1.1 Exercises, Sample Solutions - PBworksmackenziekim.pbworks.com/w/file/fetch/63434086/chap1 soln.pdf · ... SR = PQ, SP = RQ c) J, L, and K are midpoints of sides AB, ... 13. a)

GDM, Chapter 1, Sample Solutions

1.1 Exercises, Sample Solutions Copyright 2003 Pearson Education Canada Inc., Toronto, Ontario 1

1.1 Exercises, Sample Solutions 5. Equal vectors have the same magnitude and direction.

a) Opposite sides of a parallelogram are parallel and equal in length. ABDC,BCAD == b) Opposite sides of a rectangle are parallel and equal in length. The diagonals of a rectangle bisect each other. RQSP,PQSR,TRPT,TSQT ==== c) J, L, and K are midpoints of sides AB, AC, and BC, respectively. JBAJ,JBLK,AJLK,BCBK,KCJL,BKJL,LACL,LAKJ,CLKJ ========= d) Opposite sides of a regular hexagon are parallel and equal in length. ,CDBG,GEAF,GCFG,GEBG,GADG,EFCB,AFCD,ABED ========

,ABGC,FGED,CDGE,BGAF,CBDG,GAEF,ABFG,GCED ========

CBGA,DGEF ==

6. Answers may vary. Since equal vectors have the same magnitude and direction, BCABandEFDE == .

Since opposite vectors have the same magnitude and are opposite in direction, EFAB −=

and DEAB −= .

7. X is the midpoint of YZ, so XY = XZ. Also, XZandXY are opposite in direction. Therefore,

XZXY −= .

8. Answers may vary. a) 50 km/h [north] b) 12 m/s [095°] Scale 1 cm : 25 km/h. Scale: 1 cm : 4 m/s



c) 500 N [southeast] d) 2.5 m/s2 [335°] Scale: 1 cm : 250 N Scale: 1 cm : 1.25 m/s2

e) 7 m [270°] Scale: 1 cm : 3.5 m

9. a) 50 km/h [south] b) 12 m/s [275°] Scale: 1 cm : 25 km/h Scale: 1 cm : 4 cm/s

c) 500 N [northwest] d) 2.5 m/s2 [155°] Scale : 1 cm : 250 N Scale : 1 cm : 1.25 m/s2



e) 7 m [90°] Scale: 1 cm : 3.5 cm

10. a) i) False, since BCandAB 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 = 32 + 32

= 18 = 3 2

11. a) Yes, equal vectors have equal magnitudes. b) No, the vectors may not be in the same direction.

12. The fractions 46 ,

69 , … are equivalent representations of

23 so we can substitute any of these

fractions for 23 . Similarly, vectors with the same magnitude and direction are equivalent

vectors and any one of these vectors can be substituted for another.



1.2 Exercises, Sample Solutions 4. a) FEAFDBAF +=+ b) DEADDBDE +=+

AE= AE= c) FDCFEBFA +=+ d) DFBDECDA +=+ CD= BF= e) FCAFDEAF +=+ f) CEECFDEC +=+ AC= EE=

0= 5.

6. a) KRNRKN =+ b) RSKRKRRS +=+ KS= c) MRMSMN =+ d) KMNKNKKM +=+ NM=

a) b)

c) d)

e) f)



e) NMKNRSKN +=+ f) SK)RSKR(SKNMKR ++=++

KM= SKKS += KK= 0=

7. a) BADBDA += or CADCDA += b) BDCBCD += or ADCACD += c) DBCDCB += d) DBADAB += e) ABDADB += or CBDCDB += f) DCBDBC +=

8. a) xx =+ 0 b) If x is a vector with tail at A and head at B, then: BBAB0 +=+x = AB (the vector from the tail of AB to the head of BB) = x

9. Arrange vu, , and w head-to-tail. The resultant wvu ++ is the vector from the tail of u to the head of w .

10. Add the vectors in the order cba ++ )( , then in the order )( cba ++ .

cmcba +=++ )( pacba +=++ )( = n = n The same resultant is obtained regardless of the order in which we add the vectors. Hence, vector addition is associative.

11. 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) WZYZXYWX =++ b) PQRPRPPQ +=+

RQ=



c) ABCACAAB +=+ d) STUSVUVUUSST ++=++ CB= VT=

12. The vectors CAand,BC,AB are arranged head-to-tail so

CABCAB ++ is the vector from the tail of AB to the head of CA . This is the vector AA , or 0 .

13. a) DABDEBCEAC ++++ 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.

14. a) Let r be the resultant velocity of the boat. Draw a vector diagram. By the Pythagorean theorem: 22 610 +=r

= 136 Ñ11.7



The resultant speed of the boat is approximately 11.7 km/h. b) Let θ be the angle in the figure above.

tan θ = 610

θ Ñ31° The resultant path of the boat makes an angle of 90° − 31°, 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

120 = 106

x = 106120×

= 72 Pierre lands 72 m downstream.

15. In order to travel directly across the river, Pierre must steer slightly upstream, as shown in the diagram at the right.

a) sin θ = 610

θ Ñ36.9° Pierre should head at an angle of 90° − 36.9° , or 53.1° to the shore. b) Let r be the resultant velocity of the boat. By the Pythagorean theorem: 22 610 −=r

= 64 = 8 The resultant speed of the boat is 8 km/h. From part a, the width of the river is 120 m, or 0.12 km. The time it takes to cross the river is:

Time = width of riverresultant speed

812.0=

= 0.015 It will take Pierre 0.015 h, or approximately 0.9 minutes to cross the river.

16. 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) KRMKNMMKNMKR ++=++



NR= b) RN)KSRK(RKRNKS ++=++

SMRS += RM=

17. a) Let OA represent the 8-N force and OB represent the 11-N force. Complete parallelogram OACB. Then OC represents the resultant force.

b) ∠AOB = 30° AOBC is a parallelogram, so: ∠OBC = 180° − 30° = 150° In ∆OBC, use the Cosine Law to calculate r .

r 2 = 82 + 112 − 2(8)(11)cos 150°

r Ñ 18.4 The magnitude of the resultant is approximately 18.4 N. 18. a) Yes. When a and b are in the same direction, baba +and,,

lie on a straight line, and .baba +=+ This is illustrated in

the diagram at the right. b) From part a, baba +=+ when baba +and,, are in the same

direction. When baba +and,, are in opposite directions, baba +and,, lie

on a straight line, and baba +<+ . This is illustrated in the diagram

at the right. In all other cases, baba +and,, 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,



baba +>+ , or baba +<+ .

Hence, in general, baba +≤+ .

c) No. From part b, baba +≤+ for any vectors ba and .

If baba +>+ , then baba +and,, cannot form a triangle.



1.3 Exercises, Sample Solutions 4.

5. a) i) BCAB +=+ vu ii) BCAB −=− vu AC= CBAB += CBDC += DB= iii) BCAB −−=−− vu iv) ABBC −=− uv CBBA += BABC += DACD += CDBC += CA= BD= b) BCABAC += vu += Also, DCADAC += uv += Thus uvvu +=+ . This illustrates the commutative property.

a) b)

c) d)

e) f)



6. Answers may vary. a) RQTRTQ += b) STRSRT +=

QRTR −= TSRS −= c) PTTSPS += d) PTTRPR += TPTS −= TPTR −=

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, OBAOAB += CDBC −= In ∆EFO, OFEOEF += DEFA += Therefore: FAEFDECDBCAB −+−+− FADEFADECDBCCDBC −++−+−−= 0=

8. a) CGBCAB ++=++ wvu b) GCFGEF ++=−+ wvu AG= EC= c) BFCBDC ++=+− wvu b) FBGFHG ++=−− wvu DF= HB=

9. The vectors ba + and ba − are diagonals of the parallelogram 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. Therefore ba + and ba − have the same length when a and b are perpendicular. b) Diagonal OC is longer than diagonal AB when ∠BOA is acute. Therefore, ba + is longer than ba − when the angle between a and b is acute.

10. Refer to the diagram and explanation in exercise 9 above. • vu + = vu − when u and v are perpendicular.

• vu + > vu − when the angle between u and v is acute

• vu + < vu − when the angle between u and v is obtuse



11. a) If a and b are in opposite directions, then .baba +=−

If a and b are in the same direction, then .baba +<−

If a and b are not collinear, then a , b , and ba − form a triangle. Since the sum of any two sides of a triangle must be greater than the length of the third side, baba −>+ , or

.baba +<−

Therefore, for any vectors a and b , .baba +≤−

b) i) The only case that baba −≤− is when a and b are

in the same direction and ba ≥ .

ii) baba −≥− is always true.

iii) The only case that abba −≤− is when a and b are

in the same direction and ba ≤ .

iv) abba −≥− is always true.



