Vectors 3-1 VECTORS AND THEIR COMPONENTS. What Is Physics? Vectors and Scalars. Learning Objectives

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1 C H A P T E R 3 Vectors 3-1 VECTORS AND THEIR COMPONENTS Lerning Ojectives After reding this module, ou should e le to Add vectors drwing them in hed-to-til rrngements, ppling the commuttive nd ssocitive lws Sutrct vector from second one Clculte the components of vector on given coordinte sstem, showing them in drwing Given the components of vector, drw the vector nd determine its mgnitude nd orienttion Convert ngle mesures etween degrees nd rdins. Ke Ides Sclrs, such s temperture, hve mgnitude onl. The re specified numer with unit (10 C) nd oe the rules of rithmetic nd ordinr lger. Vectors, such s displcement, hve oth mgnitude nd direction (5 m, north) nd oe the rules of vector lger. Two vectors nd m e dded geometricll drwing them to common scle nd plcing them hed to til. The vector connecting the til of the first to the hed of the second is the vector sum s. To sutrct from, reverse the direction of to get ; then dd to. Vector ddition is commuttive nd oes the ssocitive lw. The (sclr) components nd of n two-dimensionl vector long the coordinte es re found dropping perpendiculr lines from the ends of onto the coordinte es. The components re given cos u nd sin u, where u is the ngle etween the positive direction of the is nd the direction of. The lgeric sign of component indictes its direction long the ssocited is. Given its components, we cn find the mgnitude nd orienttion of the vector with nd tn. Wht Is Phsics? Phsics dels with gret mn quntities tht hve oth size nd direction, nd it needs specil mthemticl lnguge the lnguge of vectors to descrie those quntities. This lnguge is lso used in engineering, the other sciences, nd even in common speech. If ou hve ever given directions such s Go five locks down this street nd then hng left, ou hve used the lnguge of vectors. In fct, nvigtion of n sort is sed on vectors, ut phsics nd engineering lso need vectors in specil ws to eplin phenomen involving rottion nd mgnetic forces, which we get to in lter chpters. In this chpter, we focus on the sic lnguge of vectors. 40 Vectors nd Sclrs A prticle moving long stright line cn move in onl two directions. We cn tke its motion to e positive in one of these directions nd negtive in the other. For prticle moving in three dimensions, however, plus sign or minus sign is no longer enough to indicte direction. Insted, we must use vector.

2 3-1 VECTORS AND THEIR COMPONENTS 41 A vector hs mgnitude s well s direction, nd vectors follow certin (vector) rules of comintion, which we emine in this chpter. A vector quntit is quntit tht hs oth mgnitude nd direction nd thus cn e represented with vector. Some phsicl quntities tht re vector quntities re displcement, velocit, nd ccelertion. You will see mn more throughout this ook, so lerning the rules of vector comintion now will help ou gretl in lter chpters. Not ll phsicl quntities involve direction. Temperture, pressure, energ, mss, nd time, for emple, do not point in the sptil sense. We cll such quntities sclrs, nd we del with them the rules of ordinr lger. A single vlue, with sign (s in temperture of 40 F), specifies sclr. The simplest vector quntit is displcement, or chnge of position. A vector tht represents displcement is clled, resonl, displcement vector. (Similrl, we hve velocit vectors nd ccelertion vectors.) If prticle chnges its position moving from A to B in Fig. 3-1, we s tht it undergoes displcement from A to B, which we represent with n rrow pointing from A to B.The rrow specifies the vector grphicll. To distinguish vector smols from other kinds of rrows in this ook, we use the outline of tringle s the rrowhed. In Fig. 3-1, the rrows from A to B, from A to B, nd from A to B hve the sme mgnitude nd direction. Thus, the specif identicl displcement vectors nd represent the sme chnge of position for the prticle. A vector cn e shifted without chnging its vlue if its length nd direction re not chnged. The displcement vector tells us nothing out the ctul pth tht the prticle tkes. In Fig. 3-1, for emple, ll three pths connecting points A nd B correspond to the sme displcement vector, tht of Fig Displcement vectors represent onl the overll effect of the motion, not the motion itself. Adding Vectors Geometricll Suppose tht, s in the vector digrm of Fig. 3-2, prticle moves from A to B nd then lter from B to C. We cn represent its overll displcement (no mtter wht its ctul pth) with two successive displcement vectors, AB nd BC. The net displcement of these two displcements is single displcement from A to C. We cll AC the vector sum (or resultnt) of the vectors AB nd BC. This sum is not the usul lgeric sum. In Fig. 3-2, we redrw the vectors of Fig. 3-2 nd relel them in the w tht we shll use from now on, nmel, with n rrow over n itlic smol, s in. If we wnt to indicte onl the mgnitude of the vector ( quntit tht lcks sign or direction), we shll use the itlic smol, s in,, nd s. (You cn use just hndwritten smol.) A smol with n overhed rrow lws implies oth properties of vector, mgnitude nd direction. We cn represent the reltion mong the three vectors in Fig. 3-2 with the vector eqution s, (3-1) which ss tht the vector s is the vector sum of vectors nd.the smol in Eq. 3-1 nd the words sum nd dd hve different menings for vectors thn the do in the usul lger ecuse the involve oth mgnitude nd direction. Figure 3-2 suggests procedure for dding two-dimensionl vectors nd geometricll. (1) On pper, sketch vector to some convenient scle nd t the proper ngle. (2) Sketch vector to the sme scle, with its til t the hed of vector, gin t the proper ngle. (3) The vector sum s is the vector tht etends from the til of to the hed of. Properties. Vector ddition, defined in this w, hs two importnt properties. First, the order of ddition does not mtter. Adding to gives the sme () Figure 3-1 () All three rrows hve the sme mgnitude nd direction nd thus represent the sme displcement. () All three pths connecting the two points correspond to the sme displcement vector. Actul pth A A B A B B () A' A" B' B" C Net displcement is the vector sum () s () To dd nd, drw them hed to til. This is the resulting vector, from til of to hed of. Figure 3-2 () AC is the vector sum of the vectors AB nd BC. () The sme vectors releled.

