Chapter 6. Additional Topics in Trigonometry. Selected Applications

Transcription

1 hpter 6 dditionl Topics in Trigonometr 6. Lw of Sines 6. Lw of osines 6. Vectors in the Plne 6. Vectors nd Dot Products 6.5 Trigonometric Form of omple Numer u u u u + Selected pplictions Tringles nd ectors he mn rel-life pplictions. The pplictions listed elow represent smll smple of the pplictions in this chpter. Flight Pth, Eercise 7, pge 5 ridge Design, Eercise 8, pge 5 Sureing, Eercise 8, pge Lndu uilding, Eercise 5, pge Velocit, Eercises 8 nd 8, pge 5 Nigtion, Eercises 89 nd 90, pge 6 Reenue, Eercises 59 nd 60, pge 6 Work, Eercise 6, pge 7 Vectors indicte quntities tht inole oth mgnitude nd direction. In hpter 6, ou will stud the opertions of ectors in the plne nd ou will lern how to represent ector opertions geometricll. You will lso lern to sole olique tringles nd write comple numers in trigonometric form. Lester Lefkowitz/Gett Imges Vectors cn e used to find the irspeed nd direction of n irplne tht will llow the irplne to mintin its groundspeed nd direction. 07

2 08 hpter 6 dditionl Topics in Trigonometr 6. Lw of Sines Introduction In hpter ou looked t techniques for soling right tringles. In this section nd the net, ou will sole olique tringles tringles tht he no right ngles. s stndrd nottion, the ngles of tringle re leled,, nd, nd their opposite sides re leled,, nd c, s shown in Figure 6.. Figure 6. c To sole n olique tringle, ou need to know the mesure of t lest one side nd the mesures of n two other prts of the tringle two sides, two ngles, or one ngle nd one side. This reks down into the following four cses. Wht ou should lern Use the Lw of Sines to sole olique tringles (S or S). Use the Lw of Sines to sole olique tringles (SS). Find res of olique tringles. Use the Lw of Sines to model nd sole rel-life prolems. Wh ou should lern it You cn use the Lw of Sines to sole rel-life prolems inoling olique tringles. For instnce, Eercise on pge 5 shows how the Lw of Sines cn e used to help determine the distnce from ot to the shoreline.. Two ngles nd n side (S or S). Two sides nd n ngle opposite one of them (SS). Three sides (SSS). Two sides nd their included ngle (SS) The first two cses cn e soled using the Lw of Sines, wheres the lst two cses require the Lw of osines (see Section 6.). Owen Frnken/oris Lw of Sines (See the proof on pge 68.) If is tringle with sides,, nd c, then sin sin c sin. STUDY TIP h Olique Tringles h Notice in Figure 6. tht ngle is the included ngle etween sides nd c, ngle is the included ngle etween sides nd c, nd ngle is the included ngle etween sides nd. c c is cute. is otuse. The Lw of Sines cn lso e written in the reciprocl form sin sin sin. c

3 Section 6. Lw of Sines 09 Emple Gien Two ngles nd One Side S For the tringle in Figure 6., 0., 8.7, nd 7. feet. Find the remining ngle nd sides. The third ngle of the tringle is the Lw of Sines, ou he Using 7. produces nd 80 sin c Emple sin c sin. 7. sin sin feet sin sin sin sin feet. sin sin 8.7 Now tr Eercise. Gien Two ngles nd One Side S pole tilts towrd the sun t n 8 ngle from the erticl, nd it csts -foot shdow. The ngle of eletion from the tip of the shdow to the top of the pole is. How tll is the pole? In Figure 6., nd So, the third ngle is the Lw of Sines, ou he sin c sin. ecuse c feet, the length of the pole is c sin sin.8 feet. sin sin 9 Now tr Eercise 5. For prctice, tr reworking Emple for pole tht tilts w from the sun under the sme conditions. = 7. ft Figure c c = ft 8.7 STUDY TIP When ou re soling tringles, creful sketch is useful s quick test for the fesiilit of n nswer. Rememer tht the longest side lies opposite the lrgest ngle, nd the shortest side lies opposite the smllest ngle. Figure 6.

4 0 hpter 6 dditionl Topics in Trigonometr The miguous se (SS) In Emples nd ou sw tht two ngles nd one side determine unique tringle. Howeer, if two sides nd one opposite ngle re gien, three possile situtions cn occur: () no such tringle eists, () one such tringle eists, or () two distinct tringles stisf the conditions. The miguous se (SS) onsider tringle in which ou re gien,, nd h sin. is cute. is cute. is cute. is cute. is otuse. is otuse. Sketch h h h Necessr condition Possile < h None h One One h < < Two None > One tringles h Emple Single- se SS For the tringle in Figure 6., inches, inches, nd. Find the remining side nd ngles. the Lw of Sines, ou he sin sin sin sin sin sin.. Now ou cn determine tht Then the remining side is gien c sin sin c Now tr Eercise. Reciprocl form Multipl ech side. Sustitute for,, nd. is cute. sin sin inches. sin sin = in. c = in. Figure 6. One solution: >

5 Section 6. Lw of Sines Emple No- se SS Show tht there is no tringle for which 5, 5, nd 85. egin mking the sketch shown in Figure 6.5. From this figure it ppers tht no tringle is formed. You cn erif this using the Lw of Sines. sin sin sin sin sin 85 sin > sin. Reciprocl form Multipl ech side. This contrdicts the fct tht So, no tringle cn e formed hing sides 5 nd 5 nd n ngle of 85. Now tr Eercise 5. = 5 h 85 = 5 Figure 6.5 No solution: < h Emple 5 Two- se SS Find two tringles for which meters, meters, nd 0.5. ecuse h sin sin meters, ou cn conclude tht there re two possile tringles (ecuse h < < ). the Lw of Sines, ou he sin sin sin sin Reciprocl form There re two ngles 6.8 nd etween 0 nd 80 whose sine is For 6.8, ou otin c For 5., ou otin c sin sin sin meters. sin sin 0.5 sin sin..9 meters. sin sin 0.5 The resulting tringles re shown in Figure 6.6. Now tr Eercise 7. STUDY TIP In Emple 5, the height h of the tringle cn e found using the formul or sin h h sin. = m = m = m = m Figure 6.6 Two solutions: h < <

6 hpter 6 dditionl Topics in Trigonometr re of n Olique Tringle The procedure used to proe the Lw of Sines leds to simple formul for the re of n olique tringle. Referring to Figure 6.7, note tht ech tringle hs height of h sin. To see this when is otuse, sustitute the reference ngle 80 for. Now the height of the tringle is gien h sin80. Using the difference formul for sine, the height is gien h sin 80 cos cos 80 sin 0 cos sin sin. onsequentl, the re of ech tringle is gien re seheight c sin c sin. similr rguments, ou cn deelop the formuls re sin c sin. sinu sin u cos cos u sin h h is cute. Figure 6.7 c is otuse. c re of n Olique Tringle The re of n tringle is one-hlf the product of the lengths of two sides times the sine of their included ngle. Tht is, re c sin sin c sin. Note tht if ngle is 90, the formul gies the re of right tringle s re c seheight. Similr results re otined for ngles nd equl to 90.