1.4 Exercises, Sample Solutions 5. a) AE2AB = b) BCABAC +=

u2= vu += 2 c) BECBCE += uv −−=

6. a) PROPOR += b) PUOPOU += OQ5.0OP += OQ2OP +=

c) XWQXOQOW ++= d) QSOQOS +=

OQ5.0OP2OQ ++= OPOQ +=

OQ5.1OP2 +=

e) OP2OQ21OA += f) QYOQOY +=

OP3OQ +=

7. a) From page 27, ba 24DC −= and ba 35OF −= . Since ba 35 − is not a scalar multiple of ba 24 − , DC and OF are not parallel. b) i) ODOEDE −= ii) OEOFEF −= )4(25.1 baba +−−−−= )25.1(35 baba −−−−= ba 65.0 −−= ba −= 5.6 iii) ODOFDF −= )4(35 baba +−−−= ba 76 −=

8. a) i) ba 42OC += ii) ba 33OD +−= iii) ba 22OE −−= iv) ba 24OF −= b) i) OCODCD −= ii) ODOEDE −= )42(33 baba +−+−= )33(22 baba +−−−−= ba −−= 5 ba 5−= iii) OEOFEF −= iv) OFOCFC −= )22(24 baba −−−−= )24(42 baba −−+= a6= ba 62 +−=



v) ODOFDF −= vi) OEOCEC −= )33(24 baba +−−−= )22(42 baba −−−+= ba 57 −= ba 64 +=

9.

10. Answers may vary.



11. a)

b) The heads of the vectors lie on a straight line. This is the line l in the diagram in part a.

12. a) BMABAM += b) DNADAN +=

AD21AB += AB

21AD +=

ADAB21 +=

b) From part a, we have:

AD21ABAM +=

ADAB21AN +=

To write AB in terms of AM and AN , eliminate AD from equations and . −2 × : ADAB2AM2 −−=−

: ADAB21AN +=

Add: AB23ANAM2 −=+−

Therefore, AN32AM

34AB −=

To write AD in terms of AM and AN , eliminate AB from equations and .

: AD21ABAM +=

−2 × : AD2ABAN2 −−=−

Add: AD23AN2AM −=−



Therefore, AN34AM

32AD +−=

13. a) OEOAOD += b) OEOBOC +=

vu += vu += 2 c) BCABAC += d) OBEOEB += vu += uv 2+−= vu −= 2

14. a) EDOEOD += b) DCODOC += uv += uvu ++= vu += vu += 2 c) CBOCOB += d) OEAD = uvu ++= 2 v= vu += 3

15. Use the results of part a to complete part b. a) u=OA b) ABOAOB += FBAFAB += vu += 2 uv += BCABAC += OEBC = vu 2+= v= CDBCBD += OACD −= uv −= u−= DECDCE += ABDE −= uv 2−−= uv −−= EODEDO += v−=EO uv −−= 2 OAEOEA += uv +=



16. a) The diagram below left shows the vectors in exercise 15a arranged tail-to-tail. The diagram below right shows the vectors in exercise 15b. 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 60°.

17. a) vu 7AR +−= vu 62BQ +−=

vu 53CP +−= vu 44DO +−= vu 35EN +−= vu 26FM +−= vu +−= 7GL 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 (−1, 7), (−2, 6), (−3, 5), (−4, 4), (−5, 3), (−6, 2), (−7, 1) 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. 18. a) Multiplying by 0: 00 =x .

Multiplying by 1: xx =1 .



b) By the definition of scalar multiplication, 00 =x . The vector x1 has the same magnitude and direction as x . Therefore, xx =1 .

19. a) In the diagram below, let a=OA .

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, am=OB and an=BC . Since OC is (m + n) times as long as OA: anm )(OC += Also: BCOBOC += = am + an Therefore, anamanm +=+ )( .

b) In the diagram below, let a=OA .

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, an=OB and )(OC anm= . 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: amn)(OC = Therefore, amnanm )()( = .

20. Consider the diagram below in which ∆DOC is similar to ∆BOA and has sides |m| times as long (m # 0).



Since OD is |m| times as long as OB and OD is oppositely directed to OB :

)(

OBOD

bam

m

+=

=

Using theTriangle Law:

bmam +=

+= CDOCOD

Therefore, m( a + b ) = m a + m b , when m # 0. 21. a) uu 33 =

= 3(1) = 3

b) vv41

41 =−

)1(41=

41=

c) The diagram for vu 32 − is below left. vu 32 − 2 = 22 + 32 − 2(2)(3)cos 120°

vu 32 − = 19



d) The diagram for vu 32 +− is above right. 232 vu +− = 22 + 32 − 2(2)(3)cos 120°

vu 32 − = 19 e) The diagram for vu 5.05.1 + is below left. vu 5.05.1 + 2 = (1.5)2 + (0.5)2 − 2(1.5)(0.5)cos 60°

vu 5.05.1 + = 7

2

f) The diagram for vu 2+− is above right. 22vu +− = 12 + 22 − 2(1)(2) cos 120°

vu 2+− = 7 22. Use the given information to construct the following diagram.

Since u and v are collinear, AC || CE. Therefore, corresponding angles are equal: ∠ACB = ∠AED



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

proportional. DEBC

ADAB = . Hence,

bn

bt

amas

= , so nt

ms = .



1.5 Exercises, Sample Solutions 5. a) i) 2u = 2[3, 2] ii) 3u = 3[3, 2]

= [6, 4] = [9, 6] iii) 5u = 5[3, 2] iv) −4u = −4[3, 2] = [15, 10] = [−12, −8] b)

c) 22 23 +=u

= 13 Therefore, u2 = 2 13 , u3 = 3 13 , u5 = 5 13 , and u4− = 4 13 .

6. v = [4, 3] 22 34 +=v

= 5

a) v3 = 3[4, 3] b) ]3,4[21

21 =v

= [12, 9] =

23,2

c) Since v = 5, the vector v2 or [8, 6] has magnitude 10.

d) Since 5=v , the vector 15 v or

53,

54 has length 1.

7. a) AB = [10 − 4, 3 − 1]

= [6, 2] BC = [6 − 10, 5 − 3] = [−4 , 2] CD = [0 − 6, 3 − 5] = [−6, −2] DA = [4 − 0, 1 − 3]



= [4, −2] b) 22 26AB += 22 2)4(BC +−=

= 40 = 20 = 102 52=

∴ 102CD = ∴ 52DA =

c) ABCD is a parallelogram since opposite sides are parallel and equal.

8. a) AB = [−1 − (−2), 7 − (−1)] = [1, 8] BC = [6 − (−1), 3 − 7] = [7 , −4] CD = [5 − 6, −5 − 3] = [−1, −8] DA = [−2 − 5, −1 − (−5)] = [−7, 4] b) 22 81AB += 22 )4(7BC −+=

= 65 65=

∴ 65CD = ∴ 65DA =

c) ABCD is a rhombus since opposite sides are parallel and all sides are equal in length.

9. a) Form any 2 vectors from the 3 given points. If the vectors are scalar multiples of each other, then the points are collinear. b) i) PQ = [2 − (−3), 4 − 1] = [5, 3] PR = [5 − (−3), 6 − 1] = [8, 5] By inspection, PQ and PR are not scalar multiples of each other. Therefore, P, Q, and R are not collinear. ii) DE = [1 − 5, −5 − 1] = [−4, −6] DF = [−3 − 5, −11 − 1] = [−8, −12] By inspection, DF = 2 DE . Therefore, D, E, and F are collinear.



10. a) ]4,2[21

21 −−=− u b) ]1,3[44 −=v

[ ]2,1 −= [ ]4,12 −= c) ]1,3[]4,2[ −+−=+ vu d) ]1,3[]4,2[ −−−=− vu [ ]3,1= [ ]5,5−= e) ]1,3[2]4,2[2 −+−−=+− vu f) ]1,3[3]4,2[232 −−−=− vu = [2, −4] + [6, −2] = [−4, 8] − [9, −3] = [8, −6] = [−13, 11]

11.

12. a) ]4,2[3]3,5[3 −+=+ ba b) ]4,2[4]3,5[242 −−=− ba = [5, 3] + [6, −12] = [10, 6] − [8, −16] = [11, −9] = [2, 22] c) ]4,2[5]3,5[353 −+−=+− ba = [−15, −9] + [10, −20] = [−5, −29]



13.

