CHAPTER 6 Introduction To Vectors - Ms. Ma's Website
CHAPTER 6Introduction to VectorsReview of Prerequisite Skills, p. 2731"2e.22"3b. 2"3d.f. 122. Find BC using the Pythagorean theorem,AC 2 5 AB 2 1 BC 2.BC 2 5 AC 2 2 AB 25 102 2 625 64BC 5 8oppositeNext, use the ratio tan A 5 adjacent .BCtan A 5AB8564533. a. To solve D ABC, find measures of the sides andangles whose values are not given: AB, /B, and/C. Find AB using the Pythagorean theorem,BC 2 5 AB 2 1 AC 2.AB 2 5 BC 2 2 AC 25 (37.0)2 2 (22.0)25 885AB 5 "8858 29.7oppositeFind /B using the ratio sin B 5 hypotenuse.ACsin B 5BC22.0537.0/B 8 36.5 /C 5 90 2 /B/C 5 90 2 36.5 /C 8 53.5 b. Find measures of the angles whose values are notgiven. Find /A using the cosine law,b2 1 c2 2 a2.cos A 52bc1. a."32c.Calculus and Vectors Solutions Manual52 1 82 2 1022(5)(8)211580/A 8 97.9 Find /B using the sine law.sin Bsin A5absin (97.9 )sin B5510sin B 8 0.5/B 8 29.7 Find /C using the sine law.sin Csin A5casin Csin (97.9 )5810sin C 8 0.8/C 8 52.4 4. Since the sum of the internal angles of a triangleequals 180 , determine the measure of /Z using/X 5 60 and /Y 5 70 ./Z 5 180 2 (/X 1 /Y)5 180 2 (60 1 70 )5 50 Find XY and YZ using the sine law.XYXY5sin Ysin ZXY65sin 70 sin 50 XZ 8 7.36YZXY5sin Xsin ZYZ65sin 60 sin 50 YZ 8 6.785. Find each angle using the cosine law.RS 2 1 RT 2 2 ST 2cos R 52(RS)(RT)42 1 72 2 5252(4)(7)56-1
8.57/R 8 44 RS 2 1 ST 2 2 RT 2cos S 52(RS)(ST)42 1 5 2 2 7252(4)(5)1525/S 8 102 RT 2 1 ST 2 2 RS 2cos T 52(RT)(ST)72 1 52 2 4252(7)(5)29535/T 8 34 6.A3.5 km5TC100 km/hA48 80 km/hTFind AC and AT using the speed of each vehicle andthe elapsed time (in hours) until it was located,distance 5 speed 3 time.1AC 5 100 km h 3 h45 25 km1AT 5 80 km h 3 h325 26 km3Find CT using the cosine law.CT 2 5 AC 2 1 AT 2 2 2(AC)(AT) cos A225 (25 km)2 1 a26 kmb322 2(25 km) a26 kmb cos 48 38 443.94 km2CT 8 21.1 km9.A70 6 kmBFind AB (the distance between the airplanes) usingthe cosine law.AB 2 5 AT 2 1 BT 2 2 2(AT)(BT)cos T5 (3.5 km)2 1 (6 km)22 2(3.5 km)(6 km) cos 70 8 33.89 km2AB 8 5.82 km7.P2 km142 Q7 kmRFind QR using the cosine law.QR 2 5 PQ 2 1 PR 2 2 2(PQ)(PR) cos P5 (2 km)2 1 (7 km)22 2(2 km)(7 km) cos 142 8 75.06 km2QR 8 8.66 km6-25 cm5 cmCBCThe pentagon can be divided into 10 congruentright triangles with height AC and base BC.10 3 /A 5 360 /A 5 36 Find AC and BC using trigonometric ratios.AC 5 AB 3 cos A5 5 cos 36 8 4.0 cmBC 5 AB 3 sin A5 5 sin 36 8 2.9 cmThe area of the pentagon is the sum of the areas ofthe 10 right triangles. Use the area of ABC todetermine the area of the pentagon.Chapter 6: Introduction to Vectors
1Areapentagon 5 10 3 (BC)(AC)215 10 3 (2.9 cm)(4.0 cm)25 59.4 cm26.1 An Introduction to Vectors,pp. 279–2811. a. False. Two vectors with the same magnitudecan have different directions, so they are not equal.b. True. Equal vectors have the same direction andthe same magnitude.c. False. Equal or opposite vectors must be paralleland have the same magnitude. If two parallel vectorshave different magnitude, they cannot be equal oropposite.d. False. Equal or opposite vectors must be paralleland have the same magnitude. Two vectors with thesame magnitude can have directions that are notparallel, so they are not equal or opposite.2. Vectors must have a magnitude and direction. Forsome scalars, it is clear what is meant by just thenumber. Other scalars are related to the magnitudeof a vector. Height is a scalar. Height is the distance (see below)from one end to the other end. No direction is given. Temperature is a scalar. Negative temperatures arebelow freezing, but this is not a direction. Weight is a vector. It is the force (see below) ofgravity acting on your mass. Mass is a scalar. There is no direction given. Area is a scalar. It is the amount space inside atwo-dimensional object. It does not havedirection. Volume is a scalar. It is the amount of space insidea three-dimensional object. No direction is given. Distance is a