7 Emple 6 Finding the re of n Olique Tringle Find the re of tringulr lot hing two sides of lengths 90 meters nd 5 meters nd n included ngle of 0. onsider 90 meters, 5 meters, nd 0, s shown in Figure 6.8. Then the re of the tringle is re sin 905sin squre meters. Section 6. Lw of Sines = 5 m 0 Figure 6.8 Now tr Eercise 9. = 90 m Emple 7 n ppliction of the Lw of Sines The course for ot rce strts t point nd proceeds in the direction S 5 W to point, then in the direction S 0 E to point, nd finll ck to, s shown in Figure 6.9. Point lies 8 kilometers directl south of point. pproimte the totl distnce of the rce course. W N S E 5 ecuse lines D nd re prllel, it follows tht D. onsequentl, tringle hs the mesures shown in Figure 6.0. For ngle, ou he Using the Lw of Sines 0 8 km ou cn let 8 nd otin nd sin 5 sin 88 c sin 0 8 sin 5 6. sin 88 c 8 sin sin 88 The totl length of the course is pproimtel Length kilometers. Now tr Eercise 7. D Figure 6.9 c Figure = 8 km 0

8 hpter 6 dditionl Topics in Trigonometr 6. Eercises See for worked-out solutions to odd-numered eercises. Voculr heck Fill in the lnks.. n tringle is one tht hs no right ngles.. Lw of Sines: c sin sin. The Lw of Sines cn e used to sole tringle for cses () ngle(s) nd side(s), which cn e denoted or, () side(s) nd ngle(s), which cn e denoted.. To find the re of n tringle, use one of the following three formuls: re,, or. In Eercises 8, use the Lw of Sines to sole the tringle. If two solutions eist, find oth , 58, 58,,.,.8.5,.8 c 5 c mm 60 in ft 0 cm c c.8 km , 8, 5 60, 9, c 0 0., 6.7,.6., 5.6, c , 8, 6 5, 6., c 5.8 0, 5, 00 0, 5, 00 76, 8, d In Eercises 9, find the re of the tringle hing the indicted ngle nd sides , 6, 0 0, 9, c 0 8 5, 67, c ,.5, c 75 5, 0, c , 6, 0 5. Height flgpole t right ngle to the horizontl is locted on slope tht mkes n ngle of with the horizontl. The flgpole csts 6-meter shdow up the slope when the ngle of eletion from the tip of the shdow to the sun is 0. () Drw tringle tht represents the prolem. Show the known quntities on the tringle nd use rile to indicte the height of the flgpole. () Write n eqution inoling the unknown quntit. (c) Find the height of the flgpole. 6. Height You re stnding 0 meters from the se of tree tht is lening 8 from the erticl w from ou. The ngle of eletion from our feet to the top of the tree is () Drw tringle tht represents the prolem. Show the known quntities on the tringle nd use rile to indicte the height of the tree. () Write n eqution inoling the unknown height of the tree. (c) Find the height of the tree.

9 Section 6. Lw of Sines 5 7. Flight Pth plne flies 500 kilometers with ering of 6 (clockwise from north) from Nples to Elgin (see figure). The plne then flies 70 kilometers from Elgin to nton. Find the ering of the flight from Elgin to nton. W N S 8. ridge Design ridge is to e uilt cross smll lke from gzeo to dock (see figure). The ering from the gzeo to the dock is S W. From tree 00 meters from the gzeo, the erings to the gzeo nd the dock re S 7 E nd S 8 E, respectiel. Find the distnce from the gzeo to the dock. Tree E nton km 500 km Not drwn to scle 00 m Elgin Nples W N N S Gzeo E. Locting Fire The ering from the Pine Kno fire tower to the olt Sttion fire tower is N 65 E, nd the two towers re 0 kilometers prt. fire spotted rngers in ech tower hs ering of N 80 E from Pine Kno nd S 70 E from olt Sttion. Find the distnce of the fire from ech tower. N W E olt Sttion S 80 0 km Pine Kno Fire Not drwn to scle. Distnce ot is siling due est prllel to the shoreline t speed of 0 miles per hour. t gien time the ering to lighthouse is S 70 E, nd 5 minutes lter the ering is S 6 E (see figure). The lighthouse is locted t the shoreline. Find the distnce from the ot to the shoreline. N 6 W E 70 d S Dock 9. Rilrod Trck Design The circulr rc of rilrod cure hs chord of length 000 feet nd centrl ngle of 0. () Drw digrm tht isull represents the prolem. Show the known quntities on the digrm nd use the riles r nd s to represent the rdius of the rc nd the length of the rc, respectiel. () Find the rdius r of the circulr rc. (c) Find the length s of the circulr rc. 0. Glide Pth pilot hs just strted on the glide pth for lnding t n irport with runw of length 9000 feet. The ngles of depression from the plne to the ends of the runw re 7.5 nd 8.8. () Drw digrm tht isull represents the prolem. () Find the ir distnce the plne must trel until touching down on the ner end of the runw. (c) Find the ground distnce the plne must trel until touching down. (d) Find the ltitude of the plne when the pilot egins the descent.. ngle of Eletion 0-meter telephone pole csts 7-meter shdow directl down slope when the ngle of eletion of the sun is (see figure). Find, the ngle of eletion of the ground. 8 0 m θ θ 7 m. Distnce The ngles of eletion nd to n irplne re eing continuousl monitored t two osertion points nd, respectiel, which re miles prt, nd the irplne is est of oth points in the sme erticl plne. () Drw digrm tht illustrtes the prolem. () Write n eqution giing the distnce d etween the plne nd point in terms of nd.