14. a) )23(22 jiu −= b) )2(33 jiv +−=− ji 46 −= ji 36 −−= c) jijivu ++−=+ 223 d) )2(23 jijivu +−−=− ji −= 5 ji 3−= e) )2(2)23(424 jijivu +−−=− f) )2(3)23(232 jijivu ++−−=+− jiji 24812 −−−= jiji 3646 +++−= ji 108 −= j7=

15. a) ]3,2[]1,4[ +=+ ba = [6, 4] ]3,2[]1,4[ −=− ba = [2, −2] b)



c) 22 46 +=+ ba 22 )2(2 −+=− ba

= 52 = 8 = 2 13 = 2 2 d) The vectors ba + and ba − are diagonals of the parallelogram formed by a and b . Since the angle between a and b is acute, the diagonal corresponding to ba + is longer than the diagonal corresponding to ba − . Thus ba + is greater than ba − .

e) If a and b are perpendicular, then ba + would be equal to ba − since ba + and

ba − would be diagonals of a rectangle. This is illustrated in the diagram below left. If the angle between a and b is obtuse, then ba + would be less than ba − as illustrated

in the diagram below right.

16. a) Let w = su + t v for some scalars s and t. [2, 8] = s[3, 0] + t[−1, 2] = [3s − t, 2t] Since the vectors are equal, their components are equal. 3s − t = 2 2t = 8, or t = 4 Substitute t = 4 in equation to determine s. 3s − 4 = 2 3s = 6 s = 2 Therefore, w = 2u + 4 v . b)



17. From exercise 16, w = 2u + 4 v . a) w = 2u + 4 v b) w = 2u + 4 v 2u = −4 v + w 4 v = −2u + w

wvu212 +−= wuv

41

21 +−=

18. a) Let w = su + t v for some scalars s and t.

[12, −1] = s[2, 1] + t[−1, 3] = [2s − t, s + 3t] Since the vectors are equal, their components are equal. 2s − t = 12 s + 3t = −1 Solve equation for s. s = −3t − 1 Substitute s = −3t − 1 in equation and solve for t. 2(−3t − 1) − t = 12 −6t − 2 − t = 12 −7t = 14 t = −2 Substitute t = −2 in equation and solve for s. s = −3(−2) − 1 = 5 Therefore, w = 5u − 2 v . b)

19. From exercise 18, w = 5u − 2 v . a) w = 5u − 2 v b) w = 5u − 2 v 5u = 2 v + w 2 v = 5u − w

wvu51

52 += wuv

21

25 −=

20. a) ]1,2[3]5,0[3 −+=+ mb

= [0, 5] + [6, −3] = [6, 2]



To obtain the other vectors in the list, subtract m = [2, −1] from the preceding vector on the list. ]1,2[]2,6[2 −−=+ mb = [4, 3] ]1,2[]3,4[ −−=+ mb = [2, 4] ]1,2[]4,2[0 −−=+ mb = [0, 5] ]1,2[]5,0[ −−=− mb = [−2, 6] ]1,2[]6,2[2 −−−=− mb = [−4, 7] ]1,2[]7,4[3 −−−=− mb = [−6, 8] b)

c) The first vector in the list is ]2,6[3 =+ mb . To obtain the other vectors in the list, repeatedly subtract m = [2, −1]. Therefore, the heads of the vectors lie on a line that passes

through the point (6, 2) and has slope 21− .

d) If b = [2, 4], then ]1,8[3 =+ mb . Thus the heads of the vectors would lie on a line that

passes through (8, 1) and has slope21− .

If b = [−1, 2], then ]1,5[3 =+ mb . Thus the heads of the vectors would lie on a line that

passes through (5, −1) and has slope21− .

21. a) ]2,1[3]1,3[232 +−−=+− vu

= [−6, 2] + [3, 6] = [−3, 8] To obtain the other vectors in the list, add vu − = [2, −3] to the preceding vector on the list.



]3,2[]8,3[2 −+−=+− vu = [−1, 5] ]3,2[]5,1[ −+−=v = [1, 2] ]3,2[]2,1[ −+=u = [3, −1] ]3,2[]1,3[2 −+−=− vu = [5, −4] ]3,2[]4,5[23 −+−=− vu = [7, −7] b)

c) The first vector in the list is ]8,3[32 −=+− vu . To obtain the other vectors in the list, repeatedly add vu − = [2, −3]. Therefore, the heads of the vectors lie on a line that passes

through the point (−3, 8) and has slope 23− .

d) Yes, since we are still repeatedly adding vu − to obtain each vector on the list.

22. a) abba +=+ LS = ba + RS = ab + = [a1, a2] + [b1, b2] = [b1, b2] + [a1, a2] = [a1 + b1, a2 + b2] = [b1 + a1, b2 + a2] = [b1 + a1, b2 + a2] LS = RS, so abba +=+ b) )()( cbacba ++=++ LS = cba ++ )( RS = )( cba ++ = ([a1, a2] + [b1, b2]) + [c1, c2] = [a1, a2] + ([b1, b2] + [c1, c2]) = [a1 + b1, a2 + b2] + [c1, c2] = [a1, a2] + [b1 + c1, b2 + c2]



= [a1 + b1 + c1, a2 + b2 + c2] = [a1 + b1 + c1, a2 + b2 + c2] LS = RS, so )()( cbacba ++=++ . c) bsasbas +=+ )( LS = )( bas + RS = bsas + = s([a1, a2] + [b1, b2]) = s[a1, a2] + s[b1, b2] = s[a1 + b1, a2 + b2] = [sa1, sa2] + [sb1, sb2] = [s(a1 + b1), s(a2 + b2)] = [sa1 + sb1, sa2 + sb2] = [sa1 + sb1, sa2 + sb2] LS = RS, so bsasbas +=+ )( . d) atasats +=+ )( LS = (s + t) a RS = atas + = (s + t)[a1, a2] = s[a1, a2] + t[a1,a2] = [(s + t)a1, (s + t)a2] = [sa1, sa2] + [ta1, ta2] = [sa1 + ta1, sa2 + ta2] = [sa1 + ta1, sa2 + ta2] LS = RS, so atasats +=+ )( . e) Since 0=+ va : av −= 0 = [0, 0] − [a1, a2] = [0 − a1, 0 − a2] = [− a1, − a2] = −[a1, a2] = – a

23. Let [2a, a] be the required vector. Since the magnitude of the vector is 4: 4)2( 22 =+ aa Square both sides and simplify. 4a2 + a2 = 16 5a2 = 16

a2 = 5

16

a = ±5

4

There are two such vectors,

5

4,5

8 and

−−5

4,5

8 .

24. vu + = [2, −1] + [x, 3] = [2 + x, 2]



Since, vu + = 5, vu + 2 = 25. (2 + x)2 + 22 = 25 4 + 4x + x2 + 4 = 25 x2 + 4x − 17 = 0

2

)17(4164 −−±−=x

2

844 ±−=

2

2124 ±−=

= 212 ±−

25. 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 bc 3= , or bac 30 += . In this situation, is not possible to express a as a linear combination of b and c .



1.6 Exercises, Sample Solutions 2. Method 1: 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 2 = 502 + 352

r Ñ 61.0 N Let θ = ∠BOA. Then the bearing of r is 270° − θ.

tan θ = 3550

θ Ñ35.0° The bearing of r is 270° − 35° = 235°. The resultant force is 61. 0 N [235°]. Method 2: Using Cartesian vectors Place the vectors on a coordinate system and represent each vector algebraically. Let 1f and 2f represent the two forces. Let r be the resultant force. 1f = [−50, 0] 2f = [0, −35] r = 1f + 2f = [−50, 0] + [0, −35] = [−50, −35] The magnitude of the resultant is: r = (−50)2 + (−35)2 Ñ 61.0 N The direction of the resultant is:



tan θ = 3550

θ Ñ35.0° Therefore, the bearing of r is 270° − 35° = 235°. The resultant force is 61. 0 N [235°].

3. Method 1: Using geometric vectors In the diagram below left, OA represents the 220 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 = 180° − 55° = 125° In ∆OAC, use the Cosine Law to calculate r .

r 2 = 2202 + 4002 − 2(220)(400)cos 125°

r Ñ 556.2 Let θ be the angle the resultant makes with the 220 N force. Use the Sine Law to determine θ.

sin θ400 =

sin 125°556.2

sin θ = 400 sin 125°

556.2

θ Ñ36.1° Therefore, the resultant force has a magnitude of 556.2 N and acts at an angle of 36.1° to the 220 N force. Method 2: Using Cartesian vectors Place the vectors on a coordinate system and represent each vector algebraically. Let 1f and 2f represent the two forces. Let r be the resultant force.



1f = [220, 0] 2f = [400 cos 55°, 400 sin 55°] Ñ [229.4, 327.7] r = 1f + 2f = [220, 0]+ [229.4, 327.7] = [449.4, 327.7] The magnitude of the resultant is: r = (449.4)2 + (327.7)2 Ñ 556.2 Let θ be the angle the resultant makes with the 220 N force.

tan θ = 327.7449.4

θ Ñ36.1° The resultant force has a magnitude of 556.2 N and acts at an angle of 36.1° to the 220 N force.