3 42 CHAPTER 3 VECTORS Strt + + Figure 3-3 The two vectors nd cn e dded in either order; see Eq Vector sum Finish You get the sme vector result for either order of dding vectors. result s dding to (Fig. 3-3); tht is, (commuttive lw). (3-2) Second, when there re more thn two vectors, we cn group them in n order s we dd them. Thus, if we wnt to dd vectors,, nd c, we cn dd nd first nd then dd their vector sum to c. We cn lso dd nd c first nd then dd tht sum to. We get the sme result either w, s shown in Fig Tht is, + c ( ) c ( c ) + (ssocitive lw). (3-3) You get the sme vector result for n order of dding the vectors. + + ( + c ) + + c + c c ( + ) +c c Figure 3-4 The three vectors,, nd c cn e grouped in n w s the re dded; see Eq Figure 3-5 The vectors nd hve the sme mgnitude nd opposite directions. The vector is vector with the sme mgnitude s ut the opposite direction (see Fig. 3-5).Adding the two vectors in Fig. 3-5 would ield ( ) 0. Thus, dding hs the effect of sutrcting. We use this propert to define the difference etween two vectors let d. Then d ( ) (vector sutrction); (3-4) tht is, we find the difference vector d dding the vector to the vector. Figure 3-6 shows how this is done geometricll. As in the usul lger, we cn move term tht includes vector smol from one side of vector eqution to the other, ut we must chnge its sign. For emple, if we re given Eq. 3-4 nd need to solve for, we cn rerrnge the eqution s d or d. () Rememer tht, lthough we hve used displcement vectors here, the rules for ddition nd sutrction hold for vectors of ll kinds, whether the represent velocities, ccelertions, or n other vector quntit. However, we cn dd onl vectors of the sme kind. For emple, we cn dd two displcements, or two velocities, ut dding displcement nd velocit mkes no sense. In the rithmetic of sclrs, tht would e like tring to dd 21 s nd 12 m. d = Note hed-to-til rrngement for ddition Checkpoint 1 () Figure 3-6 () Vectors,, nd. () To sutrct vector from vector, dd vector to vector. The mgnitudes of displcements nd re 3 m nd 4 m, respectivel, nd c. Considering vrious orienttions of nd, wht re () the mimum possile mgnitude for c nd () the minimum possile mgnitude? Components of Vectors Adding vectors geometricll cn e tedious. A neter nd esier technique involves lger ut requires tht the vectors e plced on rectngulr coordinte sstem.the nd es re usull drwn in the plne of the pge, s shown

4 3-1 VECTORS AND THEIR COMPONENTS 43 in Fig The z is comes directl out of the pge t the origin; we ignore it for now nd del onl with two-dimensionl vectors. A component of vector is the projection of the vector on n is. In Fig. 3-7, for emple, is the component of vector on (or long) the is nd is the component long the is. To find the projection of vector long n is, we drw perpendiculr lines from the two ends of the vector to the is, s shown.the projection of vector on n is is its component, nd similrl the projection on the is is the component. The process of finding the components of vector is clled resolving the vector. A component of vector hs the sme direction (long n is) s the vector. In Fig. 3-7, nd re oth positive ecuse etends in the positive direction of oth es. (Note the smll rrowheds on the components, to indicte their direction.) If we were to reverse vector, then oth components would e negtive nd their rrowheds would point towrd negtive nd. Resolving vector in Fig. 3-8 ields positive component nd negtive component. In generl, vector hs three components, lthough for the cse of Fig. 3-7 the component long the z is is zero.as Figs. 3-7 nd show, if ou shift vector without chnging its direction, its components do not chnge. Finding the Components. We cn find the components of in Fig. 3-7 geometricll from the right tringle there cos u nd sin u, (3-5) where u is the ngle tht the vector mkes with the positive direction of the is, nd is the mgnitude of. Figure 3-7c shows tht nd its nd components form right tringle. It lso shows how we cn reconstruct vector from its components we rrnge those components hed to til. Then we complete right tringle with the vector forming the hpotenuse, from the til of one component to the hed of the other component. Once vector hs een resolved into its components long set of es, the components themselves cn e used in plce of the vector. For emple, in Fig. 3-7 is given (completel determined) nd u. It cn lso e given its components nd. Both pirs of vlues contin the sme informtion. If we know vector in component nottion ( nd ) nd wnt it in mgnitude-ngle nottion ( nd u), we cn use the equtions nd tn (3-6) to trnsform it. In the more generl three-dimensionl cse, we need mgnitude nd two ngles (s,, u, nd f) or three components (,, nd z ) to specif vector. O Figure 3-7 () The components nd of vector. () The components re unchnged if the vector is shifted, s long s the mgnitude nd orienttion re mintined. (c) The components form the legs of right tringle whose hpotenuse is the mgnitude of the vector. (m) O = 5 m This is the component of the vector. () This is the component of the vector. = 7 m This is the component of the vector. Figure 3-8 The component of on the is is positive, nd tht on the is is negtive. () This is the component of the vector. The components nd the vector form right tringle. (c) (m) O Checkpoint 2 In the figure, which of the indicted methods for comining the nd components of vector re proper to determine tht vector? () () (c) (d) (e) ( f )

5 44 CHAPTER 3 VECTORS Smple Prolem 3.01 Adding vectors in drwing, orienteering In n orienteering clss, ou hve the gol of moving s fr (stright-line distnce) from se cmp s possile mking three stright-line moves. You m use the following displcements in n order (), 2.0 km due est (directl towrd the est); (), 2.0 km 30 north of est (t n ngle of 30 towrd the north from due est); (c) c, 1.0 km due west. Alterntivel, ou m sustitute either for or c for c. Wht is the gretest distnce ou cn e from se cmp t the end of the third displcement? (We re not concerned out the direction.) Resoning Using convenient scle, we drw vectors,, c,, nd c s in Fig We then mentll slide the vectors over the pge, connecting three of them t time in hed-to-til rrngements to find their vector sum d. The til of the first vector represents se cmp. The hed of the third vector represents the point t which ou stop. The vector sum d etends from the til of the first vector to the hed of the third vector. Its mgnitude d is our distnce from se cmp. Our gol here is to mimize tht se-cmp distnce. We find tht distnce d is gretest for hed-to-til rrngement of vectors,, nd c. The cn e in n c 30 Scle of km () c Figure 3-9 () Displcement vectors; three re to e used. () Your distnce from se cmp is gretest if ou undergo displcements,, nd c, in n order. order, ecuse their vector sum is the sme for n order. (Recll from Eq. 3-2 tht vectors commute.) The order shown in Fig. 3-9 is for the vector sum d ( c ). Using the scle given in Fig. 3-9, we mesure the length d of this vector sum, finding d 4.8 m. (Answer) () d = + c c This is the vector result for dding those three vectors in n order. Smple Prolem 3.02 Finding components, irplne flight A smll irplne leves n irport on n overcst d nd is lter sighted 215 km w, in direction mking n ngle of 22 est of due north. This mens tht the direction is not due north (directl towrd the north) ut is rotted 22 towrd the est from due north. How fr est nd north is the irplne from the irport when sighted? Distnce (km) km Distnce (km) Figure 3-10 A plne tkes off from n irport t the origin nd is lter sighted t P. d P KEY IDEA We re given the mgnitude (215 km) nd the ngle (22 est of due north) of vector nd need to find the components of the vector. Clcultions We drw n coordinte sstem with the positive direction of due est nd tht of due north (Fig. 3-10). For convenience, the origin is plced t the irport. (We don t hve to do this. We could shift nd mislign the coordinte sstem ut, given choice, wh mke the prolem more difficult?) The irplne s displcement d points from the origin to where the irplne is sighted. To find the components of d, we use Eq. 3-5 with u 68 ( ) d d cos u (215 km)(cos 68 ) 81 km d d sin u (215 km)(sin 68 ) 199 km km. (Answer) (Answer) Thus, the irplne is 81 km est nd km north of the irport. Additionl emples, video, nd prctice ville t WilePLUS

