1.1 Exercises, Sample Solutions

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1 DM, Chapter, Sample Solutions. Exercises, Sample Solutions 5. Equal vectors have the same magnitude and direction. a) Opposite sides of a parallelogram are parallel and equal in length. AD BC, DC AB b) Opposite sides of a rectangle are parallel and equal in length. The diagonals of a rectangle bisect each other. QT TS, PT TR, SR PQ, SP RQ c) J, L, and K are midpoints of sides AB, AC, and BC, respectively. KJ CL, KJ LA, CL LA, JL BK, JL KC, BK BC, LK AJ, LK JB, AJ JB d) Opposite sides of a regular hexagon are parallel and equal in length. ED AB, CD AF, CB EF, D A, B E, F C, AF E, B CD, ED C, F AB, EF A, D CB, AF B, E CD, ED F, C AB, EF D, A CB 6. Answers may vary. Since equal vectors have the same magnitude and direction, DE EF and AB BC. Since opposite vectors have the same magnitude and are opposite in direction, AB EF and AB DE. 7. X is the midpoint of YZ, so XY XZ. Also, XY and XZ are opposite in direction. Therefore, XY XZ. 8. Answers may vary. a) 50 km/h [north] Scale cm : 5 km/h. b) m/s [095 ] Scale: cm : 4 m/s. Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario

2 DM, Chapter, Sample Solutions c) 500 N [southeast] Scale: cm : 50 N d).5 m/s [335 ] Scale: cm :.5 m/s e) 7 m [70 ] Scale: cm : 3.5 m 9. a) 50 km/h [south] Scale: cm : 5 km/h c) 500 N [northwest] Scale : cm : 50 N b) m/s [75 ] Scale: cm : 4 cm/s d).5 m/s [55 ] Scale : cm :.5 m/s. Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario

3 DM, Chapter, Sample Solutions e) 7 m [90 ] Scale: cm : 3.5 cm 0. a) i) False, since AB and BC are not in the same direction. ii) True, since the sides of a square are equal in length. iii) False, since the magnitude of a vector is non-negative. b) Use the Pythagorean theorem. AC a) Yes, equal vectors have equal magnitudes. b) No, the vectors may not be in the same direction The fractions 6, 9, are equivalent representations of 3 so we can substitute any of these fractions for 3. Similarly, vectors with the same magnitude and direction are equivalent vectors and any one of these vectors can be substituted for another.. Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 3

4 DM, Chapter, Sample Solutions. Exercises, Sample Solutions 4. a) AF + DB AF + FE AE c) FA + EB CF + FD CD e) AF + DE AF + FC AC b) DE + DB AD + DE AE d) DA + EC BD + DF BF f) EC + FD EC + CE EE 0 5. a) b) c) d) e) f) 6. a) KN + NR KR c) MN + MS MR b) RS + KR KR + RS KS d) KM + NK NK + KM NM. Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario

5 DM, Chapter, Sample Solutions e) KN + RS KN + NM KM 7. a) DA DB + BA or DA DC + CA c) CB CD + DB e) DB DA + AB or DB DC + CB f) KR + NM + SK (KR + RS) + SK KS + SK KK 0 b) CD CB + BD or CD CA + AD d) AB AD + DB f) BC BD + DC 8. a) x + 0 x b) If x is a vector with tail at A and head at B, then: x + 0 AB + BB AB (the vector from the tail of AB to the head of BB ) x 9. Arrange u, v, and w head-to-tail. The resultant u + v + w is the vector from the tail of u to the head of w. 0. Add the vectors in the order (a + b ) + c, then in the order a + (b + c ). (a + b ) + c m + c a + (b + c ) a + p n n The same resultant is obtained regardless of the order in which we add the vectors. Hence, vector addition is associative.. Diagrams may vary. Where necessary, arrange the vectors so that they are head-to-tail. The resultant is the vector with the same tail as the first vector and the same head as the last vector. a) WX + XY + YZ WZ b) PQ + RP RP + PQ RQ. Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario

6 DM, Chapter, Sample Solutions c) AB + CA CA + AB CB d) ST + US + VU VU + US + ST VT. The vectors AB, BC, and CA are arranged head-to-tail so AB + BC + CA is the vector from the tail of AB to the head of CA. This is the vector AA, or a) AC + CE + EB + BD + DA The vectors are arranged sequentially from head-to-tail so the sum is the vector from the tail of AC to the head of DA. This is the vector AA, or 0. b) The vectors in part a have the same magnitude since the diagonals of a regular pentagon are equal in length. When the heads of the vectors are joined, 5 congruent isosceles triangles are obtained. Thus, the heads of the vectors are the vertices of a regular pentagon. 4. a) Let r be the resultant velocity of the boat. Draw a vector diagram. By the Pythagorean theorem: r Ñ.7. Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 3

7 DM, Chapter, Sample Solutions The resultant speed of the boat is approximately.7 km/h. b) Let θ be the angle in the figure above. 6 tan θ 0 θ Ñ3 The resultant path of the boat makes an angle of 90 3, or 59 with the shoreline. c) Let x be how far downstream Pierre lands. The triangle in the diagram at the right is similar to the one in part a, so: x x Pierre lands 7 m downstream. 5. In order to travel directly across the river, Pierre must steer slightly upstream, as shown in the diagram at the right. 6 a) sin θ 0 θ Ñ36.9 Pierre should head at an angle of , or 53. to the shore. b) Let r be the resultant velocity of the boat. By the Pythagorean theorem: r The resultant speed of the boat is 8 km/h. From part a, the width of the river is 0 m, or 0. km. The time it takes to cross the river is: width of river Time resultant speed It will take Pierre 0.05 h, or approximately 0.9 minutes to cross the river. 6. To add the vectors, arrange them sequentially from head-to-tail. The resultant is the vector from the tail of the first vector to the head of the last vector. a) KR + NM + MK NM + MK + KR. Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 4

8 DM, Chapter, Sample Solutions NR b) KS + RN + RK (RK + KS) + RN RS + SM RM 7. a) Let OA represent the 8-N force and OB represent the -N force. Complete parallelogram OACB. Then OC represents the resultant force. b) AOB 30 AOBC is a parallelogram, so: OBC In OBC, use the Cosine Law to calculate r. r 8 + (8)()cos 50 r Ñ 8.4 The magnitude of the resultant is approximately 8.4 N. 8. a) Yes. When a and b are in the same direction, a, b, and a + b lie on a straight line, and a + b a + b. This is illustrated in the diagram at the right. b) From part a, a + b a + b when a, b, and a + b are in the same direction. When a, b, and a + b are in opposite directions, a, b, and a + b lie on a straight line, and a + b < a + b. This is illustrated in the diagram at the right. In all other cases, a, b, and a + b form a triangle whose side lengths correspond to the magnitudes of the vectors. The sum of any two sides of a triangle must be greater than the length of the third side, or a triangle cannot be formed. Therefore,. Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 5

9 DM, Chapter, Sample Solutions a + b > a + b, or a + b < a + b. Hence, in general, a + b a + b. c) No. From part b, a + b a + b for any vectors a and b. If a + b > a + b, then a, b, and a + b cannot form a triangle.. Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 6

