CHAPTER 6 Introduction to Vectors

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1 CHAPTER 6 Introduction to Vectors Review of Prerequisite Skills, p. 73 "3 ".. e. "3. "3 d. f.. Find BC using the Pthgoren theorem, AC AB BC. BC AC AB 6 64 BC 8 Net, use the rtio tn A opposite tn A BC djcent. AB To solve DABC, find mesures of the sides nd ngles whose vlues re not given: AB, /B, nd /C. Find AB using the Pthgoren theorem, BC AB AC. AB BC AC (37.) (.) 88 AB " Find /B using the rtio sin B opposite sin B AC hpotenuse. BC. 37. /B /C 9 /B /C /C Find mesures of the ngles whose vlues re not given. Find /A using the cosine lw, cos A c. c 8 ()(8) 8 /A Find /B using the sine lw. sin B sin A sin B sin (97.9 ) sin B 8. /B Find /C using the sine lw. sin C sin A c sin C sin (97.9 ) 8 sin C 8.8 /C Since the sum of the internl ngles of tringle equls 8, determine the mesure of /Z using /X 6 nd /Y 7. /Z 8 (/X /Y) 8 (6 7 ) Find XY nd YZ using the sine lw. XY sin Y XY sin Z XY sin 7 6 sin XZ YZ sin X XY sin Z YZ sin 6 6 sin YZ Find ech ngle using the cosine lw. cos R RS RT ST (RS)(RT) 4 7 (4)(7) Clculus nd Vectors Solutions Mnul 6-

2 7 /R 8 44 cos S RS ST RT (RS)(ST) 4 7 (4)() /S 8 cos T RT ST RS (RT)(ST) 7 4 (7)() 9 3 /T A 3. km T 7 8. C km/h 48 A T 8 km/h Find AC nd AT using the speed of ech vehicle nd the elpsed time (in hours) until it ws locted, distnce speed 3 time. AC kmh 3 4 h km AT 8 kmh 3 3 h 6 3 km Find CT using the cosine lw. CT AC AT (AC)(AT) cos A ( km) 6 3 km 6 km ( km) 6 km cos km CT 8. km 9. A Find AB (the distnce etween the irplnes) using the cosine lw. AB AT BT (AT)(BT)cos T (3. km) (6 km) (3. km)(6 km) cos km AB 8.8 km 7. P km 7 km Q 4 R Find QR using the cosine lw. QR PQ PR (PQ)(PR) cos P ( km) (7 km) ( km)(7 km) cos km QR km B cm cm B C C The pentgon cn e divided into congruent right tringles with height AC nd se BC. 3 /A 36 /A 36 Find AC nd BC using trigonometric rtios. AC AB 3 cos A cos cm BC AB 3 sin A sin cm The re of the pentgon is the sum of the res of the right tringles. Use the re of ^ABC to determine the re of the pentgon. 6- Chpter 6: Introduction to Vectors

3 Are pentgon 3 (BC)(AC) 3 (.9 cm)(4. cm) 9.4 cm 6. An Introduction to Vectors, pp Flse. Two vectors with the sme mgnitude cn hve different directions, so the re not equl.. True. Equl vectors hve the sme direction nd the sme mgnitude. Flse. Equl or opposite vectors must e prllel nd hve the sme mgnitude. If two prllel vectors hve different mgnitude, the cnnot e equl or opposite. d. Flse. Equl or opposite vectors must e prllel nd hve the sme mgnitude. Two vectors with the sme mgnitude cn hve directions tht re not prllel, so the re not equl or opposite.. Vectors must hve mgnitude nd direction. For some sclrs, it is cler wht is ment just the numer. Other sclrs re relted to the mgnitude of vector. Height is sclr. Height is the distnce (see elow) from one end to the other end. No direction is given. Temperture is sclr. Negtive tempertures re elow freeing, ut this is not direction. Weight is vector. It is the force (see elow) of grvit cting on our mss. Mss is sclr. There is no direction given. Are is sclr. It is the mount spce inside two-dimensionl oject. It does not hve direction. Volume is sclr. It is the mount of spce inside three-dimensionl oject. No direction is given. Distnce is sclr. The distnce etween two points does not hve direction. Displcement is vector. Its mgnitude is relted to the sclr distnce, ut it gives direction. Speed is sclr. It is the rte of chnge of distnce ( sclr) with respect to time, ut does not give direction. Force is vector. It is push or pull in certin direction. Velocit is vector. It is the rte of chnge of displcement ( vector) with respect to time. Its mgnitude is relted to the sclr speed. 3. Answers m vr. For emple: Friction resists the motion etween two surfces in contct cting in the opposite direction of motion. A rolling ll stops due to friction which resists the direction of motion. A swinging pendulum stops due to friction resisting the swinging pendulum. 4. Answers m vr. For emple:. AD. AD BC ; AB ED CB EB AB DC ; AE CD EC AE ; DE CE EB ; ; ; AC ; & DB DA BC ; AE & EB ; EC & DE ; AB & CB. B H D E.. AB EF ut d. e. A C AB AB CD EF GH AB AB JI G 7.. km h, south. km h, west km h, northest d. km h, northwest e. 6 km h, est F AB EF 6... d. e. I J Clculus nd Vectors Solutions Mnul 6-3

4 8.. 4 km h, due south. 7 km h, southwesterl 3 km h southesterl d. km h, due est 9.. i. Flse. The hve equl mgnitude, ut opposite direction. ii. True. The hve equl mgnitude. iii. True. The se hs sides of equl length, so the vectors hve equl mgnitude. iv. True. The hve equl mgnitude nd direction.. E H To clculte BD, BE nd BH, find the lengths of their corresponding line segments BD, BE nd BH using the Pthgoren theorem. BD AB AD 3 3 BD "8 BE AB AE 3 8 BE "73 BH BD DH ("8) 8 BH "8.. The tngent vector descries Jmes s velocit t tht moment. At point A his speed is km h nd he is heding north. The tngent vector shows his velocit is km h, north.. The length of the vector represents the mgnitude of Jmes s velocit t tht point. Jmes s speed is the sme s the mgnitude of Jmes s velocit. The mgnitude of Jmes s velocit (his speed) is constnt, ut the direction of his velocit chnges t ever point. d. Point C e. This point is hlfw etween D nd A, which is of the w round the circle. Since he is running 7 8 A B F D G C km h nd the trck is km in circumference, he cn run round the trck times in one hour. Tht 7 mens ech lp tkes him 4 minutes. 8 of 4 minutes is 3. minutes. 3 f. When he hs trvelled 8 of lp, Jmes will e hlfw etween B nd C nd will e heding southwest... Find the mgnitude of AB using the distnce formul. AB "( A B ) ( B A ) "(4 ) (3 ) or 3.6. CD " AB. AB moves from A(4, ) to B(, 3) or ( B, B ) ( A 3, A ). Use this to find point D. ( D, D ) ( C 3, C ) (6 3, ) EF (3, AB ). Find point E using ( A, A ) ( B 3, B ). ( E, E ) ( F 3, F ) (3 3, ) d. GH (, 3) AB, nd moves in the opposite direction s AB. ( H, H ) ( G 3, G ). ( H, H ) ( G 3, G ) (3 3, ) (, ) 6. Vector Addition, pp Chpter 6: Introduction to Vectors

5 d... BA A c c. C A B d. c CB C A B c d. CA 3.. C C c A B B. ( + c) + ( + c) ( + ) c c + + c ( + ) + c The resultnt vectors re the sme. The order in which ou dd vectors does not mtter. A B c A c B.. PS R RQ Q RS. c. S PS PR P PQ Q + c Clculus nd Vectors Solutions Mnul S RQ RS R PQ 6. PS MR P t MS RS ST SQ TQ so ( ) ( t ) MS MQ SQ 6-