10 6 hpter 6 dditionl Topics in Trigonometr 5. Shdow Length The Lening Tower of Pis in Itl lens ecuse it ws uilt on unstle soil miture of cl, snd, nd wter. The tower is pproimtel 58.6 meters tll from its foundtion (see figure). The top of the tower lens out 5.5 meters off center. α 5.5 m β 58.6 m Snthesis True or Flse? In Eercises 7 nd 8, determine whether the sttement is true or flse. Justif our nswer. 7. If n three sides or ngles of n olique tringle re known, then the tringle cn e soled. 8. If tringle contins n otuse ngle, then it must e olique. 9. Writing n the Lw of Sines e used to sole right tringle? If so, write short prgrph eplining how to use the Lw of Sines to sole the following tringle. Is there n esier w to sole the tringle? Eplin. 50, 90, 0 θ d Not drwn to scle 0. Think out It Gien 6 nd 5, find lues of such tht the tringle hs () one solution, () two solutions, nd (c) no solution. () Find the ngle of len of the tower. () Write s function of d nd, where is the ngle of eletion to the sun. (c) Use the Lw of Sines to write n eqution for the length d of the shdow cst the tower in terms of. (d) Use grphing utilit to complete the tle. 6. Grphicl nd Numericl nlsis In the figure, nd re positie ngles. d α c () Write s function of. () Use grphing utilit to grph the function. Determine its domin nd rnge. (c) Use the result of prt () to write c s function of. (d) Use grphing utilit to grph the function in prt (c). Determine its domin nd rnge. (e) Use grphing utilit to complete the tle. Wht cn ou conclude? γ β 9 Mollweide s Formul In Eercises nd, sole the tringle. Then use one of the two forms of Mollweide s Formul to erif the results. Form Form. Sole the tringle for 5, 5, nd 6. Then use form of Mollweide s Formul to erif our solution.. Sole the tringle for, 60, nd. Then use form of Mollweide s Formul to erif our solution. Skills Reiew In Eercises nd, use the gien lues to find (if possile) the lues of the remining four trigonometric functions of... sin c cos cos c sin cos 5, sin tn 9, csc 85 c c In Eercises 5 8, write the product s sum or difference sin 8 cos 6. cos cos cos sin 8. sin sin 6 6

11 Section 6. Lw of osines 7 6. Lw of osines Introduction Two cses remin in the list of conditions needed to sole n olique tringle SSS nd SS. To use the Lw of Sines, ou must know t lest one side nd its opposite ngle. If ou re gien three sides (SSS), or two sides nd their included ngle (SS), none of the rtios in the Lw of Sines would e complete. In such cses ou cn use the Lw of osines. Lw of osines (See the proof on pge 69.) Stndrd Form c c cos c c cos c cos lterntie Form cos c c cos c c cos c Wht ou should lern Use the Lw of osines to sole olique tringles (SSS or SS). Use the Lw of osines to model nd sole rel-life prolems. Use Heron s re Formul to find res of tringles. Wh ou should lern it You cn use the Lw of osines to sole rel-life prolems inoling olique tringles. For instnce, Eercise on pge shows ou how the Lw of osines cn e used to determine the length of the gu wires tht nchor tower. Emple Three Sides of Tringle SSS Find the three ngles of the tringle shown in Figure 6.. = 8 ft c = ft = 9 ft Figure 6. It is good ide first to find the ngle opposite the longest side side in this cse. Using the lterntie form of the Lw of osines, ou find tht Gregor Schuster/zef/oris cos c c ecuse cos is negtie, ou know tht is n otuse ngle gien t this point it is simpler to use the Lw of Sines to determine. sin sin 8 sin ecuse is otuse, must e cute, ecuse tringle cn he t most one otuse ngle. So,.08 nd Now tr Eercise

12 8 hpter 6 dditionl Topics in Trigonometr Do ou see wh it ws wise to find the lrgest ngle first in Emple? Knowing the cosine of n ngle, ou cn determine whether the ngle is cute or otuse. Tht is, cos > 0 for 0 < < 90 cute cos < 0 for 90 < < 80. Otuse So, in Emple, once ou found tht ngle ws otuse, ou knew tht ngles nd were oth cute. Furthermore, if the lrgest ngle is cute, the remining two ngles re lso cute. Emple Two Sides nd the Included ngle SS Find the remining ngles nd side of the tringle shown in Figure 6.. Eplortion Wht fmilir formul do ou otin when ou use the third form of the Lw of osines c cos nd ou let 90? Wht is the reltionship etween the Lw of osines nd this formul? = 9 m 5 Figure 6. c = m STUDY TIP Use the Lw of osines to find the unknown side in the figure. c c cos 9 9 cos ecuse 5.07 meters, ou now know the rtio sin nd ou cn use the reciprocl form of the Lw of Sines to sole for. sin sin sin sin There re two ngles etween 0 nd 80 whose sine is 0.70,.7 nd For.7, For 5., 9 sin ecuse side c is the longest side of the tringle, must e the lrgest ngle of the tringle. So,.7 nd 0.. Now tr Eercise 5. When soling n olique tringle gien three sides, ou use the lterntie form of the Lw of osines to sole for n ngle. When soling n olique tringle gien two sides nd their included ngle, ou use the stndrd form of the Lw of osines to sole for n unknown side. Eplortion In Emple, suppose 5. fter soling for, which ngle would ou sole for net, or? re there two possile solutions for tht ngle? If so, how cn ou determine which ngle is the correct solution?

13 Section 6. Lw of osines 9 pplictions Emple n ppliction of the Lw of osines The pitcher s mound on women s softll field is feet from home plte nd the distnce etween the ses is 60 feet, s shown in Figure 6.. (The pitcher s mound is not hlfw etween home plte nd second se.) How fr is the pitcher s mound from first se? In tringle HPF, H 5 (line HP isects the right ngle t H), f, nd p 60. Using the Lw of osines for this SS cse, ou he h f p fp cos H cos 5º So, the pproimte distnce from the pitcher s mound to first se is h feet. Now tr Eercise ft 60 ft P h F f = ft 60 ft 5 p = 60 ft H Figure 6. Emple n ppliction of the Lw of osines ship trels 60 miles due est, then djusts its course northwrd, s shown in Figure 6.. fter treling 80 miles in the new direction, the ship is 9 miles from its point of deprture. Descrie the ering from point to point. N W S E c = 60 mi = 9 mi = 80 mi Not drwn to scle Figure 6. You he 80, 9, nd c 60; so, using the lterntie form of the Lw of osines, ou he cos c c So, rccos Therefore, the ering mesured from due north from point to point is , or N 76.5 E. Now tr Eercise

14 0 hpter 6 dditionl Topics in Trigonometr Heron s re Formul The Lw of osines cn e used to estlish the following formul for the re of tringle. This formul is clled Heron s re Formul fter the Greek mthemticin Heron (c. 00..). Heron s re Formul (See the proof on pge 70.) Gien n tringle with sides of lengths,, nd c, the re of the tringle is gien re where s ss s s c c. Emple 5 Using Heron s re Formul Find the re of tringle hing sides of lengths meters, 5 meters, nd c 7 meters. ecuse s c 68 8, Heron s re Formul ields re ss s s c squre meters. Now tr Eercise 5. You he now studied three different formuls for the re of tringle. Formuls for re of Tringle. Stndrd Formul: re h. Olique Tringle: re c sin sin c sin. Heron s re Formul: re ss s s c Eplortion n the formuls oe e used to find the re of n tpe of tringle? Eplin the dntges nd disdntges of using one formul oer nother.