10 DM, Chapter, Sample Solutions.3 Exercises, Sample Solutions 4. a) b) c) d) e) f) 5. a) i) u + v AB + BC AC iii) u v AB BC BA + CB CD + DA CA ii) u v AB BC AB + CB DC + CB DB iv) v u BC AB BC + BA BC + CD BD b) AC AB + BC u +v Also, AC AD + DC v +u Thus u + v v + u. This illustrates the commutative property..3 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario

11 DM, Chapter, Sample Solutions 6. Answers may vary. a) TQ TR + RQ b) RT RS + ST TR QR RS TS c) PS TS + PT TS TP d) PR TR + PT TR TP 7. Draw the diagonals of the hexagon to locate the center O. Since ABCDEF is a regular hexagon, the triangles in the diagram at the right are equilateral triangles. In ABO, AB AO + OB BC CD In EFO, EF EO + OF FA + DE Therefore: AB BC + CD DE + EF FA BC CD BC + CD DE + FA + DE FA 0 8. a) u + v + w AB + BC + C A c) u v + w DC + CB + BF DF b) u + v w EF + F + C EC b) u v w H + F + FB HB 9. The vectors a + b and a b are diagonals of the parallelogram K OACB formed by a and b. This is illustrated in the diagram at the right. a) If OACB is a rectangle, then the diagonals will be congruent. K Therefore a + b and a b have the same length when a and b are perpendicular. b) Diagonal OC is longer than diagonal AB when BOA is acute. Therefore, a + b is longer than a b when the angle between K a and b is acute. 0. Refer to the diagram and explanation in exercise 9 above. u + v u v when u and v are perpendicular. u + v > u v when the angle between u and v is acute u + v < u v when the angle between u and v is obtuse.3 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario

12 DM, Chapter, Sample Solutions. a) If a and b are in opposite directions, then a b a + b. If a and b are in the same direction, then a b < a + b. If a and b are not collinear, then a, b, and a b form a triangle. Since the sum of any two sides of a triangle must be greater than the length of the third side, a + b > a b, or a b < a + b. Therefore, for any vectors a and b, a b a + b. b) i) ii) The only case that a b a b is when a and b are in the same direction and a b. a b a b is always true. iii) The only case that a b b a is when a and b are in the same direction and a b. iv) a b b a is always true..3 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 3

13 DM, Chapter, Sample Solutions.4 Exercises, Sample Solutions 5. a) AB AE u c) CE CB + BE v u b) AC AB + BC u + v 6. a) OR OP + PR OP + 0.5OQ b) OU OP + PU OP + OQ c) OW OQ + QX + XW OQ + OP + 0.5OQ d) OS OQ + QS OQ + OP OP +.5OQ e) OA OQ + OP f) OY OQ + QY OQ + 3OP 7. a) From page 7, DC 4a b and OF 5a 3b. Since 5a 3b is not a scalar multiple of 4a b, DC and OF are not parallel. b) i) DE OE OD ii) EF OF OE.5a b ( a + 4b ) 5a 3b (.5a b ) 0.5a 6b 6.5a b iii) DF OF OD 5a 3b ( a + 4b ) 6 a 7b 8. a) i) OC a + 4b iii) OE a b b) i) CD OD OC 3a + 3b (a + 4b ) 5a b iii) EF OF OE 4a b ( a b ) 6a ii) OD 3a + 3b iv) OF 4a b ii) DE OE OD a b ( 3a + 3b ) a 5b iv) FC OC OF a + 4b (4a b ) a + 6b.4 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario

14 DM, Chapter, Sample Solutions v) DF OF OD 4a b ( 3a + 3b ) 7 a 5b vi) EC OC OE a + 4b ( a b ) 4a + 6b Answers may vary..4 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario

15 DM, Chapter, Sample Solutions. a) b) The heads of the vectors lie on a straight line. This is the line l in the diagram in part a.. a) AM AB + BM AB + AD b) AN AD + DN AD + AB AB + AD b) From part a, we have: AM AB + AD c AN AB + AD d To write AB in terms of AM and AN, eliminate AD from equations c and d. c: AM AB AD d: AN AB + AD 3 Add: AM + AN AB 4 Therefore, AB AM AN 3 3 To write AD in terms of AM and AN, eliminate AB from equations c and d. c: AM AB + AD d: AN AB AD 3 Add: AM AN AD.4 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 3

16 DM, Chapter, Sample Solutions 4 Therefore, AD AM + AN a) OD OA + OE u +v c) AC AB + BC u +v b) OC OB + OE u + v d) EB EO + OB v + u u v 4. a) OD OE + ED v +u u +v c) OB OC + CB u + v + u 3u + v b) OC OD + DC u +v +u u + v d) AD OE v 5. Use the results of part a to complete part b. OA u b) a) AB AF + FB v +u BC OE v CD OA u DE AB v u EO v OB OA + AB u + v AC AB + BC u + v BD BC + CD v u CE CD + DE v u DO DE + EO v u EA EO + OA v +u.4 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 4

17 DM, Chapter, Sample Solutions 6. a) The diagram below left shows the vectors in exercise 5a arranged tail-to-tail. The diagram below right shows the vectors in exercise 5b. b) In each list, the heads of the vectors are equally spaced on a circle and form the vertices of a regular hexagon. This is because in each list, the vectors have the same magnitude and the angle between any pair of vectors is a) AR u + 7v BQ u + 6v CP 3u + 5v DO 4u + 4v EN 5u + 3v FM 6u + v L 7u + v b) The heads of the vectors lie on a straight line. c) If we visualize u and v forming a parallelogram grid, the heads of the vectors in part a correspond to the points (, 7), (, 6), ( 3, 5), ( 4, 4), ( 5, 3), ( 6, ), ( 7, ) on the grid. In each ordered pair, the second coordinates in 8 more than the first coordinate, so the points lie on a straight line. 8. a) Multiplying by 0: 0 x 0. Multiplying by : x x..4 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 5

18 DM, Chapter, Sample Solutions b) By the definition of scalar multiplication, 0 x 0. The vector x has the same magnitude and direction as x. Therefore, x x. 9. a) In the diagram below, let OA a. Points B and C are constructed on OA extended such that OB is m times as long as OA and BC is n times as long as OA. Therefore, OB ma and BC na. Since OC is (m + n) times as long as OA: OC (m + n)a Also: OC OB + BC ma + na Therefore, (m + n)a ma + na. b) In the diagram below, let OA a. Points B and C are constructed on OA extended such that OB is n times as long as OA and OC is m times as long as OB. Therefore, OB na and OC m(na ). Since OC is collinear with OA and OB, and OC is m times as long as OB and OB is n times as long as OA, OC is mn times as long as OA: OC (mn)a Therefore, m(na ) (mn)a. 0. Consider the diagram below in which DOC is similar to BOA and has sides m times as long (m # 0)..4 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 6

19 DM, Chapter, Sample Solutions Since OD is m times as long as OB and OD is oppositely directed to OB : OD mob m( a + b ) Therefore, m( a + b ) m a + m b, when m # 0. Using thetriangle Law: OD OC + CD ma + mb. a) 3u 3 u 3() 3 b) v v 4 4 () 4 4 c) The diagram for u 3v is below left. u 3v + 3 ()(3)cos 0 u 3v 9.4 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 7