6 7... d. e. f. g. h See the figure in prt. for the drwn vectors. nd so cos (u), 9.. Mri s velocit is km h downstrem.. 4 km/h 7 km/h 4 km/h Mri s speed is 3 km h... f + f ship A 7 km/h f u km/h 3 km/h. The vectors form tringle with side lengths S S S S S S f, f nd f f. Find f f using the cosine lw. S S f f S f S f S S f f cos (u) S S S f f $ f S f S S f f cos (u) B u f D u C. + w Find using the Pthgoren theorem. w w w ( kmh) (8 kmh) w 89 7 Find the direction of w using the rtio tn(u) w utn kmh. nd form right tringle. Find w 8 N 8. W 7 using the Pthgoren theorem. kmh, N 8. W,, Find the ngle etween nd using the rtio tn (u) 4 utn Find AB AC using the cosine lw nd the supplement to the ngle etween nd AB AB AC AB AC. AC AB AC cos (3 ) "3 ()() AB "3 AC D A w u 8 kmh E B The digonls of prllelogrm isect ech other. So EA EC nd Therefore, EA ED EB EB EC. ED. C 6-6 Chpter 6: Introduction to Vectors

7 . P T R S Multiple pplictions of the Tringle Lw for dding vectors show tht RM TP (since oth re equl to the undrwn vector nd tht RM TM SQ ), (since oth re equl to the undrwn vector RQ ) Adding these two equtions gives RM TP SQ 6. RM TP SQ nd represent the digonls of prllelogrm with sides nd. M Since nd the onl prllelogrm with equl digonls is rectngle, the prllelogrm must lso e rectngle. 7. P G + Q M R Let point M e defined s shown. Two pplictions of the Tringle Lw for dding vectors show tht GQ GR QM RM MG MG Adding these two equtions gives GQ QM MG GR RM From the given informtion, MG nd QM GP RM (since the re opposing vectors of equl length), so GQ GP GR, s desired. Q 6.3 Multipliction of Vector Sclr, pp A vector cnnot equl sclr... 3 cm. d. 3. E N descries direction tht is towrd the north of due est ( 9 est of north), in other words 9 6 towrd the est of due north. N6 E nd ering of 6 oth descrie direction tht is 6 towrd the est of due north. So, ech is descriing the sme direction in different w. 4. Answers m vr. For emple:. cm v 6 cm v. v d. e. v v v 9 cm 3 v Clculus nd Vectors Solutions Mnul 6-7

8 d. 6. Answers m vr. For emple: d. e. 7.. m( ) n 3 m(4 ) n (3 ) m 3 nd n 4stisf the eqution, s does n multiple of the pir (3, 4). There re infinitel mn vlues possile.. d e 3 f( ) d 3e 4f d, e, nd f stisf the eqution, s does n multiple of the triple (,, ). There re infinitel mn vlues possile. 8. or nd re colliner, so where k is nonero sclr. Since k,, k cn onl e or c, 3 mc n c, 3 d e fc Yes 6-8 Chpter 6: Introduction to Vectors

9 . Two vectors re colliner if nd onl if the cn e relted sclr multiple. In this cse k. colliner. not colliner not colliner d. colliner.. is vector with length unit in the sme direction s.. is vector with length unit in the opposite direction of.. sin u 8 ().9 u from towrds 6. ` ` is positive multiple of so it points in the sme direction s, nd hs mgnitude. It is unit vector in the sme direction s. 7. A m d. 3 4 e nd mke n ngle of, so ou m find using the Pthgoren theorem. 9 " or.4 Find the direction of using the rtio tn (u) utn from towrds. Find using the cosine lw, nd the supplement to the ngle etween nd. cos ( ) "3 ()() 8.9 Find the direction of using the sine lw. sin u sin ( ) B D C AD AD c AD CD But CD c BD So AD BD CD BD c., or AD 8. PM nd so MN PN PN PM c. PQ nd so QR PR PR PQ Notice tht MN QR We cn conclude tht is prllel to nd QR MN QR MN. 9. A B C c AD E D Clculus nd Vectors Solutions Mnul 6-9

10 Answers m vr. For emple:. nd. u nd nd d. u ED v u AD v u AB CD AC v AE nd v DB AD.. Since the mgnitude of is three times the mgnitude of nd ecuse the given sum is must e in the opposite direction of n, m nd n 3 m.. Whether nd re colliner or not, m nd n will mke the given eqution true.... BE CD ( CD ) The two re therefore prllel (colliner) nd BE CD. A B E D C Appling the tringle lw for dding vectors shows tht AC AD DC The given informtion sttes tht AB 3 DC 3 AB DC B the properties of trpeoids, this gives 3 AE EC, nd since AC AE EC, the originl eqution gives AE 3 AE AD 3 AB 6.4 Properties of Vectors, pp d. AE AD 3 AB AE AD 3 AB. 3. ( + ) 4. Answers m vr. For emple: + k( + ) k k. PQ RQ (RQ SR (SR SR TS SQ RQ ) (TS PT PS PS ) (PT PT TS ) ) PQ SQ 6... EC Yes, the digonls of rectngulr prism re of equl length c c 6 6c 8.. 6i 9. 3i i 8j 4c 7i 4j 7j k (3i j k k 6i 9j 3k i 4j 3(6i k 4k j k 6i 3j 8j 6i 7k k 9j 4i 3k ) j 7k ) 9. Solve the first eqution for. 3 + c + ( + c) ( + )+ c = = + + c + (+ c) 6- Chpter 6: Introduction to Vectors

11 Sustitute into the second eqution Lstl, find 3 in terms of nd Z ( ) ( ) 3 AG BH CE DF c AG c c c c c BH Therefore, AG BH T X O Appling the tringle lw for dding vectors shows tht TY TZ ZY The given informtion sttes tht TX ZY TX ZY B the properties of trpeoids, this gives nd since TY TO OY TO OY,, the originl eqution gives TO TO TZ TX Y 3 TO TZ TX Mid-Chpter Review, pp AB BA DC AD CD CB BC DA There is not enough informtion to determine if there is vector equl to. PD DA AP. BC (prllelogrm).. RV. RV PS d. RU e. PS f. PQ 3.. Find using the cosine lw, nd the supplement to the ngle etween the vectors. cos (3)(4) 3 "3. Find u using the rtio tn u 4 3 TO 3 TZ 3 TX utn t 4 or t 4. In qudrilterl PQRS, look t ^PQR. Joining the midpoints B nd C cretes vector tht is prllel to PR nd hlf the length of PR BC. Look t ^SPR. Joining the midpoints A nd D cretes vector tht is prllel to nd hlf the length of is prllel to AD PR PR AD nd equl in length to AD. BC. Therefore, ABCD is prllelogrm. 6.. Find using the cosine lw. Note v nd the ngle etween nd is u v v u v u v u u v v. cos 6 Clculus nd Vectors Solutions Mnul 6-

12 8 (8)() u v ". Find the direction of u v using the sine lw. sin u v sin 6 u v sin u sin 6 " usin "8 8 7 u v (u v ) d. Find using the cosine lw. u u v v " (u v ) u v u v cos u v 7. Find using the cosine lw. p q p "7 q p q p q cos 6 8. m 9. BC n DC m n 4 (4)() ()() 3 BD AC. Construct prllelogrm with sides OA nd. Since the digonls isect ech other, is the digonl equl to Or nd So, And AC OC OA OB OA AB. Now OB OA AC AC OA OC OB OC. OB OA AB.. (OC OA ) Multipling gives. 3 AC CD AB 3 AD OB OA OC. BD AD BD AD AD AB BD 3 BC BC AC BC 3. The ir velocit of the irplne nd the wind velocit (W (V ir) ) hve opposite directions. V ground V ir W 46 km h due south 3... PT PT SR 4... d PS PQ RS 3 QS QS 3 QR 6. Vectors in R nd R 3, pp No, s the -coordinte is not rel numer... We first rrnge the -, -, nd -es (ech cop of the rel line) in w so tht ech pir of es re perpendiculr to ech other (i.e., the - nd -es re rrnged in their usul w to form the -plne, nd the -is psses through the origin of the -plne nd is perpendiculr to this plne). This is esiest viewed s right-hnded sstem, where, from the viewer s perspective, the positive -is points upwrd, the positive -is points out of the pge, nd the positive -is points rightwrd in the plne of the pge. Then, given point P(,, c), we locte this point s unique position moving units long the -is, then from there units prllel to the -is, nd finll c units prllel to the -is. It s ssocited unique position vector is determined drwing vector with til t the origin O(,, ) nd hed t P.. Since this position vector is unique, its coordintes re unique. Therefore 4, 3, nd c Since A nd B re rell the sme point, we cn equte their coordintes. Therefore, 3, nd c Chpter 6: Introduction to Vectors