15 Section 6. Lw of osines 6. Eercises See for worked-out solutions to odd-numered eercises. Voculr heck Fill in the lnks.. The stndrd form of the Lw of osines for cos c is.. Formul is estlished using the Lw of osines.. Three different formuls for the re of tringle re gien re, re c sin sin c sin, nd re. In Eercises 0, use the Lw of osines to sole the tringle... 6 in. in m 8.5 m mm 0 0 km 0 6 km mi.5 ft 0. ft 0. mi in. 0.8 m 5 mm , 9, 8,, c c. 50,. 08, 5, 0, c ,. 5,, 0, c 5 c , 6.., 8, 0.75, c 8 c.5 cm 5. d. d 8 cm 8 cm. d , , , 6, 0, 6., c 8 c 0 c , 6.5,.5 In Eercises 6, complete the tle soling the prllelogrm shown in the figure. (The lengths of the digonls re gien c nd d.) c d In Eercises 7 6, use Heron s re Formul to find the re of the tringle in. 5 m in. 5 m 8 in. m 9. 5, 8, c 0 0., 7, c 7 φ θ c d 0

16 hpter 6 dditionl Topics in Trigonometr...5 ft.5 ft. ft...5, 75., 0., 5, c 9 c ,.5, 6.65,.85, c. c Nigtion plne flies 80 miles from Frnklin to enterille with ering of 75 (clockwise from north). Then it flies 68 miles from enterille to Rosemont with ering of. Drw digrm tht isull represents the prolem, nd find the stright-line distnce nd ering from Rosemont to Frnklin. 8. Sureing To pproimte the length of mrsh, sureor wlks 80 meters from point to point. Then the sureor turns 80 nd wlks 0 meters to point (see figure). pproimte the length of the mrsh m.75 cm.5 cm. cm 80 m. Distnce Two ships lee port t 9.M. One trels t ering of N 5 W t miles per hour, nd the other trels t ering of S 67 W t 6 miles per hour. pproimte how fr prt the ships re t noon tht d.. Length 00-foot erticl tower is to e erected on the side of hill tht mkes 6 ngle with the horizontl (see figure). Find the length of ech of the two gu wires tht will e nchored 75 feet uphill nd downhill from the se of the tower ft 75 ft. Trusses Q is the midpoint of the line segment PR in the truss rfter shown in the figure. Wht re the lengths of the line segments PQ, QS, nd RS? R 75 ft Q 0 9. Nigtion ot rce runs long tringulr course mrked uos,, nd. The rce strts with the ots heded west for 600 meters. The other two sides of the course lie to the north of the first side, nd their lengths re 500 meters nd 800 meters. Drw digrm tht isull represents the prolem, nd find the erings for the lst two legs of the rce. 0. Streetlight Design Determine the ngle in the design of the streetlight shown in the figure. S P wning Design retrctle wning oe ptio lowers t n ngle of 50 from the eterior wll t height of 0 feet oe the ground (see figure). No direct sunlight is to enter the door when the ngle of eletion of the sun is greter thn 70. Wht is the length of the wning? 50 Sunís rs 0 ft θ Lndu uilding The Lndu uilding in mridge, Msschusetts hs tringulr-shped se. The lengths of the sides of the tringulr se re 5 feet, 57 feet, nd 90 feet. Find the re of the se of the uilding.

17 Section 6. Lw of osines 6. Geometr prking lot hs the shpe of prllelogrm (see figure). The lengths of two djcent sides re 70 meters nd 00 meters. The ngle etween the two sides is 70. Wht is the re of the prking lot? 70 m 7. Engine Design n engine hs seen-inch connecting rod fstened to crnk (see figure). () Use the Lw of osines to write n eqution giing the reltionship etween nd. () Write s function of. (Select the sign tht ields positie lues of. ) (c) Use grphing utilit to grph the function in prt (). (d) Use the grph in prt (c) to determine the totl distnce the piston moes in one ccle..5 in. 70 θ Figure for 7 Figure for 8 8. Mnufcturing In process with continuous pper, the pper psses cross three rollers of rdii inches, inches, nd 6 inches (see figure). The centers of the three-inch nd si-inch rollers re d inches prt, nd the length of the rc in contct with the pper on the four-inch roller is s inches. () Use the Lw of osines to write n eqution giing the reltionship etween d nd. () Write s function of d. (c) Write s s function of. (d) omplete the tle. 7 in. 00 m s in. in. 6 in. d (inches) (degrees) s (inches) θ d Snthesis True or Flse? In Eercises 9 5, determine whether the sttement is true or flse. Justif our nswer. 9. tringle with side lengths of 0 feet, 6 feet, nd 5 feet cn e soled using the Lw of osines. 50. Two sides nd their included ngle determine unique tringle. 5. In Heron s re Formul, s is the erge of the lengths of the three sides of the tringle. Proofs In Eercises 5 5, use the Lw of osines to proe ech of the following c cos c c c cos c c cos cos 55. Proof Use hlf-ngle formul nd the Lw of osines to show tht, for n tringle, cos ss c where s c. 56. Proof Use hlf-ngle formul nd the Lw of osines to show tht, for n tringle, sin s s where s c. 57. Writing Descrie how the Lw of osines cn e used to sole the miguous cse of the olique tringle, where feet, 0 feet, nd 0. Is the result the sme s when the Lw of Sines is used to sole the tringle? Descrie the dntges nd the disdntges of ech method. 58. Writing In Eercise 57, the Lw of osines ws used to sole tringle in the two-solution cse of SS. n the Lw of osines e used to sole the no-solution nd singlesolution cses of SS? Eplin. Skills Reiew cos c c c In Eercises 59 6, elute the epression without using clcultor. 59. rcsin 60. cos 0 6. tn 6. rcsin