20 DM, Chapter, Sample Solutions d) The diagram for u + 3v is above right. u + 3v + 3 ()(3)cos 0 u 3v 9 e) The diagram for.5u + 0.5v is below left..5u + 0.5v (.5) + (0.5) (.5)(0.5)cos u + 0.5v f) The diagram for u + v is above right. u + v + ()() cos 0 u + v 7. Use the given information to construct the following diagram. Since u and v are collinear, AC CE. Therefore, corresponding angles are equal: ACB AED.4 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 8

21 DM, Chapter, Sample Solutions Because it is a common angle: BAC DAE Since two corresponding angles of ABC and ADE are equal, the third pair of angles are equal. Hence ABC is similar to ADE. Since the corresponding sides of similar triangles are tb sa s t AB BC. Hence,, so. proportional. ma AD DE m n nb.4 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 9

22 DM, Chapter, Sample Solutions.5 Exercises, Sample Solutions 5. a) i) u [3, ] [6, 4] iii) 5 u 5[3, ] [5, 0] ii) 3 u 3[3, ] [9, 6] iv) 4 u 4[3, ] [, 8] b) c) u Therefore, u 3, 3u 3 3, 5u 5 3, and 4u v [4, 3] v a) 3v 3[4, 3] v [4,3] 3 [, 9], c) Since v 5, the vector v or [8, 6] has magnitude 0. b) 4 3 d) Since v 5, the vector 5 v or, has length a) AB [0 4, 3 ] [6, ] BC [6 0, 5 3] [ 4, ] CD [0 6, 3 5] [ 6, ] DA [4 0, 3].5 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario

23 DM, Chapter, Sample Solutions [4, ] b) AB CD 0 BC ( 4) DA 5 c) ABCD is a parallelogram since opposite sides are parallel and equal. 8. a) AB [ ( ), 7 ( )] [, 8] BC [6 ( ), 3 7] [7, 4] CD [5 6, 5 3] [, 8] DA [ 5, ( 5)] [ 7, 4] b) AB + 8 BC 7 + ( 4) CD 65 DA 65 c) ABCD is a rhombus since opposite sides are parallel and all sides are equal in length. 9. a) Form any vectors from the 3 given points. If the vectors are scalar multiples of each other, then the points are collinear. b) i) PQ [ ( 3), 4 ] [5, 3] PR [5 ( 3), 6 ] [8, 5] By inspection, PQ and PR are not scalar multiples of each other. Therefore, P, Q, and R are not collinear. ii) DE [ 5, 5 ] [ 4, 6] DF [ 3 5, ] [ 8, ] By inspection, DF DE. Therefore, D, E, and F are collinear..5 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario

24 DM, Chapter, Sample Solutions 0. a) u [, 4] [, ] c) u + v [, 4] + [3, ] [, 3] e) u + v [, 4] + [3, ] [, 4] + [6, ] [8, 6] b) 4v 4[3, ] [, 4] d) u v [, 4] [3, ] [ 5, 5] f) u 3v [, 4] 3[3, ] [ 4, 8] [9, 3] [ 3, ].. a) a + 3b [5, 3] + 3[, 4] [5, 3] + [6, ] [, 9] b) a 4b [5, 3] 4[, 4] [0, 6] [8, 6] [, ] c) 3a + 5b 3[5, 3] + 5[, 4] [ 5, 9] + [0, 0] [ 5, 9].5 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 3

25 DM, Chapter, Sample Solutions a) u (3i j ) 6i 4 j c) u + v 3i j + i + j 5i j e) 4u v 4(3i j ) (i + j ) i 8 j 4i j 8i 0 j b) 3v 3(i + j ) 6i 3 j d) u v 3i j (i + j ) i 3j f) u + 3v (3i j ) + 3(i + j ) 6i + 4 j + 6i + 3 j 7j 5. a) a + b [4,] + [, 3] [6, 4] a b [4,] [, 3] [, ] b).5 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 4

26 DM, Chapter, Sample Solutions c) a + b a b + ( ) 8 K d) The vectors a + b and a b are diagonals of the parallelogram formed by a and b. Since K the angle between a and b is acute, the diagonal corresponding to a + b is longer than the diagonal corresponding to a b. Thus a + b is greater than a b. K e) If a and b are perpendicular, then a + b would be equal to a b since a + b and a b would be diagonals of a rectangle. This is illustrated in the diagram below left. If K the angle between a and b is obtuse, then a + b would be less than a b as illustrated in the diagram below right. K 6. a) Let w s u + t v for some scalars s and t. [, 8] s[3, 0] + t[, ] [3s t, t] Since the vectors are equal, their components are equal. 3s t c t 8, or t 4 d Substitute t 4 in equation c to determine s. 3s 4 3s 6 s K Therefore, w u + 4 v. b).5 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 5

27 DM, Chapter, Sample Solutions K 7. From exercise 6, w u + 4 v. K a) w u + 4 v K u 4 v + w u v + w b) K w u + 4v K 4 v u + w v u+ w 4 K 8. a) Let w s u + t v for some scalars s and t. [, ] s[, ] + t[, 3] [s t, s + 3t] Since the vectors are equal, their components are equal. s t c s + 3t d Solve equation d for s. s 3t e Substitute s 3t in equation d and solve for t. ( 3t ) t 6t t 7t 4 t Substitute t in equation e and solve for s. s 3( ) 5 K Therefore, w 5 u v. b) K 9. From exercise 8, w 5 u v. K a) w 5 u v K 5u v + w u v+ w 5 5 b) K w 5u v K v 5u w 5 v u w 0. a) b + 3m [0, 5] + 3[, ] [0, 5] + [6, 3] [6, ].5 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 6

28 DM, Chapter, Sample Solutions To obtain the other vectors in the list, subtract m [, ] from the preceding vector on the list. b + m [6, ] [, ] [4, 3] b + m [4, 3] [, ] [, 4] b + 0m [, 4] [, ] [0, 5] b m [0, 5] [, ] [, 6] b m [, 6] [, ] [ 4, 7] b 3m [ 4, 7] [, ] [ 6, 8] b) c) The first vector in the list is b + 3m [6, ]. To obtain the other vectors in the list, repeatedly subtract m [, ]. Therefore, the heads of the vectors lie on a line that passes through the point (6, ) and has slope. d) If b [, 4], then b + 3m [8, ]. Thus the heads of the vectors would lie on a line that passes through (8, ) and has slope. If b [, ], then b + 3m [5, ]. Thus the heads of the vectors would lie on a line that passes through (5, ) and has slope.. a) u + 3v [3, ] + 3[, ] [ 6, ] + [3, 6] [ 3, 8] To obtain the other vectors in the list, add u v [, 3] to the preceding vector on the list..5 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 7