scalar. The distance between twopoints does not have direction. Displacement is a vector. Its magnitude is relatedto the scalar distance, but it gives a direction. Speed is a scalar. It is the rate of change ofdistance (a scalar) with respect to time, but doesnot give a direction. Force is a vector. It is a push or pull in a certaindirection. Velocity is a vector. It is the rate of change ofdisplacement (a vector) with respect to time. Itsmagnitude is related to the scalar speed.Calculus and Vectors Solutions Manual3. Answers may vary. For example: Friction resiststhe motion between two surfaces in contact byacting in the opposite direction of motion. A rolling ball stops due to friction which resiststhe direction of motion. A swinging pendulum stops due to frictionresisting the swinging pendulum.4. Answersmayvary.For example: a. AD 5 BC; AB5DC; AE 5 EC ; DE 5 EBb. AD5 2CB5 2CD ; AB ; AE 5 2CE;ED 5 2EB; DA 5 2BC c. AC & DB; AE & EB; EC & DE; AB & CB5.HBDEJACFIG a. AB 5 CD b. AB 5 2EF c. @ AB @ 5 @ EF @ but AB 2 EF d. GH 5 2AB e. AB 5 22JI6. a.b.c.d.e.7. a. 100 km h, southb. 50 km h, westc. 100 km h, northeastd. 25 km h, northweste. 60 km h, east6-3
8. a. 400 km h, due southb. 70 km h, southwesterlyc. 30 km h southeasterlyd. 25 km h, due east9. a. i. False. They have equal magnitude, butopposite direction.ii. True. They have equal magnitude.iii. True. The base has sides of equal length, so thevectors have equal magnitude.iv. True. They have equal magnitude and direction.b. EHFGADCBTo calculate @ BD @ , @ BE @ and @ BH @ , find the lengthsof their corresponding line segments BD, BE andBH using the Pythagorean theorem.BD 2 5 AB 2 1 AD 25 32 1 32BD 5 "18BE 2 5 AB 2 1 AE 25 32 1 82BE 5 "73BH 2 5 BD 2 1 DH 25 ("18)2 1 82BH 5 "8210. a. The tangent vector describes James’s velocityat that moment. At point A his speed is 15 km hand he is heading north. The tangent vector showshis velocity is 15 km h, north.b. The length of the vector represents the magnitudeof James’s velocity at that point. James’s speed isthe same as the magnitude of James’s velocity.c. The magnitude of James’s velocity (his speed) isconstant, but the direction of his velocity changes atevery point.d. Point Ce. This point is halfway between D and A, which is78 of the way around the circle. Since he is running 6-4 15 km h and the track is 1 km in circumference, hecan run around the track 15 times in one hour. Thatmeans each lap takes him 4 minutes. 87 of 4 minutesis 3.5 minutes.f. When he has travelled 38 of a lap, James will behalfway between B and C and will be headingsouthwest. 11. a. Find the magnitude of AB using the distanceformula. @ AB @ 5 " (xA 2 xB )2 1 (yB 2 yA )25 " (24 1 1)2 1 (3 2 2)25 "10 or 3.16 b. CD 5 AB. AB moves from A(24, 2) toB(21, 3) or (xB, yB ) 5 (xA 1 3, yA 1 1). Use thisto find point D.(xD, yD ) 5 (xC 1 3, yC 1 1)5 (26 1 3, 0 1 1)5 (23, 1) c. EF 5 AB. Find point E using(xA, yA ) 5 (xB 2 3, yB 2 1).(xE, yE ) 5 (xF 2 3, yF 2 1)5 (3 2 3, 22 2 1)5(0, 23) d. GH 5 2AB, and moves in the opposite direction as AB.(xH, yH ) 5 (xG 2 3, yG 2 1).(xH, yH ) 5 (xG 2 3, yG 2 1)5 (3 2 3, 1 2 1)5 (0, 0)6.2 Vector Addition, pp. 290–2921. a.xyx yb.–yx –yxc.yy –x–xChapter 6: Introduction to Vectors
d.c.–x–b–ya–y –x –ca–b–cA2. a. BAd.C b. 0C caAc. CB–bBa–b cB4. a.Ab(b c)aC d. CABAa (b c)b.C3. a.cbBac(a b)c(a b) cbc. The resultant vectors are the same. The order inwhich you add vectors does not matter. Aa 1 b B 1 c 5 a 1 Ab 1 c B a b c5. a. PSaR–RQQRSb.PRS–cbPS PQPQb. 0–RSa b–cSRQR–PQaCalculus and Vectors Solutions ManualPSP 6. x 1 y 5 MR 1 RS 5 MS z 1 t 5 ST 1 TQ5 SQso (x 1 y ) 1 (z 1 t ) 5 MS 1 SQ5 MQ6-5