18 hpter 6 dditionl Topics in Trigonometr 6. Vectors in the Plne Introduction Mn quntities in geometr nd phsics, such s re, time, nd temperture, cn e represented single rel numer. Other quntities, such s force nd elocit, inole oth mgnitude nd direction nd cnnot e completel chrcterized single rel numer. To represent such quntit, ou cn use directed line segment, s shown in Figure 6.5. The directed line segment PQ \ hs initil point P nd terminl point Q. Its mgnitude, or length, is denoted PQ \ nd cn e found using the Distnce Formul. P Initil point PQ Q Terminl point Figure 6.5 Figure 6.6 Wht ou should lern Represent ectors s directed line segments. Write the component forms of ectors. Perform sic ector opertions nd represent ectors grphicll. Write ectors s liner comintions of unit ectors. Find the direction ngles of ectors. Use ectors to model nd sole rel-life prolems. Wh ou should lern it Vectors re used to nlze numerous spects of eerd life. Eercise 86 on pge 5 shows ou how ectors cn e used to determine the tension in the cles of two crnes lifting n oject. Two directed line segments tht he the sme mgnitude nd direction re equilent. For emple, the directed line segments in Figure 6.6 re ll equilent. The set of ll directed line segments tht re equilent to gien directed line segment PQ \ is ector in the plne, written PQ \. Vectors re denoted lowercse, oldfce letters such s u,, nd w. Emple Equilent Directed Line Segments Let u e represented the directed line segment from P 0, 0 to Q,, nd let e represented the directed line segment from R, to S,, s shown in Figure 6.7. Show tht u. From the Distnce Formul, it follows tht PQ \ nd RS \ he the sme mgnitude. PQ \ 0 0 RS \ Moreoer, oth line segments he the sme direction, ecuse the re oth directed towrd the upper right on lines hing the sme slope. Slope of PQ \ 0 0 Slope of RS \ So, PQ \ nd RS \ he the sme mgnitude nd direction, nd it follows tht u. Now tr Eercise. Sndr ker/gett Imges (0, 0) (, ) R u P Figure 6.7 (, ) S (, ) Q

19 omponent Form of Vector The directed line segment whose initil point is the origin is often the most conenient representtie of set of equilent directed line segments. This representtie of the ector is in stndrd position. ector whose initil point is t the origin 0, 0 cn e uniquel represented the coordintes of its terminl point,. This is the component form of ector, written s,. The coordintes nd re the components of. If oth the initil point nd the terminl point lie t the origin, is the zero ector nd is denoted 0 0, 0. Section 6. Vectors in the Plne 5 omponent Form of Vector The component form of the ector with initil point P p, p nd terminl point Q q, q is gien PQ \ q p, q p,. The mgnitude (or length) of is gien q p q p. If, is unit ector. Moreoer, 0 if nd onl if is the zero ector 0. Two ectors u u, u nd, re equl if nd onl if u nd u. For instnce, in Emple, the ector u from P 0, 0 to Q, is u PQ \ 0, 0, nd the ector from R, to S, is RS \,,. TEHNOLOGY TIP You cn grph ectors with grphing utilit grphing directed line segments. onsult the user s guide for our grphing utilit for specific instructions. Emple Finding the omponent Form of Vector Find the component form nd mgnitude of the ector tht hs initil point, 7 nd terminl point, 5. Let P, 7 p, p nd Q, 5 q, q, s shown in Figure 6.8. Then, the components of, re q p 5 q p Q = (, 5) So, 5, nd the mgnitude of is P = (, 7) Now tr Eercise 5. Figure 6.8

20 6 hpter 6 dditionl Topics in Trigonometr Vector Opertions The two sic ector opertions re sclr multipliction nd ector ddition. Geometricll, the product of ector nd sclr k is the ector tht is k times s long s. If k is positie, k hs the sme direction s, nd if k is negtie, k hs the opposite direction of, s shown in Figure 6.9. To dd two ectors u nd geometricll, first position them (without chnging their lengths or directions) so tht the initil point of the second ector coincides with the terminl point of the first ector u. The sum u is the ector formed joining the initil point of the first ector u with the terminl point of the second ector, s shown in Figure 6.0. This technique is clled the prllelogrm lw for ector ddition ecuse the ector u, often clled the resultnt of ector ddition, is the digonl of prllelogrm hing u nd s its djcent sides. Figure 6.9 u + u u Figure 6.0 Definition of Vector ddition nd Sclr Multipliction Let u u, u nd, e ectors nd let k e sclr ( rel numer). Then the sum of u nd is the ector u u, u Sum nd the sclr multiple of k times u is the ector ku ku, u ku, ku. Sclr multiple u The negtie of, is, nd the difference of u nd is Negtie u + ( ) u u u dd. See Figure 6.. Figure 6. u, u. Difference To represent u geometricll, ou cn use directed line segments with the sme initil point. The difference u is the ector from the terminl point of to the terminl point of u, which is equl to u, s shown in Figure 6..

21 Section 6. Vectors in the Plne 7 The component definitions of ector ddition nd sclr multipliction re illustrted in Emple. In this emple, notice tht ech of the ector opertions cn e interpreted geometricll. Emple Vector Opertions Let, 5 nd w,, nd find ech of the following ectors... w c. w d. w (, 0) (, 5) ecuse, 5, ou he, 5, 5, 0. sketch of is shown in Figure 6... The difference of w nd is w, 5 5,. sketch of w is shown in Figure 6.. Note tht the figure shows the ector difference w s the sum w. c. The sum of nd w is w, 5,, 5,, 5 6, 8 6, 5 8,. sketch of w is shown in Figure 6.. d. The difference of nd w is w, 5,, 5,, 0 9, 9, 0,. sketch of w is shown in Figure 6.5. Note tht the figure shows the ector difference w s the sum w. Now tr Eercise Figure 6. w Figure 6. (, 5) 0 w 8 w (, ) 5 (5, ) (, ) + w Figure 6. (, 0) w w (, ) 6 Figure 6.5

22 8 hpter 6 dditionl Topics in Trigonometr Vector ddition nd sclr multipliction shre mn of the properties of ordinr rithmetic. Properties of Vector ddition nd Sclr Multipliction Let u,, nd w e ectors nd let c nd d e sclrs. Then the following properties re true.. u u. u w u w. u 0 u. u u 0 5. cdu cdu 6. c du cu du 7. cu cu c 8. u u, 0u 0 9. c c Unit Vectors In mn pplictions of ectors, it is useful to find unit ector tht hs the sme direction s gien nonzero ector. To do this, ou cn diide its length to otin u unit ector. Unit ector in direction of Note tht u is sclr multiple of. The ector u hs mgnitude of nd the sme direction s. The ector u is clled unit ector in the direction of. STUDY TIP Propert 9 cn e stted s follows: The mgnitude of the ector c is the solute lue of c times the mgnitude of. Emple Finding Unit Vector Find unit ector in the direction of, 5 nd erif tht the result hs mgnitude of. The unit ector in the direction of is, 5 5, 5 9 9, 5 9 9, This ector hs mgnitude of ecuse Now tr Eercise 7.