4. Method 1: Using geometric vectors In the diagram below left, OA represents the 20 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 = 120° OACB is a parallelogram, so: ∠OBC = 180° − 120° = 60° In ∆OCB, use the Cosine Law to calculate r .

r 2 = 202 + 302 − 2(20)(30)cos 60°


sin θ20 =

sin 60°26.5

sin θ = 20 sin 60°

26.5

θ Ñ40.8° The equilibriant is equal in magnitude but opposite in direction to the resultant. Hence, the



equilibriant has magnitude 26.5 N and acts at an angle of 180° − 40.8°, or 139.2° to the 30 N force. Method 2: Using Cartesian vectors Place the vectors on a coordinate system and represent each vector algebraically. Let 1f and 2f represent the two forces. Let r be the resultant force. 1f = [30, 0] 2f = [20 cos 120°, 20 sin 120°] Ñ [−10, 17.3] r = 1f + 2f = [30, 0] + [−10, 17.3] = [20, 17.3] The magnitude of the resultant is: r = (20)2 + (17.3)2 Ñ 26.5 Let θ be the angle the resultant makes with the 30 N force.

tan θ = 17.320

θ Ñ40.8° The equilibriant is equal in magnitude but opposite in direction to the resultant. Hence, the equilibriant has magnitude 26.5 N and acts at an angle of 180° − 40.8°, or 139.2° to the 30 N force.

5. Method 1: Using geometric vectors In the diagram below left, OA represents the unknown force and OB represents the 195 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:



∠OBC = 180° − 75° = 105° Let θ be the angle the unknown force makes with the resultant. Use the Sine Law to calculate θ.

sin θ195 =

sin 105°225

sin θ = 195 sin 105°

225

θ Ñ 56.8° Therefore, α = 180° − 105° − 56.8° = 18.2° Use the Sine Law to determine x, the magnitude of the unknown force.

x

sin 18.2° = 225

sin 105°

x = 225 sin 18.2°

sin 105°

Ñ 72.6 Therefore, the unknown force has a magnitude of 72.6 N and makes an angle of 56.8° with the resultant. Method 2: Using Cartesian vectors Place the vectors on a coordinate system and represent each vector algebraically. Let 1f and 2f represent the two forces. Let r be the resultant force. 1f = [x, 0] 2f = [195 cos 75°, 195 sin 75°] Ñ [50.5, 188.4] r = [225 cos θ, 225 sin θ]

r = 1f + 2f [225 cos θ, 225 sin θ] = [x, 0] + [50.5, 188.4] [225 cos θ, 225 sin θ] = [x + 50.5, 188.4] Since the vectors are equal, their components are equal. 225 cos θ = x + 50.5 225 sin θ = 188.4 Solve equation for θ.

sinθ = 188.4225

θ Ñ 56.8° Substitute θ Ñ 56.8° in equation and solve for x. 225 cos 56.8° = x + 50.5 x = 225 cos 56.8° − 50.5 Ñ 72.6



Therefore, the unknown force has a magnitude of 72.6 N and makes an angle of 56.8° with the resultant.

6. Method 1: 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 = 135° OCDL is a parallelogram, so: ∠OCD = 180° − 135° = 45° In ∆OCD, use the Cosine Law to calculate r .

r 2 = 40 + 302 − 2(40)(30)cos 45°


sin θ30 =

sin 45°28.3

sin θ = 30 sin 45°

28.3

θ Ñ48.6° Therefore, to keep his brothers efforts in equilibrium, Pepe should exert a force of 28.3 N at an angle of (180° − 48.6°), or 131.4° to the force exerted by Paco. Method 2: Using Cartesian vectors Place the vectors on a coordinate system and represent each vector algebraically. Let 1f and 2f represent the two forces. Let r be the resultant force. 1f = [40, 0] 2f = [30 cos 225°, 30 sin 225°] Ñ [−21.2, −21.2]



r = 1f + 2f = [40, 0] + [−21.2, −21.2] = [18.8, −21.2] The magnitude of the resultant is: r = (18.8)2 + (21.2)2 Ñ 28.4 N The direction of the resultant is:

tan θ = 21.218.8

θ Ñ 48.4° Therefore, to keep his brothers efforts in equilibrium, Pepe should exert a force of 28.3 N at an angle of (180° − 48.6°), or 131.4° to the force exerted by Paco.

7. This problem is similar to Example 4, so it is simpler to use geometric vectors. Let 1T and 2T 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 225 N weight.

Use the Sine Law to find | 1T | and | 2T |.

)2435sin(

22566sin55sin21

+==

TT

59sin

55sin2251 =T

59sin66sin225

2 =T

Ñ 215.0 Ñ 239.8 The tensions in the two ropes are 215. 0 N and 239.8 N respectively.



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.

cos 30° = T

100 tan 30° = 100m

T = 100

cos 30° m = 100 tan 30°

Ñ 115.5 Ñ 57.7 Therefore, the tension in the chain is 115.5 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 10 N directed upwards. Let 1T and 2T represent the forces in the two wires. Since the picture is hung symmetrically on the wires, | 1T | = | 2T |. Let | 1T | = | 2T | = T.

Use the Sine Law to determine T.

30sin120sin

10 T=

T = 120sin

30sin10

Ñ 5.8 N The tension in the wires is 5.8 N.



10. Method 1: 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 = 135° Since OHRW is a parallelogram: ∠OHR = 180° − 135° = 45° Use the Cosine Law to calculate r .

2r = 9002 + 1002 − 2(900)(100)cos 45° Ñ 832.3 Let ∠ROH = θ. The bearing of the plane is 135° − θ.

sin θ100 =

sin 45°832.3

sin θ = 100 sin 45°

832.3

θ Ñ4.9° Hence the bearing is 135° − 4.9° = 130.1°. The plane’s speed relative to the ground is 832.3 km/h on a bearing of 130.1°. Method 2: Using Cartesian vectors The diagram at the right represents the situation. A bearing of 135° corresponds to a direction angle of 315°. 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.



a = [900 cos 315°, 900 sin 315°] Ñ[636.4, −636.4] w = [0, 100] r = a + w = [636.4, −636.4] + [0, 100] Ñ[636.4, −536.4] The magnitude of the resultant is: r = (636.4)2 + (−536.4)2 Ñ 832.3 The bearing of the plane is 90° + θ .

tan θ = 536.4636.4

θ Ñ 40.1° Hence the bearing is 90° + 40.1° = 130.1° The plane’s speed relative to the ground is 832.3 km/h on a bearing of 130.1°.


From the given bearings: ∠WOH = 113° − 30° = 83° Since OHRW is a parallelogram: ∠OWR = 180° − 83° = 97° Use the Cosine Law to calculate r .

2r = 6002 + 802 − 2(600)(80) cos 97° Ñ 614.9 Let ∠ROH = θ. The bearing of the plane is 30° + θ.



sin θ80 =

sin 97°614.9

sin θ = 80 sin 97°

614.9

θ Ñ7.4° Hence the bearing is 30° + 7.4° = 37.4°. The plane’s speed relative to the ground is 614.9 km/h on a bearing of 37.4°. Method 2: Using Cartesian vectors The diagram at the right represents the situation. Bearings of 030° and 113° correspond to direction angles of 60° and 337° respectively. 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. a = [600 cos 60°, 600 sin 60°] Ñ[300, 519.6] w = [80 cos 337°, 80 sin 337°] Ñ[73.6, −31.3] r = a + w = [300, 519.6] + [73.6, −31.3] Ñ[373.6, 488.3] The magnitude of the resultant is: r = (373.6)2 + (488.3)2 Ñ 614.9 The bearing of the plane is 90° − θ .

tan θ = 488.3373.6

θ Ñ 52.6° Hence the bearing is 90° − 52.6° = 37.4° The plane’s speed relative to the ground is 614.9 km/h on a bearing of 37.4°.

12. 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.



From the given bearings: ∠ROW = 125° − 75° = 50° a) Let ∠HOR = θ. Use the Sine Law on ∆ORW to calculate θ.

sin θ80 =

sin 50°550

sin θ = 80 sin 50°

550

θ Ñ6.4° The heading the pilot should take is 075° − 6.4° = 68.6°. b) ∠HOW = 50° + 6.4° = 56.4°. Since OWRH is a parallelogram: ∠OHR = 180° − 56.4° =123.6° Use the Sine Law to calculate the speed of the plane relative to the ground.

80

sin 6.4° = r

sin 123.6°

r = 80 sin 123.6°

sin 6.4

Ñ 597.8 The speed of the plane relative to the ground is 597.8 km/h.

c) Time = distancespeed

= 508

597.8

Ñ 0.85 The trip will take approximately 0.85 h, or approximately 51 min.