6 3-1 VECTORS AND THEIR COMPONENTS 45 Prolem-Solving Tctics Angles, trig functions, nd inverse trig functions Tctic 1 Angles Degrees nd Rdins Angles tht re mesured reltive to the positive direction of the is re positive if the re mesured in the counterclockwise direction nd negtive if mesured clockwise. For emple, 210 nd 150 re the sme ngle. Angles m e mesured in degrees or rdins (rd).to relte the two mesures, recll tht full circle is 360 nd 2p rd.to convert, s, 40 to rdins, write 40 2 rd rd. Tctic 2 Trig Functions You need to know the definitions of the common trigonometric functions sine, cosine, nd tngent ecuse the re prt of the lnguge of science nd engineering. The re given in Fig in form tht does not depend on how the tringle is leled. You should lso e le to sketch how the trig functions vr with ngle, s in Fig. 3-12, in order to e le to judge whether clcultor result is resonle. Even knowing the signs of the functions in the vrious qudrnts cn e of help Qudrnts IV I II III IV () () cos sin Tctic 3 Inverse Trig Functions When the inverse trig functions sin 1, cos 1, nd tn 1 re tken on clcultor, ou must consider the resonleness of the nswer ou get, ecuse there is usull nother possile nswer tht the clcultor does not give. The rnge of opertion for clcultor in tking ech inverse trig function is indicted in Fig As n emple, sin hs ssocited ngles of 30 (which is displed the clcultor, since 30 flls within its rnge of opertion) nd 150. To see oth vlues, drw horizontl line through 0.5 in Fig nd note where it cuts the sine curve. How do ou distinguish correct nswer? It is the one tht seems more resonle for the given sitution tn Tctic 4 Mesuring Vector Angles The equtions for cos u nd sin u in Eq. 3-5 nd for tn u in Eq. 3-6 re vlid onl if the ngle is mesured from the positive direction of sin = cos = tn = leg opposite hpotenuse leg djcent to hpotenuse leg opposite leg djcent to Hpotenuse Leg djcent to Figure 3-11 A tringle used to define the trigonometric functions. See lso Appendi E. Leg opposite (c) Figure 3-12 Three useful curves to rememer. A clcultor s rnge of opertion for tking inverse trig functions is indicted the drker portions of the colored curves. the is. If it is mesured reltive to some other direction, then the trig functions in Eq. 3-5 m hve to e interchnged nd the rtio in Eq. 3-6 m hve to e inverted. A sfer method is to convert the ngle to one mesured from the positive direction of the is. In WilePLUS, the sstem epects ou to report n ngle of direction like this (nd positive if counterclockwise nd negtive if clockwise). Additionl emples, video, nd prctice ville t WilePLUS

7 46 CHAPTER 3 VECTORS 3-2 UNIT VECTORS, ADDING VECTORS BY COMPONENTS Lerning Ojectives After reding this module, ou should e le to Convert vector etween mgnitude-ngle nd unitvector nottions Add nd sutrct vectors in mgnitude-ngle nottion nd in unit-vector nottion Identif tht, for given vector, rotting the coordinte sstem out the origin cn chnge the vector s components ut not the vector itself. Ke Ides Unit vectors î, ĵ, nd kˆ hve mgnitudes of unit nd re in which î, ĵ, nd z kˆ re the vector components of nd directed in the positive directions of the,, nd z es,,, nd z re its sclr components. respectivel, in right-hnded coordinte sstem. We cn To dd vectors in component form, we use the rules write vector in terms of unit vectors s r r r z z z. î ĵ z kˆ, Here nd re the vectors to e dded, nd r is the vector sum. Note tht we dd components is is. The unit vectors point long es. kˆ ĵ z Figure 3.13 Unit vectors î, ĵ, nd kˆ define the directions of right-hnded coordinte sstem. î Unit Vectors A unit vector is vector tht hs mgnitude of ectl 1 nd points in prticulr direction. It lcks oth dimension nd unit. Its sole purpose is to point tht is, to specif direction. The unit vectors in the positive directions of the,, nd z es re leled î, ĵ, nd kˆ, where the ht ˆ is used insted of n overhed rrow s for other vectors (Fig. 3-13).The rrngement of es in Fig is sid to e right-hnded coordinte sstem. The sstem remins right-hnded if it is rotted rigidl. We use such coordinte sstems eclusivel in this ook. Unit vectors re ver useful for epressing other vectors; for emple, we cn epress nd of Figs. 3-7 nd 3-8 s î ĵ (3-7) nd î ĵ. (3-8) These two equtions re illustrted in Fig The quntities î nd ĵ re vectors, clled the vector components of.the quntities nd re sclrs, clled the sclr components of (or, s efore, simpl its components). This is the vector component. Figure 3-14 () The vector components of vector. () The vector components of vector. ˆj O () ˆi This is the vector component. î O ˆj () Adding Vectors Components We cn dd vectors geometricll on sketch or directl on vector-cple clcultor. A third w is to comine their components is is.