13 . From prt., A(, 3, 8), so OA (, 3, 8). Here is depiction of this vector. OA (, 3, 8) O(,, ) (, 4, ) (,, ) (4,, ) (, 4, ) (4, 4, ) (4,, ) C(4, 4, ) 4. This is not n cceptle vector in I 3 s the -coordinte is not n integer. However, since ll of the coordintes re rel numers, this is cceptle s vector in R 3.. (4, 4, ) A(4, 4, ) (, 4, ) (, 4, ) (4,, ) O(,, ) (,, ) (4,, ) ( 4,, ) B( 4, 4, ) 6.. A(,, ) is locted on the -is. B(,, ), C(,, ), nd D(,, ) re three other points on this is.. OA (,, ), the vector with til t the origin O(,, ) nd hed t A. 7.. Answers m vr. For emple: OA OC (,, ), OB (,, ), (,, ). Yes, these vectors re colliner (prllel), s the ll lie on the sme line, in this cse the -is. A generl vector ling on the -is would e of the form OA (,, ) for n rel numer. Therefore, this vector would e represented plcing the til t O, nd the hed t the point (,, ) on the -is. 8. A(,, ) E(,, 3) B(,, ) F(,, 3) D(, 3, ) (,, ) ( 4,, ) C(,, 3) O(,, ) (, 4, ) (, 4, ) ( 4, 4, ) 9.. A(3,, 4) C(,, 4) B(,, 4) Clculus nd Vectors Solutions Mnul 6-3

14 . Ever point on the plne contining points A, B, nd C hs -coordinte equl to 4. Therefore, the eqution of the plne contining these points is 4( plne prllel to the -plne through the point 4)... A(,, 3) (,, 3) (,, 3) (,, 3) O(,, ) (,, ) (,, ) (,, ) d. e. D(,, ) (,, ) (,, ) (,, ) O(,, ) (,, ) (,, ) (,, ) (,, ) (,, ) E(,, ). (,, ) (,, ) (,, ) B(,, ) (,, ) (,, ) (,, ) (,, ) (,, ) (,, ) O(,, ) (,, ) f. O(,, ) (,, ) (,, ) (,, ) (,, ) O(,, ) (,, ) C(,, ) (,, ) (,, ) F(,, ) (,, ) (,, ) (,, ) (,, ) O(,, ) (,, ).. OA (,, 3) (,, 3) (,, 3) A(,, 3) (,, 3) O(,, ) (,, ) OA (,, ) (,, ) 6-4 Chpter 6: Introduction to Vectors

15 . OB (,, ) (,, ) B(,, ) f. OF (,, ) (,, ) (,, ) O(,, ) OB (,, ) (,, ) (,, ) (,, ) (,, ) (,, ) (,, ) F(,, ) OF O(,, ) (,, ) (,, ) OC (,, ) (,, ) (,, ) (,, ) C(,, ) (,, ) d. OD (,, ) e. OE (,, ) E(,, ) (,, ) (,, ) (,, ) (,, ) (,, ) O(,, ) (,, ) (,, ) (,, ) (,, ) O(,, ) OC D(,, ) (,, ) OD (,, ) (,, ) (,, ) (,, ) O(,, ) OE.. Since P nd Q represent the sme point, we cn equte their - nd -coordintes to get the sstem of equtions c 6 Sustituting this second eqution into the first gives c 6 c So nd c.. Since P nd Q represent the sme point in R 3, the will hve the sme ssocited position vector, i.e. OP OQ. So, since these vectors re equl, the will certinl hve equl mgnitudes, i.e. OP OQ. 3. P(,, ) represents generl point on the -plne, since the -coordinte is. Similrl, Q(,, ) represents generl point in the -plne, nd R(,, ) represents generl point in the -plne. 4.. Ever point on the plne contining points M, N, nd P hs -coordinte equl to. Therefore, the eqution of the plne contining these points is (this is just the -plne).. The plne contins the origin O(,, ), nd so since it lso contins the points M, N, nd P s well, it will contin the position vectors ssocited with these points joining O (til) to the given point (hed). Tht is, the plne contins the vectors OM, ON, nd OP... A(,, ), B(, 4, ), C(, 4, ), D(,, 7), E(, 4, 7), F(,, 7). OA (,, ), OB (, 4, ), OC (, 4, ), OD (,, 7), OE (, 4, 7), OF (,, 7) Rectngle DEPF is 7 units elow the -plne. Clculus nd Vectors Solutions Mnul 6-

16 d. Ever point on the plne contining points B, C, E, nd P hs -coordinte equl to 4. Therefore, the eqution of the plne contining these points is 4 ( plne prllel to the -plne through the point 4). e. Ever point contined in rectngle BCEP hs -coordinte equl to 4, nd so is of the form (, 4,) where nd re rel numers such tht # # nd 7 # #. 6.. d. O(,, ) M(, 3, ) O(, ) e. F(,, ) P(4, ) O(,, ). D( 3, 4) f. O(, ) J(,, ) O(,, ) O(,, ) C(, 4, ) 7. The following o illustrtes the three dimensionl solid consisting of the set of ll points (,, ) such tht # #, # #, nd # #. (,, ) O(,, ) (,, ) (,, ) (,, ) (,, ) (,, ) (,, ) 6-6 Chpter 6: Introduction to Vectors

17 8. First, OP OA OB the tringle lw of vector ddition, where OB OA OP (, nd OA, ), (,, ), re drwn in stndrd position (strting from the origin O(,, )), nd OB is drwn strting from the hed of Notice tht OA OA. lies in the -plne, nd OB is perpendiculr to the -plne (so is perpendiculr to OA ). So, OP, OA, nd OB form right tringle nd, the Pthgoren theorem, Similrl, OA OP OA OB the tringle lw of vector ddition, where nd (,, ) (,, ), nd these three vectors form right tringle s well. So, OA Oviousl OB, nd so sustituting gives OP OA OB OP " 9. To find vector equivlent to OP AB (, 3, 6), where B(4,, 8), we need to move units to the right of the -coordinte for B (to 4 6), 3 units to the left of the -coordinte for B (to 3 ), nd 6 units elow the -coordinte for B (to 8 6 ). So we get the point A(6,, ). Indeed, notice tht to get from A to B (which descries vector AB ), we move units left in the -coordinte, 3 units right in the -coordinte, nd 6 units up in the -coordinte. This is equivlent to vector OP (, 3, 6). 6.6 Opertions with Algeric Vectors in R, pp A(, 3) B(, ). AB (, ) (, 3) (3, ) BA AB (3, ) (3, ) Here is sketch of these two vectors in the -coordinte plne. AB. OA "() 3 "9 AB 8.39 "3 " Also, since BA BA AB AB,? AB. A(, 3) " OB " AB BA B(, ) " A(6, ) O(, ) OA Clculus nd Vectors Solutions Mnul 6-7

18 . (, ) O(, ) (3, ) ( 3, ) (, ). The vectors with the sme mgnitude re OA nd OA OA, nd OA 3. OA "3 (4) " 4.. The i -component will e equl to the first coordinte in component form, nd so Similrl, the j 3. -component will e equl to the second coordinte in component form, nd so.. (3, ) (3, ) "(3) " "(6) "37 6 "(4) (9) "68. 4 (6, ) (4, 9) (, ) "() " 4 8. (6, ) (4, 9) (, ) "() " (, 3) (, ) ((), (3) ) (, 7). 3(4, 9) 9(, 3) (3(4) 9(), 3(9) 9(3)) (3, ) (6, ) (6, ) 3 (6) 3 (6), () 3 () 7. (, ). 3 i j, i 3(i j j ) (6 )i (i j ) (3 )j. ( 4 7i 8j ) 4(i 3( 3 ) 3i ( j j ) (i j ) 3 3 ) 3(i 3 3( ) 9i j 8.. 8j ) 3(i j ) (i i i 4j j ) (i j ) 4j " 4 " (i 3i j 3i 6j ) (i j ) 6j "3 (6) " (i j ) 3(i j ) 3 7i 7i 7j 7j "7 (7) "338 d ( 3 ) 3 3 so, from prt, 3 3 " Chpter 6: Introduction to Vectors