23 Section 6. Vectors in the Plne 9 The unit ectors, 0 nd 0, re clled the stndrd unit ectors nd re denoted i, 0 nd j 0, s shown in Figure 6.6. (Note tht the lowercse letter i is written in oldfce to distinguish it from the imginr numer i. ) These ectors cn e used to represent n ector, s follows.,, 0 0, i j The sclrs nd re clled the horizontl nd erticl components of, respectiel. The ector sum i j is clled liner comintion of the ectors i nd j. n ector in the plne cn e written s liner comintion of the stndrd unit ectors i nd j. Emple 5 Writing Liner omintion of Unit Vectors Let u e the ector with initil point, 5 nd terminl point,. Write u s liner comintion of the stndrd unit ectors i nd j. egin writing the component form of the ector u. u, 5, 8 i 8j This result is shown grphicll in Figure 6.7. Now tr Eercise 5. j = 0, i =, 0 Figure (, ) u 6 (, 5) Figure 6.7 Emple 6 Vector Opertions Let u i 8j nd i j. Find u. You could sole this prolem conerting u nd to component form. This, howeer, is not necessr. It is just s es to perform the opertions in unit ector form. u i 8j i j 6i 6j 6i j i 9j Now tr Eercise 57.

24 0 hpter 6 dditionl Topics in Trigonometr Direction ngles If u is unit ector such tht is the ngle (mesured counterclockwise) from the positie -is to u, the terminl point of u lies on the unit circle nd ou he u, cos, sin cos i sin j s shown in Figure 6.8. The ngle is the direction ngle of the ector u. Suppose tht u is unit ector with direction ngle. If i j is n ector tht mkes n ngle with the positie -is, then it hs the sme direction s u nd ou cn write cos, sin u θ = cos θ (, ) = sin θ ecuse i j cos i sin j, it follows tht the direction ngle for is determined from tn cos i sin j. sin cos Quotient identit Figure 6.8 sin cos Multipl numertor nd denomintor.. Simplif. Emple 7 Finding Direction ngles of Vectors Find the direction ngle of ech ector.. u i j. i j. The direction ngle is u (, ) tn. So, s shown in Figure The direction ngle is 5, tn. Moreoer, ecuse i j lies in Qudrnt IV, lies in Qudrnt IV nd its reference ngle is rctn 5. So, it follows tht s shown in Figure 6.0. Now tr Eercise , θ = 5 Figure 6.9 θ = (, ) Figure 6.0

25 Section 6. Vectors in the Plne pplictions of Vectors Emple 8 Finding the omponent Form of Vector Find the component form of the ector tht represents the elocit of n irplne descending t speed of 00 miles per hour t n ngle of 0 elow the horizontl, s shown in Figure 6.. The elocit ector hs mgnitude of 00 nd direction ngle of 0. 0 cos i sin j 00cos 0i 00sin 0j i 00 j 50 i 50j 50, 50 You cn check tht hs mgnitude of 00 s follows. Figure , checks. Now tr Eercise 8. Emple 9 Using Vectors to Determine Weight force of 600 pounds is required to pull ot nd triler up rmp inclined t 5 from the horizontl. Find the comined weight of the ot nd triler. sed on Figure 6., ou cn mke the following osertions. \ force of grit comined weight of ot nd triler \ force ginst rmp \ force required to moe ot up rmp 600 pounds construction, tringles WD nd re similr. So, ngle is In tringle ou he sin \ \ \ \ sin 5 5. W 5 D 5 Figure 6. So, the comined weight is pproimtel 8 pounds. (In Figure 6., note tht \ is prllel to the rmp.) Now tr Eercise 85.

26 hpter 6 dditionl Topics in Trigonometr Emple 0 Using Vectors to Find Speed nd Direction n irplne is treling t speed of 500 miles per hour with ering of 0 t fied ltitude with negligile wind elocit, s shown in Figure 6.(). s the irplne reches certin point, it encounters wind lowing with elocit of 70 miles per hour in the direction N 5 E, s shown in Figure 6.(). Wht re the resultnt speed nd direction of the irplne? STUDY TIP Recll from Section.8 tht in ir nigtion, erings re mesured in degrees clockwise from north. 0 Wind θ () Figure 6. () Using Figure 6., the elocit of the irplne (lone) is 500cos 0, sin 0 50, 50 nd the elocit of the wind is 70cos 5, sin 5 5, 5. So, the elocit of the irplne (in the wind) is 50 5, , 8.5 nd the resultnt speed of the irplne is miles per hour. Finll, if tn which implies tht is the direction ngle of the flight pth, ou he rctn So, the true direction of the irplne is 7.. Now tr Eercise 89.

27 Section 6. Vectors in the Plne 6. Eercises See for worked-out solutions to odd-numered eercises. Voculr heck Fill in the lnks.. cn e used to represent quntit tht inoles oth mgnitude nd direction.. The directed line segment PQ \ hs point P nd point Q.. The of the directed line segment PQ \ is denoted PQ \.. The set of ll directed line segments tht re equilent to gien directed line segment PQ \ is in the plne. 5. The directed line segment whose initil point is the origin is sid to e in. 6. ector tht hs mgnitude of is clled. 7. The two sic ector opertions re sclr nd ector. 8. The ector u is clled the of ector ddition. 9. The ector sum i j is clled of the ectors i nd j, nd the sclrs nd re clled the nd components of, respectiel. In Eercises nd, show tht u... (0, ) 6 (6, 5) u u (, ) (, ) (, ) (0, 0) 6 (, ) (0, 5) In Eercises, find the component form nd the mgnitude of the ector... (, ) (, ) Initil Point 9. 5, 0. 7, 0.,. (, ) 5 (, ) Terminl Point, 5 7 0, 5,,, 5 5 In Eercises 8, use the figure to sketch grph of the specified ector. To print n enlrged cop of the grph, go to the wesite 5 (, ) 5 (, ) (, ) 5 (, ) 6 (, 5) 5 5 (, ).. u 5. u 6. u 7. u 8. u u

28 hpter 6 dditionl Topics in Trigonometr In Eercises 9, use the figure to sketch grph of the specified ector. To print n enlrged cop of the grph, go to the wesite 9. u 0.. u.. u. u In Eercises 5 0, find () u, () u, (c) u, nd (d) u. 5. u,, 7, u 6, 8,, u 0, 5,, 9 u i j, i j 0. In Eercises, use the figure nd write the ector in terms of the other two ectors.. w.. u. In Eercises 5, find unit ector in the direction of the gien ector. 5. u 6, 0 6. u 0, 7., 8., 9., , 0. i j. w i j. w j. w i u u w In Eercises 5 50, find the ector with the gien mgnitude nd the sme direction s u. Mgnitude Direction u 5, 6 u, u 5,,, 0 u i j, i j Mgnitude Direction u i j u i j u i u 5j In Eercises 5 5, the initil nd terminl points of ector re gien. Write the ector s liner comintion of the stndrd unit ectors i nd j. Initil Point Terminl Point 5.,, ,, 6 5., 5, 5. 6, 0, In Eercises 55 60, find the component form of nd sketch the specified ector opertions geometricll, where u i j nd w i j. 55. u 56. w 57. u w 58. u w 59. u w 60. u w In Eercises 6 66, find the mgnitude nd direction ngle of the ector cos 0i sin 0j 8cos 5i sin 5j 6. 6i 6j 6. i 7j 65. i 5j 66. i 5j In Eercises 67 7, find the component form of gien its mgnitude nd the ngle it mkes with the positie -is. Sketch. Mgnitude ngle in the direction i j 7. in the direction i j In Eercises 7 76, find the component form of the sum of u nd with direction ngles u nd. Mgnitude ngle 7. u 5 u