8 3-2 UNIT VECTORS, ADDING VECTORS BY COMPONENTS 47 To strt, consider the sttement r, (3-9) which ss tht the vector r is the sme s the vector ( ). Thus, ech component of r must e the sme s the corresponding component of ( ) r (3-10) r (3-11) r z z z. (3-12) In other words, two vectors must e equl if their corresponding components re equl. Equtions 3-9 to 3-12 tell us tht to dd vectors nd, we must (1) resolve the vectors into their sclr components; (2) comine these sclr components, is is, to get the components of the sum r ; nd (3) comine the components of r to get r itself. We hve choice in step 3. We cn epress r in unit-vector nottion or in mgnitude-ngle nottion. This procedure for dding vectors components lso pplies to vector sutrctions. Recll tht sutrction such s d cn e rewritten s n ddition d ( ). To sutrct, we dd nd components, to get d, d, nd d z z z, where d d î d ĵ d z kˆ. (3-13) Checkpoint 3 () In the figure here, wht re the signs of the components of nd d d 1 2? () Wht re the signs of the components of d 1 nd d 2? (c) Wht re the signs of the nd components of? d 1 d 2 d 1 d 2 Vectors nd the Lws of Phsics So fr, in ever figure tht includes coordinte sstem, the nd es re prllel to the edges of the ook pge. Thus, when vector is included, its compo- nents nd re lso prllel to the edges (s in Fig. 3-15).The onl reson for tht orienttion of the es is tht it looks proper ; there is no deeper reson. We could, insted, rotte the es (ut not the vector ) through n ngle f s in Fig. 3-15, in which cse the components would hve new vlues, cll them nd. Since there re n infinite numer of choices of f, there re n infinite numer of different pirs of components for. Which then is the right pir of components? The nswer is tht the re ll equll vlid ecuse ech pir (with its es) just gives us different w of descriing the sme vector ; ll produce the sme mgnitude nd direction for the vector. In Fig we hve (3-14) nd u u f. (3-15) The point is tht we hve gret freedom in choosing coordinte sstem, ecuse the reltions mong vectors do not depend on the loction of the origin or on the orienttion of the es.this is lso true of the reltions of phsics; the re ll independent of the choice of coordinte sstem. Add to tht the simplicit nd richness of the lnguge of vectors nd ou cn see wh the lws of phsics re lmost lws presented in tht lnguge one eqution, like Eq. 3-9, cn represent three (or even more) reltions, like Eqs. 3-10, 3-11, nd ' ' O O ' ' () Rotting the es chnges the components ut not the vector. () Figure 3-15 () The vector nd its components. () The sme vector, with the es of the coordinte sstem rotted through n ngle f. φ '

9 48 CHAPTER 3 VECTORS Smple Prolem 3.03 Serching through hedge mze A hedge mze is mze formed tll rows of hedge. After entering, ou serch for the center point nd then for the eit. Figure 3-16 shows the entrnce to such mze nd the first two choices we mke t the junctions we encounter in moving from point i to point c. We undergo three displcements s indicted in the overhed view of Fig d m 1 40 d m 2 30 d m 3 0, where the lst segment is prllel to the superimposed is. When we rech point c, wht re the mgnitude nd ngle of our net displcement dnet from point i? KEY IDEAS (1) To find the net displcement d net, we need to sum the three individul displcement vectors dnet d 1 d 2 d 3. (2) To do this, we first evlute this sum for the components lone, d net, d l d 2 d 3, (3-16) nd then the components lone, d net, d 1 d 2 d 3. (3-17) (3) Finll, we construct dnet from its nd components. Clcultions To evlute Eqs nd 3-17, we find the nd components of ech displcement. As n emple, the components for the first displcement re shown in Fig. 3-16c.We drw similr digrms for the other two displcements nd then we ppl the prt of Eq. 3-5 to ech displcement, using ngles reltive to the positive direction of the is d l (6.00 m) cos m d 2 (8.00 m) cos ( 60 ) 4.00 m d 3 (5.00 m) cos m. Eqution 3-16 then gives us d net, 4.60 m 4.00 m 5.00 m m. Similrl, to evlute Eq. 3-17, we ppl the prt of Eq. 3-5 to ech displcement d l (6.00 m) sin 40 = 3.86 m d 2 (8.00 m) sin ( 60 ) = 6.93 m d 3 (5.00 m) sin 0 0 m. Eqution 3-17 then gives us d net, 3.86 m 6.93 m 0 m 3.07 m. Net we use these components of dnet to construct the vector s shown in Fig. 3-16d the components re in hed-totil rrngement nd form the legs of right tringle, nd i d 1 u 1 u 2 Three vectors d2 First vector d 1 d 1 i () c i Net vector () d net, d 3 c d 1 (c) d net, d net c (d) Figure 3-16 () Three displcements through hedge mze. () The displcement vectors. (c) The first displcement vector nd its components. (d) The net displcement vector nd its components.

10 3-2 UNIT VECTORS, ADDING VECTORS BY COMPONENTS 49 the vector forms the hpotenuse. We find the mgnitude nd ngle of dnet with Eq The mgnitude is d net 2d 2 net, d 2 net, (3-18) 2(13.60 m) 2 ( 3.07 m) m. (Answer) To find the ngle (mesured from the positive direction of ), we tke n inverse tngent tn d 1 net, (3-19) d net, tn 3.07 m (Answer) m The ngle is negtive ecuse it is mesured clockwise from positive. We must lws e lert when we tke n inverse tngent on clcultor. The nswer it displs is mthemticll correct ut it m not e the correct nswer for the phsicl sitution. In those cses, we hve to dd 180 to the displed nswer, to reverse the vector. To check, we lws need to drw the vector nd its components s we did in Fig. 3-16d. In our phsicl sitution, the figure shows us tht 12.7 is resonle nswer, wheres is clerl not. We cn see ll this on the grph of tngent versus ngle in Fig. 3-12c. In our mze prolem, the rgument of the inverse tngent is 3.07/13.60, or On the grph drw horizontl line through tht vlue on the verticl is. The line cuts through the drker plotted rnch t 12.7 nd lso through the lighter rnch t 167. The first cut is wht clcultor displs. Smple Prolem 3.04 Adding vectors, unit-vector components Figure 3-17 shows the following three vectors (4.2 m)î (1.5 m)ĵ, ( 1.6 m)î (2.9 m)ĵ, nd c ( 3.7 m)ĵ. Wht is their vector sum r which is lso shown? () Then rrnge the net components hed to til. 2.6i ˆ c To dd these vectors, find their net component nd their net component. () r r 2.3ĵ This is the result of the ddition. Figure 3-17 Vector r is the vector sum of the other three vectors. KEY IDEA We cn dd the three vectors components, is is, nd then comine the components to write the vector sum r. Clcultions For the is, we dd the components of,, nd c, to get the component of the vector sum r r c 4.2 m 1.6 m m. Similrl, for the is, r c 1.5 m 2.9 m 3.7 m 2.3 m. We then comine these components of r to write the vector in unit-vector nottion r (2.6 m)î (2.3 m)ĵ, (Answer) where (2.6 m)î is the vector component of r long the is nd (2.3 m)ĵ is tht long the is. Figure 3-17 shows one w to rrnge these vector components to form r. (Cn ou sketch the other w?) We cn lso nswer the question giving the mgnitude nd n ngle for r. From Eq. 3-6, the mgnitude is r 2(2.6 m) 2 nd the ngle (mesured from the direction) is tn m ( 2.3 m) m 2.6 m 41, where the minus sign mens clockwise. (Answer) (Answer) Additionl emples, video, nd prctice ville t WilePLUS