13. Since a distance of 80 km was flown in 20 minutes, the speed of the plane relative to the

ground was

h31

km80 , or 240 km/h.

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:

tan 10° = 240w

w = 240 tan 10° Ñ 42.3 The speed of the wind is 42.3 km/h.



1.7 Exercises, Sample Solutions 6. Use the formula

vuvu ⋅=θcos .

a) 2222 1540

]1,5[]4,0[cos++

⋅=θ b) 2222 2)1()2(3

]2,1[]2,3[cos+−−+

−⋅−=θ

= 0 + 44 26

= −3 − 4

13 5

= 126

= −765

θ Ñ 78.7° θ Ñ 150.3°

c) 2222 )5()2()1(4

]5,2[]1,4[cos−+−−+

−−⋅−=θ d) 2222 )4(236

]4,2[]3,6[cos−++

−⋅=θ

= −8 + 517 29

= 12 − 12

45 20

= −3493

= 0

θ Ñ 97.8° θ = 90°

7. vu ⋅ is positive so the angle between the vectors is acute.

8. a) AB = [−2 − (−1), 1 − 0] = [−1, 1] AC = [1 − (−1), 4 − 0] = [2 , 4] BC = [1 − (−2), 4 − 1] = [3, 3]

cos ∠A = ACAB

ACAB ⋅ cos ∠B = BCBA

BCBA ⋅

2222 42)1()1(

]4,2[]1,1[

++−

⋅−= 2222 33)1(1

]3,3[]1,1[

+−+

⋅−=

= −2 + 4

2 20 =

3 − 32 18

= 240

= 0

∠A Ñ 71.6° ∠B = 90°



∠C = 180° − ∠A − ∠B = 180° − 71.6° − 90° = 18.4° b) PQ = [8 − 2, 3 − 6] = [6, −3] PR = [−4 − 2, 0 − 6] = [−6 , −6] QR = [−4 − 8, 0 − 3] = [−12, −3]

cos ∠P = PRPQ

PRPQ ⋅ cos ∠Q = QRQP

QRQP ⋅

2222 )6()6()3(6

]6,6[]3,6[

−+−−+

−−⋅−= 2222 )3()12()3()6(

]3,12[]3,6[

−+−−+−

−−⋅−=

= −36 + 18

45 72 =

72 − 945 153

= −183240

= 636885

∠P Ñ 108.4° ∠Q Ñ 40.6° ∠R = 180° − ∠P − ∠Q = 180° − 108.4° − 40.6° = 31.0°

9. a) See diagram below.

b) Quadrilateral ABCD Use the diagram in part a. AB = [9, 3], DC = [9, 3], BC = [1, −3], AD = [1, −3] ADAB ⋅ = [9, 3]⋅[1, −3] = 9 − 9 = 0



BCBA ⋅ = [–9, –3]⋅[1, −3] = –9 + 9 = 0 DCDA ⋅ = [–1, 3]⋅[9, 3] = –9 + 9 = 0 CBCD ⋅ = [–9, –3]⋅[–1, 3] = 9 – 9 = 0 Therefore, ABCD is a rectangle. Quadrilateral PQRS Use the diagram in part a. PQ = [12, −4], SR = [12, −4], QR = [−2, −7], PS = [−2, −7] Since opposite sides are equal, PQRS is a parallelogram. PQ ⋅ PS = [12, −4]⋅[−2, −7] = −24 + 28 = 4 Therefore, ∠P ≠ 90°. Hence PQRS is a not rectangle.

10. The vertices of ∆PQR are P(−3, −2), Q(−1, 4), and R(5, 2). PQ = [−1 − (−3), 4 − (−2)] = [2, 6] PR = [5 − (−3), 2 − (−2)] = [8 , 4] QR = [5 − (−1), 2 − 4] = [6, −2] a) i) PQ ⋅ PR = [2, 6]⋅ [8, 4] ii) QP ⋅ QR = [−2, −6]⋅ [6 , −2] = 16 + 24 = −12 + 12 = 40 = 0 iii) RQ ⋅ RP = [−6, 2]⋅ [−8, −4] = 48 − 8 = 40 b) The geometric definition of the dot product is θcosbaba =⋅ , 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.



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.

11. The vertices of parallelogram ABCD are A(2, 0), B(6, 1), C(8, 4), and D(4, 3). AB = [6 − 2, 1 − 0] = [4, 1] Since ABCD is a parallelogram, DC = AB = [4, 1]. AD = [4 − 2, 3 − 0] = [2, 3] Since ABCD is a parallelogram, BC = AD = [2, 3]. a) i) AB ⋅ AD = [4, 1]⋅[2, 3] ii) CB ⋅ CD = [−2, −3]⋅[−4, −1] = 8 + 3 = 8 + 3 = 11 = 11 iii) BA ⋅ BC = [−4, −1]⋅[2, 3] iv) DA ⋅ DC = [−2, −3]⋅[4, 1] = −8 − 3 = −8 − 3 = −11 = −11 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 θcosbaba =⋅ .

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 Adjacent angles in a parallelogram are supplementary, so ∠B = 180° − ∠A. Also opposite sides are equal, so AD = BC. Clearly AB = BA. Therefore equation becomes:



− BA ⋅ BC = −| AB || AD | cos (180° − ∠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.

12. 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 = [−2 − (−5), 1 − 5] = [3, −4] AC = [7 − (−5), 8 − 5] = [12, 3] BC = [7 − (−2), 8 − 1] = [9, 7] AB ⋅ AC = [3, −4]⋅[12, 3] = 36 − 12 ≠ 0 AB ⋅ BC = [3, −4]⋅ [9, 7] = 27 − 28 ≠ 0 AC ⋅ BC = [12, 3]⋅[9, 7] = 108 − 21 ≠ 0 Therefore, A, B, and C do not form a right angle. ii) JK = [5 − (−3), 0 − (−4)] = [8, 4] JL = [2 − (−3), 6 − (−4)] = [5, 10] KL = [2 − 5, 6 − 0] = [−3, 6] JK ⋅ JL = [8, 4]⋅[5, 10] = 40 + 40 ≠ 0 JL ⋅ KL = [5, 10]⋅[−3, 6] = −15 + 60 ≠ 0 JK ⋅ KL = [8, 4]⋅[−3, 6] = −24 + 24 = 0 Therefore, J, K, and L form a right angle.



13. a) Form any 2 vectors from the 3 given points, and find the angle between the vectors. If the angle is 0° or 180°, the points are collinear. b) i) DE = [2 − (−4), 3 − 7] = [6, −4] DF = [8 − (−4), −1 − 7] = [12, −8]

cos ∠D = DFDE

DFDE ⋅

2222 )8(12)4(6

]8,12[]4,6[

−+−+

−⋅−=

= 72 + 3252 208

= 10410816

= 1 ∠D = 0° Therefore, D, E, and F are collinear. ii) RS = [4 − 7, 1 − 2] = [−3, −1] RT = [−4 − 7, −2 − 2] = [−11, −4]

cos ∠R = RTRS

RTRS ⋅

2222 )4()11()1()3(

]4,11[]1,3[

−+−−+−

−−⋅−−=

= 33 + 410 137

= 371370

∠D Ñ 1.5° Therefore, R, S, and T are not collinear.

14. Suppose the square is ABCD and AB = [5, 2]. Since BC is perpendicular to AB and AB has

slope52 , BC has slope

25− . Furthermore

AB = BC, so BC = [−2, 5] or [2, −5]. Thus the



other three sides of the square are either [5, 2], [2, −5], [2, −5] or [5, 2], [−2, 5], [−2, 5].

15. There are 2 cases to consider. Case 1. [4, 2] represents the width of the rectangle.

The slope of the line segment representing the width is 42 , so the slope of the line segment

representing the length is 24− . 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, 2], [4, −8], [4, −8] or [4, 2], [−4, 8], [−4, 8]. Case 2. [4, 2] represents the length of the rectangle. Since the length is double the width, the vector representing the width is either [−1, 2] or [1, −2]. Therefore, vectors that could represent the other 3 sides of the rectangle are [4, 2], [−1, 2], [−1, 2] or [4, 2], [1, −2], [1, −2].

16. a) In ∆OBC:

cos θ = OCOB

OC = OB cos θ = θcosb

Area of rectangle OAED = (OA)(OD) = (OA)(OC) since OD = OC = θcosba

= ba ⋅ 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 ba .

c) When θ = 45°, BC = OC = OD, and so D, B, and E are collinear.

17. a) See diagram for part c. b) i) wvu )( ⋅ = ([2, 0]⋅[2, 1]) w ii) uwv )( ⋅ = ([2, 1]⋅[1, 2])u = (4 + 0) w = (2 + 2)u = 4 w or [4, 8] = 4u or [8, 0] iii) vuw )( ⋅ = ([1, 2]⋅[2, 0]) v = (2 + 0) v = 2 v or [4, 2] c) The expressions in part b are scalar multiples of u , v , and w .