29 DM, Chapter, Sample Solutions b) u + v [ 3, 8] + [, 3] [, 5] v [, 5] + [, 3] [, ] u [, ] + [, 3] [3, ] u v [3, ] + [, 3] [5, 4] 3u v [5, 4] + [, 3] [7, 7] c) The first vector in the list is u + 3v [ 3, 8]. To obtain the other vectors in the list, repeatedly add u v [, 3]. Therefore, the heads of the vectors lie on a line that passes 3 through the point ( 3, 8) and has slope. d) Yes, since we are still repeatedly adding u v to obtain each vector on the list.. a) a + b b + a LS a + b [a, a] + [b, b] [a + b, a + b] [b + a, b + a] LS RS, so a + b b + a b) (a + b ) + c a + (b + c ) LS (a + b ) + c ([a, a] + [b, b]) + [c, c] [a + b, a + b] + [c, c] RS b + a [b, b] + [a, a] [b + a, b + a] RS a + (b + c ) [a, a] + ([b, b] + [c, c]) [a, a] + [b + c, b + c].5 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 8

30 DM, Chapter, Sample Solutions [a + b + c, a + b + c] LS RS, so (a + b ) + c a + (b + c ). [a + b + c, a + b + c] c) s (a + b ) sa + sb LS s (a + b ) s([a, a] + [b, b]) s[a + b, a + b] [s(a + b), s(a + b)] [sa + sb, sa + sb] LS RS, so s (a + b ) sa + sb. RS sa + sb s[a, a] + s[b, b] [sa, sa] + [sb, sb] [sa + sb, sa + sb] d) ( s + t )a sa + ta LS (s + t) a (s + t)[a, a] [(s + t)a, (s + t)a] [sa + ta, sa + ta] LS RS, so ( s + t )a sa + ta. RS sa + ta s[a, a] + t[a,a] [sa, sa] + [ta, ta] [sa + ta, sa + ta] e) Since a + v 0 : v 0 a [0, 0] [a, a] [0 a, 0 a] [ a, a] [a, a] K a 3. Let [a, a] be the required vector. Since the magnitude of the vector is 4: (a ) + a 4 Square both sides and simplify. 4a + a 6 5a 6 6 a 5 4 a± ,, There are two such vectors, and u + v [, ] + [x, 3] [ + x, ].5 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 9

31 DM, Chapter, Sample Solutions Since, u + v 5, u + v 5. ( + x) x + x x + 4x ± 6 4( 7) x 4 ± 84 4 ± ± 5. It is not possible to express a as a linear combination of b and c if c is collinear to b. For example, consider the diagram at the right, where c 3b, or c 0a + 3b. In this situation, is not possible to express a as a linear combination of b and c..5 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 0

32 DM, Chapter, Sample Solutions.6 Exercises, Sample Solutions. Method : Using geometric vectors In the diagram below left, OB represents the 50 N force and OC represents the 35 N force. Complete rectangle OBAC as shown in the diagram below right. Let OA represent the resultant force In OBC, use the Pythagorean Theorem to calculate r. r r Ñ 6.0 N Let θ BOA. Then the bearing of r is 70 θ. 35 tan θ 50 θ Ñ35.0 The bearing of r is The resultant force is 6. 0 N [35 ]. Method : Using Cartesian vectors Place the vectors on a coordinate system and represent each vector algebraically. Let f and f represent the two forces. Let r be the resultant force. f [ 50, 0] f [0, 35] r f + f [ 50, 0] + [0, 35] [ 50, 35] The magnitude of the resultant is: r ( 50) + ( 35) Ñ 6.0 N The direction of the resultant is:.6 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario

33 DM, Chapter, Sample Solutions 35 tan θ 50 θ Ñ35.0 Therefore, the bearing of r is The resultant force is 6. 0 N [35 ]. 3. Method : Using geometric vectors In the diagram below left, OA represents the 0 N force and OB represents the 400 N force. Complete parallelogram OACB as shown in the diagram below right. Let OC represent the resultant force BOA 55 OACB is a parallelogram, so: OAC In OAC, use the Cosine Law to calculate r. r (0)(400)cos 5 r Ñ 556. Let θ be the angle the resultant makes with the 0 N force. Use the Sine Law to determine θ. sin θ sin sin 5 sin θ 556. θ Ñ36. Therefore, the resultant force has a magnitude of 556. N and acts at an angle of 36. to the 0 N force. Method : Using Cartesian vectors Place the vectors on a coordinate system and represent each vector algebraically. Let f and f represent the two forces. Let r be the resultant force..6 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario

34 DM, Chapter, Sample Solutions f [0, 0] f [400 cos 55, 400 sin 55 ] Ñ [9.4, 37.7] r f + f [0, 0]+ [9.4, 37.7] [449.4, 37.7] The magnitude of the resultant is: r (449.4) + (37.7) Ñ 556. Let θ be the angle the resultant makes with the 0 N force tan θ θ Ñ36. The resultant force has a magnitude of 556. N and acts at an angle of 36. to the 0 N force. 4. Method : Using geometric vectors In the diagram below left, OA represents the 0 N force and OB represents the 30 N force. Complete parallelogram OACB as shown in the diagram below right. Let OC represent the resultant force and e represent the equilibriant. BOA 0 OACB is a parallelogram, so: OBC In OCB, use the Cosine Law to calculate r. r (0)(30)cos 60 r Ñ 6.5 Let θ be the angle the resultant makes with the 30 N force. Use the Sine Law to determine θ. sin θ sin sin 60 sin θ 6.5 θ Ñ40.8 The equilibriant is equal in magnitude but opposite in direction to the resultant. Hence, the.6 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 3

35 DM, Chapter, Sample Solutions equilibriant has magnitude 6.5 N and acts at an angle of , or 39. to the 30 N force. Method : Using Cartesian vectors Place the vectors on a coordinate system and represent each vector algebraically. Let f and f represent the two forces. Let r be the resultant force. f [30, 0] f [0 cos 0, 0 sin 0 ] Ñ [ 0, 7.3] r f + f [30, 0] + [ 0, 7.3] [0, 7.3] The magnitude of the resultant is: r (0) + (7.3) Ñ 6.5 Let θ be the angle the resultant makes with the 30 N force. 7.3 tan θ 0 θ Ñ40.8 The equilibriant is equal in magnitude but opposite in direction to the resultant. Hence, the equilibriant has magnitude 6.5 N and acts at an angle of , or 39. to the 30 N force. 5. Method : Using geometric vectors In the diagram below left, OA represents the unknown force and OB represents the 95 N force. Complete parallelogram OACB as shown in the diagram below right. Let OC represent the resultant force BOA 75 OACB is a parallelogram, so:.6 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 4

36 DM, Chapter, Sample Solutions OBC Let θ be the angle the unknown force makes with the resultant. Use the Sine Law to calculate θ. sin θ sin sin 05 sin θ 5 θ Ñ 56.8 Therefore, α Use the Sine Law to determine x, the magnitude of the unknown force. 5 x sin 8. sin 05 5 sin 8. x sin 05 Ñ 7.6 Therefore, the unknown force has a magnitude of 7.6 N and makes an angle of 56.8 with the resultant. Method : Using Cartesian vectors Place the vectors on a coordinate system and represent each vector algebraically. Let f and f represent the two forces. Let r be the resultant force. f [x, 0] f [95 cos 75, 95 sin 75 ] Ñ [50.5, 88.4] r [5 cos θ, 5 sin θ] r f + f [5 cos θ, 5 sin θ] [x, 0] + [50.5, 88.4] [5 cos θ, 5 sin θ] [x , 88.4] Since the vectors are equal, their components are equal. 5 cos θ x c d 5 sin θ 88.4 Solve equation d for θ sinθ 5 θ Ñ 56.8 Substitute θ Ñ 56.8 in equation c and solve for x. 5 cos 56.8 x x 5 cos Ñ Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 5