29 Section 6. Vectors in the Plne Mgnitude ngle In Eercises 77 nd 78, use the Lw of osines to find the ngle etween the ectors. (ssume 0 } } 80. ) u u 0 50 u 5 50 i j, i j, In Eercises 79 nd 80, grph the ectors nd the resultnt of the ectors. Find the mgnitude nd direction of the resultnt w i j w i j 00 Resultnt Force In Eercises 8 nd 8, find the ngle etween the forces gien the mgnitude of their resultnt. (Hint: Write force s ector in the direction of the positie -is nd force s ector t n ngle with the positie -is.) Force Force Resultnt Force 8. 5 pounds 60 pounds 90 pounds pounds 000 pounds 750 pounds 8. Velocit ll is thrown with n initil elocit of 70 feet per second, t n ngle of 0 with the horizontl (see figure). Find the erticl nd horizontl components of the elocit. u u u 70 ft sec Velocit gun with muzzle elocit of 00 feet per second is fired t n ngle of with the horizontl. Find the erticl nd horizontl components of the elocit. 85. Tension Use the figure to determine the tension in ech cle supporting the lod. in. 0 in. 0 in. 86. Tension The crnes shown in the figure re lifting n oject tht weighs 0,0 pounds. Find the tension in the cle of ech crne Numericl nd Grphicl nlsis loded rge is eing towed two tugots, nd the mgnitude of the resultnt is 6000 pounds directed long the is of the rge (see figure). Ech tow line mkes n ngle of degrees with the is of the rge l () Write the resultnt tension T of ech tow line s function of. Determine the domin of the function. () Use grphing utilit to complete the tle. T θ θ (c) Use grphing utilit to grph the tension function. (d) Eplin wh the tension increses s increses.

30 6 hpter 6 dditionl Topics in Trigonometr 88. Numericl nd Grphicl nlsis To crr 00-pound clindricl weight, two people lift on the ends of short ropes tht re tied to n eelet on the top center of the clinder. Ech rope mkes n ngle of degrees with the erticl (see figure). θ θ (c) Write the elocit of the jet reltie to the ir s ector in component form. (d) Wht is the speed of the jet with respect to the ground? (e) Wht is the true direction of the jet? 9. Numericl nd Grphicl nlsis Forces with mgnitudes of 50 newtons nd 0 newtons ct on hook (see figure). 50 newtons 00 l () Write the tension T of ech rope s function of. Determine the domin of the function. () Use grphing utilit to complete the tle. θ 0 newtons (c) Use grphing utilit to grph the tension function. (d) Eplin wh the tension increses s increses. 89. Nigtion n irplne is fling in the direction 8 with n irspeed of 860 kilometers per hour. ecuse of the wind, its groundspeed nd direction re, respectiel, 800 kilometers per hour nd 0. Find the direction nd speed of the wind. Wind T kilometers per hour 860 kilometers per hour 90. Nigtion commercil jet is fling from Mimi to Settle. The jet s elocit with respect to the ir is 580 miles per hour, nd its ering is. The wind, t the ltitude of the plne, is lowing from the southwest with elocit of 60 miles per hour. () Drw figure tht gies isul representtion of the prolem. () Write the elocit of the wind s ector in component form. W N S E () Find the direction nd mgnitude of the resultnt of the forces when () Write the mgnitude M of the resultnt nd the direction of the resultnt s functions of, where (c) Use grphing utilit to complete the tle. M (d) Use grphing utilit to grph the two functions. (e) Eplin wh one function decreses for incresing, wheres the other doesn t. 9. Numericl nd Grphicl nlsis tetherll weighing pound is pulled outwrd from the pole horizontl force u until the rope mkes n ngle of degrees with the pole (see figure). θ Tension l () Write the tension T in the rope nd the mgnitude of u s functions of. Determine the domins of the functions. u

31 Section 6. Vectors in the Plne 7 () Use grphing utilit to complete the tle. (c) Use grphing utilit to grph the two functions for (d) ompre T nd u s increses. Snthesis T u True or Flse? In Eercises 9 96, determine whether the sttement is true or flse. Justif our nswer. 9. If u nd he the sme mgnitude nd direction, then u. 9. If u is unit ector in the direction of, then u. 95. If i j 0, then. 96. If u i j is unit ector, then. True or Flse? In Eercises 97 0, use the figure to determine whether the sttement is true or flse. Justif our nswer. 05. Think out It onsider two forces of equl mgnitude cting on point. () If the mgnitude of the resultnt is the sum of the mgnitudes of the two forces, mke conjecture out the ngle etween the forces. () If the resultnt of the forces is 0, mke conjecture out the ngle etween the forces. (c) n the mgnitude of the resultnt e greter thn the sum of the mgnitudes of the two forces? Eplin. 06. Grphicl Resoning onsider two forces F 0, u nd c 97. d 98. c s 99. u c 00. w s 0. w d 0. d 0 0. u t 0. t w d t s w F 5cos, sin. () Find F F s function of. () Use grphing utilit to grph the function for 0 <. (c) Use the grph in prt () to determine the rnge of the function. Wht is its mimum, nd for wht lue of does it occur? Wht is its minimum, nd for wht lue of does it occur? (d) Eplin wh the mgnitude of the resultnt is neer Proof Proe tht cos i sin j is unit ector for n lue of. 08. Technolog Write progrm for our grphing utilit tht grphs two ectors nd their difference gien the ectors in component form. In Eercises 09 nd 0, use the progrm in Eercise 08 to find the difference of the ectors shown in the grph Skills Reiew In Eercises 6, simplif the epression s 5 t 5 s 50t (, 6) (, 5) (5, ) 6 8 In Eercises 7 0, use the trigonometric sustitution to write the lgeric epression s trigonometric function of, where 0 < < / , 9, 00,, 7 sin 7 sec 0 cot csc (9, ) In Eercises, sole the eqution.. cos cos 0. sin sin 0. sec 0. cos cot cos 0 00 ( 0, 70) (80, 80) (0, 60) ( 00, 0) 50 50