18. If a and b are perpendicular, then ba ⋅ = 0.

a) [k, −2]⋅[−1, 2] = 0 b) [−3, 4]⋅[5, k] = 0 −k − 4 = 0 −15 + 4k = 0

k = −4 k = 154

19. a) )( cba +⋅ = [2, −3]⋅([−1, 4] + [5, 2]) b) =⋅+ cba )( ([2, −3] + [−1, 4])⋅[5, 2] = [2, −3]⋅[4, 6] = [1, 1]⋅[5, 2] = 8 − 18 = 5 + 2 = −10 = 7 c) =+⋅+ )()( caba ([2, −3] + [−1, 4])⋅([2, −3] + [5, 2]) = [1, 1]⋅[7, −1] = 7 − 1 = 6 d) =−⋅+ )25()32( baba (2[2, −3] + 3[−1, 4])⋅(5[2, −3] − 2[−1, 4]) = [1, 6]⋅[12, −23] = 12 − 138 = −126

20. a) The magnitude of the force in the direction of the displacement is the horizontal component of the force, θcosF .

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 θcosF

= F θcosd

= dF ⋅

21. Work = dF ⋅ = F θcosd

= 30(100)cos 30° Ñ 2598.1 The work done in moving the wagon 100 m is approximately 2598.1 joules.



22.

23. a) To draw a different rectangle, construct C so that ∠OCA = 90°. Then construct D so that OC = OD.

b) i) a) In ∆OAC:

cos θ = OCOA

OC = OA cos θ = θcosa Area of rectangle OBED = (OB)(OD) = (OB)(OC) since OD = OC = θcosab

= ba ⋅ 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.



1.8 Exercises, Sample Solutions 1. a) cabacba ⋅+⋅=+⋅ )( b) baaabaa ⋅+⋅=+⋅ )(

baa ⋅+= 2 c) vuuuvuu ⋅+⋅=+⋅ 2)2( d) vuuuvuu ⋅−⋅=−⋅ 96)32(3

vuu ⋅+= 22 vuu ⋅−= 96 2

2. a) bbaabababa ⋅+−⋅+=−⋅+ )()()()( bbbaabaa ⋅−⋅−⋅+⋅= bbbabaaa ⋅−⋅−⋅+⋅=

22 ba −=

b) bbaabababa 2)()()2()( ⋅−+⋅−=+⋅− bbbaabaa ⋅−⋅+⋅−⋅= 22 bbbabaaa ⋅−⋅+⋅−⋅= 22

22 2bbaa −⋅+=

c) bbaabababa 2)4()4()2()4( ⋅++⋅+=+⋅+ bbbaabaa ⋅+⋅+⋅+⋅= 284 bbbabaaa ⋅+⋅+⋅+⋅= 284

22 294 bbaa +⋅+=

d) bbaabababa 2)32(3)32()23()32( ⋅+−⋅+=−⋅+ bbbaabaa ⋅−⋅−⋅+⋅= 6496 bbbabaaa ⋅−⋅−⋅+⋅= 6496

22 656 bbaa −⋅+=

3. a) The dot product does not satisfy the associative law because the expressions cba ⋅⋅ )( and

)( cba ⋅⋅ have no meaning. • cba ⋅⋅ )( has no meaning because ba ⋅ is a scalar and we cannot take its dot product

with vector c . • )( cba ⋅⋅ also has no meaning because it too involves the dot product of vector ( a ) with

a scalar ( cb ⋅ ).



b) No meaning can be given to the expression cba ⋅⋅ since cba ⋅⋅ )( and )( cba ⋅⋅ have no meaning.

4. Proof that 00 =⋅a Geometric proof Cartesian proof

θcos00 aa =⋅ Let a = ],[ 21 aa

= 0 ]0,0[],[0 21 ⋅=⋅ aaa )0()0( 21 aa += = 0 Proof that || aua =⋅ , where u is a unit vector in the direction of a

If u is a unit vector in the direction of a , then u = aa ||1 .

Geometric proof Cartesian proof

aa

aua||

1⋅=⋅ Let a = ],[ 21 aa

= aaa

⋅||

1 aa

aua||

1⋅=⋅

= 2

||1 aa

= aaa

⋅||

1

= || a = ],[],[121212

22

1

aaaaaa

⋅+

= 2

22

1

22

21

aa

aa

+

+

= 22

21 aa +

= || a

5. a) If cbca ⋅=⋅ then: 0=⋅−⋅ cbca 0)( =−⋅ bac 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:

• c = 0 • 0=− ba or ba = • )( cab −⊥

Hence if cbca ⋅=⋅ , it does not necessarily follow that ba = .



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, cbca ⋅=⋅ . However, ba ≠ .

b) cbca

⋅⋅ is not the same as

bcac for real numbers so it cannot be written as

ba . If

θ1 is the angle between a and c and θ2 is the angle between b and c , then

2

1

cos

cos

θθ

cb

cacbca =

⋅⋅

2

1

cos

cos

θθ

b

a=

6. a) CBOCOB += BCABAC +=

OAOC += OAOC −= ac += ac −=

b) If 0)()( =−⋅+ acac , then ACOB ⋅ = 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 Dot products ab = ba abba ⋅=⋅ a(b + c) = ab + ac cabacba ⋅+⋅=+⋅ )(

aa = a 2aaa =⋅

(ka)b = a(kb) = k(ab) )()( bakbkabak ⋅=⋅=⋅ a(0) = 0 00 =⋅a These are properties of real numbers that do not have corresponding properties for dot products: (ab)c = a(bc), a(1) = a.

8. a) bbaabababa ⋅++⋅+=+⋅+ )()()()( (Distributive property) bbbaabaa ⋅+⋅+⋅+⋅= bbbabaaa ⋅+⋅+⋅+⋅= (Commutative property) bbbaaa ⋅+⋅+⋅= 2



b) bbbaaababa ⋅+⋅+⋅=+⋅+ 2)()(

22 cos2 bbaa ++= θ

222 2 bbaaba +⋅+=+

22 cos2 bbaa ++= θ

c) 222

cos2 bbaaba ++=+ θ

If a and b are perpendicular, then cos θ = 0 and θcos2 ba = 0.

Therefore, 222

baba +=+ .

This is the Pythagorean theorem.

9. aaaaaaa

⋅⋅=↓

aa

a

= 2

2

a= This is reasonable as illustrated in the diagram at the right.

10. a) bbbbaba

⋅⋅=↓ b) a

aaabab

⋅⋅=↓

]4,8[]4,8[]4,8[]4,8[]4,6[ −

−⋅−−⋅= ]4,6[

]4,6[]4,6[]4,6[]4,8[

⋅⋅−=

= 48 − 1664 + 16 [8, −4] =

48 − 1636 + 16 [6, 4]

= 3280 [8, −4] =

3252 [6, 4]

= 25 [8, −4] =

813 [6, 4]

=

−

58,

516 =

1332,

1348

c)

−=↓

58,

516ba

=↓

1332,

1348ab

= [3.2, −1.6] Ñ [3.7, 2.5]



d) Since abba ↓≠↓ , vector projection is not commutative.

11. Use the formula bbbbaba

⋅⋅=↓ .

a) ba ↓ ]3,2[]3,2[]3,2[]3,2[]0,3[

⋅⋅=

= 6 − 04 + 9 [2,3]

= 613 [2, 3]

=

1318,

1312

b) ]4,5[]4,5[]4,5[

]4,5[]5,4[ −

−⋅−

−⋅=↓ ba

= −20 + 2025 + 16 [−5, 4]

= 0[8, −4] = [0, 0] or 0

c) ]1,3[]1,3[]1,3[

]1,3[]2,4[

⋅

⋅−−=↓ ba

= −12 − 2

9 + 1 [3, 1]

= −1410 [3, 1]

= −75 [3, 1]

=

−−

57,

521



d) ]2,6[]2,6[]2,6[]2,6[]3,2[

⋅⋅−=↓ ba

= 12 − 636 + 4 [6, 2]

= 640 [6, 2]

= 320 [6, 2]

=

103,

109 or ji

103

109 +

12. PQ = [−1 − (−4), 6 − 0]

= [3, 6] PR = [3 − (−4), 4 − 0] = [7, 4] QR = [3 − (−1), 4 − 6] = [4, −2] a) See diagram in part b.