19 D(4, ) B( 4, 4) 4 A( 8, ) C(, ) F( 7, ) H(6, ) 4 G(, ) E(, 4) 6 8 so oviousl we will hve OA BC. (It turns out tht their common mgnitude is "6 3 "4.).. C( 4, ) B(6, 6) A(, 3). AB (4, 4) (8, ) CD (4, ) (4, ) (, ) EF (, 4) (7, ) (, 4) GH (6, 4) (6, ) (, ) (, ). AB "4 " CD " 4 " EF "(6) 4 " GH 8 7. " ".. B the prllelogrm lw of vector ddition, OC OA OB (6, 3) (, 6) (7, 3) For the other vectors, BA OA OB (6, 3) (, 6) BC (, OC 9) OB (7, 3) (, 6). OA (6, 3) (6, BC 3),. AB (6, 6) (, 3) (4, 3) AB "4 3 " AC (4, ) (, 3) (6, 8) AC "(6) 8 " CB (6, 6) (4, ) (, ) CB " () " CB 8.8 AC, AB Since CB AC AB, the tringle is right tringle... A(, ) C(, 8) B(7, ) Clculus nd Vectors Solutions Mnul 6-9

20 . X( 6, ) A(, ) C(, 8) Z(, 4) B(7, ) Sustituting this into the lst eqution ove, we cn now solve for. () 4 4 So nd C(, ) B( 6, 9) D(8, ) Y(4, 8) A(, 3) As first possiilit for the fourth verte, there is X(, ). From the sketch in prt., we see tht we. Becuse ABCD is rectngle, we will hve would then hve BC AD CX BA (, ) (6, 9) (8, ) (, 3) (, 8) ( 7, ()) ( 6, 9) (6, 8) (8, 4) So, nd i.e., So B similr resoning for the other.. Since nd points lelled in the sketch in prt., PA PA 7, PB C(, 7). X(6, )., (, ) (, ) AY CB PB (, ), ( (), ) (7, 8) (, ) (, ) (, ) (, ), this mens tht ( ) So Finll, () Y(4, 8). 4 BZ AC ( 7, ()) ( (), 8 ) (3, 6) 7 3 So P,. 6 So Z(, 4). In conclusion, the three possile. This point Q on the -is will e of the form loctions for fourth verte in prllelogrm Q(, ) for some rel numer. Resoning with vertices A, B, nd C re X(6, ), Y(4, 8), similrl to prt., we hve nd Z(, 4). QA (, ) (, ) 3.. 3(, ) (, 3) (, 33) QB (, ) (3 (), 3 (3)) (, 33) (, ) (, ) (3, 3 ) (, 33) (, So since QA ) 3 QB 3 33, So 7 nd. ( ) ( ). (, ) 3(6, ) (6, 4) 4 4 ( 8, ) (6, 4) To solve for, use So Q, Chpter 6: Introduction to Vectors

21 6. QP is in the direction opposite to nd QP OP OQ PQ, (, 9) (, ) (9, 4) QP "9 4 "68 4 A unit vector in the direction of QP is u 4 QP 9 4, 4 Indeed, is oviousl in the sme direction s (since u u 4 is positive sclr multiple of QP QP ), nd notice tht u Å Å O, P, nd R cn e thought of s the vertices of tringle. PR OR OP (8, ) (7, 4) PR (, ) () () OR 66 (8) () OP 6 (7) 4 6 OR B the cosine lw, the ngle,, etween nd OP u stisfies cos u OR OP PR OR? OP !6?! ucos!6?! So the ngle etween OR nd is out.. We found the vector in prt., so RP nd RP PR PR OP 8.86 (, ) PR (, ) 66 Also, the prllelogrm lw of vector ddition, OQ OR OP (8, ) (7, 4) OQ (, 3) () 3 74 Plcing RP (, ) nd OQ (, 3) with their tils t the origin, tringle is formed joining the heds of these two vectors. The third side of this tringle is the vector v RP OQ (, ) (, 3) v (6, ) 6 6 Now resoning similr to prt., the cosine lw implies tht the ngle, u, etween RP nd OQ stisfies cos u RP OQ v RP? OQ !66?!74 ucos !66?! So the ngle etween RP nd OQ is out 3.4. However, since we re discussing the digonls of prllelogrm OPQR here, it would lso hve een pproprite to report the supplement of this ngle, or out , s the ngle etween these vectors. 6.7 Opertions with Vectors in R 3, pp OA. OA i j 4k. OB "() 4 " OB (3, 4, 4) "3 4 (4) " c (, 3, 3) (,, 4) (, 8, ) ( (), 3 8, (3) 4 ) (, 3, ) ` 3 c ` " (3) 3 Clculus nd Vectors Solutions Mnul 6-

22 4.. OP OA OB ((3), 4, ()) (, 6, ). OA "(3) 4 3 OB " () 3 OP "() AB OB OA (,, ) (3, 4, ) ( (3), 4, () ) (,, 3) AB AB " () (3) " represents the vector from the tip of to the tip of OB OA It is the difference etween the two vectors.... (, 4, ) (, 3, ) (,, ) (, 4, ) (, 6, 4) (,, ) ( (), 4 6, (4) ) (, 3, 3). 3 (, 4, ) 3(, 3, ) (,, ) (, 8, ) (3, 9, 6) (,, ) ( 3, 8 9, 6 ) (7, 6, 8) 3 (, 4, ) (, 3, ) 3(,, ),, (, 3, ) (6, 3, ) (6), 3 3, () 3 d. 3,, 3 3 3(, 4, ) (, 3, ) 3(,, ) (3,, 3) (,, ) (6, 3, ) (3 6, 3, 3 ) (, 3, 3) 6.. p q (i j (. p q i (i j )i k ) (i j ( 3i )i k k ( )j j k ) ( )k ) (i j ( k )j j k ) ( )k p q (i (4i j j k ) (4 d. p (4i q 9i )i k (i ) (i j 3j ( j (i 3k )j j k ) k ) ( )k (4 9i )i k j ) (i k ) 7.. m 3j ( n 3k )j j (i k j k ) ) ( )k (i ( m 4i. m n j ())i k ) 3k ()j (i j k ( )k ) n "4 (i () k (3) ) (i ( i j ())i k j j "6 k 8. ) ( )k m m n 3n " (i (4i k " k ) 3(i 8.4 ) (4 i (6))i (6i j 3j 4k 3j 3j k 6k ) ) ( 6)k m d. m 3n "() (i k 3 ) i 4 k " m 8. "() () " 8.8 i 3i j i 6j k 8j 7k k Divide on oth sides to get: i 4j k Plug this eqution into the first given eqution: i 4j k i i j j k ( i )i k (i 9.. The vectors OA, OB j ( 6k 4)j 4j k ) ( )k, nd OC represent the -plne, -plne, nd -plne, respectivel. The re lso the vector from the origin to points (,, ), (,,c), nd (,, c), respectivel.. OA OB i OC i j OA i j k j ck ck " OB " c OB " c 6- Chpter 6: Introduction to Vectors