7. a. 2x b. y c. x 1 y d. 2x 1 y e. x 1 y 1 z f. 2x 2 y g. 2x 1 y 1 z h. 2x 2z8. a.11.wa wuyByuy –xCux –yxxDAyb. See the figure in part a. for the drawn vectors. 0 y 2 x 0 2 5 0 y 0 2 1 0 x 0 2 2 20 y 0 0 2x 0 cos (u) and 0 2x 0 5 0 x 0 , so 0 y 2 x 0 2 5 0 x 2 y 0 29. a. Maria’s velocity is 11 km h downstream.b.4 km/h11 km/h7 km/hc.4 km/h7 km/h3 km/hMaria’s speed is 3 km h.10. a.f1 f2f1ushipf2 Find 0 a 1 w 0 using the Pythagorean theorem. 2 0 a 1 w 0 5 0 a 02 1 0 w 025 (150 km h)2 1 (80 km h)25 28 900 0 a 1 w 0 2 5 170 Find the direction of a 1 w using the ratio 0w0tan(u) 5 0a080 km hu 5 tan21150 km h8 N 28.1 W a 1 w 5 170 km h, N 28.1 W 12. x , y , and x 1 y form a right triangle. Find 0 x 1 y 0 using the Pythagorean theorem. 0 x 1 y 02 5 0 x 02 1 0 y 025 72 1 2425 625 0 x 1 y 0 5 25 Find the angle between x and x 1 y using the ratio 0y0tan (u) 5 0x024u 5 tan2178 73.7 13. Find @ AB 1 AC @ using the cosine law and thesupplement to the angle between AB and AC. @ AB 1 AC @ 2 5 @ AB @ 2 1 @ AC @ 2 2 2 @ AB @ @ AC @ cos (30 )"35 12 1 12 2 2(1)(1)25 2 2 "3 @ AB 1 AC @ 8 0.5214.DCb. The vectors form a triangle with side lengthsSSSSSS@ f1 @ , @ f2 @ and @ f1 1 f2 @ . Find @ f1 1 f2 @ using thecosine law.@ f1 1 f @ 2 5 @ f1 @ 2 1 @ f2 @ 2 2 2 @ f1 @ @ f2 @ cos (u)2SSSSS6-6SSSES@ f1 1 f2 0 5 @ f1 @ 2 1 @ f2 @ 2 2 2 @ f1 @ @ f2 @ cos (u)SaSSABThe diagonals of a parallelogram bisect each other. 5 2EB. So EA 5 2EC and ED Therefore, EA 1 EB 1 EC 1 ED 5 0.Chapter 6: Introduction to Vectors
15. PTabMRbaQ6.3 Multiplication of a Vector by aScalar, pp. 298–3011. A vector cannot equal a scalar.2. a.SMultiple applications of the Triangle Law foradding vectors show that RM 1 b 5 a 1 TP (since both are equal to theundrawn vector TM ), and that RM 1 a 5 b 1 SQ (since both are equal to theundrawn vector RQ )Adding these two equations gives 2 RM 1 a 1 b 5 a 1 b 1 TP1SQ 2 RM 5 TP 1 SQ 16. a 1 b and a 2 b represent the diagonals of a parallelogram with sides a and b.3 cmb.9 cmc.2 cmd.6 cma ba–bba Since @ a 1 b @ 5 @ a 2 b @ and the only parallelogramwith equal diagonals is a rectangle, the parallelogrammust also be a rectangle.17.P3. E25 N describes a direction that is 25 toward thenorth of due east (90 east of north), in other words90 2 25 5 65 toward the east of due north. N65 Eand “a bearing of 65 ” both describe a direction thatis 65 toward the east of due north. So, each isdescribing the same direction in a different way.4. Answers may vary. For example:GQM vRLet point M be defined as shown. Two applicationsof theLaw for adding vectors show that Triangle GQ 1 QM 1 MG 5 0 GR 1 RM 1 MG 5 0Adding these two equations gives GQ 1 QM 1 2 MG 1 GR 1 RM 5 0From the given information,2 MG5GP and QM 1 RM 5 0 (since they are opposing vectors ofequal length), so GQ 1 GP 1 GR 5 0, as desired.a. 2vb.c.1 v2d.e. 22vCalculus and Vectors Solutions Manual2 2 v31 v0v06-7
5. a.b.c.y–3b3byd.aabyx 3ybb2a 3be.x–b2a – 3b–b–bb.–yx–yc.–2x y–x–xyd.–x–x–2x – y–y 3 c 5 2a , b 5 a2 mc 1 nb 5 0 3 m(2a ) 1 n a a b 5 02 m(4a ) 1 n (3a ) 5 0m 5 3 and n 5 24 satisfy the equation, as does anymultiple of the pair (3, 24). There are infinitelymany values possible. 3 c 5 2a , b 5 ab.2 da 1 eb 1 f c 5 0 3 da 1 ea a b 1 f(2a ) 5 02 2da 1 3ea 1 4f a 5 0d 5 2, e 5 0, and f 5 21 satisfy the equation, asdoes any multiple of the triple (2, 0, 21). There areinfinitely many values possible.8.or7. a.–yx – 3y a and b are collinear, so a 5 kb, where k is a nonzero scalar. Since 0 a 0 5 @ b @ , k can only be 21 or 1.9.ba4a6. Answers may vary. For example:aa.aab–2b2aYes6-8Chapter 6: Introduction to Vectors
10. Two vectors are collinear if and only if they can be related by a scalar multiple. In this case a 2 kba. collinearb. not collinearc. not collineard. collinear1 11. a. 0 x 0 x is a vector with length 1 unit in the same direction as x .1 b. 2 0 x 0 x is a vector with length 1 unit in the opposite direction of x .12.aabbb232 13. a. 2 a31 b. a31 c. 0 a 032 d. 0 a 034 e. a3 14. x and y make an angle of 90 , so you may find 0 2x 1 y 0 using the Pythagorean theorem. 0 2x 1 y 0 2 5 0 2x 0 2 1 0 y 0 22252 11 0 2x 1 y 0 5 "5 or 2.24 Find the direction of 2x 1 y using the ratio 0y0tan (u) 5 0 2x 01u 5 tan212 8 26.6 from x towards 2x 1 y 15. Find 0 2x 1 y 0 using the cosine law, and the supplement to the angle between x and y . 