11 50 CHAPTER 3 VECTORS 3-3 MULTIPLYING VECTORS Lerning Ojectives After reding this module, ou should e le to Multipl vectors sclrs Identif tht multipling vector sclr gives vector, tking the dot (or sclr) product of two vectors gives sclr, nd tking the cross (or vector) product gives new vector tht is perpendiculr to the originl two Find the dot product of two vectors in mgnitude-ngle nottion nd in unit-vector nottion Find the ngle etween two vectors tking their dot product in oth mgnitude-ngle nottion nd unit-vector nottion. Ke Ides The product of sclr s nd vector v is new vector whose mgnitude is sv nd whose direction is the sme s tht of v if s is positive, nd opposite tht of v if s is negtive. To divide v s, multipl v 1/s. The sclr (or dot) product of two vectors nd is written nd is the sclr quntit given cos, in which is the ngle etween the directions of nd. A sclr product is the product of the mgnitude of one vector nd the sclr component of the second vector long the direction of the first vector. In unit-vector nottion, ( î ĵ z kˆ ) ( î ĵ z kˆ ), which m e epnded ccording to the distriutive lw. Note tht Given two vectors, use dot product to find how much of one vector lies long the other vector Find the cross product of two vectors in mgnitudengle nd unit-vector nottions Use the right-hnd rule to find the direction of the vector tht results from cross product In nested products, where one product is uried inside nother, follow the norml lgeric procedure strting with the innermost product nd working outwrd. The vector (or cross) product of two vectors nd is written nd is vector c whose mgnitude c is given c sin, in which is the smller of the ngles etween the directions of nd. The direction of c is perpendiculr to the plne defined nd nd is given right-hnd rule, s shown in Fig Note tht ( ). In unit-vector nottion, ( î ĵ z kˆ ) ( î ĵ z kˆ ), which we m epnd with the distriutive lw. In nested products, where one product is uried inside nother, follow the norml lgeric procedure strting with the innermost product nd working outwrd. Multipling Vectors* There re three ws in which vectors cn e multiplied, ut none is ectl like the usul lgeric multipliction. As ou red this mteril, keep in mind tht vector-cple clcultor will help ou multipl vectors onl if ou understnd the sic rules of tht multipliction. Multipling Vector Sclr If we multipl vector sclr s, we get new vector. Its mgnitude is the product of the mgnitude of nd the solute vlue of s. Its direction is the direction of if s is positive ut the opposite direction if s is negtive. To divide s, we multipl 1/s. Multipling Vector Vector There re two ws to multipl vector vector one w produces sclr (clled the sclr product), nd the other produces new vector (clled the vector product). (Students commonl confuse the two ws.) *This mteril will not e emploed until lter (Chpter 7 for sclr products nd Chpter 11 for vector products), nd so our instructor m wish to postpone it.

12 3-3 M ULTIPLYING VECTORS 51 The Sclr Product The sclr product of the vectors nd in Fig is written s nd defined to e cos f, (3-20) where is the mgnitude of, is the mgnitude of, nd is the ngle etween nd (or, more properl, etween the directions of nd ). There re ctull two such ngles nd 360. Either cn e used in Eq. 3-20, ecuse their cosines re the sme. Note tht there re onl sclrs on the right side of Eq (including the vlue of cos ). Thus on the left side represents sclr quntit. Becuse of the nottion, is lso known s the dot product nd is spoken s dot. A dot product cn e regrded s the product of two quntities (1) the mgnitude of one of the vectors nd (2) the sclr component of the second vector long the direction of the first vector. For emple, in Fig. 3-18, hs sclr component cos long the direction of ; note tht perpendiculr dropped from the hed of onto determines tht component. Similrl, hs sclr component cos long the direction of. If the ngle etween two vectors is 0, the component of one vector long the other is mimum, nd so lso is the dot product of the vectors. If, insted, is 90, the component of one vector long the other is zero, nd so is the dot product. Eqution 3-20 cn e rewritten s follows to emphsize the components ( cos f)() ()( cos f). (3-21) The commuttive lw pplies to sclr product, so we cn write. When two vectors re in unit-vector nottion, we write their dot product s ( î ĵ z kˆ ) ( î ĵ z kˆ ), (3-22) which we cn epnd ccording to the distriutive lw Ech vector component of the first vector is to e dotted with ech vector component of the second vector. B doing so, we cn show tht z z. (3-23) φ () Component of long direction of is cos φ Figure 3-18 () Two vectors nd, with n ngle f etween them. () Ech vector hs component long the direction of the other vector. Multipling these gives the dot product. Or multipling these gives the dot product. φ Component of long direction of is cos φ ()

13 52 CHAPTER 3 VECTORS Checkpoint 4 Vectors nd hve mgnitudes of 3 units nd 4 units, respectivel.wht is the ngle etween the directions of C nd D if C D equls () zero, () 12 units, nd (c) 12 units? C D The Vector Product The vector product of nd, written, produces third vector c whose mgnitude is c sin f, (3-24) where f is the smller of the two ngles etween nd. (You must use the smller of the two ngles etween the vectors ecuse sin f nd sin(360 f) differ in lgeric sign.) Becuse of the nottion, is lso known s the cross product, nd in speech it is cross. If nd re prllel or ntiprllel, 0. The mgnitude of, which cn e written s, is mimum when nd re perpendiculr to ech other. The direction of c is perpendiculr to the plne tht contins nd. Figure 3-19 shows how to determine the direction of c with wht is known s right-hnd rule. Plce the vectors nd til to til without ltering their orienttions, nd imgine line tht is perpendiculr to their plne where the meet. Pretend to plce our right hnd round tht line in such w tht our fingers would sweep into through the smller ngle etween them. Your outstretched thum points in the direction of c. The order of the vector multipliction is importnt. In Fig. 3-19, we re determining the direction of c, so the fingers re plced to sweep into through the smller ngle. The thum ends up in the opposite direction from previousl, nd so it must e tht c c ; tht is, ( ). (3-25) In other words, the commuttive lw does not ppl to vector product. In unit-vector nottion, we write ( î ĵ z kˆ ) ( î ĵ z kˆ ), (3-26) which cn e epnded ccording to the distriutive lw; tht is, ech component of the first vector is to e crossed with ech component of the second vector. The cross products of unit vectors re given in Appendi E (see Products of Vectors ). For emple, in the epnsion of Eq. 3-26, we hve î î ( î î) 0, ecuse the two unit vectors î nd î re prllel nd thus hve zero cross product. Similrl, we hve î ĵ ( î ĵ) kˆ. In the lst step we used Eq to evlute the mgnitude of î ĵ s unit. (These vectors î nd ĵ ech hve mgnitude of unit, nd the ngle etween them is 90.) Also, we used the right-hnd rule to get the direction of î ĵ s eing in the positive direction of the z is (thus in the direction of kˆ ).