23 d. AB (,, c) (,, ) (,, c) AB is direction vector from A to B... OA "() (6) 3 "49 7. d. AB " 9 " 8.49 e. BA OA OB (, 6, 3) (3, 4, ) (,, 9) f.. OB AB "(3) OB OA (4) "69 3 (3, 4, ) (, 6, 3) (3 (), 4 (6), 3) (,, 9) BA "() () (9) " 8.49 B(3,, 7) BC C(7, 3, ) DC D(4,, 3) AB A(, 3, ) AD In order to show tht ABCD is prllelogrm, we must show tht AB DC or BC AD. This will show the hve the sme direction, thus the opposite sides re prllel. B showing the vectors re equl the will hve the sme mgnitude, impling the opposite sides hving congruen AB (3,, 7) (, 3, ) (3, 3, 7 ) DC (3, 4, ) (7, 3, ) (4,, 3) (7 4, 3, 3) (3, Thus AB 4, ) DC. Do the clcultions for the other pir s check. BC (7, 3, ) (3,, 7) (7 3, 3 (), 7) AD (4,, ) (4,, 3) (, 3, ) (4, 3, 3 ) So BC (4,, AD ). We hve shown AB DC nd BC AD, so ABCD is prllelogrm.. (,, c) (,, c) (, 6, c) (,, c) (,, c) (,, c) (,, c c c) ( 3, 4, c) (,, ) 3 ; 3 ; 3 4; 7 c ; c OB O OA OC. V (,, ), the origin V end point of OA V 3 end point of OB (,, ) V 4 end V OA point of OB OC (, 4, ) (,, ) (,, ) (, 4, ) (, 4, ) V 6 OA OC (, 6, 6) (,, ) (,, ) (,, ) V 7 OB OC (, 7, 4) (, 4, ) (,, ) (, 4, ) V 8 OA OB (, OC 9, ) (,, ) (, 9, ) ( V 7 ) (, 9, ) (,, ) 4. An point on the -is hs -coordinte nd -coordinte. The -coordinte of ech of A nd B is 3, so the -component of the distnce from the desired point is the sme for ech of A nd B. The -component of the distnce from the desired point will e for ech of A nd B, (). So, the -coordinte of the desired point hs to e hlfw etween the -coordintes of A nd B. The desired point is (,, ). A B C Clculus nd Vectors Solutions Mnul 6-3

24 . A To solve this prolem, we must first consider the tringle formed,, nd. We will use their mgnitudes to solve for ngle A, which will e used to solve for in the tringle formed, nd,. Using the cosine lw, we see tht: cos (A) Now, consider the tringle formed, nd,. Using the cosine lw gin: cos (A) ( ) ( ) ( ) "9 or Liner Comintions nd Spnning Sets, pp The re colliner, thus liner comintion is not pplicle.. It is not possile to use in spnning set. Therefore, the remining vectors onl spn R. 3. The set of vectors spnned (, ) is m(, ). If we let m, then m(, ) (, ). 4. i spns the set m(,, ). This is n vector long the -is. Emples: (,, ), (,,. As in question, it is not possile to use ) in spnning set. + C B 6. (, ), (, )6, (, 4), (, )6, (, ), (3, 6)6 re ll the possile spnning sets for R with vectors i ( 6i 8j 3 4( j 6j c ) 4i k 8k 4 i 6 6j c 4k 8i 8j c ) ( c 4j 4j 4 ) i k k 4 4i 4c j 8k ( 6i 3 j j 4i j c k i 3j k )4( 43j k 4k 8i 4j c ) k ( j c ) k. ( 4 8c ) 4c i 3i j 8j j 6k k 4i j 8k 3 (3 6 9c ) 3c i 4i j j j 6k k 3i 9j 6k ( 4 8c ) 3i 7i 8j 3j k 3 (3 6 9c ) 4k 4i j k 8. (,, ), (,, )6: (,, ) (,, ) (,, ) (3, 4, ) 3(,, ) 4(,, ) (,, ), (,, )6 (,, ) (,, ) 3(,, ) (3, 4, ) 3(,, ) (,, ) 9.. It is the set of vectors in the -plne.. (, 4, ) (,, ) 4(,, ) B prt. the vector is not in the -plne. There is no comintion tht would produce numer other thn for the -component. d. It would still onl spn the -plne. There would e no need for tht vector.. Looking t the -component: 3c The -component: 6 c The -component: c 3c c c Chpter 6: Introduction to Vectors

25 Sustituting this into the first nd second eqution: (, 34) (, 3) (, ) Looking t the -component: Looking t the -component: 34 3 Sustituting in : Sustituting into -component eqution: (8) (, 34) (, 3) 8(, ).. (, ) (, ) (, ) Sustitute in : Sustitute this ck into the first eqution:. Using the formuls in prt : For (, 3): (, 3) (, ) 4(, ) For (4, ): (4, ) 9(, ) 4(, ) For (4, ) (4, ) 7(, ) 8(, ) 3. Tr: (,, 3) (4,, ) (4,, 6) components: components: Sustitute in : Clculus nd Vectors Solutions Mnul Sustitute this result into the -components: 4 3 Check sustituting into -components: Therefore: (,, 3) (4,, ) (4,, 6) for n nd. The do not lie on the sme plne.. (, 3, 4) (,, ) (3, 4, 7) components: 3 3 components: 3 4 Sustitute in : 9 4 Check with components: Since there eists n nd to form liner comintion of of the vectors to form the third, the lie on the sme plne. 3(, 3, 4) (, 4. Let vector, ) (3, 4, nd 7) (, 3, 4) (, 3, ) (vectors from the origin to points A nd B, respectivel). To determine, we let c (vector from origin to C) e liner comintion of nd. (, 3, 4) (, 3, ) (, 6, ) components: components: Sustitute in : Sustitute in in component eqution: 6 components: 4 Sustitute in nd : m, n 3. Non-prllel vectors cnnot e equl, unless their mgnitudes equl. 6. Answers m vr. For emple: Tr liner comintions of the vectors such tht 6-

26 the component equls. Then clculte wht p nd q would equl. (4,, 7) (,, 6) (6,, ) So p 6nd q (4,, ) (,, 6) (,, ) So p nd q (4,, 7) (,, 6) 3 3 3, 3, So p 3 nd q As in question, non-prllel vectors. Their mgnitudes must e gin to mke the equlit true. m m 3 (m )(m 3) m, 3 m m 6 (m )(m 3) m, 3 So, when m 3, their sum will e. Review Eercise, pp flse; Let then: ( ). true;, nd c oth represent the lengths of the digonl of prllelogrm, the first with sides nd nd the second with sides nd c ; since oth prllelogrms hve s side nd digonls of equl length. true; Sutrcting from oth sides shows tht c c d. true; Drw the prllelogrm formed nd SW. FW nd RS RF re the opposite sides of prllelogrm nd must e equl. e. true; The distriutive lw for sclrs f. flse; Let nd let Then, nd ut c c ( d d. ) c d c c c so c d.. Sustitute the given vlues of,, nd into the epression 3 ( 3 ( 3 3 4c ) c 3 ( 3 3c ) ) 4 c 4 6 8c 6 9 9c c c 9c 3 8c. Simplif the epression efore sustituting the given vlues of nd 3( 4, (4, 6 ) 6 3 ( ) ) XY 3 3c OY OX (,, ) (,, ) (,, ) (4 (), 4, 8 ) (, 3, 6) XY "( ) ( ) ( ) "() (3) (6) " "49 7. The components of unit vector in the sme direction s XY re 7(, 3, 6) ( 7, 3 7, 6 7). 4.. The position vector OP is equivlent to OP YX (,, ) (,, ) (,, ) (,, 6 ) (6, 3, 6). YX "(6) (3) (6) "8 9 The components of unit vector in the sme direction s re. MN YX NM (,, ) (,, ) (,, ) ( 8, 3, ) (6,, 3) YX. 9(6, 3, 6) ( 3, 3, 3) 6-6 Chpter 6: Introduction to Vectors