0 2x 1 y 0 2 5 0 2x 0 2 1 0 y 0 2 2 20 2x 0 0 y 0 cos (150 )sin u 8 (1)122.91 u 8 9.9 from x towards y 1 16. b 5 a0a0 1 @b@ 5 a 0a0 1 @b@ 5 0 a 00a0 @b@ 5 1 b is a positive multiple of a , so it points in the same direction as a and has magnitude 1. It is a unit vector in the same direction as a .17.Am55 22 1 12 2 2(2)(1)cADCCalculus and Vectors Solutions ManualBD AD 5 c 1 CD AD 5 b 1 BD 2 AD 5 c 1 b 1 CD 1 BDBut CD 1 BD 5 0. 1 1 So 2 AD 5 c 1 b, or AD 5 2 c 1 2 b. 18. PM 5 a and PN 5 bso MN 5 PN 2 PM2a 5b PQ 5 2a and PR 5 2bso QR 5 PR 2 PQ 5 2b 2 2aNotice that 2MN 5 2b 2 2a5 QR We can conclude that QR is parallel to MN and @ QR @ 5 2 @ MN @ .19.AB2"32 0 2x 1 y 0 8 2.91 Find the direction of 2x 1 y using the sine law.sin usin (150 ) 5 0y00 2x 1 y 0bECD6-9
Answers mayvary. For example: a. u 5 AB and v 5 CD b. u 5 AD and v 5 AE c. u 5 AC and v 5 DB d. u 5 ED and v 5 AD 20. a. Since the magnitude of x is three times the magnitude of y and because the given sum is 0, mx must be in the opposite direction of ny and0 n 0 5 30 m 0. b. Whether x and y are collinear or not, m 5 0 andgiven equation true.n 5 0 will make the 21. a. CD 5 b 2 a b. BE 5 2b 22a 5 2(b 2 a ) 5 2CDThe two are thereforeparallel (collinear) and @ BE @ 5 2 @ CD @22.ABED AC 5 AE 1 EC , the original equation gives 33AE 1 AE 5 AD 1 AB22 53AE 5 AD 1 AB22 23AE 5 AD 1 AB556.4 Properties of Vectors, pp. 306–3071. a. 0b. 1 c. 0d. 16-10a bbbb aa3.cb(b c)(a b)a(a b) c a (b c) a b c4. Answers may vary. For example:CAE 5 EC, and since abApplying the triangle law for adding vectors showsthat AC 5 AD 1 DCThe given information states that 2AB 5 DC3 3AB 5 DC2By the properties of trapezoids, this gives32 2. a 1 b 5 b 1 aak(a b)kb a bka 5. PQ 5 RQ 1SR 1TS 1 PT 5 (RQ 1 SR ) 1 (TS 1 PT )5 (SR1 RQ) 1 (PT 1 TS ) 5 SQ 1 PS 5 PS 1 SQ5 PQ6. a. ECb. 0c. Yes, the diagonals of a rectangular prism are ofequal length 7. 5 3a 2 6b 2 15c 2 6a 1 12b 2 6c 2 a 1 3b 2 3c 5 24a 1 9b 2 24c 8. a. 5 6i 28j 1 2k 1 6i 2 9j 1 3k 5 12i 2 17j 1 5k b. 5 3i 2 4j 1 k 2 10i 1 15j 2 5k 5 27i 1 11j 2 4k c. 5 2(3i 2 4j 1 k 1 6i 2 9j 13k) 23(26i18j22k114i221j1 7k ) 5 26i 1 13j 2 7k 9. Solve the first equation for x . 1 3 x5 a2 y22Chapter 6: Introduction to Vectors
Substitute into the second equation. 1 3 6b 5 2 a a 2 y b 1 5y221 12 by5 a11313 Lastly, find x in terms of a and b. 1 3 1 12 x5 a2 a a1bb22 13135 18 5 a2 b1313 10. a 5 x 2 y 2 1 5 y 1 z 2 (b 1 z )33 2 2 5 y2 z2b33 2 5 (y 2 z ) 2 b3 2 5 b2b31 52 b 3 11. a. AG 5 a 1 b 1 c BH 5 2a 1 b 1 c CE 5 2a 2 b 1 c DF 5 a 2 b 1 c 2 22b. @ AG @ 5 0 a 0 1 @ b @ 1 0 c 0 2 25 0 2a 0 2 1 @ b @ 1 0 c 0 2 5 @ BH @ 2Therefore, @ AG @ 5 @ BH @ T12.Z XOYApplying the triangle law for adding vectorsshows that TY 5 TZ 1 ZYThe given information states that TX 5 2 ZY 1TX 5 ZY2By the properties of trapezoids, this gives 12 TO 5 OY, and since TY 5 TO 1 OY, theoriginal equation gives 11TO 1 TO 5 TZ 1 TX22Calculus and Vectors Solutions Manual 31TO 5 TZ 1 TX22 21TO 5 TZ 1 TX33Mid-Chapter Review, pp. 308–309 1. a. AB 5 DC BA 5 CD AD 5 BC CB 5 DAThere is not enough informationto determine if there is a vector equal to AP.b. @ PD @ 5 @ DA @ 5 @ BC @ (parallelogram) 2. a. RV b. RV c. PS d. RU e. PS f. PQ 3. a. Find @ a 1 b @ using the cosine law, and thesupplement to the angle between the vectors. @ a 1 b @ 2 5 0 a 0 2 1 @ b @ 2 2 2 0 a 0 @ b @ cos 60 15 32 1 42 2 2(3)(4)253 @ a 1 b @ 5 "3b. Find u usingthe ratio @b@tan u 5 0a04534u 5 tan2138 53 4. t 5 4 or t 5 245. In quadrilateral PQRS, look at PQR. Joining themidpoints B and C creates a vector BC that is parallel to PR and half the length of PR. Look at SPR. Joining the midpoints A and D creates a vector AD that is parallel to PR and half the length of PR. BC is parallel to AD and equal in length to AD.Therefore, ABCD is a parallelogram. 6. a. Find 0 u 2 v 0 using the cosine law. Note 0 2v 0 5 0 v 0 and the angle between u and 2v is 120 . 2 2 2 0 u 2 v 0 5 0 u 0 1 0 2v 0 2 2 0 u 0 0 2v 0 cos 60 6-11