14 3-3 M ULTIPLYING VECTORS 53 Continuing to epnd Eq. 3-26, ou cn show tht ( z z ) î ( z z ) ĵ ( ) kˆ. (3-27) A determinnt (Appendi E) or vector-cple clcultor cn lso e used. To check whether n z coordinte sstem is right-hnded coordinte sstem, use the right-hnd rule for the cross product î ĵ kˆ with tht sstem. If our fingers sweep î (positive direction of ) into ĵ (positive direction of ) with the outstretched thum pointing in the positive direction of z (not the negtive direction), then the sstem is right-hnded. Checkpoint 5 Vectors nd hve mgnitudes of 3 units nd 4 units, respectivel. Wht is the ngle etween the directions of C nd D if the mgnitude of the vector product C D is () zero nd () 12 units? C D A c () c () Figure 3-19 Illustrtion of the right-hnd rule for vector products. () Sweep vector into vector with the fingers of our right hnd. Your outstretched thum shows the direction of vector. () Showing tht c is the reverse of.

15 54 CHAPTER 3 VECTORS Smple Prolem 3.05 Angle etween two vectors using dot products Wht is the ngle etween 3.0 î 4.0 ĵ nd 2.0î 3.0 kˆ? (Cution Although mn of the following steps cn e pssed with vector-cple clcultor, ou will lern more out sclr products if, t lest here, ou use these steps.) KEY IDEA The ngle etween the directions of two vectors is included in the definition of their sclr product (Eq. 3-20) cos f. (3-28) Clcultions In Eq. 3-28, is the mgnitude of, or ( 4.0) , nd is the mgnitude of, or 2( 2.0) (3-29) (3-30) We cn seprtel evlute the left side of Eq writing the vectors in unit-vector nottion nd using the distriutive lw (3.0î 4.0 ĵ) ( 2.0î 3.0 kˆ ) (3.0 î) ( 2.0 î) (3.0 î) (3.0 kˆ ) ( 4.0 ĵ) ( 2.0 î) ( 4.0 ĵ) (3.0 kˆ ). We net ppl Eq to ech term in this lst epression. The ngle etween the unit vectors in the first term ( î nd î) is 0, nd in the other terms it is 90.We then hve (6.0)(1) (9.0)(0) (8.0)(0) (12)(0) 6.0. Sustituting this result nd the results of Eqs nd 3-30 into Eq ields 6.0 (5.00)(3.61) cos f, so cos (5.00)(3.61) (Answer) Smple Prolem 3.06 Cross product, right-hnd rule In Fig. 3-20, vector lies in the plne, hs mgnitude of 18 units, nd points in direction 250 from the positive direction of the is. Also, vector hs mgnitude of 12 units nd points in the positive direction of the z is.wht is the vector product c? KEY IDEA When we hve two vectors in mgnitude-ngle nottion, we find the mgnitude of their cross product with Eq nd the direction of their cross product with the right-hnd rule of Fig Clcultions For the mgnitude we write c sin f (18)(12)(sin 90 ) 216. (Answer) To determine the direction in Fig. 3-20, imgine plcing the fingers of our right hnd round line perpendiculr to the plne of nd (the line on which c is shown) such tht our fingers sweep into. Your outstretched thum then z Sweep into. Figure 3-20 Vector c (in the plne) is the vector (or cross) product of vectors nd. c = gives the direction of c.thus, s shown in the figure, c lies in the plne. Becuse its direction is perpendiculr to the direction of ( cross product lws gives perpendiculr vector), it is t n ngle of (Answer) from the positive direction of the is. This is the resulting vector, perpendiculr to oth nd. Smple Prolem 3.07 Cross product, unit-vector nottion If 3î 4ĵ nd 2î 3 kˆ, wht is c? KEY IDEA When two vectors re in unit-vector nottion, we cn find their cross product using the distriutive lw. Clcultions Here we write c (3î 4) ĵ ( 2î 3 kˆ ) 3î ( 2 î) 3î 3kˆ ( 4 ĵ) ( 2 î) ( 4 ĵ) 3 kˆ.

16 REVIEW & SUMMARY 55 We net evlute ech term with Eq. 3-24, finding the direction with the right-hnd rule. For the first term here, the ngle f etween the two vectors eing crossed is 0. For the other terms, f is 90.We find c 6(0) 9( ĵ) 8( kˆ ) 12î 12î 9 ĵ 8 kˆ. (Answer) This vector c is perpendiculr to oth nd, fct ou cn check showing tht c = 0 nd c = 0; tht is, there is no component of c long the direction of either or. In generl A cross product gives perpendiculr vector, two perpendiculr vectors hve zero dot product, nd two vectors long the sme is hve zero cross product. Additionl emples, video, nd prctice ville t WilePLUS Review & Summr Sclrs nd Vectors Sclrs, such s temperture, hve mgnitude onl. The re specified numer with unit (10 C) nd oe the rules of rithmetic nd ordinr lger. Vectors, such s displcement, hve oth mgnitude nd direction (5 m, north) nd oe the rules of vector lger. Adding Vectors Geometricll Two vectors nd m e dded geometricll drwing them to common scle nd plcing them hed to til. The vector connecting the til of the first to the hed of the second is the vector sum s. To sutrct from, reverse the direction of to get ; then dd to.vector ddition is commuttive nd oes the ssocitive lw ( ) c ( c ). Components of Vector The (sclr) components nd of n two-dimensionl vector long the coordinte es re found dropping perpendiculr lines from the ends of onto the coordinte es. The components re given cos u nd sin u, (3-5) where u is the ngle etween the positive direction of the is nd the direction of. The lgeric sign of component indictes its direction long the ssocited is. Given its components, we cn find the mgnitude nd orienttion (direction) of the vector using nd tn (3-6) Unit-Vector Nottion Unit vectors î, ĵ, nd kˆ hve mgnitudes of unit nd re directed in the positive directions of the,, nd z es, respectivel, in right-hnded coordinte sstem (s defined the vector products of the unit vectors). We cn write vector in terms of unit vectors s î ĵ z kˆ, (3-7) in which î, ĵ, nd z kˆ re the vector components of nd,, nd z re its sclr components. (3-2) (3-3) To dd vectors in com- Adding Vectors in Component Form ponent form, we use the rules r r r z z z. (3-10 to 3-12) Here nd re the vectors to e dded, nd r is the vector sum. Note tht we dd components is is.we cn then epress the sum in unit-vector nottion or mgnitude-ngle nottion. Product of Sclr nd Vector The product of sclr s nd vector v is new vector whose mgnitude is sv nd whose direction is the sme s tht of v if s is positive, nd opposite tht of v if s is negtive. (The negtive sign reverses the vector.) To divide v s, multipl v 1/s. The Sclr Product The sclr (or dot) product of two vectors nd is written nd is the sclr quntit given cos f, (3-20) in which f is the ngle etween the directions of nd. A sclr product is the product of the mgnitude of one vector nd the sclr component of the second vector long the direction of the first vector. Note tht, which mens tht the sclr product oes the commuttive lw. In unit-vector nottion, ( î ĵ z kˆ ) ( î ĵ z kˆ ), (3-22) which m e epnded ccording to the distriutive lw. The Vector Product The vector (or cross) product of two vectors nd is written nd is vector c whose mgnitude c is given c sin f, (3-24) in which f is the smller of the ngles etween the directions of nd. The direction of c is perpendiculr to the plne defined nd nd is given right-hnd rule, s shown in Fig Note tht ( ), which mens tht the vector product does not oe the commuttive lw. In unit-vector nottion, ( î ĵ z kˆ ) ( î ĵ z kˆ ), (3-26) which we m epnd with the distriutive lw.