27 "49 7 The components of the unit vector with the opposite direction to MN re 7(6,, 3) ( 6 7, 7, 3 7) 6.. The two digonls cn e found clculting OA OB nd OA OB. O OA OB (3,, 6) (6, 6, ) (3 6, 6, 6 ) OA OB (3, 8, 8) (3,, 6) (6, 6, ) (3 (6), 6, 6 ()) (9, 4, 4). To determine the ngle etween the sides of the prllelogrm, clculte nd OA OB OA, OB, nd ppl the cosine lw. 7.. NM "(6) () (3) "49 7 "76 "9 cos u 8.98 u A "3 AB "( ()) ( ) (3 ) "(3) () () "9 4 "4 OA OB OA "(3) () (6) OB "(6) (6) () OA OB "(9) (4) (4) cos u OA OB OA OB OA OB cos u (7) ("9) ("3) (7)("9) B OA + OB BC "(3 ) (3 ) (4 3) "() (3) (7) " 9 49 "9 CA "( 3) ( 3) ( (4)) "(4) () () "6 4 "4 Tringle ABC is right tringle if nd onl if AB CA BC. AB CA ("4) ("4) 4 4 BC 9 ("9) 9 So tringle ABC is right tringle.. Are of tringle For tringle ABC the longest side BC h. is the hpotenuse, so AB nd CA re the se nd height of the tringle. Are ( AB )( CA ) "4"4 "63 3 or. "7 Perimeter of tringle equls the sum of the sides. Perimeter AB BC CA "4 "9 " d. The fourth verte D is the hed of the digonl vector from A. To find tke. AB AD AB AC ( (),, 3 ) (3,, ) AC (3 (), 3, 4 ) (4,, ) AD AB AC (3 4,, ()) (7,, 3) So the fourth verte is D( 7,, (3)) or D(6,, ). Clculus nd Vectors Solutions Mnul 6-7

28 8.. + c c. Since the vectors nd re perpendiculr,. So, (4) (3) 6 9 " 9. Epress r s liner comintion of p nd q : Solve for nd : r p q (, ) (, 7) (3, ) (, ) (, 7) (3, ) (, ) ( 3, 7 ) Solve the sstem of equtions: 3 7 Use the method of elimintion: 3() 3(7 ) B sustitution, 3 Therefore Epress q (, 7) 3 (3, ) (, s liner comintion of p ) nd r. Solve for nd : q p r (3, ) (, 7) (, ) (3, ) (, 7) (, ) ( 3, ) (, 7 ) Solve the sstem of equtions: 3 7 Use the method of elimintion: (3) ( ) B sustitution, 3 Therefore Epress p 3(, 7) 3(, ) (3, s liner comintion of q ) nd r. Solve for nd : p q r (, 7) (3, ) (, ) (, 7) (3, ) (, ) (, 7) (3, ) Solve the sstem of equtions: 3 7 Use the method of elimintion: () (3 ) B sustitution, Therefore 3(3, ) (, ) (, 7).. Let P(,, ) e point equidistnt from A nd B. Then PA PB. ( ) ( ()) ( 3) ( ) ( ) ( (3)) (,, ) nd (, 3, ) clerl stisf the eqution nd re equidistnt from A nd B... (4, 3, ) (,, 4) (6, 8, c) 3(7, c, 4) (4, 3, ) (,, 8) 3, 4, c (4, 3, ) 8, 4 3c, c Solve the equtions: i ii. c c c iii c 3 4 3(8) (, 3c, ) 6-8 Chpter 6: Introduction to Vectors

29 . (3,, 4),, (3,, c) c, 3 c, (3,, 4) (,, ) (3,, c) (c, 3c, ) (3,, 4) ( 3 c, 3c, c) Solve the sstem of equtions: 3c 4 c Use the method of elimintion: (4) ( c) 8 c 3c 3 3c c B sustitution, 8 Solve the eqution: 3 3 c 3 (8) 3 () Find AB, BC, Test " " "(3) () " AB, BC, CA CA AB "( ) ( ()) ( ) "() (3) () BC "(4 ) ( ) ( ) "() (4) () CA "(4 ) ( ()) ( ) in the Pthgoren theorem: AB CA (") (") BC (") So tringle ABC is right tringle.. Yes, P(,, 3), Q(, 4, 6), nd R(,, 3) re colliner ecuse: P (, 4, 6) Q (, 4, 6) R (, 4, 6) 3.. Find AB, BC, CA AB "( 3) ( ) ( 4) "9 BC 3 "( ) ( ) (3 ) CA "( 3) ( ) (3 4) Test in the Pthgoren theorem: BC CA A"6B A"3B 6 3 AB 9 (3) 9 So tringle ABC is right tringle.. Since tringle ABC is right tringle, 6 cos/abc Å DA, BC nd EB, ED. DC, AB nd CE, EA AD DC AC, But AC DB Therefore, AD DC DB.. C(3,, ); P(3, 4, ); E(, 4, ); F(, 4, ). DB (3, 4, ) (3, 4, ) CF ( 3, 4, ) (3, 4, ) D P O "() () () "() () () "6 "() () () "3 AB, BC, X u CA B Clculus nd Vectors Solutions Mnul 6-9

30 OD DP the Pthgoren theorem Thus ODPB is squre nd cos u, so the ngle etween the vectors is 9. d. E P O u 8 cos 3! d + e 3 d e Use the cosine lw to evlute d e d e d e d e cos u (3) () 3 (3)() cos 9 3 "3 d e " d d e e Use the cosine lw to evlute d e d e d e d e cos u (3) () 3 (3)() cos "3 d e "8. e d 8.83 (d e ) d e u A OA 3, OP " u 8 (m/poa) A: 4 km/h Let represent the ir speed of the irplne nd let W represent the velocit of the wind. In one hour, the plne will trvel A W kilometers. Becuse A nd W mke right ngle, use the Pthgoren theorem: A W A W A W " km So in 3 hours, the plne will trvel 3(4.3)km km. A 7 tn u W A 4 (4) () utn The direction of the irplne is S4. W. 8.. An pir of nonero, noncolliner vectors will spn R. To show tht (, 3) nd (3, ) re noncolliner, show tht there does not eist n numer k such tht k(, 3) (3, ). Solve the sstem of equtions: k 3 3k Solving oth equtions gives two different vlues 3 for k, nd 3, so (, 3) nd (3, ) re noncolliner nd thus spn R. (33, 79) m(, 3) n(3, ) (33, 79) (m, 3m) (3n, n) (33, 79) (m 3n, 3m n) Solve the sstem of equtions: 33 m 3n 79 3m n Use the method of elimintion: 3(33) 3(m 3n) (79) (3m n) 969 6m 9n 9 6m n 6 n B sustitution, m 77. W: km/h 6-3 Chpter 6: Introduction to Vectors

31 9.. Find nd such tht (, 9, 4) (, 3, ) (3,, 4) (, 9, 4) (, 3, ) (3,, 4) (, 9, 4) ( 3, 3, 4) i. 3 ii. 9 3 iii. 4 4 Use the method of elimintion with i. nd iii. (4) ( 4) B sustitution,. lies in the plne determined nd c ecuse it cn e written s liner comintion of nd. If vector is in the spn of nd c then cn e written s liner comintion of c., nd c. Find m nd n such tht (3, 36, 3) m(, 3, ) n(3,, 4) (m, 3m, m) (3n, n, 4n) (m 3n, 3m n, m 4n) Solve the sstem of equtions: 3 m 3n 36 3m n 3 m 4n Use the method of elimintion: (3) (m 4n) 46 m 8n 3 m 3n 33 n 3 n B sustitution,. So, vector m is in the spn of nd c... (,, 4) (, 4, 4) (4,, 4) (4, 4, 4). (,, 4) (, 4, 4) (4, 4, 4) (4,, 4) (,, ) (, 4, ) (4,, ) (4, 4, ) PO (4, 4, 4) so, OP PO (4, 4, 4) (4, 4, 4) (,, 4) (, 4, 4) (4,, 4) (4, 4, 4) (,, ) (, 4, ) (4,, ) (4, 4, ) The vector PQ from P(4, 4, 4) to Q(, 4, ) cn e written s PQ (4,, 4). d. (,, 4) (, 4, 4) (4,, 4) (4, 4, 4) (,, ) (, 4, ) (4,, ) (4, 4, ) (,, ) (, 4, ) (4,, ) (4, 4, ) The vector with the coordintes (4, 4, ).. ( 4 c ) 3 c ( c ) 3 3( 3 3c c ) 4(,, ) 3(,, 3) (,, 3) (4, 4, 4) (6, 3, 9) (,, 3) (, 7, ) 7 Clculus nd Vectors Solutions Mnul 6-3