15 8 1 10 2 2(8)(10)a b2 0 u 2 v 0 5 2"21 b. Find the direction of u 2 v using the sine law.sin usin 60 5 0 2v 00u 2 v05sin u 5sin 60 "215u 5 sin21"288 71 11 (u 1 v )c. (u 1 v ) 50u 1 v02"21 d. Find 0 5u 1 2v 0 using the cosine law. 0 5u 1 2v 0 2 5 0 5u 0 2 1 0 2v 0 2 2 2 0 5u 0 0 2v 0 cos 120 15 402 1 202 2 2(40)(20)a2 b2 0 5u 1 2v 0 5 20"7 7. Find 0 2p 2 q 0 using the cosine law. 0 2p 2 q 0 2 5 0 2p 0 2 1 0 2q 0 2 2 2 0 2p 0 0 2q 0 cos 60 15 22 1 12 2 2(2)(1)a b 5 32 8. 0 m 1 n 0 5 0 m 02 0 n 0 9. BC 5 2y DC 5 x BD 5 2x 2 y AC 5 x 2 y 10. Construct a parallelogram with sides OA and OC .Since the diagonals bisect each other, 2OB is the diagonal equal to OA 1 OC. Or OB 5 OA 1 ABand AB 5 12 AC. So, OB 5 OA 1 12 AC. And AC 5 OC 2 OA . Now OB 5 OA 1 12 ( OC 2 OA) Multiplying by 2 gives 2OB 5 OA 1 OC. 11. AC 1 CD 5 AD 3x 2 y 1 2y 5 AD 3x 1 y 5 AD AB 1 BD 5 AD x 1 BD 5 3x 1 y BD 5 2x 1 y AB 1 BC 5 AC x 1 BC 5 3x 2 y BC 5 2x 2 y 12. The air velocityof the airplane (Vair ) and the wind velocity (W ) have opposite directions.2 2 Vground 5 Vair 2 W5 460 km h due south6-12 13. a. PTb. PT c. SR14. a. 1 a133a bbb.32–2 bc.a–bad.12a12 b 15. PS 5 PQ1 QS 53b2a RS 5 QS 2 QR 5 23a6.5 Vectors in R 2 and R 3, pp. 316–3181. No, as the y-coordinate is not a real number.2. a. We first arrange the x-, y-, and z-axes (each acopy of the real line) in a way so that each pair ofaxes are perpendicular to each other (i.e., the x- andy-axes are arranged in their usual way to form thexy-plane, and the z-axis passes through the origin ofthe xy-plane and is perpendicular to this plane).This is easiest viewed as a “right-handed system,”where, from the viewer’s perspective, the positivez-axis points upward, the positive x-axis points outof the page, and the positive y-axis points rightwardin the plane of the page. Then, given point P(a, b, c),we locate this point’s unique position by moving aunits along the x-axis, then from there b unitsparallel to the y-axis, and finally c units parallel tothe z-axis. It’s associated unique position vector isdetermined by drawing a vector with tail at theorigin O(0, 0, 0) and head at P.b. Since this position vector is unique, itscoordinates are unique. Therefore a 5 24, b 5 23,and c 5 28.3. a. Since A and B are really the same point, wecan equate their coordinates. Therefore a 5 5,b 5 23, and c 5 8.Chapter 6: Introduction to Vectors
zb. From part a., A(5, 23, 8), so OA 5 (5, 23, 8).Here is a depiction of this vector.zOA (5, –3, 8)(0, 4, 0)yO(0, 0, 0)(0, 0, –2)(4, 0, 0)y(4, 0, –2)x4. This is not an acceptable vector in I 3 as thez-coordinate is not an integer. However, since all ofthe coordinates are real numbers, this is acceptableas a vector in R 3.5.zO(0, 0, 0)(0, –4, –2)(0, –4, 0)y(4, –4, 0)(0, 4, –2)(4, 4, 0)C(4, 4, –2)x6. a. A(0, 21, 0) is located on the y-axis.B(0, 22, 0), C(0, 2, 0), and D(0, 10, 0) are threeother points on this axis. b. OA 5 (0, 21, 0), the vector with tail at theorigin O(0, 0, 0) and head at A.7. a. Answers may vary.For example: OA 5 (0, 0, 1), OB 5 (0, 0, 21),OC 5 (0, 0, 25)b. Yes, these vectors are collinear (parallel), as theyall lie on the same line, in this case the z-axis.c. A general vectorlying on the z-axis would be of the form OA 5 (0, 0, a) for any real number a.Therefore, this vector would be represented byplacing the tail at O, and the head at the point(0, 0, a) on the z-axis.z8.(0, 0, –2)F(0, 2, 3)(4, 0, 0)xA(4, –4, –2)E(2, 0, 3)(4, 0, –2)B(0, –2, 0)z(–4, 0, 2) B(–4, 4, 2)(–4, 0, 0)(–4, 4, 0)y(0, 4, 2)D(2, 3, 0)C(0, 0, –3)x(0, 0, 2)O(0, 0, 0)yA(1, 0, 0)z9. a.(0, 4, 0)yxxA(3, 2, –4)Calculus and Vectors Solutions ManualC(0, 1, –4)B(1, 1, –4)6-13