37 DM, Chapter, Sample Solutions Therefore, the unknown force has a magnitude of 7.6 N and makes an angle of 56.8 with the resultant. 6. Method : Using geometric vectors In the diagram below left, OC represents the 40 N force applied by Paco and OL represents the 30 N force applied by Louis. Complete parallelogram OCDL as shown in the diagram below right. Let r represent the resultant of forces applied by Paco and Louis, and e represent the force applied by Pepe. LOC 35 OCDL is a parallelogram, so: OCD In OCD, use the Cosine Law to calculate r. r (40)(30)cos 45 r Ñ 8.3 Let θ be the angle the resultant makes with the 40 N force. Use the Sine Law to determine θ. sin θ sin sin 45 sin θ 8.3 θ Ñ48.6 Therefore, to keep his brothers efforts in equilibrium, Pepe should exert a force of 8.3 N at an angle of ( ), or 3.4 to the force exerted by Paco. Method : Using Cartesian vectors Place the vectors on a coordinate system and represent each vector algebraically. Let f and f represent the two forces. Let r be the resultant force. f [40, 0] f [30 cos 5, 30 sin 5 ] Ñ [.,.].6 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 6

38 DM, Chapter, Sample Solutions r f + f [40, 0] + [.,.] [8.8,.] The magnitude of the resultant is: r (8.8) + (.) Ñ 8.4 N The direction of the resultant is:. tan θ 8.8 θ Ñ 48.4 Therefore, to keep his brothers efforts in equilibrium, Pepe should exert a force of 8.3 N at an angle of ( ), or 3.4 to the force exerted by Paco. 7. This problem is similar to Example 4, so it is simpler to use geometric vectors. Let T and T represent the forces in the two ropes respectively. Draw a vector diagram and the corresponding triangle diagram. Since the forces are in equilibrium, the resultant of the forces in the two ropes is equal and opposite to the force exerted by the 5 N weight. Use the Sine Law to find T and T. T sin 55 D T sin 66 D 5 sin(35 D + 4 D ) 5 sin 55 D 5 sin 66 D T sin 59 D sin 59 D Ñ 5.0 Ñ 39.8 The tensions in the two ropes are 5. 0 N and 39.8 N respectively. T.6 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 7

39 DM, Chapter, Sample Solutions 8. The simpler approach is to use geometric vectors. The diagram, below left, shows the forces acting on the child. Let m represent the force exerted by the mother and T represent the force in the chain. The weight of the child is the equilibriant of these two forces. Use the triangle diagram, above right. m 00 tan 30 cos T 00 T m 00 tan 30 cos 30 Ñ 5.5 Ñ 57.7 Therefore, the tension in the chain is 5.5 N. The magnitude of the force exerted by the mother is 57.7 N. 9. The simpler approach is to use geometric vectors. Draw a diagram. The forces are in equilibrium. The forces in the wires are along the wires directed downwards. To keep the system in equilibrium, the nail exerts a force of 0 N directed upwards. Let T and T represent the forces in the two wires. Since the picture is hung symmetrically on the wires, T T. Let T T T. Use the Sine Law to determine T. 0 T D sin 0 sin 30 D 0 sin 30 D T sin 0 D Ñ 5.8 N The tension in the wires is 5.8 N..6 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 8

40 DM, Chapter, Sample Solutions 0. Method : Using geometric vectors Let OW represent the velocity of the wind. Let OH represent the velocity of the plane in still air. Complete parallelogram OWRH, where OR represents the velocity of the plane relative to the ground. From the given bearing: WOH 35 Since OHRW is a parallelogram: OHR Use the Cosine Law to calculate r. r (900)(00)cos 45 Ñ 83.3 Let ROH θ. The bearing of the plane is 35 θ. sin θ sin sin 45 sin θ 83.3 θ Ñ4.9 Hence the bearing is The plane s speed relative to the ground is 83.3 km/h on a bearing of 30.. Method : Using Cartesian vectors The diagram at the right represents the situation. A bearing of 35 corresponds to a direction angle of 35. Let a represent the velocity of the plane in still air. Let w represent the velocity of the wind. Let r represent the velocity of the plane relative to the ground..6 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 9

41 DM, Chapter, Sample Solutions a [900 cos 35, 900 sin 35 ] Ñ[636.4, 636.4] w [0, 00] r a + w [636.4, 636.4] + [0, 00] Ñ[636.4, 536.4] The magnitude of the resultant is: r (636.4) + ( 536.4) Ñ 83.3 The bearing of the plane is 90 + θ tan θ θ Ñ 40. Hence the bearing is The plane s speed relative to the ground is 83.3 km/h on a bearing of Method : Using geometric vectors Let OW represent the velocity of the wind. Let OH represent the velocity of the plane in still air. Complete parallelogram OWRH, where OR represents the velocity of the plane relative to the ground. From the given bearings: WOH Since OHRW is a parallelogram: OWR Use the Cosine Law to calculate r. r (600)(80) cos 97 Ñ 64.9 Let ROH θ. The bearing of the plane is 30 + θ..6 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 0

42 DM, Chapter, Sample Solutions sin θ sin sin 97 sin θ 64.9 θ Ñ7.4 Hence the bearing is The plane s speed relative to the ground is 64.9 km/h on a bearing of Method : Using Cartesian vectors The diagram at the right represents the situation. Bearings of 030 and 3 correspond to direction angles of 60 and 337 respectively. Let Let Let a represent the velocity of the plane in still air. w represent the velocity of the wind. r represent the velocity of the plane relative to the ground. a [600 cos 60, 600 sin 60 ] Ñ[300, 59.6] w [80 cos 337, 80 sin 337 ] Ñ[73.6, 3.3] r a + w [300, 59.6] + [73.6, 3.3] Ñ[373.6, 488.3] The magnitude of the resultant is: r (373.6) + (488.3) Ñ 64.9 The bearing of the plane is 90 θ tan θ θ Ñ 5.6 Hence the bearing is The plane s speed relative to the ground is 64.9 km/h on a bearing of The simpler approach is to use geometric vectors. Let OW represent the velocity of the wind. Let OR represent the velocity of the plane relative to the ground. Complete parallelogram OWRH, where OH represents the velocity of the plane in still air..6 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario

43 DM, Chapter, Sample Solutions From the given bearings: ROW a) Let HOR θ. Use the Sine Law on ORW to calculate θ. sin θ sin sin 50 sin θ 550 θ Ñ6.4 The heading the pilot should take is b) HOW Since OWRH is a parallelogram: OHR Use the Sine Law to calculate the speed of the plane relative to the ground. r 80 sin 6.4 sin sin 3.6 r sin 6.4 Ñ The speed of the plane relative to the ground is km/h. distance c) Time speed Ñ 0.85 The trip will take approximately 0.85 h, or approximately 5 min..6 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario

44 DM, Chapter, Sample Solutions 3. Since a distance of 80 km was flown in 0 minutes, the speed of the plane relative to the 80 km, or 40 km/h. ground was h 3 Draw a diagram. Let OW represent the velocity of the wind. Let OH represent the velocity of the plane in still air. Let OR represent the velocity of the plane relative to the ground Complete parallelogram OWRH. OHR is a right triangle, so: w tan 0 40 w 40 tan 0 Ñ 4.3 The speed of the wind is 4.3 km/h..6 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 3

45 DM, Chapter, Sample Solutions.7 Exercises, Sample Solutions u v 6. Use the formula cosθ. uv [0, 4] [5,] a) cosθ θ Ñ ( ) ( ) θ Ñ 50.3 [4, ] [, 5] c) cosθ [3, ] [, ] b) cosθ 4 + ( ) ( ) + ( 5) θ Ñ 97.8 [6, 3] [, 4] d) cosθ ( 4) θ u v is positive so the angle between the vectors is acute. 8. a) AB [ ( ), 0] [, ] AC [ ( ), 4 0] [, 4] BC [ ( ), 4 ] [3, 3] cos A AB AC cos B AB AC [,] [, 4] ( ) + () A Ñ BA BC BA BC [, ] [3, 3] + ( ) B 90.7 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario

46 DM, Chapter, Sample Solutions C 80 A B b) PQ [8, 3 6] [6, 3] PR [ 4, 0 6] [ 6, 6] QR [ 4 8, 0 3] [, 3] cos P PQ PR PQ PR cos Q [6, 3] [ 6, 6] 6 + ( 3) ( 6) + ( 6) P Ñ 08.4 QP QR QP QR [ 6, 3] [, 3] ( 6) + ( 3) ( ) + ( 3) Q Ñ 40.6 R 80 P Q a) See diagram below. b) Quadrilateral ABCD Use the diagram in part a. AB [9, 3], DC [9, 3], BC [, 3], AD [, 3] AB AD [9, 3] [, 3] Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario

47 DM, Chapter, Sample Solutions BA BC [ 9, 3] [, 3] DA DC [, 3] [9, 3] CD CB [ 9, 3] [, 3] Therefore, ABCD is a rectangle. Quadrilateral PQRS Use the diagram in part a. PQ [, 4], SR [, 4], QR [, 7], PS [, 7] Since opposite sides are equal, PQRS is a parallelogram. PQ PS [, 4] [, 7] Therefore, P 90. Hence PQRS is a not rectangle. 0. The vertices of PQR are P( 3, ), Q(, 4), and R(5, ). PQ [ ( 3), 4 ( )] [, 6] PR [5 ( 3), ( )] [8, 4] QR [5 ( ), 4] [6, ] ii) QP QR [, 6] [6, ] a) i) PQ PR [, 6] [8, 4] iii) RQ RP [ 6, ] [ 8, 4] b) The geometric definition of the dot product is a b a b cosθ, so: PQ PR PQ PR cos P RQ RP RQ RP cos R From the diagram, PQR is isosceles with PQ RQ and P R. Therefore, PQ PR and RQ RP are equal..7 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 3

48 DM, Chapter, Sample Solutions c) From the geometric definition of the dot product: PQ PR PQ PR cos P QP QR QP QR cos Q RQ RP RQ RP cos R If PQR is equilateral, PQ RQ PR and P Q R. Therefore, PQ PR, QP QR, and RQ RP will be equal.. The vertices of parallelogram ABCD are A(, 0), B(6, ), C(8, 4), and D(4, 3). AB [6, 0] [4, ] Since ABCD is a parallelogram, DC AB [4, ]. AD [4, 3 0] [, 3] Since ABCD is a parallelogram, BC AD [, 3]. a) i) AB AD [4, ] [, 3] 8+3 iii) BA BC [ 4, ] [, 3] 8 3 ii) CB CD [, 3] [ 4, ] 8+3 iv) DA DC [, 3] [4, ] 8 3 b) There are two relationships: AB AD CB CD and BA BC DA DC. AB AD BA BC and CB CD DA DC. c) The geometric definition of a dot product is a b a b cosθ. From part b, we have AB AD CB CD and BA BC DA DC. AB AD AB AD cos A CB CD CB CD cos C In any parallelogram ABCD, opposite sides are equal so AB DC and AD BC. Opposite angles are also equal, so A C. Hence AB AD and CB CD are equal. Similarly, BA BC and DA DC are equal. From part b, we also have AB AD BA BC AB AD AB AD cos A BA BC BA BC cos B c Adjacent angles in a parallelogram are supplementary, so B 80 A. Also opposite sides are equal, so AD BC. Clearly AB BA. Therefore equation c becomes:.7 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 4

49 DM, Chapter, Sample Solutions BA BC AB AD cos (80 A) AB BC ( cos A) AB AD cos A Hence AB AD BA BC. Similarly, CB CD DA DC. d) If the parallelogram is a rectangle, AB AD CB CD BA BC DA DC 0.. a) Three vectors can be formed from the 3 points. If the dot product of any pair of vectors is 0, then the points form a right angle. b) i) AB [ ( 5), 5] [3, 4] AC [7 ( 5), 8 5] [, 3] BC [7 ( ), 8 ] [9, 7] AB AC [3, 4] [, 3] 36 0 AB BC [3, 4] [9, 7] AC BC [, 3] [9, 7] 08 0 Therefore, A, B, and C do not form a right angle. ii) JK [5 ( 3), 0 ( 4)] [8, 4] JL [ ( 3), 6 ( 4)] [5, 0] KL [ 5, 6 0] [ 3, 6] JK JL [8, 4] [5, 0] JL KL [5, 0] [ 3, 6] JK KL [8, 4] [ 3, 6] Therefore, J, K, and L form a right angle..7 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 5

50 DM, Chapter, Sample Solutions 3. a) Form any vectors from the 3 given points, and find the angle between the vectors. If the angle is 0 or 80, the points are collinear. b) i) DE [ ( 4), 3 7] [6, 4] DF [8 ( 4), 7] [, 8] cos D DE DF DE DF [6, 4] [, 8] 6 + ( 4) + ( 8) D 0 Therefore, D, E, and F are collinear. ii) RS [4 7, ] [ 3, ] RT [ 4 7, ] [, 4] cos R RS RT RS RT [ 3, ] [, 4] ( 3) + ( ) ( ) + ( 4) D Ñ.5 Therefore, R, S, and T are not collinear. 4. Suppose the square is ABCD and AB [5, ]. Since BC is perpendicular to AB and AB has 5 slope, BC has slope. Furthermore 5 AB BC, so BC [, 5] or [, 5]. Thus the.7 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 6

