0, ! 25]

! f) v = [!650, 0]

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 .

" 24.100% The angle of the resultant with the west direction is ! tan!1 ! !51.429o . (The resultant is pointing #$ 19.219 &'

SW.)

MHR • Calculus and Vectors 12 Solutions 693

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%$

2 2 = 6 + 2 = 40 =" 6.3 a + b 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 .

MHR • Calculus and Vectors 12 Solutions 694

Chapter 7 Section 1 Question 12 Page 368 a)

 a (7, 2)

 b

b) a + b =! 10

! ! a ! b = #"!8, 4%$

2 2 = (!8) + 4 = 80 =" 8.9

a + b 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 .

MHR • Calculus and Vectors 12 Solutions 695

Chapter 7 Section 1 Question 13 Page 368

! ! "! a) L.S. = (u + v) + w !u ,u # !v ,v # !w , w # = (" 1 2 $ + " 1 2 $) + " 1 2 $ !u v , u v # !w , w # = (" 1 + 1 2 + 2 $) + " 1 2 $ !u v w , u v w # = " 1 + 1 + 1 2 + 2 + 2 $ ! ! "! R.S. = u + (v + w) !u ,u # !v ,v # !w , w # = " 1 2 $ + (" 1 2 $ + " 1 2 $) !u ,u # !v w , v w # = " 1 2 $ + " 1 + 1 2 + 2 $ !u v w , u v w # = " 1 + 1 + 1 2 + 2 + 2 $ L.S. = R.S.

! ! "! ! ! "! Therefore, u + v + w = u + v + w . ( ) ( )

! ! b) L.S. = k (u + v) k !u , u # !v , v # = (" 1 2 $ + " 1 2 $) k !u v , u v # = " 1 + 1 2 + 2 $ = !k u + v , k u + v # " ( 1 1 ) ( 2 2 )$ !ku kv , ku kv # = " 1 + 1 2 + 2 $ ! ! R.S. = ku + kv k !u , u # k !v , v # = " 1 2 $ + " 1 2 $ !ku , ku # !kv , kv # = " 1 2 $ + " 1 2 $ !ku kv , ku kv # = " 1 + 1 2 + 2 $ L.S. = R.S.

! ! ! ! Therefore, k u + v = ku + kv . ( )

MHR • Calculus and Vectors 12 Solutions 696

! ! c) L.S. = u + v !u , u # !v , v # = " 1 2 $ + " 1 2 $ !u v , u v # = " 1 + 1 2 + 2 $ ! ! R.S. = v + u !v , v # !u , u # = " 1 2 $ + " 1 2 $ !v u , v u # = " 1 + 1 2 + 2 $ = !u + v , u + v # " 1 1 2 2 $

! ! ! ! Therefore, u + v = v + u .

! d) L.S.=(k + m)u k m !u , u # = ( + )" 1 2 $ = ! k + m u , k + m u # "( ) 1 ( ) 2 $ !ku mu , ku mu # = " 1 + 1 2 + 2 $ ! ! R.S. = ku + mu k !u , u # m!u , u # = " 1 2 $ + " 1 2 $ !ku , ku # !mu , mu # = " 1 2 $ + " 1 2 $ = !ku + mu , ku + mu # " 1 1 2 2 $

! ! ! 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$ # %

MHR • Calculus and Vectors 12 Solutions 697

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.

" 125.506 % The angle of the resultant with the east direction is tan!1 ! 14.376o . (The resultant is #$ 489.682&' 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

120º 200 N 120 N 60º 60º

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]. 2 2 The magnitude of the resultant is !40 + 277.13 ! 280.0 . ( ) ( ) " 277.13% The angle of the resultant is tan!1 ! !81.79o . #$ !40 &'

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.

MHR • Calculus and Vectors 12 Solutions 698

Chapter 7 Section 1 Question 18 Page 369

Draw a diagram.

550 N Jason

120º 608 N 150º 90º Sam

700 N Nick

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 .

" !223.7% The resultant angle is tan!1 ! !33.9o . #$ 333 &'

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

MHR • Calculus and Vectors 12 Solutions 699

Chapter 7 Section 1 Question 19 Page 369

12 000 kg·m/s2

15º 15 000 kg·m/s2

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 .

" 3105.8 % The resultant angle is tan!1 ! 6.66o . #$ 26 591.1&'

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

2Ω

4Ω

5Ω

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 . + ! = "

" !3% b) The angle of the resultant is tan!1 ! !36.9o . #$ 4 &'

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

MHR • Calculus and Vectors 12 Solutions 700

Chapter 7 Section 1 Question 21 Page 369

Draw a diagram.

! h 500 θ km/h 15 km/h 12 32º

42º

!500 500 $ The resultant vector must be cos 42o , sin 42o . # 12 12 & " % The current vector is [15cos 58º, 15sin 58º]. The heading vector is the difference between the resultant vector and the current vector. ! !500 o 500 o $ h = # cos 42 , sin 42 & ' [15cos 58º, 15sin 58º] " 12 12 % " [23.016, 15.160] ! h = (23.016)2 + (15.160)2

" 27.560

# 15.160 & ! = tan"1 $% 23.016'( ! 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.

MHR • Calculus and Vectors 12 Solutions 701

Chapter 7 Section 1 Question 22 Page 369 a) Solve for x. x2 + (3x)2 = 9 10x2 = 81 Square both sides. 81 x2 = 10 9 x = ± 10 9 10 x = ± 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 = 0 5x2 % 2x % 7 = 0 (5x % 7)(x +1) = 0

7 x = or x = !1 5

MHR • Calculus and Vectors 12 Solutions 702

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.

! ! v1 v2

O ! ! v1 + v2

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 = 9 x ! 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 2 y = 3(x ! 7) ! 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.

MHR • Calculus and Vectors 12 Solutions 703

Chapter 7 Section 1 Question 25 Page 369

Method 1: o o Use a trigonometric identity. Let !A = 40 and !B = 20 . ! A + B$ ! A ' B$ sin A + sin B = 2sin cos "# 2 %& "# 2 %& sin 40° + sin 20° = 2sin30°cos10°

! 1$ = 2 sin 90° '10° "# 2%& ( ) = sin80°

Method 2: Start with an equilateral triangle.

40º 20º 1 1 y

60º 80º 100º 60º x 1 – x

Using the sine law in the two smaller triangles, you get y 1 = sin60o sin80o 1! x = sin 20o x = o sin 40

From the second and last fractions, sin 40o x = sin80o

From the third and last actions, sin 20o 1! x = sin 40o

Adding these two equation together, sin 40o sin 20o 1 = + sin80o sin80o sin80o = sin 40o + sin 20o

MHR • Calculus and Vectors 12 Solutions 704

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

!" " g ! h = 5000(4500)cos180o d) 22 500 000 # "

Chapter 7 Section 2 Question 2 Page 376

! ! a) u !v = 6(10)cos30o " 52.0

! ! b) s!t = 30(15)cos120o = "225

!" !" c) f ! g = 5.8(13.4)cos180o # "77.7

! ! q r 4.0(6.1)cos90o d) ! = = 0

! ! e) a !b = 850(400)cos58o " 180 172.5

!" !" f) m! p = 16(2)cos153o # "28.5

MHR • Calculus and Vectors 12 Solutions 705

Chapter 7 Section 2 Question 3 Page 376

Answers may vary. For example: ! ! ! ! ! ! ! a) u !(ku + v) = k (u !u) + u !v ! 2 ! ! = k u + u !v

! ! ! ! ! ! ! b) (ku ! v)"(lv) = kl (u "v) ! l (v "v) ! ! ! 2 = kl u "v ! l v ( )

! ! ! ! ! ! ! ! ! ! ! ! c) (u ! v)"(u ! v) = u "u ! u "v ! v "u + v "v ! 2 ! ! ! 2 = u ! 2u "v + v

! ! "! ! ! "! ! ! ! "! ! ! d) u + v ! w + x = u ! w + u ! x + v ! w + v ! x ( ) ( )

Chapter 7 Section 2 Question 4 Page 376

! ! a) 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) v ! w 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).

MHR • Calculus and Vectors 12 Solutions 706

c) This expression has meaning. v ! w 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

MHR • Calculus and Vectors 12 Solutions 707

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.

MHR • Calculus and Vectors 12 Solutions 708

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 u and v are perpendicular, then u !v = 0 .

"!2, 5$# % "!k, 4$# = 0 ! ! u and v are perpendicular if k = –10. 2k + 20 = 0

k = &10

Chapter 7 Section 2 Question 12 Page 376

! ! ! ! If u and v are perpendicular, then u !v = 0 . "!k, 3$# % "!k, 2k $# = 0 k 2 + 6k = 0 k(k + 6) = 0 k = 0 or k = 6

! ! u and 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.

MHR • Calculus and Vectors 12 Solutions 709

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 a = [3, ! 5] and b = [2, 1]. ! ! 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 .

! !1 3 " ! c) Let u , and a !1, 3" . = # $ = % & %2 2 & ! ! Since a = 2u , the vectors are collinear.

2 2 ! ! 1$ ! 3$ u = # & + # & " 2% " 2 % 1 3 = + 4 4 = 1 ! ! Since u = 1, u is a unit vector.

MHR • Calculus and Vectors 12 Solutions 710

! ! L.S. = a !u "1 3 $ = "1, 3$ ! & , ' # % 2 2 #& %' ( 1+ ( 3+ = 1* - + 3* - ) 2, ) 2 , 1 3 = + 2 2 = 2 ! R.S. = a

2 = 12 + ( 3) = 1+ 3 2

! ! ! 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 a = [a1, a2 ] and b = [b1, b2 ] so that a !b = 0 . ! ! Show a ! b . ! ! Since a !b = 0 ,

a1b1 + a2b2 = 0

a1b1 = !a2b2 b a 2 = ! 1 b a 1 2

! ! This statement says that the slope of b is the negative reciprocal of the slope of a . ! ! Therefore, a ! b .

Proof 2 ( ! ) ! ! ! ! Let a = [a1, a2 ] and b = [b1, b2 ] so that a ! b . ! ! Show that a !b = 0 . ! ! Since a ! b , consider the slopes of these two vectors; the slopes must be negative reciprocals of each other.

MHR • Calculus and Vectors 12 Solutions 711

b a 2 = ! 1 b1 a2

a1b1 = !a2b2 a b + a b = 0 1 1 2 2

! ! Therefore, a !b = 0 .

! ! b) Let a = [a1, a2 ] and b = [b1,b2 ] be any two vectors. ! ! L.S. = a !b "a , a $ "b , b $ = # 1 2 % ! # 1 2 % = a b + a b !1 !1 2 2 R.S. = b! a "b , b $ "a , a $ = # 1 2 % ! # 1 2 %

= b1a1 + b2a2 = a b + a b commutuative property for multiplication in " 1 1 2 2 ! ! ! ! Therefore, L.S. = R.S. and a !b = b ! a .

! 2 2 c) Let u = [u1, u2 ] where u1 + u2 =1 . ! ! Let a = ku !ku ,ku # = " 1 2 $ ! ! L.S. = a !u "ku , ku $ "u , u $ = # 1 2 % ! # 1 2 % 2 2 = ku1 + ku2 ! R.S. = a "ku , ku $ = # 1 2 % 2 2 ku ku = ( 1 ) + ( 2 )

2 2 2 2 = k u1 + k u2 k 2 u 2 u 2 = ( 1 + 2 ) = k 2 (1) = k

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

MHR • Calculus and Vectors 12 Solutions 712

Chapter 7 Section 2 Question 15 Page 377

! ! ! Let three vectors be a = [3, 0], b = [0, 5], and c = [0, 7]. ! ! ! ! ! ! In this case, a !b = 0 and a !c = 0 , 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 x2 + y2 = a2 .