17 56 CHAPTER 3 VECTORS Questions 1 Cn the sum of the mgnitudes of two vectors ever e equl to the mgnitude of the sum of the sme two vectors? If no, wh not? If es, when? 2 The two vectors shown in Fig lie in n plne. Wht re the signs of the nd components, respectivel, of () d 1 d2, () d 1 d2, nd (c) d 2 d1? 3 Being prt of the Gtors, the Universit of Florid golfing tem must pl on putting green with n lligtor pit. Figure 3-22 shows n overhed view of one putting chllenge of the tem; n coordinte sstem is superimposed. Tem memers must putt from the origin to the hole, which is t coordintes (8 m, 12 m), ut the cn putt the golf ll using onl one or more of the following displcements, one or more times (8 m)î (6 m)ĵ, d2 (6 m)ĵ, (8 m)î. d 1 The pit is t coordintes (8 m, 6 m). If tem memer putts the ll into or through the pit, the memer is utomticll trnsferred to Florid Stte Universit, the rch rivl. Wht sequence of displcements should tem memer use to void the pit nd the school trnsfer? 4 Eqution 3-2 shows tht the ddition of two vectors nd is commuttive. Does tht men sutrction is commuttive, so tht? 5 Which of the rrngements of es in Fig cn e leled right-hnded coordinte sstem? As usul, ech is lel indictes the positive side of the is. z z ( ) (d ) z z ( ) (e ) Figure 3-23 Question 5. d 2 d 1 d 3 Figure 3-21 Question 2. Hole Gtor pit Figure 3-22 Question 3. z (c ) z ( f ) 6 Descrie two vectors nd such tht () c nd c; () ; (c) c nd 2 2 c 2. 7 If d ( c), does () ( d) c ( ), () ( ) d c, nd (c) c ( d)? 8 If c, must equl c? 9 If F q( v B ) nd v is perpendiculr to B, then wht is the direction of B in the three situtions shown in Fig when constnt q is () positive nd () negtive? F F z v z z v (1) (2) (3) Figure 3-24 Question Figure 3-25 shows vector nd four other vectors tht hve the sme D B mgnitude ut differ in orienttion. () Which of those other four vectors hve the sme dot product with A? () A Which hve negtive dot product with A? C 11 In gme held within threedimensionl E mze, ou must move Figure 3-25 Question 10. our gme piece from strt, t z coordintes (0, 0, 0), to finish, t coordintes ( 2 cm, 4 cm, 4 cm). The gme piece cn undergo onl the displcements (in centimeters) given elow. If, long the w, the gme piece lnds t coordintes ( 5 cm, 1 cm, 1 cm) or (5 cm, 2 cm, 1 cm), ou lose the gme. Which displcements nd in wht sequence will get our gme piece to finish? p 7î 2ĵ 3kˆ r 2î 3ĵ 2kˆ q 2î ĵ 4kˆ s 3î 5ĵ 3 kˆ. 12 The nd components of four vectors,, c, nd d re given elow. For which vectors will our clcultor give ou the correct ngle u when ou use it to find u with Eq. 3-6? Answer first emining Fig. 3-12, nd then check our nswers with our clcultor. 3 3 c 3 c d 3 d Which of the following re correct (meningful) vector epressions? Wht is wrong with n incorrect epression? () A ( B C ) (f) A ( B C) () A ( B C ) (g) 5 A (c) A ( B C ) (h) 5 ( B C) (d) A ( B C ) (i) 5 ( B C) (e) A ( B C ) (j) ( A B ) ( B C ) A F v