51 DM, Chapter, Sample Solutions other three sides of the square are either [5, ], [, 5], [, 5] or [5, ], [, 5], [, 5]. 5. There are cases to consider. Case. [4, ] represents the width of the rectangle. The slope of the line segment representing the width is, so the slope of the line segment 4 4. Since the length is double the width, the vector representing the length is either [4, 8] or [ 4, 8]. Therefore, vectors that could represent the other 3 sides of the rectangle are [4, ], [4, 8], [4, 8] or [4, ], [ 4, 8], [ 4, 8]. Case. [4, ] represents the length of the rectangle. Since the length is double the width, the vector representing the width is either [, ] or [, ]. Therefore, vectors that could represent the other 3 sides of the rectangle are [4, ], [, ], [, ] or [4, ], [, ], [, ]. representing the length is 6. a) In OBC: OC cos θ OB OC OB cos θ b cos θ Area of rectangle OAED (OA)(OD) (OA)(OC) since OD OC a b cosθ a b b) When θ 90, rectangle OAED cannot be constructed. When θ 0, a and b are parallel, and the area of the rectangle OAED is a b. c) When θ 45, BC OC OD, and so D, B, and E are collinear. 7. a) See diagram for part c. ii) (v w)u ([, ] [, ]) u b) i) (u v ) w ([, 0] [, ]) w (4 + 0) w ( + ) u 4 w or [4, 8] 4 u or [8, 0] K iii) ( w u )v ([, ] [, 0]) v ( + 0) v v or [4, ] c) The expressions in part b are scalar multiples of u, v, and w..7 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 7

52 DM, Chapter, Sample Solutions 8. If a and b are perpendicular, then a b 0. a) [k, ] [, ] 0 k 4 0 b) [ 3, 4] [5, k] k 0 5 k 4 k 4 9. a) a (b + c ) [, 3] ([, 4] + [5, ]) b) (a + b ) c ([, 3] + [, 4]) [5, ] [, 3] [4, 6] [, ] [5, ] c) (a + b ) (a + c ) ([, 3] + [, 4]) ([, 3] + [5, ]) [, ] [7, ] 7 6 d) (a + 3b ) (5a b ) ([, 3] + 3[, 4]) (5[, 3] [, 4]) [, 6] [, 3] a) The magnitude of the force in the direction of the displacement is the horizontal component of the force, F cosθ. b) The work done is the product of the magnitude of the displacement and the magnitude of the force in the direction of the displacement. Thus: Work d F cosθ F d cosθ F d. Work F d F d cosθ 30(00)cos 30 Ñ 598. The work done in moving the wagon 00 m is approximately 598. joules..7 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 8

53 DM, Chapter, Sample Solutions. 3. a) To draw a different rectangle, construct C so that OCA 90. Then construct D so that OC OD. b) i) a) In OAC: OC cos θ OA OC OA cos θ a cosθ Area of rectangle OBED (OB)(OD) (OB)(OC) since OD OC b a cosθ a b ii) When θ 90 or θ 0, OAC and rectangle OBED cannot be constructed. iii) When θ 45, DA CA BE, and so D, A, and E are collinear..7 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 9

54 DM, Chapter, Sample Solutions.8 Exercises, Sample Solutions. a) a (b + c ) a b + a c c) u (u + v ) u u + u v u + u v b) a (a + b ) a a + a b a + a b d) 3u (u 3v ) 6u u 9u v 6 u 9u v. a) (a + b ) (a b ) (a + b ) a (a + b ) b a a + b a a b b b a a + a b a b b b a b b) (a b ) (a + b ) (a b ) a + (a b ) b a a b a + a b b b a a a b + a b b b a + a b b c) (4a + b ) (a + b ) (4a + b ) a + (4a + b ) b 4a a + b a + 8a b + b b 4a a + a b + 8a b + b b 4 a + 9a b + b d) (a + 3b ) (3a b ) (a + 3b ) 3a (a + 3b ) b 6a a + 9b a 4a b 6b b 6a a + 9a b 4a b 6b b 6 a + 5a b 6 b 3. a) The dot product does not satisfy the associative law because the expressions (a b ) c and a (b c ) have no meaning. ( a b ) c has no meaning because a b is a scalar and we cannot take its dot product with vector c. a (b c ) also has no meaning because it too involves the dot product of vector ( a ) with a scalar ( b c )..8 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario

55 DM, Chapter, Sample Solutions b) No meaning can be given to the expression a b c since (a b ) c and a (b c ) have no meaning. 4. Proof that a 0 0 eometric proof a 0 a 0 cosθ 0 Cartesian proof Let a [a, a ] a 0 [a, a ] [0, 0] a (0) + a (0) 0 Proof that a u a, where u is a unit vector in the direction of a If u is a unit vector in the direction of a, then u a. a eometric proof Cartesian proof a u a a Let a [a, a ] a a a a u a a a a a a a a a a [a, a ] [a, a ] a + a a + a a + a a + a a 5. a) If a c b c then: a c b c 0 c (a b ) 0 Two vectors have a dot product of 0 if one of the vectors is 0 or if the vectors are perpendicular. Therefore, there are three possibilities: c0 a b 0 or a b b (a c ) Hence if a c b c, it does not necessarily follow that a b..8 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario

56 DM, Chapter, Sample Solutions Less formally, we can also argue as follows: Suppose a, b, and c are the side vectors of an equilateral triangle. Since the magnitudes of the vectors are equal and the angle between the vectors are equal, a c b c. However, a b. ac a c a for real numbers so it cannot be written as. If b) is not the same as bc b c b K θ is the angle between a and c and θ is the angle between b and c, then a c cosθ a c b c b c cosθ a cosθ b cosθ AC AB + BC 6. a) OB OC + CB OC + OA OC OA c+a c a b) If (c + a ) (c a ) 0, then OB AC 0. Therefore, diagonals OB and OC are perpendicular, so parallelogram OABC is a rhombus. 7. These properties are common to dot products and the products of real numbers. Real numbers Dotproducts ab ba a b b a a (b + c ) a b + a c a(b + c) ab + ac aa a a a a (ka)b a(kb) k(ab) (ka ) b a kb k (a b ) K a(0) 0 a 0 0 These are properties of real numbers that do not have corresponding properties for dot products: (ab)c a(bc), a() a. 8. a) (a + b) (a + b ) (a + b ) a + (a + b ) b a a + b a + a b + b b a a + a b + a b + b b a a + a b + b b (Distributive property) (Commutative property).8 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 3

57 DM, Chapter, Sample Solutions b) (a + b) (a + b ) a a + a b + b b a + a b cosθ + b a + b a + a b + b a + a b cosθ + b c) a + b a + a b cosθ + b If a and b are perpendicular, then cos θ 0 and a b cosθ 0. Therefore, a + b a + b. This is the Pythagorean theorem. a a 9. a a a a a a a a a This is reasonable as illustrated in the diagram at the right. a b 0. a) a b b b b [6, 4] [8, 4] [8, 4] [8, 4] [8, 4] b a b) b a a a a [8, 4] [6, 4] [6, 4] [6, 4] [6, 4] [8, 4] [6, 4] 3 80 [8, 4] 3 5 [6, 4] 5 [8, 4] 8 3 [6, 4] 6 8, , c) a b, 5 5 [3.,.6] 48 3 b a, 3 3 Ñ [3.7,.5].8 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 4