C(x, a2 ! x2 )

A(–a, 0) B(a, 0)

!!!" !!!" 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 !!!" 2 !!!" 2 = CO – OA !!!" !!!" = 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.

MHR • Calculus and Vectors 12 Solutions 713

Chapter 7 Section 2 Question 17 Page 377

! ! "! Let u = [1, 2], v = [3, 4], w = [!1, ! 4], and k = 2 .

! ! 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 . ( )

MHR • Calculus and Vectors 12 Solutions 714

! ! "! 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 Page 377

! ! "! Let u = [u1, u2 ], v = [v1, v2 ], and w = [w1, w2 ].

! ! a) L.S. = (ku)!v k "u , u $ "v , v $ = ( # 1 2 %)! # 1 2 % "ku , ku $ "v , v $ = # 1 2 % ! # 1 2 %

= ku1v1 + ku2v2 ! ! M.S. = k (u !v) k "u , u $ "v , v $ = (# 1 2 % ! # 1 2 %) k u v u v = ( 1 1 + 2 2 )

= ku1v1 + ku2v2 ! ! R.S. = u !(kv) "u , u $ k "v , v $ = # 1 2 % !( # 1 2 %) "u , u $ "kv , kv $ = # 1 2 % ! # 1 2 % = ku v + ku v 1 1 2 2

! ! ! ! ! ! Clearly all three sides are equal and ku !v = k u !v = u ! kv . ( ) ( ) ( )

MHR • Calculus and Vectors 12 Solutions 715

! ! "! b) L.S. = u !(v + w) = "u u $ ! "v , v $ + "w , w $ # 1, 2 % (# 1 2 % # 1 2 %) = "u u $ ! "v + w , v + w $ # 1, 2 % # 1 1 2 2 %

= u1(v1 + w1 ) + u2 (v2 + w2 ) = u v + u w + u v + u w !1 !1 !1 "!1 2 2 2 2 R.S. = u !v + u ! w = "u u $ ! "v , v $ + "u u $ ! "w , w $ # 1, 2 % # 1 2 % # 1, 2 % # 1 2 %

= u1v1 + u2v2 + u1w1 + u2 w2 = u v + u w + u v + u w 1 1 1 1 2 2 2 2

! ! "! ! ! ! "! Therefore, u ! v + w = u !v + u ! w . ( )

! ! "! c) L.S. = (u + v)! w "u , u $ "v , v $ "w , w $ = (# 1 2 % + # 1 2 %)! # 1 2 % "u v , u v $ "w , w $ = # 1 + 1 2 + 2 % ! # 1 2 %

= (u1 + v1 )w1 + (u2 + v2 )w2 = u w + v w + u w + v w !1 "!1 !1 "!1 2 2 2 2 R.S. = u ! w + v ! w "u , u $ "w , w $ "v , v $ "w , w $ = # 1 2 % ! # 1 2 % + # 1 2 % ! # 1 2 %

= u1w1 + u2 w2 + v1w1 + v2 w2 = u w + v w + u w + v w 1 1 1 1 2 2 2 2

! ! "! ! "! ! "! 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.

MHR • Calculus and Vectors 12 Solutions 716

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 = 0 5x + 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# . " $ " $

!!!" !!!" 2 OP !OQ = " 2 2 5x + 5y = " 2 2 y = "x " 10

2 The set of points Q form a line perpendicular to OP and have a y-intercept of ! . 10 c) Let Q(x, y) be any point on the locus. !!!" !!!" OQ = !x, y# and OP = !5, 5# . " $ " $

!!!" !!!" 3 OP !OQ = " 2 3 5x + 5y = " 2 3 y = "x " 10

3 The set of points Q form a line perpendicular to OP and have a y-intercept of ! . 10

MHR • Calculus and Vectors 12 Solutions 717

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 y = ax2 + 5; where a < 0 . Substitute (3, 0) to determine a. 0 = a(3)2 + 5 5 a = ! 9 5 The equation of the parabola is y = ! x2 + 5 . 9 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.

MHR • Calculus and Vectors 12 Solutions 718

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.

Chapter 7 Section 3 Question 2 Page 384

!" " a) 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.

MHR • Calculus and Vectors 12 Solutions 719

Chapter 7 Section 3 Question 3 Page 384

!" " p "q a) cos! = !" " p q

#7, 8% " #4, 3% = $ & $ & $#7, 8&% $#4, 3&% 7(4) + 8(3) = 72 + 82 42 + 32 52 = 53.1507 ( 52 + ! = cos'1 )* 53.1507,- # 11.9º

! ! r "s b) cos! = ! ! r s

$#2, # 8& " $6, #1& = % ' % ' %$#2, # 8'& %$6, #1'& #2(6) + (–8)(–1) = (–2)2 + (–8)2 62 + (–1)2 #4 = 50.1597 ( #4 + ! = cos#1 )* 50.1597,- " 94.6º ! ! t "u c) cos! = ! ! t u

$#7, 2& " $6, 11& = % ' % ' %$#7, 2'& %$6, 11'& #7(6) + 2(11) = (#7)2 + 22 62 +112 #20 = 91.2195 ( #20 + ! = cos#1 )* 91.2195,- " 102.7º

MHR • Calculus and Vectors 12 Solutions 720

! "! e" f d) cos! = ! "! 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

Chapter 7 Section 3 Question 4 Page 384

" % ! ! 1 ! a) proj! u = u cos! $ ! v' v $ ' # v & " 1 !% = 10cos 25o v #$ 18 &' " % ! 1 ! " 0.50v or 9.1$ ! v' $ ' # v &

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

MHR • Calculus and Vectors 12 Solutions 721

! ! b) proj! u u cos v = ! = 7cos 110o " "2.4

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

! ! c) proj! u u cos v = ! = 20cos 90o = 0

The projection has zero magnitude.

Chapter 7 Section 3 Question 5 Page 384

! ! ! " a !b% ! a) proj! a ! ! b b = $ ' # b!b&

" *)6, (1,+ ! *)11, 5,+% = $ ' *)11, 5,+ # *)11, 5,+ ! *)11, 5,+ & " 61 % = )11, 5+ #$ 146&' * , " )4.6, 2.1+ * ,

" !" " " c ! d % !" b) proj!" c " !" d d = $ ' # c ! d &

" )(2, 7+* ! )(,4, 3+* % = $ ' )(,4, 3+* # )(,4, 3+* ! )(,4, 3+*& " 13% = (,4, 3* #$ 25&' ) + = (,2.08, 1.56* ) +

" !" " " e! f % !" c) proj!" e !" !" f f = $ ' # f ! f &

" *)(2, ( 5,+ ! *)(5, 1,+% = $ ' *)(5, 1,+ # *)(5, 1,+ ! *)(5, 1,+ & " 5 % = )(5, 1+ #$ 26&' * , # )(1.0, 0.2+ * , MHR • Calculus and Vectors 12 Solutions 722

"! ! "! " g ! h% ! d) proj! g ! ! h h = $ ' # h! h &

" *)10, ( 3,+ ! *)4, ( 4,+% = $ ' *)4, ( 4,+ # *)4, ( 4,+ ! *)4, ( 4,+ & " 52% = )4, ( 4+ #$ 32&' * , = )6.5, ( 6.5+ * ,

Chapter 7 Section 3 Question 6 Page 385

!" " !" " 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

! 1 a) u = "!6, 1$# "!6, 1$# 1 = "!6, 1$# 37 ! 6 1 # = % , & 37 37 " $

!" ! 6 1 $ b) f = 25# , & " 37 37 % ! 150 25 $ = # , & " 37 37 %

MHR • Calculus and Vectors 12 Solutions 723

! c) s = "!11, 0$# "! ! W = F %s ! 150 25 # = & , ' % "!11, 0$# " 37 37 $ ( 150 + ( 25 + = * - (11) + * - (0) ) 37 , ) 37 , # 271.3 J

Chapter 7 Section 3 Question 8 Page 385

!" " W = F s cos! !" 150 = F (8)cos 20o !" 150 F = 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$ . # % # % # % A

B C

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

MHR • Calculus and Vectors 12 Solutions 724

!!!" !!!" AB!CB cos B = !!!" !!!" AB CB

#"1, " 8% ! #5, 1% = $ & $ & $#"1, " 8&% $#5, 1&% "1(5) + ("8)(1) = ("1)2 + ("8)2 52 +12 "13 # 41.1096 ( "13 + 'B = cos"1 )* 41.1096,- # 108.4

Angle B is approximately 108.4 º.

!!!" !!!" AC! BC cosC = !!!" !!!" 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 ( 39 + 'C = cos"1 )* 55.1543,- # 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°.

MHR • Calculus and Vectors 12 Solutions 725

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$ . # % # % # % # % (2, 3) (5, 3) ! u ! ! v (0, 0) (3, 0)

"5, 3$ & "1, ' 3$ cos! = # % # % #"5, 3%$ #"1, ' 3%$ 5(1) + 3('3) = 52 + 32 12 + ('3)2 '4 ! 18.4391 ( '4 + ! = cos'1 )* 18.4391,- ! 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

Q(–6, 4) R(4, 3)

P(–2, 1) S

a) Method 1: !!!" !!!" !!" OS = OP + PS !!!" !!!" = OP + QR = #"!2, 1%$ + #"10,–1%$ = "8, 0$ # %

S has coordinates (8, 0).

MHR • Calculus and Vectors 12 Solutions 726

Method 2: Let S have coordinates (x, y). Since RS and PQ are parallel, their slopes are the same. y ! 3 4 ! 1 = x – 4 –6 ! (!2)

3 = – 4 3x + 4y = 24 1

Since PS and QR are parallel, their slopes are the same. y ! 1 3! 4 = x + 2 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).

!!!" !!!" QP "QR b) cos!PQR = !!!" !!!" QP QR

$4, # 3& " $10, #1& = % ' % ' %$4, # 3'& %$10, #1'& 4(10) + (#3)(–1) = 42 + (#3)2 102 + (–1)2 44 # 50.2494

MHR • Calculus and Vectors 12 Solutions 727

# 44 & !PQR = cos"1 $% 50.2494'( ! 28.9

The interior angles are 29° and 180º – 29º = 151°. c) Let θ be the angle between the diagonals PR and QS.

!!!" !!!" PR "QS cos! = !!!" !!!" PR QS

#6, 2% " #14, ' 4% = $ & $ & $#6, 2&% $#14, ' 4&% 6(14) + 2('4) = 62 + 22 142 + ('4)2 76 # 92.0869 ( 76 + ! = cos'1 )* 92.0869,- # 34.4

Angle θ is approximately 34º.

The angles between the diagonals are 34° and 180º – 34º = 146°.