32 8 hpter 6 dditionl Topics in Trigonometr 6. Vectors nd Dot Products The Dot Product of Two Vectors So fr ou he studied two ector opertions ector ddition nd multipliction sclr ech of which ields nother ector. In this section, ou will stud third ector opertion, the dot product. This product ields sclr, rther thn ector. Definition of Dot Product The dot product of u u, u nd, is gien u u u. Properties of the Dot Product (See the proofs on pge 7.) Let u,, nd w e ectors in the plne or in spce nd let c e sclr. Wht ou should lern Find the dot product of two ectors nd use properties of the dot product. Find ngles etween ectors nd determine whether two ectors re orthogonl. Write ectors s sums of two ector components. Use ectors to find the work done force. Wh ou should lern it You cn use the dot product of two ectors to sole rel-life prolems inoling two ector quntities. For instnce, Eercise 6 on pge 6 shows ou how the dot product cn e used to find the force necessr to keep truck from rolling down hill..... u u 0 0 u w u u w 5. cu cu u c Emple Finding Dot Products Find ech dot product.., 5,.,, c. 0,, ln Thornton/Gett Imges.. c., 5, 5 8 5,, 0 0,, Now tr Eercise. In Emple, e sure ou see tht the dot product of two ectors is sclr ( rel numer), not ector. Moreoer, notice tht the dot product cn e positie, zero, or negtie.

33 Emple Using Properties of Dot Products Let u,,,, nd w,. Find ech dot product.. u w. u egin finding the dot product of u nd... u,, u w,, 8 u u 8 Notice tht the product in prt () is ector, wheres the product in prt () is sclr. n ou see wh? Now tr Eercise 9. Section 6. Vectors nd Dot Products 9 Emple Dot Product nd Mgnitude The dot product of u with itself is 5. Wht is the mgnitude of u? ecuse u u u 5, it follows tht u u u 5. Now tr Eercise. The ngle etween Two Vectors The ngle etween two nonzero ectors is the ngle, 0, etween their respectie stndrd position ectors, s shown in Figure 6.. This ngle cn e found using the dot product. (Note tht the ngle etween the zero ector nd nother ector is not defined.) ngle etween Two Vectors (See the proof on pge 7.) If is the ngle etween two nonzero ectors u nd, then cos u u. u u θ Origin Figure 6.

34 0 hpter 6 dditionl Topics in Trigonometr Emple Finding the ngle etween Two Vectors Find the ngle etween u, nd, 5. cos u This implies tht the ngle etween the two ectors is rccos u,, 5,, s shown in Figure 6.5. Rewriting the epression for the ngle etween two ectors in the form u u cos Now tr Eercise 7. lterntie form of dot product produces n lterntie w to clculte the dot product. From this form, ou cn see tht ecuse u nd re lws positie, u nd cos will lws he the sme sign. Figure 6.6 shows the fie possile orienttions of two ectors. TEHNOLOGY TIP The grphing utilit progrm Finding the ngle etween Two Vectors, found t this tetook s Online Stud enter, grphs two ectors u, nd c, d in stndrd position nd finds the mesure of the ngle etween them. Use the progrm to erif Emple. 6 5 =, 5. θ u =, 5 6 Figure 6.5 u θ < < cos < cos < 0 cos 0 Opposite direction Otuse ngle 90 ngle Figure 6.6 u θ u θ u θ 0 < < 0 < cos < cute ngle 0 u cos Sme direction Definition of Orthogonl Vectors The ectors u nd re orthogonl if u 0. The terms orthogonl nd perpendiculr men essentill the sme thing meeting t right ngles. Een though the ngle etween the zero ector nd nother ector is not defined, it is conenient to etend the definition of orthogonlit to include the zero ector. In other words, the zero ector is orthogonl to eer ector u ecuse 0 u 0.

35 Section 6. Vectors nd Dot Products Emple 5 Determining Orthogonl Vectors re the ectors u, nd 6, orthogonl? egin finding the dot product of the two ectors. u, 6, 6 0 ecuse the dot product is 0, the two ectors re orthogonl, s shown in Figure 6.7. = 6, u =, Figure 6.7 Now tr Eercise 5. Finding Vector omponents You he lred seen pplictions in which two ectors re dded to produce resultnt ector. Mn pplictions in phsics nd engineering pose the reerse prolem decomposing gien ector into the sum of two ector components. onsider ot on n inclined rmp, s shown in Figure 6.8. The force F due to grit pulls the ot down the rmp nd ginst the rmp. These two orthogonl forces, nd w, re ector components of F. Tht is, F w w. w Vector components of F The negtie of component w represents the force needed to keep the ot from rolling down the rmp, nd w represents the force tht the tires must withstnd ginst the rmp. procedure for finding nd is shown on the net pge. w w w F Figure 6.8 w

36 hpter 6 dditionl Topics in Trigonometr Definition of Vector omponents Let u nd e nonzero ectors such tht u w w w where nd re orthogonl nd is prllel to (or sclr multiple of), s shown in Figure 6.9. The ectors w nd w re clled ector components of u. The ector w is the projection of u onto nd is denoted The ector w w w proj u. w is gien w u w. w u u w θ θ w w is cute. Figure 6.9 is otuse From the definition of ector components, ou cn see tht it is es to find the component w once ou he found the projection of u onto. To find the projection, ou cn use the dot product, s follows. So, nd u w w c w u c w c u c w c 0 w proj u c u. w is sclr multiple of. Tke dot product of ech side with. w nd re orthogonl. Projection of u onto Let u nd e nonzero ectors. The projection of u onto is gien proj u u.

37 Section 6. Vectors nd Dot Products Emple 6 Decomposing Vector into omponents Find the projection of u, 5 onto 6,. Then write u s the sum of two orthogonl ectors, one of which is proj u. The projection of u onto is w proj u u 8 0 6, 6 5, 5 s shown in Figure 6.0. The other component, w, is w = 6, 5 6 So, w u w, 5 6 5, 5 9 5, 7 5. u w w 6 5, 5 9 7, 5 5, 5. Now tr Eercise 5. 5 Figure 6.0 w u =, 5 Emple 7 Finding Force 00-pound crt sits on rmp inclined t 0, s shown in Figure 6.. Wht force is required to keep the crt from rolling down the rmp? ecuse the force due to grit is erticl nd downwrd, ou cn represent the grittionl force the ector F 00j. Force due to grit To find the force required to keep the crt from rolling down the rmp, project F onto unit ector in the direction of the rmp, s follows. cos 0i sin 0j i j Therefore, the projection of F onto is w proj F F F i j. Unit ector long rmp The mgnitude of this force is 00, nd therefore force of 00 pounds is required to keep the crt from rolling down the rmp. Now tr Eercise 6. 0 w Figure 6. F