b. Every point on the plane containing points A, B,and C has z-coordinate equal to 24. Therefore, theequation of the plane containing these points isz 5 24 (a plane parallel to the xy-plane throughthe point z 5 24).10. a.A(1, 2, 3)(0, 0, 3)zd.(1, 0, 1)z(0, 2, 0)y(1, 0, 0)(1, 2, 0)(1, 0, 0)ze.(0, –1, 1)x(0, 0, 1)E(1, –1, 1)(1, 0, 1)(0, –1, 0)zb.(–2, 0, 0)(–2, 0, 1)B(–2, 1, 1)(1, 0, 0)x(0, 1, 1)(0, 1, 0)xyzf.(1, 0, 0)O(0, 0, 0)(0, –1, 0)(1, –1, 0)zc.C(1, –2, 1) (0, –2, 0)xF(1, –1, –1)(0, 0, 1)yy(0, 0, –1)(1, 0, –1) 11. a. OA 5 (1, 2, 3)O(0, 0, 0)(1, 0, 0)O(0, 0, 0)(0, –1, –1)(0, –2, 1)(1, –2, 0)yO(0, 0, 0)(1, –1, 0)(–2, 1, 0)(0, 0, 1)y(0, 1, 0)(1, 1, 0)xO(0, 0, 0)(0, 1, 1)O(0, 0, 0)(0, 2, 3)(1, 0, 3)D(1, 1, 1)(0, 0, 1)(0, 0, 3)(1, 0, 1)zA(1, 2, 3)(0, 2, 3)(1, 0, 3)O(0, 0, 0)(1, 0, 0)OA(0, 2, 0)y(1, 2, 0)x6-14Chapter 6: Introduction to Vectors
b. OB 5 (22, 1, 1)f. OF 5 (1, 21, 21)z (–2, 0, 1)(–2, 0, 0)zB(–2, 1, 1)OB(1, 0, 0)(0, –1, 0)(1, –1, 0)(–2, 1, 0)(0, 0, 1)(0, 1, 1)y(0, 1, 0)(0, –1, –1)F(1, –1, –1)O(0, 0, 0)xx c. OC 5 (1, 22, 1)z(0, –2, 1)(0, –2, 0) (1, 0, 1)C(1, –2, 1)(0, 0, 1)OCyO(0, 0, 0)(1, –2, 0)(1, 0, 0)x d. OD 5 (1, 1, 1)zD(1, 1, 1)(0, 0, 1)(1, 0, 1)y(0, 1, 0)(1, 1, 0)(1, 0, 0)x(0, 1, 1)ODO(0, 0, 0) e. OE 5 (1, 21, 1)z(0, –1, 1)E(1, –1, 1)(0, 0, 1)(1, 0, 1)(0, –1, 0)O(0, 0, 0)(1, –1, 0)x(1, 0, 0)yO(0, 0, 0)(0, 0, –1)OF(1, 0, –1)12. a. Since P and Q represent the same point,we can equate their y- and z-coordinates to get thesystem of equationsa2c56a 5 11Substituting this second equation into the first gives11 2 c 5 6c55So a 5 11 and c 5 5.b. Since P and Q represent the same point in R 3,they will have the same associated position vector,i.e. OP 5 OQ. So, since these vectors are equal,they will certainly have equal magnitudes, i.e. @ OP @ 5 @ OQ @ .13. P(x, y, 0) represents a general point on thexy-plane, since the z-coordinate is 0. Similarly,Q(x, 0, z) represents a general point in the xz-plane,and R(0, y, z) represents a general point in theyz-plane.14. a. Every point on the plane containing points M,N, and P has y-coordinate equal to 0. Therefore, theequation of the plane containing these points isy 5 0 (this is just the xz-plane).b. The plane y 5 0 contains the origin O(0, 0, 0),and so since it also contains the points M, N, and Pas well, it will contain the position vectors associatedwith these points joining O (tail) to the given point(head). That is, the plane y 5 0 contains the vectors OM, ON, and OP.15. a. A(22, 0, 0), B(22, 4, 0), C(0, 4, 0),D(0, 0, 27), E(0, 4, 27), F(22, 0, 27) yOECalculus and Vectors Solutions Manual b. OA 5 (22, 0, 0), OB 5 (22, 4, 0), OC 5 (0, 4, 0), OD 5 (0, 0, 27), OE 5 (0, 4, 27), OF 5 (22, 0, 27)c. Rectangle DEPF is 7 units below the xy-plane.6-15
d. Every point on the plane containing points B, C,E, and P has y-coordinate equal to 4. Therefore, theequation of the plane containing these points isy 5 4 (a plane parallel to the xz-plane through thepoint y 5 4).e. Every point contained in rectangle BCEP hasy-coordinate equal to 4, and so is of the form(x, 4, z) where x and z are real numbers such that22 # x # 0 and 27 # z # 0.16. a.yzd.O(0, 0, 0)yM(–1, 3, –2)xze.F(0, 0, 5)O(0, 0)xO(0, 0, 0)P(4, –2)xyb.yD(–3, 4)zf.O(0, 0)xJ(–2, –2, 0)yO(0, 0, 0)xzc.C(2, 4, 5)O(0, 0, 0)y17. The following box illustrates the three dimensionalsolid consisting of the set of all points (x, y, z) suchthat 0 # x # 1, 0 # y # 1, and 0 # z # 1.zO(0, 0, 0)x(0, 0, 1)(1, 1, 1)(1, 0, 1)(1, 0, 0)x6-16(0, 1, 1)y(0, 1, 0)(1, 1, 0)Chapter 6: Introduction to Vectors
18. First, OP 5 OA 1 OB by the triangle law of5 (5, 210, 0),OAvectoraddition,where OB 5 (0, 0, 210), OP and OA are drawn instandard position (starting from the origin O(0, 0, 0)),and OB is drawn starting from the head of OA .Notice that OA lies in the xy-plane, and OB isperpendicular to the xy-plane (so is perpendicular toOA ). So, OP, OA, and OB form a right triangleand, 22by the Pythagorean theorem, @ OP @ 5 @ OA @ 1 @ OB @ 2 Similarly, OA 5 a 1 b by the triangle law of vectoraddition, where a 5 (5, 0, 0) and b 5 (0, 210, 0), and these three vectors form aright triangle as well. So, @ OA @ 2 5 0 a 0 2 1 @ b @ 