b) i) PQPQPQPQPRPQPR

⋅⋅=↓ ii) RQ

RQRQRQRPRQRP

⋅⋅=↓

]6,3[]6,3[]6,3[]6,3[]4,7[

⋅⋅= ]2,4[

]2,4[]2,4[]2,4[]4,7[ −

−⋅−−⋅−−=

= 21 + 249 + 36 [3, 6] =

28 − 816 + 4 [−4, 2]

= 4545 [3, 6] =

2020 [−4, 2]

= [3, 6] = [−4, 2] = PQ = RQ

iii) PRPRPRPRPQPRPQ

⋅⋅=↓ ii) PR

PRPRPRQRPRQR

⋅⋅=↓

]4,7[]4,7[]4,7[]4,7[]6,3[

⋅⋅= ]4,7[

]4,7[]4,7[]4,7[]2,4[

⋅⋅−=

= 21 + 2449 + 16 [7, 4] =

28 − 849 + 16 [7, 4]

= 4565 [7, 4] =

2065 [7, 4]



= 913 [7, 4] =

413 [7, 4]

PR139= PR

134=

c) PR134PR

139PRQRPRPQ +=↓+↓

= PR

13. Use the formula bbbbaba

⋅⋅=↓

bb

ba

= 2

cosθ

bba

=↓ 27

60cos)7)(4(

= 1449 b

= 27 b

Ñ 0.3b

14. Use the formula vvvvuvu

⋅⋅=↓

vv

vu

= 2

cosθ

vvu

=↓ 211

135cos)11)(8(



= −44 2

121 v

= −4 211 v

Ñ −0.5 v

15. ba ↓ is a scalar multiple of b , where the scalar is

⋅⋅bbba .

a) Therefore, 0=↓ ba when 0=⋅ ba , that is, when a is perpendicular to b .This is illustrated in the diagram, below left.

b) Therefore, ba ↓ is undefined when 0=⋅ bb . This occurs when b is the zero vector. This is illustrated on the diagram, above right.

16. a) bba ↓↓ )( Since ba ↓ is a scalar multiple of b , it is collinear to b . Therefore, projecting ba ↓ perpendicularly on b gives ba ↓ . b) )( bab ↓↓ Use the diagram in part a. Since ba ↓ is a scalar multiple of b , it is collinear to b . Therefore, projecting b perpendicularly on ba ↓ gives b . c) aba ↓↓ )( aba ↓↓ )( is the projection of ba ↓ on a . It is a scalar multiple of a , k a , as shown in the diagram at the right. d) )( baa ↓↓ Use the diagram in part a. )( baa ↓↓ is the projection of a on ba ↓ . It is ba ↓ as illustrated in the diagram.



17. a) It is possible to have abba ↓=↓ when a and b are perpendicular. Then 0=↓=↓ abba . When ba = , clearly abba ↓=↓ . b) Since cb ↓ is collinear to c , projecting a perpendicularly on c or on cb ↓ produces the same vector. This vector is ca ↓ .

18. Let u = [a, b]. Then u lies on a line with slope ab . Since v is perpendicular to u , it lies on a

line with slope ba− . If v is twice as long as u , then v = [−2b, 2a] or v = [2b, −2a]. Hence

there are 2 cases to consider. Case 1. v = [−2b, 2a] Since the sum of u and v is [6, 8]: [a, b] + [−2b, 2a] = [6, 8] [a − 2b, b + 2a] = [6, 8] Since the vectors are equal, their components are equal. a − 2b = 6 2a + b = 8 Eliminate a. Eliminate b. −2 × : −2a + 4b = −12 : a − 2b = 6 : 2a + b = 8____ 2 × : 4a + 2b = 16 Add: 5b = −4 Add: 5a = 22

b = 54− a =

522

Thus u =

−

54,

522 and v =

544,

58 .

Case 2. v = [2b,− 2a] Since the sum of u and v is [6, 8]: [a, b] + [2b, −2a] = [6, 8] [a + 2b, b − 2a] = [6, 8] Since the vectors are equal, their components are equal. a + 2b = 6 −2a + b = 8 Eliminate a. Eliminate b. 2 × : 2a + 4b = 12 : a + 2b = 6 : −2a + b = 8____ −2 × : 4a − 2b = −16 Add: 5b = 20 Add: 5a = −10 b = 4 a = −2 Thus u = [−2, 4] and v = [8, 4].



19. a) )()()()( 41

41

41

41 22

babababababa −⋅−−+⋅+=−−+

)2)2 (41(4

1 bbbaaabbbaaa ⋅+⋅−⋅−⋅+⋅+⋅=

2222

41

21

41

41

21

41 bbabba aa −⋅+−+⋅+=

ba ⋅=

b) A similar equation for the product xy in algebra is xy = 14 (x + y)2 − 14 (x − y)2

14 (x + y)2 + 14 (x − y)2 = 14 (x2 + 2xy + y2) − 14 (x2 − 2xy + y2)

= 14 x2 + 12 xy + 14 y2 − 14 x2 + 12 xy − 14 y2

= xy


Chapter 1 Review Exercises, Sample Solutions Copyright 2003 Pearson Education Canada Inc., Toronto, Ontario 1

Chapter 1 Review Exercises, Sample Solutions 1. Three examples of scalar quantities are distance, speed, and mass. Three examples of vector

quantities are displacement, velocity, and weight.

2. Answers may vary. a) 20 N [east] b) 24 m/s [135°] Scale : 1 cm : 10 N Scale: 1 cm : 8 m/s

3. Answers may vary. a) Equal vectors have the same magnitude and direction. BCAD,ABDC,OBDO,OCAO ==== b) Opposite vectors have the same magnitude but are in opposite directions. CBAD,BADC,BODO,COAO −=−=−=−=

4. a) AEHAHE += b) CFGCGF += c) HGDHDG += d) GCDGDC +=

5. a) PQPRPG =+ b) RBRQRA =+

c) FERSCDEFRSCD ++=−+ d) SFQBDRFSQBDR ++=−+

FE)DECD( ++= DRDR2DR ++=

FECE += = 4 DR CEGC += GE=

6. AADACDBCAB =+++ 0=



7.

8. Answers may vary. a) ADFAFD += b) DBEDEB += ADAF +−= DBDE +−= AFAD −= DEDB −= c) ABCACB += b) BEABAE += ABAC +−= BEBA +−= ACAB −= BABE −=

9. Answers may vary.

10. a) ED2AC = b) CDACAD += = u2 CDED2 += = vu +2 c) BAEBEA += DECD −−= uv −−=

11. DGCDCG += BMCBCM += CBCD

21+= CDCB

21+=

12. a) i) u3 = 3[−1, 2] ii) u2 = 2[−1, 2]

= [−3, 6] = [−2, 4]



iii) u− = −[−1, 2] iv) u4− = −4[−1, 2] = [1, −2] = [4, −8] b) u = (−1)2 + 22

= 5 Therefore, u3 = 3 5 , u2 = 2 5 , u− = 5 , and u4 = 4 5 .

13. a) AB = [7 − 3, 4 − 2] BC = [−1 − 7, 10 − 4] CA = [3 − (−1), 2 − 10] = [4, 2] = [−8, 6] = [4, −8] b) | AB | = 42 + 22 | BC | = (−8)2 + 62 | CA | = 42 + (−8)2 = 20 = 100 = 80 | BC |2 = | AB |2 + | CA |2 since 100 = 20 + 80. Therefore, ∆ABC is a scalene, right triangle.

14. a) Let w = su + t v for some scalars s and t. [4, 8] = s[0, 2] + t[1, −3] = [t, 2s − 3t] Since the vectors are equal, their components are equal. t = 4 2s − 3t = 8 Substitute t = 4 in equation to determine s. 2s −3(4) = 8 2s = 20 s = 10 Therefore, w = 10u + 4 v . b)



15. Method 1: Using geometric vectors In the diagram below left, OA and OB represent the two 90 N forces. Complete parallelogram OACB as shown in the diagram below right. Let OC represent the resultant force.

∠BOA = 48° OACB is a parallelogram, so: ∠OAC = 180° − 48° = 132° Use the Cosine Law to calculate r .

r 2 = 902 + 902 − 2(90)(90)cos 132°

r Ñ 164.4 Let θ = ∠COA. Use the Sine Law to determine θ.

sin θ90 =

sin 132°164.4

sin θ = 90 sin 132°

164.4

θ Ñ24° The resultant force has a magnitude of 164.4 N and acts at an angle of 24° with each force. Therefore, a force of 164.4 N at an angle of 180° to the resultant must be applied to the object to create equilibrium. Method 2: Using Cartesian vectors Place the vectors on a coordinate grid as shown in the diagram at the right. Let 1f and 2f represent the two forces. Let r be the resultant force. 1f = [90, 0] 2f = [90 cos 48°, 90 sin 48°] Ñ[60.2, 66.9] r = 1f + 2f = [90, 0] + [60.2, 66.9] = [150.2, 66.9] The magnitude of the resultant is:



r = (150.2)2 + (66.9)2 Ñ 164.4 The direction of the resultant is:

tan θ = 66.9150.2

The resultant force has a magnitude of 164.4 N and acts at an angle of 24° with each force. Therefore, a force of 164.4 N at an angle of 180° to the resultant must be applied to the object to create equilibrium.