Chapter 7 Section 3 Question 12 Page 385

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

! Use i = [1, 0] as a vector along the x-axis. !!!" " PQ "i cos! = !!!" " PQ i

$4, # 4& " $1, 0& = % ' % ' %$4, # 4'& (1) 4(1) + (–4)(0) = 42 + (–4)2 4 # 5.6569

MHR • Calculus and Vectors 12 Solutions 728

# 4 & ! = cos"1 $% 5.6569'( ! 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

! ! ! proju v = v cos! ! ! u = v cos! " ! u ! ! v u cos! = ! u ! ! v #u = ! u

Chapter 7 Section 3 Question 14 Page 385

! ! " 1 ! # proj! v = v cos! $ ! u % u $ u % & ' ! ! " 1 ! # u = v cos! $ ! u %( ! $ u % u & ' ! ! v u cos! ! = ! ! u u u ! ! v )u ! = ! ! u u )u

Chapter 7 Section 3 Question 15 Page 385

! ! 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.

MHR • Calculus and Vectors 12 Solutions 729

Chapter 7 Section 3 Question 16 Page 385

sin 10º 10º cos 10º

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. ! u = [cos 10º, sin 10º]

10º

! v = [0, –9800]

! ! ! v !u proj! v = ! u u

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

MHR • Calculus and Vectors 12 Solutions 730

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].

! ! ! v !u proj! 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. 2 2 x2 + (30 587.5) = ( 30 0002 +18 0002 ) 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.

MHR • Calculus and Vectors 12 Solutions 731

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 d1 = [1, 1] and d 2 = [1, !1]. " !!" " !!" " " % !!" " " % !!" i ! d1 j ! d1 !!" !!" proj i = $ !!" !!" ' d1 proj j = $ !!" !!" ' d1 d1 d1 # d1 ! d1 & # d1 ! d1 &

" )(1, 0+* ! )(1, 1+*% " )(0, 1+* ! )(1,1+*% = $ ' )(1, 1+* = $ ' )(1, 1+* # )(1, 1+* ! )(1, 1+* & # )(1, 1+* ! )(1, 1+*& " 1% " 1% = (1, 1* = (1, 1* #$ 2&' ) + #$ 2&' ) + (1 1 * (1 1 * = , = , ,2 2 - ,2 2 - ) + ) +

MHR • Calculus and Vectors 12 Solutions 732

" !!" " !!" " " % !!" " " % !!" i ! d2 j ! d2 !!" !!" proj i = $ !!" !!" ' d2 proj j = $ !!" !!" ' d2 d2 d2 # d2 ! d2 & # d2 ! d2 &

" )(1, 0+* ! )(1, ,1+* % " )(0, 1+* ! )(1, ,1+* % = $ ' )(1, ,1+* = $ ' )(1, ,1+* # )(1, ,1+* ! )(1, ,1+*& # )(1, ,1+* ! )(1, ,1+*& " 1% " ,1% = (1, ,1* = (1, ,1* #$ 2&' ) + #$ 2 &' ) + (1 1 * ( 1 1 * = , , = , , -2 2 . - 2 2 . ) + ) +

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.

MHR • Calculus and Vectors 12 Solutions 733

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.

MHR • Calculus and Vectors 12 Solutions 734

Chapter 7 Section 3 Question 25 Page 385

! ! ! " a !b% ! a) proj! a ! ! b b = $ ' # b!b&

" )(6, 5+* ! )(1, 3+*% = $ ' )(1, 3+* # )(1, 3+* ! )(1, 3+* & " 21% = (1, 3* #$ 10&' ) + = (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].

Chapter 7 Section 3 Question 26 Page 385

"! ! "! " F !u% ! proj! F ! ! u u = $ ' # u !u &

" )(25, 18+* ! )(2, 5+*% = $ ' )(2, 5+* # )(2, 5+* ! )(2, 5+* & " 140% = (2, 5* #$ 29 &' ) + (280 700 * = , , - ) 29 29 + # (9.7, 24.1* ) +

As in question 25, find the perpendicular component vector by subtraction. !280 700 $ !445 '178$ [25,18] – # , & = # , & " 29 29 % " 29 29 % [15.3, 6.1] ! '

Use the dot product to check perpendicularity. !280 700" !445 178" % , & # % , $ & = 0 ' 29 29 ( ' 29 29 ( !280 700" !445 178" The perpendicular component vectors are $ , % and $ , # % . & 29 29 ' & 29 29 '

MHR • Calculus and Vectors 12 Solutions 735

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 u !v = 0 ; the two projection vectors are 0 if the vectors u and v are perpendicular.

These two situations are demonstrated in the diagrams below. ! u ! u ! v ! v

! ! ! ! b) proju v = projv u ! ! v cos! = u cos! ! ! v = u or cos! = 0 ! ! ! ! This is true whenever u and v have the same magnitude or if the vectors u and v are perpendicular. This first situation is demonstrated in the diagram below.

! u

! v

Chapter 7 Section 3 Question 28 Page 385

There are three situations to consider.

Case 1: Suppose the “3” side has length 60. 1 Then the “4” side has length 80 and the area is (60 cm)(80 cm) = 2400 cm2. 2

MHR • Calculus and Vectors 12 Solutions 736

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,

5x 4x

60 3x

4x 5x = 60 3x 2 12x ! 300x = 0 x(x ! 25) = 0 x = 0, 25 1 Therefore, x = 25 and the area is (125 cm)(60 cm) = 3750 cm2. 2 The maximum possible area is Case 3 which is 3750 cm2.

Chapter 7 Section 3 Question 29 Page 385

1471 538(un ) f (un ) = log (un ) + n u un 1471 = n + 538 n ( ) = 1471+ 538 = 2009

MHR • Calculus and Vectors 12 Solutions 737

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)

Q(–4, 1, 3)

P(2, 2, –5)

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 x2 + y2 . For P, d = 22 + 32 = 13.

2 For Q, d = (!4) +12 = 17.

2 For R, d = 62 + !2 = 40. ( )

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

MHR • Calculus and Vectors 12 Solutions 738

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)

(–2, 0, 4)

Chapter 7 Section 4 Question 4 Page 399

! a) u = (!1)2 + 52 + (!2)2

= 30

MHR • Calculus and Vectors 12 Solutions 739

! 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) 3i ! 5 j + 2k b) ! 3i ! 6 j + 9k c) 5i ! 7k

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 v = !2u .

Chapter 7 Section 4 Question 8 Page 399

"4,1,!7$ = 42 +12 + (!7)2 # % = 66 1 ! Unit vectors are of the form ± ! v . v

! 4 1 7 " ! 4 1 7 " Therefore, the unit vectors are $ , , # % and $# , # , % . & 66 66 66 ' & 66 66 66 '

MHR • Calculus and Vectors 12 Solutions 740

Chapter 7 Section 4 Question 9 Page 399

! ! ! ! u and v are collinear if and only if u kv for some k . = !" a) [a, 3, 6]= k [!8, 12, b] 1 Comparing the y-coordinates gives k = . 4 1 1 So, a = !8 6 = b 4 ( ) 4 a = !2 b = 24 b) [a, 2, 0]= k [!3, ! 6, ! b] 1 Comparing the y-coordinates gives k = ! . 3 1 1 So, a = ! (!3) 0 = ! (!b) 3 3 a = 1 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]

MHR • Calculus and Vectors 12 Solutions 741

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

MHR • Calculus and Vectors 12 Solutions 742

Chapter 7 Section 4 Question 12 Page 399

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

Chapter 7 Section 4 Question 13 Page 399

!!!" !!!" !!!" DE = OE ! OD !!!" #"!4, 2, 6%$ = #"3, 3, 1%$ ! OD !!!" OD = #"3, 3, 1%$ ! #"!4, 2, 6%$ !!!" OD = "7, 1, ! 5$ # %

Chapter 7 Section 4 Question 14 Page 399

!!!" a) AB = #"0 ! 0, 4 ! (!3), ! 4 ! 2%$ = "0, 7, ! 6$ # %

!!!" b) CD = #"!3! 4, ! 3! 5, 5! 0%$ = "!7, ! 8, 5$ # %

Chapter 7 Section 4 Question 15 Page 400

! a) 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$ # %

MHR • Calculus and Vectors 12 Solutions 743

! ! ! 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

Chapter 7 Section 4 Question 16 Page 400

!" " g " h cos! = !" " 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.

MHR • Calculus and Vectors 12 Solutions 744

Chapter 7 Section 4 Question 17 Page 400 a) Let [x, y, z] be a vector orthogonal to [3, –1, 4]. x, y, z ! 3,"1, 4 = 0 [ ] [ ] 3x " y + 4z = 0 Select any values for x and z and solve for y. !5 Let x = 2 and y = 1, then z = . 4 7 Let x = –3 and y = 5, then z = . 2

" 5 % " 7 % Two orthogonal vectors are 2,1, ! and !3,5, . $ 4 ' $ 2 ' # & # & b) Let [x, y, z] be a vector orthogonal to [–4, –9, 3]. x, y, z ! "4, " 9, 3 = 0 [ ] [ ] "4x " 9y + 3z = 0

Select any values for x and y and solve for z. Let x = 1 and y = 2. !4(1) ! 9(2) + 3z = 0 3z = 22 22 z = 3

Let x = –3 and y = 2. !4(!3) ! 9(2) + 3z = 0 3z = 6 z = 2

! 22 $ Two orthogonal vectors are 1,2, and [–3, 2, 2]. # 3 & " %

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.

MHR • Calculus and Vectors 12 Solutions 745

Chapter 7 Section 4 Question 19 Page 400

! 1 ! Unit vectors parallel to u have the form ± ! u . u ! a) a = 52 + (!3)2 + 22

= 38

! ! 5 3 2 " ! 5 3 2 " The unit vectors parallel to a are $ , # , % and $# , , # % . & 38 38 38 ' & 38 38 38 '

!!!" !!!" !!!" b) PQ = OQ ! OP = #"5+ 7, ! 2 ! 8, ! 2 ! 3%$ = #"12, !10, ! 5%$ !!!" PQ = 122 + (!10)2 + (!5)2

= 269

!!!" ! 12 10 5 " ! 12 10 5 " The unit vectors parallel to PQ are $ , # , # % and $# , , % . & 269 269 269 ' & 269 269 269 '

! c) u = [5, 6, ! 3] ! 2 u = 52 + 62 + (!3) = 70 ! 5 6 3 5 6 3 The unit vectors parallel to u are u1 = i + j ! k and u 2 = ! i ! j + k . 70 70 70 70 70 70

!" d) f = #"!3, ! 2, ! 9%$ !" f = (!3)2 + (!2)2 + (!9)2

= 94

!" 3 2 9 3 2 9 The unit vectors parallel to f are f = ! i ! j ! k and f = i + j + k . 1 94 94 94 2 94 94 94

MHR • Calculus and Vectors 12 Solutions 746

Chapter 7 Section 4 Question 20 Page 400

! ! ! ! L.S. = ku R.S. = ku1 + ku2 + ku 3 k !u , u , u # = " 1 2 3 $ k !u , 0, 0# !0, u , 0# !0, 0, u # = (" 1 $ + " 2 $ + " 3 $) ! ! ! = k !u i + u j + u 3 k # " 1 2 $ ! ! ! = ku i + ku j + ku k !1 2 3! ! = ku1 !1, 0, 0# + ku2 !0, 1, 0# + ku !0, 0, 1# " $ " $ 3 " $ ! ! ! = ku1 + ku2 + ku 3

Therefore, L.S. = R.S.