25 25 1 1005 125 Obviously @ OB @ 2 5 100, and so substituting gives@ OP @ 2 5 @ OA @ 2 1 @ OB @ 2 6.6 Operations with AlgebraicVectors in R 2, pp. 324–326yA(–1, 3)yABB(2, 5)A(–1, 3)BAx 5 125 1 1005 225 @ OP @ 5 "2255 15 19. To find a vector AB equivalent toOP 5 (22, 3, 6), where B(4, 22, 8), we need tomove 2 units to the right of the x-coordinate for B(to 4 1 2 5 6), 3 units to the left of the y-coordinatefor B (to 22 2 3 5 25), and 6 units below thez-coordinate for B (to 8 2 6 5 2). So we get thepoint A(6, 25, 2). Indeed, notice that to get from Ato B (which describes vector AB ), we move 2 unitsleft in the x-coordinate, 3 units right in they-coordinate, and 6 units up in the z-coordinate.This is equivalent to vector OP 5 (22, 3, 6).1. a. AB 5 (2, 5) 2 (21, 3)5 (3, 2) BA 5 2AB5 2 (3, 2)5 (23, 22)Here is a sketch of these two vectors in thexy-coordinate plane.B(2, 5)xCalculus and Vectors Solutions Manualb. 0 OA 0 5 "(21)2 1 325 "108 3.16 0 OB 0 5 "22 1 52 5 "29 8 5.39c. 0 AB 0 5 "32 1 225 "138 3.61 BA52AB,Also,since 0BA 0 5 0 2AB 0 5 0 21 0 ? 0 AB 05 0 AB 0 5 "138 3.612.30252015105O(0, 0)–9 –6 –3 0–5–10yA(6, 10)xOA3 6 96-17
a.2015105O(0, 0)–12 –9 –6 –3 0(–3, –5) –5–10–15–20(–12, –20)y(12, 20)(3, 5)x3 6 9 12b. The vectors with the same magnitude are 11OA and 2 OA,22 2OA and 22OA 3. @ OA @ 5 "32 1 (24)25 "255 54. a. The i -component will be equal to the firstcoordinate in componentform, and so a 5 23. Similarly, the j -component will be equal to thesecond coordinate in component form, and so b 5 5.b. 0 ( 23, b)0 5 0 ( 23, 5)05 "(23)2 1 525 "348 5.83 5. a. 0 a 0 5 "(260)2 1 1125 "37215 61 @ b @ 5 "(240)2 1 (29)25 "1681 5 41 b. a 1 b 5 (260, 11) 1 (240, 29)5 (2100, 2) @ a 1 b @ 5 "(2100)2 1 225 "10 004 8 100.02 a 2 b 5 (260, 11) 2 (240, 29)5 (220, 20) @ a 2 b @ 5 "(220)2 1 20 25 "8008 28.286. a. 2(22, 3) 1 (2, 1) 5 (2(22) 1 2, 2(3) 1 1)5 (22, 7)b. 23(4, 29) 2 9(2, 3)5 (23(4) 2 9(2), 23(29) 2 9(3))5 (230, 0)6-1812c. 2 (6, 22) 1 (6, 15)2321215 a2 (6) 1 (6), 2 (22) 1 (15)b23235 (1, 11) 7. x 5 2i 2 j, y 5 2i 1 5j a. 3x 2 y 5 3(2i 2 j ) 2 (2i 1 5j ) 5 (6 1 1)i 1 (23 2 5)j 5 7i 2 8j b. 2 (x 1 2y ) 1 3(2x 2 3y ) 5 24x 2 11y 5 24(2i 2 j ) 2 11(2i 1 5j ) 5 3i 2 51j c. 2(x 1 3y ) 2 3(y 1 5x ) 5 213x 1 3y 5 213(2i 2 j ) 1 3(2i 1 5j ) 5 229i 1 28j 8. a. x 1 y 5 (2i 2 j ) 1 (2i 1 5j ) 5 i 1 4j 0x 1 y 0 5 @ i 1 4j @5 "12 1 425 "178 4.12 b. x 2 y 5 (2i 2 j ) 2 (2i 1 5j ) 5 3i 2 6j 0x 2 y 0 5 @ 3i 2 6j @5 "32 1 (26)25 "458 6.71 c. 2x 2 3y 5 2(2i 2 j ) 2 3(2i 1 5j ) 5 7i 2 17j 0 2x 2 3y 0 5 @ 7i 2 17j @5 "72 1 (217)25 "3388 18.38 d. 0 3y 2 2x 0 5 0 2 (2x 2 3y ) 0 5 0 21 0 0 2x 2 3y 0 5 0 2x 2 3y 0so, from part c., 0 3y 2 2x 0 5 0 2x 2 3y 05 "3388 18.38Chapter 6: Introduction to Vectors
9.B(–4, 4)A(–8, 2)8642so obviously we will have @ OA @ 5 @ BC @ .(It turns out that their common magnitude is"62 1 32 5 "45.)11. a.y yD(4, 5)xC(2, 1)–8 –6 –4 –2 0 2 4 6 8–2F(–7, 0)H(6, –2)–4 G(1, –2)E(–1, –4)–6–8 C(–4, 11)B(6, 6)A(2, 3)x a. AB 5 (24, 4) 2 (28, 2)5 (4, 2) CD 5 (4, 5) 2 (2, 1)5 (2, 4) EF 5 (27, 0) 2 (21, 24)5 (26, 4) GH 5 (6, 22) 2 (1, 22)5 (5, 0) b. @ AB @ 5 "42 1 225 "20 8 4.47@ CD @ 5 "22 1 425 "208 4.47 @ EF @ 5 "(26)2 1 425 "528 7.21 @ GH @ 5 "52 1 025 "255510. a. By the parallelogram law of vector addition, OC 5 OA 1 OB5 (6, 3) 1 (11, 26)5 (17, 23)For the other vectors, BA 5 OA 2 OB5 (6, 3) 2 (11, 26)5 (25, 9) BC 5 OC 2 OB5 (17, 23) 2 (11, 26)5 (6, 3) b. OA 5 (6, 3) 5 BC,Calculus and Vectors Solutions Manual b. AB 5 (6, 6) 2 (2, 3)5 (4, 3) @ AB @ 5 "42 1 325 "25 55AC 5 (24, 11) 2 (2, 3)5 (26, 8) @ AC @ 5 "(26)2 1 825 "100 5 10CB 5 (6, 6) 2 (24, 11)5 (10, 25) @ CB @ 5 "102 1 (25)25 "125 8 11.18 @c. CB @ 2 5 125 @ AC @ 2 5 100, @ AB @ 2 5 25Since @ CB @ 2 5 @ AC @ 2 1 @ AB @ 2, the triangle is a righttriangle.12. a.y C(2, 8)A(–1, 2)xB(7, –2)6-19