32 .. AB ecuse it is the dimeter of the circle. " or 4.47 "8 or If A, B, nd C re vertices of right tringle, then So, tringle ABC is right tringle. 3.. FL. MK JK FG JM GH HL c d. e. 4. " A( 3, 4) BC "( 3) ( (4)) CA "( (3)) ( 4) BC CA AB BC CA A"B A"8B "() (4) "(8) (4) B(3, 4) 8 AB HJ IH HG IK KJ GF IH FG FJ HK GF c IJ c C(, ).. " the Pthgoren theorem. " the Pthgoren theorem "4 9 the Pthgoren theorem 6. Cse If nd c re colliner, then is lso colliner with oth nd But is perpendiculr to nd c, so c. 4c is perpendiculr to 4c. Cse If nd re not colliner, then spnning sets, nd c c spn plne in nd is in tht plne. If R 3, is perpendiculr to 4c nd c, then it is perpendiculr to the plne nd ll vectors in the plne. So, is perpendiculr to 4c. Chpter 6 Test, p Let P e the til of nd let Q e the hed of The vector sums 3 ( c nd 3( c ) c. )4 4 cn e depicted s in the digrm elow, using the tringle lw of ddition. We see tht PQ ( c ) ( ) c. This is the ssocitive propert for vector ddition. P ( + ) PQ= ( + ) + c = + ( + c).. AB. AB (6 (), 7 3, 3 ()) (8, 4, 8) BA "8 ()AB 4 8 (8, 4, 8); BA AB ; unit vector in direction of BA BA BA (8, 4, 8) 3 3, 3 ( + c) 3. Let PQ, QR, nd s in the digrm elow. Note tht RS QS, 6 nd tht tringle PQR nd tringle PRS shre ngle u. c Q 6-3 Chpter 6: Introduction to Vectors

33 P + B the cosine lw: cos u nd cos u. Hence, ( ) " " "(3) ("7) "(3) "9 u R Q S 4.. We hve 3 nd 3. Multipling the first eqution nd the second eqution ields: nd Adding these equtions, we hve: Sustituting this into the first eqution ields: Simplifing, we hve: 3(. 3 ) 3... First, conduct sclr multipliction on the third vector, ielding: (,, c) (,, ) (6, 3, ) (3,, c). Now, ech of the three components corresponds to n eqution. First, 6 3, which implies. Second, 3. Sustituting nd simplifing ields. Third, c so.. nd c, c. spn R, ecuse n vector (, ) in R cn e written s liner comintion of nd. These two vectors re not multiples of ech other.. First, conduct sclr multipliction on the vectors, ielding: (p, 3p) (3q, q) (3, 9). Now, ech component corresponds to n eqution. First, p 3q 3. Second, 3p q 9. Multipling the second eqution 3 nd dding the result to the first eqution ields: 7p 4, which implies p. Sustituting this into the first eqution nd simplifing ields 6.. m nc q 3. (,, 9) m(3,, 4) n(,, 3) (,, 9) (3m, m, 4m) (n, n, 3n) Ech of the three components corresponds to n eqution. First, 3m n. Second, m n. Third, 9 4m 3n. Multipling the first eqution nd dding the result to the second eqution ields m. Sustituting m into the first eqution ields n 7. Since m nd n lso solves the third component s eqution, 7 m nc for m nd n 7. Hence, cn e written s liner comintion of nd. r mp nq c. (6,, 4) m(, 3, 4) n(4,, 6) (6,, 4) (m, 3m, 4m) (4n, n, 6n) Ech of the three components corresponds to n eqution. First, 6 m 4n. Second, 3m n. Third, 4 4m 6n. Multipling the first eqution nd dding the result to the third eqution ields n 4. Sustituting n 4 into the first eqution ields m. We hve tht n 4 nd m is the unique solution to the first nd third equtions, ut n 4 nd m does not solve the second eqution. Hence, this sstem of equtions hs no solution, nd cnnot e written s liner comintion of nd In other words, r p r q. does not lie in the plne determined nd 7. nd p q. hve mgnitudes of nd, respectivel, nd hve n ngle of etween them, s depicted in the picture elow. Clculus nd Vectors Solutions Mnul 6-33

34 Since 6 is the complement of 3 cn e depicted s elow. 3 + u 3 6 B the cosine lw: cos "3 or 3.6 The direction of 3 is u, the ngle from. This cn e computed from the sine lw: 3 sin 6 sin u sin u sin 6 3 usin (4) sin 6 "3 reltive to 8. DE u 8 DE CE 73.9 CD Also, BA BA CA CB Thus, DE BA usin sin Chpter 6: Introduction to Vectors

35 CHAPTER 6 Introduction to Vectors Review of Prerequisite Skills, p. 73 "3 ".. e. "3. "3 d. f.. Find BC using the Pthgoren theorem, AC AB BC. BC AC AB 6 64 BC 8 Net, use the rtio tn A opposite tn A BC djcent. AB To solve DABC, find mesures of the sides nd ngles whose vlues re not given: AB, /B, nd /C. Find AB using the Pthgoren theorem, BC AB AC. AB BC AC (37.) (.) 88 AB " Find /B using the rtio sin B opposite sin B AC hpotenuse. BC. 37. /B /C 9 /B /C /C Find mesures of the ngles whose vlues re not given. Find /A using the cosine lw, cos A c. c 8 ()(8) 8 /A Find /B using the sine lw. sin B sin A sin B sin (97.9 ) sin B 8. /B Find /C using the sine lw. sin C sin A c sin C sin (97.9 ) 8 sin C 8.8 /C Since the sum of the internl ngles of tringle equls 8, determine the mesure of /Z using /X 6 nd /Y 7. /Z 8 (/X /Y) 8 (6 7 ) Find XY nd YZ using the sine lw. XY sin Y XY sin Z XY sin 7 6 sin XZ YZ sin X XY sin Z YZ sin 6 6 sin YZ Find ech ngle using the cosine lw. cos R RS RT ST (RS)(RT) 4 7 (4)(7) Clculus nd Vectors Solutions Mnul 6-

36 7 /R 8 44 cos S RS ST RT (RS)(ST) 4 7 (4)() /S 8 cos T RT ST RS (RT)(ST) 7 4 (7)() 9 3 /T A 3. km T 7 8. C km/h 48 A T 8 km/h Find AC nd AT using the speed of ech vehicle nd the elpsed time (in hours) until it ws locted, distnce speed 3 time. AC kmh 3 4 h km AT 8 kmh 3 3 h 6 3 km Find CT using the cosine lw. CT AC AT (AC)(AT) cos A ( km) 6 3 km 6 km ( km) 6 km cos km CT 8. km 9. A Find AB (the distnce etween the irplnes) using the cosine lw. AB AT BT (AT)(BT)cos T (3. km) (6 km) (3. km)(6 km) cos km AB 8.8 km 7. P km 7 km Q 4 R Find QR using the cosine lw. QR PQ PR (PQ)(PR) cos P ( km) (7 km) ( km)(7 km) cos km QR km B cm cm B C C The pentgon cn e divided into congruent right tringles with height AC nd se BC. 3 /A 36 /A 36 Find AC nd BC using trigonometric rtios. AC AB 3 cos A cos cm BC AB 3 sin A sin cm The re of the pentgon is the sum of the res of the right tringles. Use the re of ^ABC to determine the re of the pentgon. 6- Chpter 6: Introduction to Vectors