Chapter 7 Section 4 Question 21 Page 400

! ! L.S. = u + v !u , u , u # !v , v , v # = " 1 2 3 $ + " 1 2 3 $ !u , 0, 0# !0, u , 0# !0, 0, u # !v , 0, 0# !0, v , 0# !0, 0, v # = " 1 $ + " 2 $ + " 3 $ + " 1 $ + " 2 $ + " 3 $ ! ! ! ! ! ! = u i + u j + u k + v i + v j + v k 1 2 ! 3 1 ! 2 3 ! = (u1 + v1 )i + (u2 + v2 ) j + (u3 + v3 )k (u v ) !1, 0, 0# (u v ) !0, 1, 0# (u v ) !0, 0, 1# = 1 + 1 " $ + 2 + 2 " $ + 3 + 3 " $ !u v , 0, 0# !0, u v , 0# !0, 0, u v # = " 1 + 1 $ + " 2 + 2 $ + " 3 + 3 $ !u v , u v , u v # = " 1 + 1 2 + 2 3 + 3 $ R.S. = !u + v , u + v , u + v # " 1 1 2 2 3 3 $

Therefore, L.S. = R.S.

Chapter 7 Section 4 Question 22 Page 400

!!!!" !!!" !!!" P P = P O + OP 1 2 !!1!" !!!"2 = OP + P O !!!"2 !!1!" = OP2 ! OP1 "x , y , z $ "x , y , z $ = # 2 2 2 % ! # 1 1 1 % = "x ! x , y ! y , z ! z $ # 2 1 2 1 2 1 %

MHR • Calculus and Vectors 12 Solutions 747

Chapter 7 Section 4 Question 23 Page 400

!!!" !!!" !!!" 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. x2 + y2 + z2 = 0

x2 + y2 + z2 = 0

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.

MHR • Calculus and Vectors 12 Solutions 748

Chapter 7 Section 4 Question 25 Page 400

!!!" !!!" !!!" a) AB = #"!5, 5, 0%$ AC = #"!5, 0, 5%$ BC = #"0, ! 5, 5%$ !!!" !!!" !!!" AB = (–5)2 + 52 + 02 AC = (–5)2 + 02 + 52 BC = 02 + (–5)2 + 52

= 50 = 50 = 50

!!!" !!!" !!!" Since AB = BC = AC , ! ABC is an equilateral triangle.

!5 5 5$ b) Draw a diagram. It appears that the closest point to the origin is the centroid of the triangle , , . #3 3 3& " % z

C(0, 0, 5)

B(0, 5, 0) y

A(5, 0, 0) x

In chapter 6, it was shown that the centroid of ΔABC is defined by the position vector !!!" 1 !!!" !!!" !!!" OX = OA + OB + OC . 3( ) ! 5 5 5 " The point on the interior of the triangle closest to the origin is the point # , , $ . % 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. !" R = [25, 0, 0]+ [0, 35, 0]+ [0, 0, 40] = [25, 35, 40] !" R = 252 + 352 + 402

= 3450 # 58.7

The magnitude of the resultant is about 58.7 N.

MHR • Calculus and Vectors 12 Solutions 749

! ! u "v b) cos! = ! ! 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.

Chapter 7 Section 4 Question 27 Page 401

! a = [2, ! 3, 5] ! b = [6, 3, !1] ! c = [!5, !1, 7]

! ! ! ! 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

MHR • Calculus and Vectors 12 Solutions 750

Chapter 7 Section 4 Question 28 Page 401

! ! L.S. = u !v "u ,u ,u $ "v ,v ,v $ = # 1 2 3 % ! # 1 2 3 % ! ! ! ! ! ! = u i + u j + u k ! v i + v j + v k ( 1 2 3 ) ( 1 2 3 ) ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! = u v i !i + u v i ! j + u v i ! k + u v j !i + +u v j ! j + +u v j ! k + +u v k !i + +u v k ! j + u v k ! k Step 1 1 1! ! 1 2 1 3 ! ! 2 1 2 2 ! ! 2 3 3 1 3 2 3 3 = u1v1 i !i + 0 + 0 + 0 + u2v2 j ! j + 0 + 0 + 0 + u3v3 k ! k Step 2

= u1v1 + u2v2 + u3v3 R.S. = u v + u v + u v 1 1 2 2 3 3

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 i ! j = i ! k = j !i = j ! k = k !i = k ! j = 0 (perpendicular vectors) ! ! ! ! ! ! and i !i = j ! j = k ! k =1 (parallel unit vectors) .

Chapter 7 Section 4 Question 29 Page 401

! ! ! ! u ! v if and only if u "v = 0 . a) [4, 1, 3]!["1, 5, k]= 0 "4 + 5 + 3k = 0 3k = "1 1 k = " 3 b) [k, 3, 6]![5, k, 8]= 0 5k + 3k + 48 = 0 8k = "48 k = "6 c) "!11, 3, 2k $# % "!k, 4, k $# = 0 11k +12 + 2k 2 = 0 2k 2 +11k +12 = 0 (2k + 3)(k + 4) = 0

3 k = ! or k = ! 4 2

MHR • Calculus and Vectors 12 Solutions 751

Chapter 7 Section 4 Question 30 Page 401

! ! ! ! ! ! i ! j = #"1, 0, 0%$ ! #"0, 1, 0%$ j ! k = #"0, 1, 0%$ ! #"0, 0, 1%$ i ! k = #"1, 0, 0%$ ! #"0, 0, 1%$ = 1(0) + 0(1) + 0(0) = 0(0) +1(0) + 0(1) = 1(0) + 0(0) + 0(1) = 0 = 0 = 0

Since i ! j = j ! k = i ! k = 0 , then i , j , and k are mutually orthogonal.

Chapter 7 Section 4 Question 31 Page 401

! ! a) i) L.S. = u + v !u , u , u # !v , v , v # = " 1 2 3 $ + " 1 2 3 $ !u v , u v , u v # = " 1 + 1 2 + 2 3 + 3 $ !v u , v u , v u # = " 1 + 1 2 + 2 3 + 3 $ !v , v , v # !u , u , u # = " 1 2 3 $ + " 1 2 3 $ ! ! R.S. = v + u = !v , v , v # + !u , u , u # " 1 2 3 $ " 1 2 3 $ Therefore, L.S. = R.S.

It is proven that u + v = v + u .

! ! ii) L.S. = u !v "u , u , u $ "v , v , v $ = # 1 2 3 % ! # 1 2 3 %

= u1v1 + u2v2 + u3v3 = v u + v u + v u !1 !1 2 2 3 3 R.S. = v !u "v , v , v $ "u , u , u $ = # 1 2 3 % ! # 1 2 3 % = v u + v u + v u 1 1 2 2 3 3 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.

MHR • Calculus and Vectors 12 Solutions 752

Chapter 7 Section 4 Question 32 Page 401

! ! ! Let a = [a1, a2 , a3 ], b = [b1, b2 , b3 ], and c = [c1, c2 , c3 ].

! ! ! L.S. = (a + b) + c !a , a , a # !b , b , b # !c , c , c # = (" 1 2 3 $ + " 1 2 3 $) + " 1 2 3 $ !a b , a b , a b # !c , c , c # = " 1 + 1 2 + 2 3 + 3 $ + " 1 2 3 $ !(a b ) c , (a b ) c , (a b ) c # = " 1 + 1 + 1 2 + 2 + 2 3 + 3 + 3 $ !a (b c ), a (b c ), a (b c )# = " 1 + 1 + 1 2 + 2 + 2 3 + 3 + 3 $ ! ! ! R.S. = a + (b + c) !a , a , a # !b , b , b # !c , c , c # = " 1 2 3 $ + (" 1 2 3 $ + " 1 2 3 $) !a , a , a # !b c , b c , b c # = " 1 2 3 $ + " 1 + 1 2 + 2 3 + 3 $ = !a + (b + c ), a + (b + c ), a + (b + c )# " 1 1 1 2 2 2 3 3 3 $ 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

! ! ! Let a = [a1, a2 , a3 ], b = [b1, b2 , b3 ], and c = [c1, c2 , c3 ].

! ! a) L.S. = k (a + b) k !a , a , a # !b , b , b # = (" 1 2 3 $ + " 1 2 3 $) k !a b , a b , a b # = " 1 + 1 2 + 2 3 + 3 $ !k(a b), k(a b ), k(a b )# = " 1 + 2 + 2 3 + 3 $ !ka kb , ka kb , ka kb # = " 1 + 1 2 + 2 3 + 3 $ ! ! R.S. = ka + kb k !a , a , a # k !b , b , b # = " 1 2 3 $ + " 1 2 3 $ !ka , ka , ka # !kb , kb , kb # = " 1 2 3 $ + " 1 2 3 $ = !ka + kb , ka + kb , ka + kb # " 1 1 2 2 3 3 $ Therefore, L.S. = R.S.

! ! ! ! It has been proven that k (a + b)= ka + kb .

MHR • Calculus and Vectors 12 Solutions 753

! ! ! b) L.S. = a !(b + c) "a , a , a $ "b , b , b $ "c , c , c $ = # 1 2 3 % !(# 1 2 3 % + # 1 2 3 %) = "a ,a a $ ! "b + c , b + c , b + c $ # 1 2, 3 % # 1 1 2 2 3 3 %

= a1(b1 + c1 ) + a2 (b2 + c2 ) + a3(b3 + c3 ) = a b + a c + a b + a c + a b + a c !1 !1 !1 !1 2 2 2 2 3 3 3 3 R.S. = a !b + a !c "a , a , a $ "b , b , b $ "a , a , a $ "c , c , c $ = # 1 2 3 % ! # 1 2 3 % + # 1 2 3 % ! # 1 2 3 %

= a1b1 + a2b2 + a3b3 + a1c1 + a2c2 + a3c3 = a b + a c + a b + a c + a b + a c 1 1 1 1 2 2 2 2 3 3 3 3 Therefore, L.S. = R.S.

! ! ! ! ! ! ! It has been proven that a !(b + c)= a !b + a !c .

! ! c) L.S. = k (a !b) k "a , a , a $ "b , b , b $ = (# 1 2 3 % ! # 1 2 3 %) k a b a b a b = ( 1 1 + 2 2 + 3 3 )

= k(a1b1 ) + k(a2b2 ) + k(a3b3 )

= ka1b1 + ka2b2 + ka3b3 ! ! M.S. = ka !(b) = k "a , a , a $ ! "b , b , b $ ( # 1 2, 3 %) # 1 2 3 % "ka , ka , ka $ "b , b , b $ = # 1 2 3 % ! # 1 2 3 %

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

= ka1b1 + ka2b2 + ka3b3 ! ! R.S. = a !(kb) = "a , a , a $ ! k "b , b , b $ # 1 2, 3 % ( # 1 2 3 %) = "a , a , a $ ! "kb , kb , kb $ # 1 2, 3 % # 1 2 3 %

= a1(kb1 ) + a2 (kb2 ) + a3(kb3 )

= a1kb1 + a2kb2 + a3kb3 = ka b + ka b + ka b 1 1 2 2 3 3 Therefore, L.S. = M.S = R.S.

! ! ! ! ! ! It has been proven that k (a !b)= (ka)!b = a !(kb).