b.X(–6, 12)yC(2, 8)Z(10, 4)A(–1, 2)xB(7, –2)Y(4, –8)c. As a first possibility for the fourth vertex, there isX(x1, x2 ). From the sketch in part b., we see that wewould then have CX 5 BA(x1 2 2, x2 2 8) 5 (21 2 7, 2 2 (22))5 (28, 4)x1 2 2 5 28x2 2 8 5 4So X(26, 12). By similar reasoning for the otherpoints labelled in the sketch in part b.,AY 5 CB(y1 2 (21), y2 2 2) 5 (7 2 2, 22 2 8)5 (5, 210)y1 1 1 5 5y2 2 2 5 210So Y(4, 28). Finally, BZ 5 AC(z1 2 7, z2 2 (22)) 5 (2 2 (21), 8 2 2)5 (3, 6)z1 2 7 5 3z2 1 2 5 6So Z(10, 4). In conclusion, the three possiblelocations for a fourth vertex in a parallelogramwith vertices A, B, and C are X(26, 12), Y(4, 28),and Z(10, 4).13. a. 3(x, 1) 2 5(2, 3y) 5 (11, 33)(3x 2 5(2), 3 2 5(3y)) 5 (11, 33)(3x 2 10, 3 2 15y) 5 (11, 33)3x 2 10 5 113 2 15y 5 33So x 5 7 and y 5 22.b. 22(x, x 1 y) 2 3(6, y) 5 (6, 4)(22x 2 18, 22x 2 5y) 5 (6, 4)22x 2 18 5 622x 2 5y 5 4To solve for x, use22x 2 18 5 6x 5 2126-20Substituting this into the last equation above, wecan now solve for y.22(212) 2 5y 5 4y54So x 5 212 and y 5 4.14. a.yC(x, y)B(–6, 9)D(8, 11)A(2, 3)xb. Because ABCD is a rectangle,we will have BC 5 AD(x, y) 2 (26, 9) 5 (8, 11) 2 (2, 3)(x 1 6, y 2 9) 5 (6, 8)x1656y2958So, x 5 0 and y 5 17, i.e., C(0, 17). 15. a. Since 0 PA 0 5 0 PB 0, and PA 5 (5, 0) 2 (a, 0)5 (5 2 a, 0), PB 5 (0, 2) 2 (a, 0)5 (2a, 2),this means that(5 2 a)2 5 (2a)2 1 2225 2 10a 1 a 2 5 a 2 1 410a 5 2121a51021So Pa , 0b.10b. This point Q on the y-axis will be of the formQ(0, b) for some real number b. Reasoningsimilarlyto part a., we have QA 5 (5, 0) 2 (0, b)5 (5, 2b) QB 5 (0, 2) 2 (0, b)5 (0, 2 2 b) So since @ QA @ 5 @ QB @ ,(2b)2 1 52 5 (2 2 b)2b 2 1 25 5 4 2 4b 1 b 24b 5 22121b52421So Qa0, 2 b.4Chapter 6: Introducti
6.1 An Introduction to Vectors, pp. 279-281 1. a.False. Two vectors with the same magnitude can have different directions, so they are not equal. b. True. Equal vectors have the same direction and the same magnitude. c. False. Equal or opposite vectors must be parallel and have the same magnitude. If two parallel vectors
Part One: Heir of Ash Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26 Chapter 27 Chapter 28 Chapter 29 Chapter 30 .
I think of atomic vectors as “just the data” Atomic vectors are the building blocks for augmented vectors Augmented vectors Augmented vectors are atomic vectors with additional attributes attach
Chapter 6 139 Vectors and Scalars (ii) Vectors Addition is Associative: i.e. a b c a b c where . a , b . and . c . are any three vectors. (iii) O is the identity in vectors addition: Fig.9. For every vector . a O a Where . O. is the zero vector. Remarks: Non-parallel vectors are not added or subtracted by the .
TO KILL A MOCKINGBIRD. Contents Dedication Epigraph Part One Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Part Two Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18. Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26
Two nonparallel vectors always define a plane, and the angle is the angle between the vectors measured in that plane. Note that if both a and b are unit vectors, then kakkbk 1, and ab cos . So, in general if you want to find the cosine of the angle between two vectors a and b, first compute the unit vectors aˆ and bˆ in the directions of a .
Draw vectors on your map from point to point along the trip through NYC in different colors. North Vectors-Red South Vectors-Blue East Vectors-Green West Vectors-Yellow Site Address Penn Station 33 rd St and 7th Ave Empire State Building 34th St and 5th Ave NY NY Library 41st and 5th Ave .
DEDICATION PART ONE Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 PART TWO Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 .
Anatomy Fig 1. Upper limb venous anatomy [1] Vessel Selection Right arm preferable to left (as the catheter is more likely to advance into the correct vessel), vessel selection in order: 1. Basilic 2. Brachial 3. Cephalic Pre-procedure Patient information and consent Purpose of procedure, risks, benefits, alternatives. Line care: Consider using local patient information leaflet as available .