37 Are pentgon 3 (BC)(AC) 3 (.9 cm)(4. cm) 9.4 cm 6. An Introduction to Vectors, pp Flse. Two vectors with the sme mgnitude cn hve different directions, so the re not equl.. True. Equl vectors hve the sme direction nd the sme mgnitude. Flse. Equl or opposite vectors must e prllel nd hve the sme mgnitude. If two prllel vectors hve different mgnitude, the cnnot e equl or opposite. d. Flse. Equl or opposite vectors must e prllel nd hve the sme mgnitude. Two vectors with the sme mgnitude cn hve directions tht re not prllel, so the re not equl or opposite.. Vectors must hve mgnitude nd direction. For some sclrs, it is cler wht is ment just the numer. Other sclrs re relted to the mgnitude of vector. Height is sclr. Height is the distnce (see elow) from one end to the other end. No direction is given. Temperture is sclr. Negtive tempertures re elow freeing, ut this is not direction. Weight is vector. It is the force (see elow) of grvit cting on our mss. Mss is sclr. There is no direction given. Are is sclr. It is the mount spce inside two-dimensionl oject. It does not hve direction. Volume is sclr. It is the mount of spce inside three-dimensionl oject. No direction is given. Distnce is sclr. The distnce etween two points does not hve direction. Displcement is vector. Its mgnitude is relted to the sclr distnce, ut it gives direction. Speed is sclr. It is the rte of chnge of distnce ( sclr) with respect to time, ut does not give direction. Force is vector. It is push or pull in certin direction. Velocit is vector. It is the rte of chnge of displcement ( vector) with respect to time. Its mgnitude is relted to the sclr speed. 3. Answers m vr. For emple: Friction resists the motion etween two surfces in contct cting in the opposite direction of motion. A rolling ll stops due to friction which resists the direction of motion. A swinging pendulum stops due to friction resisting the swinging pendulum. 4. Answers m vr. For emple:. AD. AD BC ; AB ED CB EB AB DC ; AE CD EC AE ; DE CE EB ; ; ; AC ; & DB DA BC ; AE & EB ; EC & DE ; AB & CB. B H D E.. AB EF ut d. e. A C AB AB CD EF GH AB AB JI G 7.. km h, south. km h, west km h, northest d. km h, northwest e. 6 km h, est F AB EF 6... d. e. I J Clculus nd Vectors Solutions Mnul 6-3

38 Are pentgon 3 (BC)(AC) 3 (.9 cm)(4. cm) 9.4 cm 6. An Introduction to Vectors, pp Flse. Two vectors with the sme mgnitude cn hve different directions, so the re not equl.. True. Equl vectors hve the sme direction nd the sme mgnitude. Flse. Equl or opposite vectors must e prllel nd hve the sme mgnitude. If two prllel vectors hve different mgnitude, the cnnot e equl or opposite. d. Flse. Equl or opposite vectors must e prllel nd hve the sme mgnitude. Two vectors with the sme mgnitude cn hve directions tht re not prllel, so the re not equl or opposite.. Vectors must hve mgnitude nd direction. For some sclrs, it is cler wht is ment just the numer. Other sclrs re relted to the mgnitude of vector. Height is sclr. Height is the distnce (see elow) from one end to the other end. No direction is given. Temperture is sclr. Negtive tempertures re elow freeing, ut this is not direction. Weight is vector. It is the force (see elow) of grvit cting on our mss. Mss is sclr. There is no direction given. Are is sclr. It is the mount spce inside two-dimensionl oject. It does not hve direction. Volume is sclr. It is the mount of spce inside three-dimensionl oject. No direction is given. Distnce is sclr. The distnce etween two points does not hve direction. Displcement is vector. Its mgnitude is relted to the sclr distnce, ut it gives direction. Speed is sclr. It is the rte of chnge of distnce ( sclr) with respect to time, ut does not give direction. Force is vector. It is push or pull in certin direction. Velocit is vector. It is the rte of chnge of displcement ( vector) with respect to time. Its mgnitude is relted to the sclr speed. 3. Answers m vr. For emple: Friction resists the motion etween two surfces in contct cting in the opposite direction of motion. A rolling ll stops due to friction which resists the direction of motion. A swinging pendulum stops due to friction resisting the swinging pendulum. 4. Answers m vr. For emple:. AD. AD BC ; AB ED CB EB AB DC ; AE CD EC AE ; DE CE EB ; ; ; AC ; & DB DA BC ; AE & EB ; EC & DE ; AB & CB. B H D E.. AB EF ut d. e. A C AB AB CD EF GH AB AB JI G 7.. km h, south. km h, west km h, northest d. km h, northwest e. 6 km h, est F AB EF 6... d. e. I J Clculus nd Vectors Solutions Mnul 6-3

39 8.. 4 km h, due south. 7 km h, southwesterl 3 km h southesterl d. km h, due est 9.. i. Flse. The hve equl mgnitude, ut opposite direction. ii. True. The hve equl mgnitude. iii. True. The se hs sides of equl length, so the vectors hve equl mgnitude. iv. True. The hve equl mgnitude nd direction.. E H To clculte BD, BE nd BH, find the lengths of their corresponding line segments BD, BE nd BH using the Pthgoren theorem. BD AB AD 3 3 BD "8 BE AB AE 3 8 BE "73 BH BD DH ("8) 8 BH "8.. The tngent vector descries Jmes s velocit t tht moment. At point A his speed is km h nd he is heding north. The tngent vector shows his velocit is km h, north.. The length of the vector represents the mgnitude of Jmes s velocit t tht point. Jmes s speed is the sme s the mgnitude of Jmes s velocit. The mgnitude of Jmes s velocit (his speed) is constnt, ut the direction of his velocit chnges t ever point. d. Point C e. This point is hlfw etween D nd A, which is of the w round the circle. Since he is running 7 8 A B F D G C km h nd the trck is km in circumference, he cn run round the trck times in one hour. Tht 7 mens ech lp tkes him 4 minutes. 8 of 4 minutes is 3. minutes. 3 f. When he hs trvelled 8 of lp, Jmes will e hlfw etween B nd C nd will e heding southwest... Find the mgnitude of AB using the distnce formul. AB "( A B ) ( B A ) "(4 ) (3 ) or 3.6. CD " AB. AB moves from A(4, ) to B(, 3) or ( B, B ) ( A 3, A ). Use this to find point D. ( D, D ) ( C 3, C ) (6 3, ) EF (3, AB ). Find point E using ( A, A ) ( B 3, B ). ( E, E ) ( F 3, F ) (3 3, ) d. GH (, 3) AB, nd moves in the opposite direction s AB. ( H, H ) ( G 3, G ). ( H, H ) ( G 3, G ) (3 3, ) (, ) 6. Vector Addition, pp Chpter 6: Introduction to Vectors

40 8.. 4 km h, due south. 7 km h, southwesterl 3 km h southesterl d. km h, due est 9.. i. Flse. The hve equl mgnitude, ut opposite direction. ii. True. The hve equl mgnitude. iii. True. The se hs sides of equl length, so the vectors hve equl mgnitude. iv. True. The hve equl mgnitude nd direction.. E H To clculte BD, BE nd BH, find the lengths of their corresponding line segments BD, BE nd BH using the Pthgoren theorem. BD AB AD 3 3 BD "8 BE AB AE 3 8 BE "73 BH BD DH ("8) 8 BH "8.. The tngent vector descries Jmes s velocit t tht moment. At point A his speed is km h nd he is heding north. The tngent vector shows his velocit is km h, north.. The length of the vector represents the mgnitude of Jmes s velocit t tht point. Jmes s speed is the sme s the mgnitude of Jmes s velocit. The mgnitude of Jmes s velocit (his speed) is constnt, ut the direction of his velocit chnges t ever point. d. Point C e. This point is hlfw etween D nd A, which is of the w round the circle. Since he is running 7 8 A B F D G C km h nd the trck is km in circumference, he cn run round the trck times in one hour. Tht 7 mens ech lp tkes him 4 minutes. 8 of 4 minutes is 3. minutes. 3 f. When he hs trvelled 8 of lp, Jmes will e hlfw etween B nd C nd will e heding southwest... Find the mgnitude of AB using the distnce formul. AB "( A B ) ( B A ) "(4 ) (3 ) or 3.6. CD " AB. AB moves from A(4, ) to B(, 3) or ( B, B ) ( A 3, A ). Use this to find point D. ( D, D ) ( C 3, C ) (6 3, ) EF (3, AB ). Find point E using ( A, A ) ( B 3, B ). ( E, E ) ( F 3, F ) (3 3, ) d. GH (, 3) AB, nd moves in the opposite direction s AB. ( H, H ) ( G 3, G ). ( H, H ) ( G 3, G ) (3 3, ) (, ) 6. Vector Addition, pp Chpter 6: Introduction to Vectors

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