MHR • Calculus and Vectors 12 Solutions 754

Chapter 7 Section 4 Question 34 Page 401

! ! ! " u !v % ! proj! u ! ! v v = $ ' # v !v &

" )(3, 4, 7+* ! )(1, 2, 3+*% = $ ' )(1, 2, 3+* # )(1, 2, 3+* ! )(1, 2, 3+* & " 32% = (1, 2, 3* #$ 14&' ) + (16 32 48* = , , , 7 7 7 - ) +

! !16 32 48" The component vector for the v direction is # , , $ . % 7 7 7 & The component vector in the perpendicular direction can be found by vector subtraction. !16 32 48" "5 !4 1 # [3, 4, 7] – # , , $ = $ , , % % 7 7 7 & &7 7 7 '

Check perpendicularity using the dot product. "5 !4 1 % ( 5+ ( 4+ ( 1+ [3, 4, 7] · $ , , ' = 3* - + 4* ! - + 7* - #7 7 7 & ) 7, ) 7, ) 7, 0 = 7 = 0 !16 32 48" "5 !4 1 # Therefore, the two perpendicular components are # , , $ and $ , , % . % 7 7 7 & &7 7 7 '

Chapter 7 Section 4 Question 35 Page 401

14º

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].

MHR • Calculus and Vectors 12 Solutions 755

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].

2 2 2 The magnitude of the final ground velocity is 200 + 20 + 0 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. z

553 m

x 10º 120º 60º 9º

y

! 553 $ The first person is located at , 0, 0 . # tan 10o & " % ! 553 553 $ The second person is located at cos 120o , sin 120o , 0 . # tan 9o tan 9o & " % ! 553 553 $ ! 553 $ The displacement vector is cos 120o , sin 120o , 0 ' , 0, 0 "# tan9o tan 9o %& "# tan 10o %& ( 553 553 553 + cos 120o , sin 120o , 0 = * o ' o o - ) tan 9 tan 10 tan 9 , ! ('4882.0, 3023.7. 0+ ) , 2 2 " 553 553 % " 553 % The distance between these points is cos 120o ! + sin 120o + 02 ! 5742.5. #$ tan 9o tan 10o &' #$ tan 9o &'

The distance between the two people is 5742.5 m.

MHR • Calculus and Vectors 12 Solutions 756

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 ! 9.8t 2 $ # %

! ii) s(t) = "6t, 18t cos 35o , 18t sin 35o ! 9.8t 2 $ # % 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 o h!(t) = 30sin 20 "19.6t

The maximum height occurs when h!(t) = 0. 30sin 20o !19.6t = 0 30sin 20o t = 19.6 t ! 0.52

! s(0.52) " !1.04, 14.66, 2.69# " $

ii) h(t) = 18tsin 35º – 9.8t2 o h!(t) = 18sin 35 "19.6t

The maximum height occurs when h!(t) = 0. 18sin 35o !19.6t = 0 18sin 35o t = 19.6 t ! 0.53

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

MHR • Calculus and Vectors 12 Solutions 757

Chapter 7 Section 4 Question 38 Page 402 a)

x

!!" 5 fN R 1 30º

x = 5tan30º ! 2.9 fN !!" R = 52 + 2.92 1 # 5.8

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

z 20º 5 fN

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).

!" R 1.8 fN 20º 5.8 fN

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.

MHR • Calculus and Vectors 12 Solutions 758

Chapter 7 Section 4 Question 39 Page 402

z E(–4, 2, 0) G

D(2, 1, 3)

F(6, –2, 4)

y

x

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).

MHR • Calculus and Vectors 12 Solutions 759

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 x2 + y2 =100 and y2 + z2 =100 . 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 v = [w, x, y, z] 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 v = [w, x, y, z] 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.

MHR • Calculus and Vectors 12 Solutions 760

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 11 x = 2 ! "11 % v = , ! 4, 1 $ 2 ' # &

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 11 x = k 2 ! "11k % v = , ! 4k, k $ 2 ' # &

! This last example shows that there is a vector v for every choice of parameter k. Therefore, there are infinitely many solutions.

MHR • Calculus and Vectors 12 Solutions 761

Chapter 7 Section 4 Question 43 Page 402

! ! a "b a) cos! = ! ! 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 ( 12 + ! = cos'1 )* 17.1464,- " 45.6º ! ! b"c cos! = ! ! 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 ( 45 + ! = cos#1 )* 51.2348,- " 28.6º ! ! a "c cos! = ! ! 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 ( 40 + ! = cos'1 )* 41.8330,- " 17.0º

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

MHR • Calculus and Vectors 12 Solutions 762

!" " d "e b) cos! = !" " 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 ( '6 + ! = cos'1 )* 27.9285,- # 102.4º

! "! e" f cos! = ! "! 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 ( #8 + ! = cos#1 )* 35.6931,- # 103.0º

!" !" d " f cos! = !" !" 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 ( 34 + ! = cos'1 )* 38.3406,- # 27.5º

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

MHR • Calculus and Vectors 12 Solutions 763

Chapter 7 Section 4 Question 44 Page 402

Proof 1 ( ! ) ! ! Suppose u ! v . ! ! ! ! Then u !v = 0 and v !u = 0 . ! ! 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 u !v = 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 ! ! 4u "v = 0 ! ! u "v = 0

! ! ! ! Therefore, u !v = 0 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.

MHR • Calculus and Vectors 12 Solutions 764

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

2 1 The probability is or or about 0.28%. 720 360

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

2 5 ! 3 5 5 3 5 5 ± (–5) ! 4(1)(–5) + x = 2 2 1 4 2 Q R S 5 ± 45 P = 2 5 ± 3 5 = 2

QR 3 RQ 3 Then, = = RS ! 5+ 3 5 $ QP " 5! 3 5 % # & ' 4 1! $ ' " 2 % # 2 & 6 6 = = 5+ 3 5 ' 8 2 ! 5+ 3 5 6 3+ 3 5 6 3+ 3 5 = ( = ( '3+ 3 5 3+ 3 5 !3+ 3 5 3+ 3 5 6(3) 1+ 5 6(3) 1+ 5 = ( ) = ( ) '9 + 45 !9 + 45 1+ 5 1+ 5 = = 2 2

MHR • Calculus and Vectors 12 Solutions 765

Chapter 7 Section 5 The Cross Product and Its Properties

Chapter 7 Section 5 Question 1 Page 410

! ! ! ! a) u ! v = u v (sin")nˆ = 60(80)sin55o nˆ " 3931.9nˆ

! ! ! ! b) u ! v = u v (sin")nˆ = 112(128)sin 164o nˆ " 3951.5nˆ

Chapter 7 Section 5 Question 2 Page 410

! ! a) 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% $ &

MHR • Calculus and Vectors 12 Solutions 766

Chapter 7 Section 5 Question 3 Page 411

! ! a) 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 u and 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 is orthogonal to u and 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% $ &

! ! ! ! Use the dot product to check if u ! v is orthogonal to u and v . ! ! ! 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 is orthogonal to u and 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% $ &

! ! ! ! Use the dot product to check if u ! v is orthogonal to u and v .

MHR • Calculus and Vectors 12 Solutions 767

! ! ! 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 is orthogonal to u and v .

Chapter 7 Section 5 Question 4 Page 411

!" " !" " !" " The area of the parallelogram defined by p and q is p " q = p q sin! .

!" " 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

The area of the parallelogram is about 615.2 square units.

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).

MHR • Calculus and Vectors 12 Solutions 768

Chapter 7 Section 5 Question 6 Page 411

! ! a) Let u = [!3, 4, 7] and v = [2, 8, 3]. ! ! 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 ! 2 ! 2 ! ! 2 u v ! u "v = 74(77) ! 472 ( ) = 3489

! ! ! 2 ! 2 ! ! 2 Therefore, it has been verified that u ! v = u v " (u #v) .

! ! b) Let u and v be any two vectors. ! ! L.S. = u ! v ! ! = u v sin"

! 2 ! 2 ! ! 2 R.S. = u v # (u $v) ! 2 ! 2 ! ! 2 = u v # ( u v cos") ! 2 ! 2 ! 2 ! 2 = u v # u v cos2 "

! 2 ! 2 = u v (1# cos2 ")

! 2 ! 2 = u v sin2 " sin2 " + cos2 " = 1 ! ! = u v sin"

Therefore, L.S. = R.S.

MHR • Calculus and Vectors 12 Solutions 769

Chapter 7 Section 5 Question 7 Page 411

! ! ! a) 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

MHR • Calculus and Vectors 12 Solutions 770

Chapter 7 Section 5 Question 8 Page 411

! "! "! ! Use c ! d and 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 "! 1 ! ! p = u ! v 654 ( ) # 13 14 17 % = ' , , ( 654 654 654 $ &

! ! ! 13 14 17 " A vector orthogonal to both u and v is # , , $ . % 654 654 654 &

Chapter 7 Section 5 Question 10 Page 411

!!!" !!!" !!!" !!!" !!!" !!!" a) PQ = OQ ! OP and SR = OR ! OS = #"1, 6, 2%$ ! #"0, 2, 5%$ = #"7, 4, 2%$ ! #"6, 0, 5%$ = "1, 4, ! 3$ = "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.

MHR • Calculus and Vectors 12 Solutions 771

!!!" !!" 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

The area of the parallelogram is about 32.2 square units. 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

!" g = 42 + 52 + 22

= 45 ! h = (–2)2 + 62 +12

= 41 !" " g h 4(–2) 5(6) 2(1) ! = + + = 24

MHR • Calculus and Vectors 12 Solutions 772

Using the cross product: ! ! g " h sin! = ! ! g h 1269 = 45 41

! = 56.0º

Using the dot product: !" " g " h cos! = !" " g h

24 = 45 41

! # 56.0°

Chapter 7 Section 5 Question 12 Page 411

! ! ! ! ! ! The area of the parallelogram is a "b = a b sin! where a and b are two adjacent sides and θ is the angle between the sides. 85 = 10(9) sin! 85 sin! = 90 # 85& ! = sin"1 ± $% 90'(

o o ! ! 70.8 and 109.2

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$ # %

MHR • Calculus and Vectors 12 Solutions 773

! ! "! 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 a "b = a b sin! where a and 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 ! = 0o or 180o 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% $ &

MHR • Calculus and Vectors 12 Solutions 774

! ! ! ! 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 a = [a1, a2 , a3 ], b = [b1, b2 , b3 ], and c = [c1, c2 , c3 ]. ! ! ! L.S. = a ! (b + c) "a , a , a $ "b , b , b $ "c , c , c $ = # 1 2 3 % ! (# 1 2 3 % + # 1 2 3 %) "a , a , a $ "b c , b c , b c $ = # 1 2 3 % ! # 1 + 1 2 + 2 3 + 3 % "a (b c ) (b c )a , a (b c ) (b c )a , a (b c ) (b c )a $ = # 2 3 + 3 & 2 + 2 3 3 1 + 1 & 3 + 3 1 1 2 + 2 & 1 + 1 2 % "a b a c a b a c , a b a c a b a c , a b a c a b a c $ = # 2 3 + 2 3 & 3 2 & 3 2 3 1 + 3 1 & 1 3 & 1 3 1 2 + 1 2 & 2 1 & 2 1 % ! ! ! ! R.S. = a ! b + a ! c "a , a , a $ "b , b , b $ "a , a , a $ "c , c , c $ = # 1 2 3 % ! # 1 2 3 % + # 1 2 3 % ! # 1 2 3 % "a b a b , a b a b , a b a b $ "a c a c , a c a c , a c a c $ = # 2 3 & 3 2 3 1 & 1 3 1 2 & 2 1 % + # 2 3 & 3 2 3 1 & 1 3 1 2 & 2 1 % = "a b + a c & a b & a c , a b + a c & a b & a c , a b + a c & a b & a c $ # 2 3 2 3 3 2 3 2 3 1 3 1 1 3 1 3 1 2 1 2 2 1 2 1 % 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% $ &