From the given bearings: ∠WOH = 220° − 120° = 100° Since OHRW is a parallelogram: ∠OHR = 180° − 100° = 80° Use the Cosine Law to calculate r .

2r = 550 + 502 − 2(550)(50)cos 80° Ñ 543.6 Let ∠ROH = θ. The bearing of the plane is 120° + θ.

sin θ50 =

sin 80°543.6

sin θ = 50 sin 80°

543.6

θ Ñ5.2° Hence the bearing is 120° + 5.2° = 125.2°. The plane’s speed relative to the ground is 543.6 km/h on a bearing of 125.2°.



Method 2: Using Cartesian vectors The diagram at the right represents the situation. Bearings of 120° and 220° correspond to direction angles of 330° and 230° respectively. 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. a = [550 cos 330°, 550 sin 330°] w = [50 cos 230°, 50 sin 230°] r = a + w = [550 cos 330° + 50 cos 230°, 550 sin 330° + 50 sin 230°] Ñ[444.2, −313.3] The magnitude of the resultant is: r = (444.2)2 + (−313.3)2 Ñ 543.6 The bearing of the plane is 90° + θ .

tan θ = 313.3444.2


17. Let r be the resultant velocity of the boat. 22 25 −=r

= 21 Ñ4.6 The resultant speed of the boat is approximately 4.6 m/s. Since the river is 70 m wide, the time taken to cross the river,

is 7021

s, or 15.3 s.

To head directly across the river, the boat must take the heading shown in the diagram.

sin θ = 25

θ Ñ 23.6° Therefore, the boat’s heading is 90° − 23.5°, or 66.4° to the shoreline.



18. Use the formulababa ⋅=θcos .

a) 22 342

]3,4[]0,2[cos+

⋅=θ b) 2222 531)2(

]5,3[]1,2[cos++−

⋅−=θ

= 810 =

−6 + 55 34

θ Ñ 36.9° = −15 34

θ Ñ 94.4°

c) 2222 )3()1()2(4

]3,1[]2,4[cos−+−−+

−−⋅−=θ d) 2222 )1()2(62

]1,2[]6,2[cos−+−+

−−⋅=θ

= −4 + 620 10

= −4 − 6

40 5

= 2200

= −10200

θ Ñ 81.9° θ = 135°

19. AB = [8 − (−3), 1 − 5] = [11, −4] AC= [−2 − (−3), −1 − 5] = [1 , −6] BC = [−2 − 8, −1 − 1] = [−10, −2]

cos ∠A = ACAB

ACAB ⋅

2222 )6(1)4(11

]6,1[]4,11[

−+−+

−⋅−=

37137

2411+=

37137

25=

Ñ 60.6°

cos ∠B = BCBA

BCBA ⋅

2222 )2()10(4)11(

]2,10[]4,11[

−+−+−

−−⋅−=



1041378110 −=

37137

102=

Ñ 31.3° ∠C = 180° − ∠A − ∠B = 180° − 60.6° − 31.3° = 88.1°

20. a) )( cba +⋅ = [−1, 3]⋅([4, 2] + [−2, −1]) b) =⋅− cba )( ([−1, 3] −[4, 2])⋅[−2, −1] = [−1, 3]⋅[2, 1] = [−5, 1]⋅[−2, −1] = −2 + 3 = 10 − 1 = 1 = 9 c) =−⋅+ )3()2( baca (2[−1, 3] +[−2, −1])⋅([−1, 3] − 3[4, 2]) = [−4, 5]⋅[−13, −3] = 52 − 15 = 37

21. a) cabacba ⋅+⋅=+⋅ )( = [−1, 3]⋅[4, 2] + [−1, 3]⋅[−2, −1] = −4 + 6 + 2 − 3 = 1 b) cbcacba ⋅−⋅=⋅− )( = [−1, 3]⋅[−2, −1] − [4, 2]⋅[−2, −1] = 2 − 3 − (−8 −2) = 9 c) bcacbaaabaca ⋅−⋅+⋅−⋅=−⋅+ 362)3()2( = 2[−1, 3]⋅[−1, 3] − 6[−1, 3]⋅[4, 2] + [−2, −1]⋅[−1, 3] − 3[−2, −1]⋅[4, 2] = 2(1 + 9) − 6(−4 + 6) + 2 − 3 − 3(−8 − 2) = 37

22. a) vvuvvuuuvuvu ⋅+⋅+⋅+⋅=+⋅+ 362)2()3(

22 372 vvuu +⋅+=

b) bbabbaaababa ⋅−⋅+⋅−⋅=−⋅+ 1612129)43()43(

22 169 ba −=



23. a)

b) bbbbaba

⋅⋅=↓ a

aaabab

⋅⋅=↓

bb

ba

= 2

cosθ a

a

ab

= 2

cosθ

b

= 23

)150)(cos3)(5( a

= 25

)150)(cos5)(3(

Ñ−1.44b Ñ −0.52 a


Chapter 1, Self-Test, Sample Solutions Copyright 2003 Pearson Education Canada Inc., Toronto, Ontario 1

Chapter 1 Self-Test, Sample Solutions 1. a) ADCDDACD +=− b) ABPBCPDCAD =+++

CDBC += BD= c) BCDCCBDC +=− DCAD += AC=

2. ADCDCBABDACDBCAB +++=−+− )ADCB()CDAB( +++=

)BCCB()BAAB( +++= 00 += 0=

3. Method 1: 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 = 140° − 40° = 100° Since OHRW is a parallelogram: ∠OHR = 180° − 100° = 80°



Use the Cosine Law to calculate r .

2r = 4202 + 402 − 2(420)(40)cos 80° Ñ 414.9 Let ∠ROH = θ. The bearing of the plane is 140° − θ.

sin θ40 =

sin 80°414.9

sin θ = 40 sin 80°

414.9

θ Ñ5.5° Hence the bearing is 140° − 5.5° = 134.5° The plane’s speed relative to the ground is 414.9 km/h on a bearing of 134.5°. Method 2: Using Cartesian vectors The diagram at the right represents the situation. Bearings of 140° and 40° correspond to direction angles of 310° and 50° respectively. 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. a = [420 cos 310°, 420 sin 310°] w = [40 cos 50°, 40 sin 50°] r = a + w = [420 cos 310° + 40 cos 50°, 420 sin 310° + 40 sin 50°] Ñ[295.7, −291.1] The magnitude of the resultant is: r = (295.7)2 + (−291.1)2 Ñ 414.9 The bearing of the plane is 90° + θ .

tan θ = 291.1295.7


4. a) The vector bb1− has unit length and is opposite in direction to b .

b = 12 + 22

= 5



Therefore:

]2,1[5

11 −=− bb

−−=5

2,5

1

b) a = 42 + 62

= 52 = 2 13 Therefore, the required vector has magnitude 2 13 . Since this vector makes an angle of 60° with the positive x-axis, its components are [2 13 cos 60°, 2 13 sin 60°] Ñ [3.61, 6.24].

c) bbbbaba

⋅⋅=↓

]2,1[]2,1[]2,1[]2,1[]6,4[

⋅⋅=

= 4 + 12 1 + 4 [1,2]

= 16 5 [1, 2]

=

532,

516

5. ba − = [k, 2] − [7, 6]

= [k − 7, −4] Since ba − = 5:

(k − 7)2 + (−4)2 = 5 Square both sides and simplify. k2 − 14k + 49 + 16 = 25 k2 − 14k + 40 = 0 (k − 10)(k − 4) = 0 k = 10 or k = 4

6. We are given that AB = a and BC =b . Since DE and EF are parallel to AB and BC respectively, DE is a scalar multiple of AB and EF is a scalar multiple of BC . Thus, we can write DE = k a and EF = l b .



BCABAC += = a + b

EDFEFD += = −l b − k a = −k a − l b Since AC is parallel to FD : AC = s FD Substitute and in . a + b = s(−k a − l b ) a + b = −sk a − slb Equate the coefficients of a and b to obtain: 1 = −sk and 1 = −sl

s = − 1k s = −

1l

Therefore −1k = −

1l , or k = l.

7. a) See diagram at right. b) The heads of the vectors lie on a straight line.

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1.1 Exercises, Sample Solutions - PBworksmackenziekim.pbworks.com/w/file/fetch/63434086/chap1 soln.pdf · ... SR = PQ, SP = RQ c) J, L, and K are midpoints of sides AB, ... 13. a)