58 DM, Chapter, Sample Solutions d) Since a b b a, vector projection is not commutative. a b. Use the formula a b b. b b [3,0] [, 3] [, 3] a) a b [, 3] [, 3] [,3] 6 3 [, 3] 8, 3 3 [4, 5] [ 5, 4] [ 5, 4] b) a b [ 5, 4] [ 5, 4] [ 5, 4] 0[8, 4] [0, 0] or 0 [ 4, ] [3, ] [3, ] c) a b [ 3, ] [ 3, ] 9 + [3, ] 4 0 [3, ] 7 5 [3, ] 7, Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 5

59 DM, Chapter, Sample Solutions [, 3] [6, ] [6, ] d) a b [ 6, ] [ 6, ] [6, ] 6 40 [6, ] 3 0 [6, ] i+ j, or PQ [ ( 4), 6 0] [3, 6] PR [3 ( 4), 4 0] [7, 4] QR [3 ( ), 4 6] [4, ] a) See diagram in part b. PR PQ PQ b) i) PR PQ PQ PQ [7, 4] [3, 6] [3, 6] [3, 6] [3, 6] [3, 6] [3, 6] [3, 6] PQ PQ PR PR iii) PQ PR PR PR [3,6] [7, 4] [7, 4] [ 7, 4 ] [ 7, 4 ] RP RQ RQ ii) RP RQ RQ RQ [ 7, 4] [ 4, ] [ 4, ] [ 4,] [ 4, ] [ 4, ] 0 0 [ 4, ] [ 4, ] RQ QR PR PR ii) QR PR PR PR [4, ] [7, 4] [7, 4] [ 7, 4 ] [ 7, 4 ] [7, 4] [7, 4] [7, 4] 0 65 [7, 4].8 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 6

60 DM, Chapter, Sample Solutions 4 3 [7, 4] 9 3 [7, 4] 9 PR 3 4 PR PR + PR 3 3 PR c) PQ PR + QR PR a b 3. Use the formula a b b b b a b cosθ b D (4)(7) cos 60 b a b b 7 b Ñ 0.3 b u v 4. Use the formula u v v v v u v cosθ v (8)() cos 35 D v u v b v.8 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 7

61 DM, Chapter, Sample Solutions 44 v 4 v Ñ 0.5 v a b 5. a b is a scalar multiple of b, where the scalar is. b b a) Therefore, a b 0 when a b 0, that is, when a is perpendicular to b.this is illustrated in the diagram, below left. b) Therefore, a b is undefined when b b 0. This occurs when b is the zero vector. This is illustrated on the diagram, above right. 6. a) (a b ) b Since a b is a scalar multiple of b, it is collinear to b. Therefore, projecting a b perpendicularly on b gives a b. b) b (a b ) Use the diagram in part a. Since a b is a scalar multiple of b, it is collinear to b. Therefore, projecting b perpendicularly on a b gives b. c) (a b ) a (a b ) a is the projection of a b on a. It is a scalar multiple of a, k a, as shown in the diagram at the right. d) a (a b ) Use the diagram in part a. a (a b ) is the projection of a on a b. It is a b as illustrated in the diagram..8 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 8

62 DM, Chapter, Sample Solutions 7. a) It is possible to have a b b a when a and b are perpendicular. Then K a b b a 0. When a b, clearly a b b a. b) Since b c is collinear to c, projecting a perpendicularly on c or on b c produces the same vector. This vector is a c. b 8. Let u [a, b]. Then u lies on a line with slope. Since v is perpendicular to u, it lies on a a a line with slope. If v is twice as long as u, then v [ b, a] or v [b, a]. Hence b there are cases to consider. Case. v [ b, a] Since the sum of u and v is [6, 8]: [a, b] + [ b, a] [6, 8] [a b, b + a] [6, 8] Since the vectors are equal, their components are equal. a b 6 c a + b 8 d Eliminate b. Eliminate a. c: a + 4b c: a b 6 d: a + b 8 d: 4a + b 6 Add: 5b 4 Add: 5a 4 b a Thus u, and v, Case. v [b, a] Since the sum of u and v is [6, 8]: [a, b] + [b, a] [6, 8] [a + b, b a] [6, 8] Since the vectors are equal, their components are equal. a + b 6 c a + b 8 d Eliminate b. Eliminate a. c: a + 4b c: a + b 6 d: a + b 8 d: 4a b 6 Add: 5b 0 Add: 5a 0 b4 a Thus u [, 4] and v [8, 4]..8 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 9

63 DM, Chapter, Sample Solutions 9. a) 4 a + b 4 a b 4 (a + b ) (a + b ) 4 (a b ) (a b ) 4 (a a + a b + b b ) 4 (a a a b + b b ) 4 a + a b + 4 b 4 a + a b 4 b K a b b) A similar equation for the product xy in algebra is xy 4 (x + y) 4 (x y) 4 (x + y) + 4 (x y) 4 (x + xy + y ) 4 (x xy + y ) 4 x + xy + 4 y 4 x + xy 4 y xy.8 Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 0

64 DM, Chapter, Sample Solutions Chapter Review Exercises, Sample Solutions. Three examples of scalar quantities are distance, speed, and mass. Three examples of vector quantities are displacement, velocity, and weight.. Answers may vary. a) 0 N [east] Scale : cm : 0 N b) 4 m/s [35 ] Scale: cm : 8 m/s 3. Answers may vary. a) Equal vectors have the same magnitude and direction. AO OC, DO OB, DC AB, AD BC b) Opposite vectors have the same magnitude but are in opposite directions. AO CO, DO BO, DC BA, AD CB 4. a) HE HA + AE c) D DH + H b) F C + CF d) DC D + C 5. a) P + PR PQ b) RA + RQ RB c) CD + RS EF CD + RS + FE d) DR + QB FS DR + QB + SF (CD + DE) + FE DR + DR + DR CE + FE C + CE E 4 DR 6. AB + BC + CD + DA AA 0 Chapter Review Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario

65 DM, Chapter, Sample Solutions Answers may vary. a) FD FA + AD AF + AD AD AF b) EB ED + DB DE + DB DB DE c) CB CA + AB AC + AB AB AC b) AE AB + BE BA + BE BE BA 9. Answers may vary. 0. a) AC ED u b) AD AC + CD ED + CD u + v c) EA EB + BA CD DE v u. C CD + D CD + CB. a) i) 3u 3[, ] [ 3, 6] CM CB + BM CB + CD ii) u [, ] [, 4] Chapter Review Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario

66 DM, Chapter, Sample Solutions iii) u [, ] [, ] b) u iv) 4u 4[, ] [4, 8] ( ) + 5 Therefore, 3u 3 5, u 5, u 5, and 4u a) AB [7 3, 4 ] [4, ] BC [ 7, 0 4] [ 8, 6] CA [3 ( ), 0] [4, 8] b) AB 4 + BC ( 8) + 6 CA 4 + ( 8) BC AB + CA since Therefore, ABC is a scalene, right triangle. 4. a) K Let w s u + t v for some scalars s and t. [4, 8] s[0, ] + t[, 3] [t, s 3t] Since the vectors are equal, their components are equal. t4 c s 3t 8 d Substitute t 4 in equation d to determine s. s 3(4) 8 s 0 s 0 K Therefore, w 0 u + 4 v. b) Chapter Review Exercises, Sample Solutions Copyright 003 Pearson Education Canada Inc., Toronto, Ontario 3

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