MHR • Calculus and Vectors 12 Solutions 775

! ! ! ! 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 a = [a1, a2 , a3 ], b = [b1, b2 , b3 ], and c = [c1, c2 , c3 ]. ! ! ! L.S. = (a + b) ! c "a , a , a $ "b , b , b $ "c , c , c $ = (# 1 2 3 % + # 1 2 3 %) ! # 1 2 3 % "a b , a b , a b $ "c , c , c $ = # 1 + 1 2 + 2 3 + 3 % ! # 1 2 3 % "(a b )c c (a b ), (a b )c c (a b ), (a b )c c (a b)$ = # 2 + 2 3 & 2 3 + 3 3 + 3 1 & 3 1 + 1 1 + 1 2 & 1 2 + % "a c b c a c b c , a c b c a c b c , a c b c a c b c $ = # 2 3 + 2 3 & 3 2 & 3 2 3 1 + 3 1 & 1 3 & 1 3 1 2 + 1 2 & 2 1 & 2 1 % ! ! ! ! R.S. = a ! c + b ! c "a , a , a $ "c , c , c $ "b , b , b $ "c , c , c $ = # 1 2 3 % ! # 1 2 3 % + # 1 2 3 % ! # 1 2 3 % "a c a c , a c a c , a c a c $ "b c b c , b c b c , b c b c $ = # 2 3 & 3 2 3 1 & 1 3 1 2 & 2 1 % + # 2 3 & 3 2 3 1 & 1 3 1 2 & 2 1 % = "a c + b c & a c & b c , a c + b c & a c & b c , a c + b c & a c & b c $ # 2 3 2 3 3 2 3 2 3 1 3 1 1 3 1 3 1 2 1 2 2 1 2 1 % 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% $ &

MHR • Calculus and Vectors 12 Solutions 776

! ! 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 u = [u1, u2 , u3 ] and v = [v1, v2 , v3 ]. ! ! L.S. = k (u ! v) k "u , u , u $ "v , v , v $ = (# 1 2 3 % ! # 1 2 3 %) k "u v v u , u v v u , u v v u $ = # 2 3 & 2 3 3 1 & 3 1 1 2 & 1 2 % "k(u v v u ), k(u v v u ), k(u v v u )$ = # 2 3 & 2 3 3 1 & 3 1 1 2 & 1 2 % "ku v kv u , ku v kv u , ku v kv u $ = # 2 3 & 2 3 3 1 & 3 1 1 2 & 1 2 % ! ! M.S. = (ku) ! v k "u , u , u $ "v , v , v $ = ( # 1 2 3 %) ! # 1 2 3 % "ku , ku , ku $ "v , v , v $ = # 1 2 3 % ! # 1 2 3 % "ku v kv u , ku v kv u , ku v kv u $ = # 2 3 & 2 3 3 1 & 3 1 1 2 & 1 2 % ! ! R.S. = u ! (kv) "u , u , u $ k "v , v , v $ = # 1 2 3 % ! # 1 2 3 % ( ) "u , u , u $ "kv , kv , kv $ = # 1 2 3 % ! # 1 2 3 % = "ku v & kv u , ku v & kv u , ku v & kv u $ # 2 3 2 3 3 1 3 1 1 2 1 2 % Therefore, L.S. = M.S. = R.S.

! ! ! ! ! ! Therefore, k (u ! v)= (ku)! v = u !(kv).

MHR • Calculus and Vectors 12 Solutions 777

Chapter 7 Section 5 Question 18 Page 412

No. Choose three collinear vectors. ! ! ! Let a = [1, 2, 3], b = [2, 4, 6], and c = [3, 6, 9]. ! ! 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 a !b = a ! c but b " c . Any three collinear vectors will show this.

Chapter 7 Section 5 Question 19 Page 412

! ! ! ! ! ! ! ! a ! b = a b sin" and a #b = a b cos". ! ! ! ! Therefore, a ! b < a #b if sin" < cos".

Examining the related functions y = sin! and y = cos! 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.

! ! a "b The easy test is to determine the measure of θ is cos! = ! ! . a b

MHR • Calculus and Vectors 12 Solutions 778

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

"1 # "5 & ! = cos % ( $ 6 6 ' o ! 146.4

! ! ! ! In this case, a !b > a "b . c) [2, 1, 1]![3, 1, 2]= 9 [2, 1, 1] = 6 [3, 1, 2] = 14

"1 # 9 $ ! = cos % & ' 6 14 ( !10.9o

! ! ! ! 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

! ! ! 2 ! 2 ! ! 2 a) From question 6, it is known that 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). ! 2 ! 2 ! ! 2 u v ! (u "v) # 0 ! 2 ! 2 ! ! 2 u v # (u "v) ! ! ! ! u v # u "v ! ! ! ! [Note: Actually, the stronger result, u v ! u "v , is true.]

MHR • Calculus and Vectors 12 Solutions 779

b) Proof 1 ( ! ) ! ! ! ! Let u and v be collinear. Then u = kv . ! ! ! 2 ! 2 ! ! 2 u ! v = u v " (u #v) ! ! ! 2 ! 2 ! ! 2 kv ! v = u v " (u #v) ! 2 ! 2 ! ! 2 ! ! ! 0 = u v " (u #v) u ! v = 0 for collinear vectors. ! 2 ! 2 ! ! 2 0 = u v " (u #v) ! ! 2 ! 2 ! 2 (u #v) = u v ! ! ! ! u v u v Take the positive square root of both sides. # =

Proof 2 ( ! ) ! ! ! ! Let u !v = u v ! ! ! ! u !v = u v ! ! 2 ! 2 ! 2 (u !v) = u v Square both sides ! 2 ! 2 ! ! 2 0 = u v " (u !v) ! 2 ! 2 ! ! 2 0 = u v " (u !v) ! ! 0 = u # v (From question 6.) ! ! 0 = u v sin$

! ! If sin! = 0 , then ! = 0o and u and v are collinear. ! ! ! ! ! ! If u = 0 or v = 0 , then u = 0 or v = 0 and the vectors are (trivially) collinear.

! ! ! ! ! ! Therefore, u and v be collinear if and only if ( ! ) u !v = u v .

Chapter 7 Section 5 Question 21 Page 412

! ! ! Vectors a, b, and c are position vectors for the three vertices of the triangle. ! ! ! ! Two of the sides of the triangle can be represented by b ! a and c ! a . ! ! ! ! ! ! ! ! The area of the parallelogram defined by b ! a and c ! a is b ! a " c ! a . ( ) ( ) The area of the triangle is one-half of this expression.

MHR • Calculus and Vectors 12 Solutions 780

1 ! ! ! ! !Area = b " a # c " a 2 ( ) ( ) 1 ! ! ! ! ! ! ! ! = b # c " b # a " a # c + a # a Distributive property 2 1 ! ! ! ! ! ! ! ! ! = b # c " b # a " a # c a # a = 0 2 1 ! ! ! ! ! ! ! ! ! ! = b # c + a # b + c # a a # b = "b # a 2 1 ! ! ! ! ! ! = a # b + b # c + c # a Commutative property 2

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 Since j ! (i ! k) = j ! (" j) = 0 ! ! ! ! ! "! ! ! ! ! ! ! and (i ! j) ! k = k ! k = 0, and and ( j ! i) ! k = ("k) ! k = 0, ! ! ! ! ! ! ! ! ! ! ! ! i ! j ! k = i ! j ! k j ! i ! k = j ! i ! k ( ) ( ) ( ) ( )

! ! ! b) Let a = [a1, a2 , a3 ], b = [b1, b2 , b3 ], and c = [c1, c2 , c3 ]. ! ! ! L.S. = a ! (b ! c) "a , a , a $ "b , b , b $ "c , c , c $ = # 1 2 3 % ! (# 1 2 3 % ! # 1 2 3 %) "a , a , a $ "b c b c , b c b c , b c b c $ = # 1 2 3 % ! # 2 3 & 3 2 3 1 & 1 3 1 2 & 2 1 % "a (b c b c ) a (b c b c ) a (b c b c ) a (b c b c ), a (b c b c ) a (b c b c )$ = # 2 1 2 & 2 1 & 3 3 1 & 1 3 3 2 3 & 3 2 & 1 1 2 & 2 1 1 3 1 & 1 3 & 2 2 3 & 3 2 % "a b c a b c a b c a b c , a b c a b c a b c a b c , a b c a b c a b c a b c $ = # 2 1 2 & 2 2 1 & 3 3 1 + 3 1 3 3 2 3 & 3 3 2 & 1 1 2 + 1 2 1 1 3 1 & 1 1 3 & 2 2 3 + 2 3 2 % ! ! ! ! ! ! R.S. = (a !c)b " (a !b)c #a , a , a % #c , c , c % #b , b , b % #a , a , a % #b , b , b % #c , c , c % = ($ 1 2 3 & ! $ 1 2 3 &)$ 1 2 3 & " ($ 1 2 3 & ! $ 1 2 3 &)$ 1 2 3 & (a c a c a c ) #b , b , b % (a b a b a b ) #c , c , c % = 1 1 + 2 2 + 3 3 $ 1 2 3 & " 1 1 + 2 2 + 3 3 $ 1 2 3 & #a b c a b c a b c , a b c a b c a b c , a b c a b c a b c % = $ 1 1 1 + 2 1 2 + 3 1 3 1 2 1 + 2 2 2 + 3 2 3 1 3 1 + 2 3 2 + 3 3 3 & #a b c a b c a b c , a b c a b c a b c , a b c a b c a b c % " $ 1 1 1 + 2 2 1 + 3 3 1 1 1 2 + 2 2 2 + 3 3 2 1 1 3 + 2 2 3 + 3 3 3 & = #a b c " a b c " a b c + a b c , a b c " a b c " a b c + a b c , a b c " a b c " a b c + a b c % $ 2 1 2 2 2 1 3 3 1 3 1 3 3 2 3 3 3 2 1 1 2 1 2 1 1 3 1 1 1 3 2 2 3 2 3 2 & Therefore, L.S. = R.S.

! ! ! ! ! ! ! ! ! Therefore, a ! b ! c = a !c b " a !b c . ( ) ( ) ( )

MHR • Calculus and Vectors 12 Solutions 781

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 A = x y ! x y + x y ! x y + x y ! x y + x y ! x y + x y ! x y 2 1 2 2 1 2 3 3 2 3 4 4 3 4 5 5 4 5 1 1 5 1 Therefore, A = 0(7) ! 2(5) + 2(6) ! 5(7) + 5(4) ! 6(6) + 6(2) !1(4) +1(5) ! 0(2) 2 1 = !36 2 = 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).

30(100) The probability it will pass through (success) is ! 0.223or 22.3%. (127)(106)

MHR • Calculus and Vectors 12 Solutions 782

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.]

Chapter 7 Section 6 Question 2 Page 418

! ! ! " u !v % ! a) proj! 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, 2, 7* #$ 6(6) + 2(2) + 7(7)&' ) + 48 = )(6, 2, 7+* 89

! 48 proj! u = [6, 2, 7] v 89 48 = 62 + 22 + 72 89 48 89 = 89 " 5.1

! ! ! " u !v % ! b) proj! 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, 7, 6+ #$ 3(3) + 7(7) + 6(6) &' * , 35 = *)3, 7, 6,+ 94

MHR • Calculus and Vectors 12 Solutions 783

! 35 proj! u = [3, 7, 6] v 94 35 = 32 + 72 + 62 94 35 94 = 94 " 3.6

! ! ! " u !v % ! c) proj! 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, 1, ( 8+ #$ 6(6) +1(1) + (–8)(–8) &' * , (43 = *)6, 1, ( 8,+ 101

! !43 proj! u = "6, 1, ! 8$ v 101 # % 43 = 62 +12 + (–8)2 101 43 101 = 101 " 4.3

! ! ! " u !v % ! d) proj! 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, 1, 0+ #$ 9(9) +1(1) + 0(0) &' * , 9 = *)9, 1, 0,+ 82

MHR • Calculus and Vectors 12 Solutions 784

! 9 proj! u = [9, 1, 0] v 82 9 = 92 +12 + 02 82 9 82 = 82 " 1.0

Chapter 7 Section 6 Question 3 Page 418

!" " a) To calculate the work done against gravity, use only the vertical components of F and 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

The work done against gravity is 120 J.

!" " W = F !s = #"3, 5, 12%$ ! #"2, 0, 10%$ = 126

The work done in the direction of travel is 126 J.

!" " c) Wgravity = F !s

= #"0, 0, 12%$ ! #"0, 0, 6%$ = 72

The work done against gravity is 72 J.

MHR • Calculus and Vectors 12 Solutions 785

!" " W = F !s = #"3, 5, 12%$ ! #"2, 1, 6%$ = 83

The work done in the direction of travel is 83 J.

Chapter 7 Section 6 Question 4 Page 418

! ! ! ! ! ! a) 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

MHR • Calculus and Vectors 12 Solutions 786

Chapter 7 Section 6 Question 5 Page 418

!" " " a) 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

The volume is 0 cubic units.

!" " " 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

The volume is 8 cubic units.

MHR • Calculus and Vectors 12 Solutions 787

Chapter 7 Section 6 Question 6 Page 418

! ! ! ! ! """! ! """! The area of the parallelogram defined by a and b is a !b where a = AB and b = AC . 1 ! ! For triangle !ABC , calculate a !b . 2 !!!" 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

1 A = (71.83) 2 # 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("!) 65(0.6) x = 40 x = 0.975

The girl should sit 0.975 m form the balance point.

MHR • Calculus and Vectors 12 Solutions 788

Chapter 7 Section 6 Question 9 Page 418

! ! 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$ # %

! ! 2 "! "! 2 2 2 a) u ! v " (w# w) = ["1, " 5, 4] " (%$6, 2, 1'& # %$6, 2, 1'&) 2 = ( ("1)2 + ("5)2 + 42 ) " (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

MHR • Calculus and Vectors 12 Solutions 789

Chapter 7 Section 6 Question 10 Page 418 a)

D

B E ! b θ A ! C F a ! ! 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. 2 ! ! 2 ! ! 2 Therefore, DF = a!b + a "b .

! ! 2 ! ! 2 ! 2 ! 2 c) The proof or verification of a!b + a "b = a b can be done geometrically or using coordinates. First show a proof using coordinates. ! ! Let a = [a1, a2 , a3 ] and b = [b1, b2 , b3 ]. ! ! 2 ! ! 2 L.S. = a ! b + a "b

2 2 #a , a , a % #b , b , b % #a , a , a % #b , b , b % = $ 1 2 3 & ! $ 1 2 3 & + $ 1 2 3 & " $ 1 2 3 & 2 2 #a b a b , a b a b , a b a b % a b a b a b = $ 2 3 ' 3 2 3 1 ' 1 3 1 2 ' 2 1 & + 1 1 + 2 2 + 3 3 2 2 2 2 = (a2b3 ' a3b2 ) + (a3b1 ' a1b3 ) + (a1b2 ' a2b1 ) + (a1b1 + a2b2 + a3b3 ) 2 2 2 2 2 2 2 2 2 2 2 2 = a2 b3 ' 2a2b3a3b2 + a3 b2 + a3 b1 ' 2a1b3a3b1 + a1 b3 + a1 b2 ' 2a2b1a1b2 + a2 b1 2 2 2 2 2 2 + a1 b1 + a2 b2 + a3 b3 + 2a1b1a2b2 + 2a2b2a3b3 + 2a1b1a3b3 = a 2b 2 + a 2b 2 + a 2b 2 + a 2b 2 + a 2b 2 + a 2b 2 + a 2b 2 + a 2b 2 + a 2b 2 1 1 1 2 1 3 2 1 2 2 2 3 3 1 3 2 3 3

MHR • Calculus and Vectors 12 Solutions 790

! 2 ! 2 R.S. = a b

2 2 !a , a , a # !b , b , b # = " 1 2 3 $ " 1 2 3 $ 2 2 2 2 2 2 = (a1 + a2 + a3 )(b1 + b2 + b3 ) = a 2b 2 + a 2b 2 + a 2b 2 + a 2b 2 + a 2b 2 + a 2b 2 + a 2b 2 + a 2b 2 + a 2b 2 1 1 1 2 1 3 2 1 2 2 2 3 3 1 3 2 3 3 Therefore, L.S. = R.S.

Now show a proof using geometric vectors. ! ! 2 ! ! 2 L.S. = a ! b + a "b

! ! 2 ! ! 2 = a b sin# + a b cos#

! 2 ! 2 ! 2 ! 2 = a b sin2 # + a b cos2 #

! 2 ! 2 = a b (sin2 # + cos2 #)

! 2 ! 2 = a b (1)

! 2 ! 2 = a b

! 2 ! 2 R.S. = a b

Therefore, L.S. = R.S.

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"

! o o ! total = (0.75)10sin90 + 0.35(5)sin(#90 )

# 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).

MHR • Calculus and Vectors 12 Solutions 791

Chapter 7 Section 6 Question 12 Page 419

! ! "! ! = r F sin" 10 = 0.20(80)sin" 10 sin" = (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 torque ! engine speed Horsepower = 5252

• 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.

! "! e! f = 0 if e and f are collinear and the angle between them is 0° or 180°. ! "! ! !" e ! f = 0 if e and f are perpendicular and the angle between them is 90°.

MHR • Calculus and Vectors 12 Solutions 792

Chapter 7 Section 6 Question 15 Page 419

! ! "! ! = r F sin" "! 100 = 0.30 F sin 40o "! 100 F = (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 ! c

! a

1 ! ! ! b) V = a b c 2 1 = 14 20 180 2 = 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.

MHR • Calculus and Vectors 12 Solutions 793

Chapter 7 Section 4 Question 17 Page 419

! ! Let u = [u1, u2 , u3 ] and v = [v1, v2 , v3 ] ! ! "! L.S. = u ! v " w ! ! ! ! = u ! v "(ku + mv) = #u , u , u % ! #v , v , v % " #k #u , u , u % + m#v , v , v %% $ 1 2 3 & $ 1 2 3 & $ $ 1 2 3 & $ 1 2 3 && #u v u v , u v u v , u v u v % #ku mv , ku mv , ku mv % = $ 2 3 ' 3 2 3 1 ' 1 3 1 2 ' 2 1 & " $ 1 + 1 2 + 2 3 + 3 &

= (u2v3 ' u3v2 )(ku1 + mv1 ) + (u3v1 ' u1v3 )(ku2 + mv2 ) + (u1v2 ' u2v1 )(ku3 + mv3 )

= ku1u2v3 + mu2v1v3 ' ku1u3v2 ' mu3v1v2 + ku2u3v1 + mu3v1v2 ' ku1u2v3 ' mu1v2v3

+ ku1u3v2 + mu1v2v3 ' ku2u3v1 ' mu2v1v3 = 0 R.S. = 0

Therefore, L.S. = R.S.

Chapter 7 Section 6 Question 18 Page 419

For an informal proof, draw a diagram. ! ! a !b

! c ! a ! b ! ! ! (a !b)! c

! ! ! Suppose a and b lie in a plane and c is a vector mot in this plane. ! ! ! ! a !b is perpendicular to both a and 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 a and b . Therefore (a !b)! c lies in the plane containing a and b .

! ! ! ! ! For a more formal proof, show that the volume of the parallelepiped formed by (a !b)! c , a, and b has ! ! ! ! ! volume zero; i.e., (a !b)! c , a, and b are coplanar vectors.

! ! ! Let a = [a1, a2 , a3 ], b = [b1, b2 , b3 ], and c = [c1, c2 , c3 ].

MHR • Calculus and Vectors 12 Solutions 794

! ! ! ! ! V = ((a ! b) ! c)" a ! b = #a , a , a % ! #b , b , b % ! #c , c , c % " #a , a , a % ! #b , b , b % (($ 1 2 3 & $ 1 2 3 &) $ 1 2 3 &) $ 1 2 3 & $ 1 2 3 & #a b a b , a b a b , a b a b % #c , c , c % #a b a b , a b a b , a b a b % = $ 2 3 ' 3 2 3 1 ' 1 3 1 2 ' 2 1 & ! $ 1 2 3 & " $ 2 3 ' 3 2 3 1 ' 1 3 1 2 ' 2 1 & #(a b a b)c c (a b a b ), (a b a b)c c (a b a b ), (a b a b )c c (a b a b )% $ 3 1 ' 1 3 ' 2 1 2 ' 2 1 1 2 ' 2 1 ' 3 2 3 ' 3 2 2 3 ' 3 2 2 ' 1 3 1 ' 1 3 & = #a b a b , a b a b , a b a b % "$ 2 3 ' 3 2 3 1 ' 1 3 1 2 ' 2 1 & #a b c a b c a b c a b c , a b c a b c a b c a b c , a b c a b c a b c a b c % $ 3 1 3 ' 1 3 3 ' 1 2 2 + 2 1 2 1 2 1 ' 2 1 1 ' 2 3 3 + 3 2 3 2 3 2 ' 3 2 2 ' 3 1 1 + 1 3 1 & = #a b a b , a b a b , a b a b % "$ 2 3 ' 3 2 3 1 ' 1 3 1 2 ' 2 1 & (a b c ' a b c ' a b c + a b c )(a b ' a b ) + (a b c ' a b c ' a b c + a b c )(a b ' a b ) = 3 1 3 1 3 3 1 2 2 2 1 2 2 3 3 2 1 2 1 2 1 1 2 3 3 3 2 3 3 1 1 3 +(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 u, v, and w are mutually orthogonal, they could be considered to form axes for 3-space in the same way ! ! ! as do i, j, and 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 i, j, and k are the ! ! ! ! ! ! original vectors, then the new vectors are i + j = [1, 1, 0], j + k = [0, 1, 1], and i + k = [1, 0, 1].

! ! ! ! (i + j)!( j + k) = #"1, 1, 0%$ ! #"0, 1, 1%$ = 1 & 0

! ! ! ! Therefore, i + j is not perpendicular to j + k .

MHR • Calculus and Vectors 12 Solutions 795

Chapter 7 Section 6 Question 20 Page 419

Draw a regularA pentagon ABCDE. The interior angles are 540º ÷ 5 = 108º.

B E G H K I J

C D

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. HA DE = AE AD HA 1 = 1 x 1 HA = x

ΔDHE is isosceles, so DE = DH = 1.

Therefore, 1 1+ = x x x2 ! x !1= 0 1± 5 x = (Quadratic formula) 2

1+ 5 Since x must be positive, x = cm. 2

MHR • Calculus and Vectors 12 Solutions 796

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, …

MHR • Calculus and Vectors 12 Solutions 797

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 ! 1 ! Unit vectors collinear with v have the form ± ! v v

! 2 1 " ! 2 1 " The required unit vectors are $# , % and $ , # % . & 5 5 ' & 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 = 3

x = !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$ # %

MHR • Calculus and Vectors 12 Solutions 798

! ! 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]. 2 2 The magnitude of the resultant is 203.82 + 263.36 ! 333.02 . ( ) ( ) " 263.36% The angle of the resultant with the east direction is tan!1 ! 52.3o . #$ 203.82&'

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

Chapter 7 Review Question 5 Page 420

! ! a) u !v = 5(–6) + 2(7) = "16

! ! b) u !v = (–3)(3) + 2(7) = 5

MHR • Calculus and Vectors 12 Solutions 799

! ! c) u !v = 3(4) + 2(–6) = 0

Chapter 7 Review Question 6 Page 420

! ! The vectors in part c) are orthogonal since u !v = 0 .

Chapter 7 Review Question 7 Page 420

! ! a "b cos! = ! ! a b 20 = 5.2(7.3) $ 20 ' ! = cos#1 %& 5.2(7.3)() o " 58.2

Chapter 7 Review Question 8 Page 420

! ! a "b a) cos! = ! ! a b

$6, # 5& " $7, 2& = % ' % ' %$6, # 5'& %$7, 2'& 6(7) + (–5)(2) = 62 + (–5)2 72 + 22 32 = 56.8595 ( 32 + ! = cos#1 )* 56.8595,- o " 55.8

MHR • Calculus and Vectors 12 Solutions 800

!" " p "q b) cos! = !" " p q

$#9, # 4& " $7, # 3& = % ' % ' %$#9, # 4'& %$7, # 3'& –9(7) + (–4)(–3) = (–9)2 + (–4)2 72 + (–3)2 #51 = 75.0067 ( #51 + ! = cos#1 )* 75.0067,- o # 132.8

Chapter 7 Review Question 9 Page 420

! ! a) proj! u u cos v = ! = 56cos 125o " "32.1 The projection has magnitude 32.1 and has direction opposite to v .

! ! ! " u !v % ! b) proj! u ! ! v v = $ ' # v !v &

" )(7, 1+* ! )(9, , 3+* % = $ ' )(9, , 3+* # )(9, , 3+* ! )(9, , 3+*& " 60% = (9, , 3* #$ 90&' ) + " (6, , 2* ) +

MHR • Calculus and Vectors 12 Solutions 801

Chapter 7 Review Question 10 Page 420

!" !" a) 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

Chapter 7 Review Question 12 Page 420

!!!" !!!" !!!" a) 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

MHR • Calculus and Vectors 12 Solutions 802

Chapter 7 Review Question 13 Page 420

! ! ! a) 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]![k, " 4, 5]= 0 6k " 4 + 40 = 0 6k = "36 k = "6

Chapter 7 Review Question 15 Page 420

! ! ! ! a) u ! v = u v (sin")nˆ = 200(350)sin 110o nˆ 65 778.5nˆ "

! ! b) u ! v = $#4, 1, " 3&% ! $#3, 7, 8&% = $#1(8) " 7(–3), – 3(3) " 8(4), 4(7) " 3(1)&% = #29, " 41, 25% $ &

MHR • Calculus and Vectors 12 Solutions 803

Chapter 7 Review Question 16 Page 420

! ! 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 a = [1, 0, 0], b = [1, 1, 1], and c = [0, 0, 1]. ! ! ! 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 .

Chapter 7 Review Question 18 Page 420

! ! "! a) ! = 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.)

MHR • Calculus and Vectors 12 Solutions 804

Chapter 7 Review Question 19 Page 420

! ! ! " u !v % ! proj! 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, ( 8, 9+ #$ 4(4) + (–8)(–8) + 9(9)&' * , (21 = *)4, ( 8, 9,+ 161

! !21 proj! u = "4, ! 8, 9$ v 161 # % 21 = 42 + (!8)2 + 92 161 21 161 = 161 " 1.7

MHR • Calculus and Vectors 12 Solutions 805

Chapter 7 Practice Test

Chapter 7 Practice Test Question 1 Page 422

B is the best answer.

[3, ! 4]"[6, k]= 0 18 ! 4k = 0 4k =18 k = 4.5

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

MHR • Calculus and Vectors 12 Solutions 806

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

! ! u "v b) cos! = ! ! u v 37 = $#8, 3&% $#2, 7&% 37 = 82 + 32 22 + 72 37 = 62.2013 ( 37 + ! = cos'1 )* 62.2013,- o " 53.5

! ! ! " v !u% ! c) proj! v ! ! u u = $ ' # u !u& " 37 % = $ ' )(8, 3+* # )(8, 3+* ! )(8, 3+*& " 37 % = (8, 3* #$ 8(8) + 3(3)&' ) + 37 = (8, 3* 73 ) + " (4.1, 1.5* ) +

MHR • Calculus and Vectors 12 Solutions 807

! ! ! ! ! d) No. The projections are not equal. Proj u is in the direction of v while the direction of proju v is in the ! v 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

MHR • Calculus and Vectors 12 Solutions 808

! ! 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 u ! v = 0 .

! ! b) The vectors in part d are orthogonal since u !v = 0 .

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. z

(0, 1, 4)

(0, 4, 3)

(0, 8, 1) y

x

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 y2 + z2 =100 .

MHR • Calculus and Vectors 12 Solutions 809

Chapter 7 Practice Test Question 12 Page 423

! a) u = 22 + 42 + (–9)2

= 101 ! 1 ! Unit vectors collinear with u have the form ± ! u . u

! ! 2 4 9 " ! 2 4 9 " The two unit vectors collinear with u are $ , ,# % and $# , # , % . & 101 101 101' & 101 101 101'

! ! ! ! b) Choose any vector v that is perpendicular to u . You want a vector that makes the dot product u !v = 0 . ! Choose a convenient vector such as v = [2, !1, 0]. ! v = 22 + (–1)2 + 02

= 5 ! ! 2 1 " ! 2 1 " Therefore two possible unit vectors collinear with u are $ , # , 0% and $# , , 0% . & 5 5 ' & 5 5 '

! ! ! c) The choice of vector v was somewhat arbitrary. You just need the dot product u !v = 0 . ! In general, if v = [v1, v2 , v3 ], then 2v1 + 4v2 ! 9v3 = 0 . For every choice of v1 and v2 , there will be a ! ! (unique) value for v3 . 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. 1 ! ! A = a " b ! 2 1 ! ! = a b sin# 2 1 = (14)(17)sin 24o 2 " 48.4

The area of the triangle is approximately 48.4 square units.

MHR • Calculus and Vectors 12 Solutions 810

Chapter 7 Practice Test Question 14 Page 423

! "! ! "! One vector perpendicular to c and d is 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

! 54 34 33 " ! 54 34 33 " The required unit vectors are $# , # , # % 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

The area is 17 square units.

! ! 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.

MHR • Calculus and Vectors 12 Solutions 811

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.

MHR • Calculus and Vectors 12 Solutions 812

Chapter 7 Practice Test Question 18 Page 423

!" " If you assume the force is parallel to the ramp, then the angle between F and s is 0o . !" " 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

z

[–12, 0, 0]

[0, 450, 95.65] 12º y

x

MHR • Calculus and Vectors 12 Solutions 813

Chapter 7 Practice Test Question 20 Page 423

Draw a diagram of the situation.

120 N

8 8sin 12º 10º 12º

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. o o W!" = (120sin 22 )(8sin 12 ) g # 74.8

The work against gravity is approximately 74.8 J.

o The mechanical work done pulling the cart up the ramp is 120(8)cos 10 or approximately 945.4 J.

Chapter 7 Practice Test Question 21 Page 423

Draw a diagram.

125 N

65º 70 N

40º

In component form, the vectors are !70cos 40o , 70sin 40o # and !125cos 65o , 125sin 65o # . " $ " $ The resultant vector is the vector sum.

MHR • Calculus and Vectors 12 Solutions 814

!" R !70cos 40o 125cos 65o , 70sin 40o 125sin 65o # = " + + $ # "!106.45, 158.28$# !" R = 106.52 +158.32

# 190.7 ' 158.28* % = tan&1 () 106.45+, o # 56.1 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. ! ! ! " u !v % ! proj! u ! ! v v = $ ' # v !v & " (7, 9* ! (6, 2*% ) + ) + 6, 2 = $ ' )( +* # )(6, 2+* ! )(6, 2+*& 60 = (6, 2* 40 ) + = (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 a and b of the parallelogram. ! ! A = a ! b ! ! A = a b sin" 200 = 90(30)sin" 200 sin" = 2700 $ 2 ' " = sin#1 %& 27() o " " 4.2

The interior angles of the parallelogram are 4.2º and 180º – 4.2º = 175.8º.

MHR • Calculus and Vectors 12 Solutions 815

Chapter 7 Practice Test Question 24 Page 423

Use the formula for torque. ! ! "! ! = r F sin" "! 175 = 0.25 F sin 78o "! 175 F = 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 u ! v = 0 .

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.

4 N 5 N 20º

θ θ 20 N

Resolve the forces along the surface of the ramp. 20sin! = 5+ 4cos20o 5+ 4cos20o sin! = 20

o "1 # 5+ 4cos20 & ! = sin % ( $ 20 ' o ! ! 26.0

MHR • Calculus and Vectors 12 Solutions 816

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 u, v, and w . These expressions should be either equal or opposite to each other.

Chapter 7 Practice Test Question 28 Page 423

! ! ! Let a = [a1,a2 ], b = [b1, b2 ], and c = [c1,c2 ]. ! ! ! a) L.S. = (a + b) + c !a ,a # !b , b # !c ,c # = (" 1 2 $ + " 1 2 $) + " 1 2 $ !a b , a b # !c ,c # = " 1 + 1 2 + 2 $ + " 1 2 $ !(a b ) c , (a b ) c # = " 1 + 1 + 1 2 + 2 + 2 $ ! ! ! R.S. = a + (b + c) !a ,a # !b , b # !c ,c # = " 1 2 $ + (" 1 2 $ + " 1 2 $) !a ,a # !b c . b c # = " 1 2 $ + " 1 + 1 2 + 2 $ !a (b c ), a (b c )# = " 1 + 1 + 1 2 + 2 + 2 $ = !(a + b ) + c , (a + b ) + c # Associative property for real numbers " 1 1 1 2 2 2 $

Therefore, L.S. = R.S.

MHR • Calculus and Vectors 12 Solutions 817

! ! b) L.S. = k (a + b) k !a ,a # !b , b # = (" 1 2 $ + " 1 2 $) k !a b , a b # = " 1 + 1 2 + 2 $ !k(a b ), k(a b )# = " 1 + 1 2 + 2 $ !ka kb , ka kb # Distributive property for real numbers = " 1 + 1 2 + 2 $ ! ! R.S. = ka + kb k !a ,a # k !b , b # = " 1 2 $ + " 1 2 $ !ka ,ka # !kb . kb # = " 1 2 $ + " 1 2 $ = !ka + kb , ka + kb # " 1 1 2 2 $

Therefore, L.S. = R.S.

MHR • Calculus and Vectors 12 Solutions 818