18 PROBLEMS 57 Prolems SSM Tutoring prolem ville (t instructor s discretion) in WilePLUS nd WeAssign Worked-out solution ville in Student Solutions Mnul WWW Worked-out solution is t Numer of dots indictes level of prolem difficult ILW Interctive solution is t Additionl informtion ville in The Fling Circus of Phsics nd t flingcircusofphsics.com http// Module 3-1 Vectors nd Their Components 1 SSM Wht re () the component nd () the component of vector in the plne if its direction is 250 counterclockwise from the positive direction of the is nd its mgnitude is 7.3 m? r 2 A displcement vector r in the plne is 15 m long nd directed t ngle u 30 in Fig Determine () the component nd () the component of the vector. Figure 3-26 Prolem 2. 3 SSM The component of vector A is 25.0 m nd the component is 40.0 m. () Wht is the mgnitude of A? () Wht is the ngle etween the direction of A nd the positive direction of? 4 Epress the following ngles in rdins () 20.0, () 50.0, (c) 100. Convert the following ngles to degrees (d) rd, (e) 2.10 rd, (f) 7.70 rd. 5 A ship sets out to sil to point 120 km due north. An unepected storm lows the ship to point 100 km due est of its strting point. () How fr nd () in wht direction must it now sil to rech its originl destintion? 6 In Fig. 3-27, hev piece of mchiner is rised sliding it distnce d 12.5 m long plnk oriented t ngle u 20.0 to the horizontl. How fr is it moved () verticll nd () horizontll? 7 Consider two displcements, one of mgnitude 3 m nd nother Figure 3-27 Prolem 6. of mgnitude 4 m. Show how the displcement vectors m e comined to get resultnt displcement of mgnitude () 7 m, () 1 m, nd (c) 5 m. Module 3-2 Unit Vectors, Adding Vectors Components 8 A person wlks in the following pttern 3.1 km north, then 2.4 km west, nd finll 5.2 km south. () Sketch the vector digrm tht represents this motion. () How fr nd (c) in wht direction would ird fl in stright line from the sme strting point to the sme finl point? 9 Two vectors re given (4.0 m)î (3.0 m)ĵ (1.0 m)kˆ nd ( 1.0 m)î (1.0 m)ĵ (4.0 m)kˆ. In unit-vector nottion, find (), (), nd (c) third vector c such tht c Find the (), (), nd (c) z components of the sum r of the displcements c nd d whose components in meters re c 7.4, c 3.8, c z 6.1; d 4.4, d 2.0, d z SSM () In unit-vector nottion, wht is the sum if (4.0 m) î (3.0 m) ĵ nd ( 13.0 m) î (7.0 m) ĵ? Wht re the () mgnitude nd (c) direction of? d 12 A cr is driven est for distnce of 50 km, then north for 30 km, nd then in direction 30 est of north for 25 km. Sketch the vector digrm nd determine () the mgnitude nd () the ngle of the cr s totl displcement from its strting point. 13 A person desires to rech point tht is 3.40 km from her present loction nd in direction tht is 35.0 north of est. However, she must trvel long streets tht re oriented either north south or est west. Wht is the minimum distnce she could trvel to rech her destintion? 14 You re to mke four stright-line moves over flt desert floor, strting t the origin of n coordinte sstem nd ending t the coordintes ( 140 m, 30 m). The component nd component of our moves re the following, respectivel, in meters (20 nd 60), then ( nd 70), then ( 20 nd c ), then ( 60 nd 70). Wht re () component nd () component c? Wht re (c) the mgnitude nd (d) the ngle (reltive to the positive direction of the is) of the overll displcement? 15 SSM ILW WWW The two vec- tors nd in Fig hve equl mgnitudes of 10.0 m nd the ngles re 1 30 nd Find the () nd () components of their 2 vector sum r, (c) the mgnitude of r, nd (d) the ngle r mkes with the positive direction of the is. 16 For the displcement vectors 1 (3.0 m)î (4.0 m)ĵ nd O (5.0 m)î ( 2.0 m)ĵ, give in Figure 3-28 Prolem 15. () unit-vector nottion, nd s () mgnitude nd (c) n ngle (reltive to ). Now give î in (d) unit-vector nottion, nd s (e) mgnitude nd (f) n ngle. 17 ILW Three vectors,, nd c ech hve mgnitude of 50 m nd lie in n plne. Their directions reltive to the positive direction of the is re 30, 195, nd 315, respectivel.wht re () the mgnitude nd () the ngle of the vector c, nd (c) the mgnitude nd (d) the ngle of c? Wht re the (e) mgnitude nd (f) ngle of fourth vector d such tht ( ) (c d ) 0? 18 In the sum A B C, vector A hs mgnitude of 12.0 m nd is ngled 40.0 counterclockwise from the direction, nd vector C hs mgnitude of 15.0 m nd is ngled 20.0 counterclockwise from the direction. Wht re () the mgnitude nd () the ngle (reltive to ) of B? 19 In gme of lwn chess, where pieces re moved etween the centers of squres tht re ech 1.00 m on edge, knight is moved in the following w (1) two squres forwrd, one squre rightwrd; (2) two squres leftwrd, one squre forwrd; (3) two squres forwrd, one squre leftwrd. Wht re () the mgnitude nd () the ngle (reltive to forwrd ) of the knight s overll displcement for the series of three moves?

19 58 CHAPTER 3 VECTORS 20 An eplorer is cught in whiteout (in which the snowfll is so thick tht the ground cnnot e distinguished from the sk) while returning to se cmp. He ws supposed to trvel due north for 5.6 km, ut when the snow clers, he discovers tht he ctull trveled 7.8 km t 50 north of due est. () How fr nd () in wht direction must he now trvel to rech se cmp? 21 An nt, crzed the Sun on hot Tes fternoon, drts over n plne scrtched in the dirt. The nd components of four consecutive drts re the following, ll in centimeters (30.0, 40.0), (, 70.0), ( 20.0, c ), ( 80.0, 70.0). The overll displcement of the four drts hs the components ( 140, 20.0). Wht re () nd () c? Wht re the (c) mgnitude nd (d) ngle (reltive to the positive direction of the is) of the overll displcement? 22 () Wht is the sum of the following four vectors in unitvector nottion? For tht sum, wht re () the mgnitude, (c) the ngle in degrees, nd (d) the ngle in rdins? E 6.00 m t rd G 4.00 m t 1.20 rd 23 If is dded to C 3.0î 4.0ĵ, the result is vector in the positive direction of the is, with mgnitude equl to tht of C. Wht is the mgnitude of B? 24 Vector A, which is directed long n is, is to e dded to vector B, which hs mgnitude of 7.0 m.the sum is third vector tht is directed long the is, with mgnitude tht is 3.0 times tht of A.Wht is tht mgnitude of A? 25 Osis B is 25 km due est of osis A. Strting from osis A, cmel wlks 24 km in direction 15 south of est nd then wlks 8.0 km due north. How fr is the cmel then from osis B? 26 Wht is the sum of the following four vectors in () unitvector nottion, nd s () mgnitude nd (c) n ngle? B F 5.00 m t 75.0 H 6.00 m t 210 A (2.00 m)î (3.00 m)ĵ B 4.00 m, t 65.0 C ( 4.00 m)î ( 6.00 m)ĵ D 5.00 m, t If d1 d 2 5d 3, d 1 d 2 3d 3, nd d3 2î 4ĵ, then wht re, in unit-vector nottion, () d1 nd () d2? 28 Two eetles run cross flt snd, strting t the sme point. Beetle 1 runs 0.50 m due est, then 0.80 m t 30 north of due est. Beetle 2 lso mkes two runs; the first is 1.6 m t 40 est of due north. Wht must e () the mgnitude nd () the direction of its second run if it is to end up t the new loction of eetle 1? 29 Tpicl ckrd nts often crete network of chemicl trils for guidnce. Etending outwrd from the nest, tril rnches (ifurctes) repetedl, with 60 etween the rnches. If roming nt chnces upon tril, it cn tell the w to the nest t n rnch point If it is moving w from the nest, it hs two choices of pth requiring smll turn in its trvel direction, either 30 leftwrd or 30 rightwrd. If it is moving towrd the nest, it hs onl one such choice. Figure 3-29 shows tpicl nt tril, with lettered stright sections of 2.0 cm length nd smmetric ifurction of 60. Pth v is prllel to the is. Wht re the () mgnitude nd () ngle (reltive to the positive direction of the superimposed is) of n nt s displcement from the nest (find it in the figure) if the nt enters the tril t point A? Wht re the (c) mgnitude nd (d) ngle if it enters t point B? 30 e f Here re two vectors d (4.0 m)î (3.0 m)ĵ nd (6.0 m)î (8.0 m)ĵ. Wht re () the mgnitude nd () the ngle (reltive to î) of? Wht re (c) the mgnitude nd (d) the ngle of? Wht re (e) the mgnitude nd (f) the ngle of ; (g) the mgnitude nd (h) the ngle of ; nd (i) the mgnitude nd (j) the ngle of? (k) Wht is the ngle etween the directions of nd? 31 In Fig. 3-30, vector with mgnitude of 17.0 m is directed t ngle 56.0 counterclockwise from the is. Wht re the components () nd () of the vector? A second coordinte sstem is inclined ngle 18.0 with respect to the first. Wht re the components (c) nd (d) in this primed coordinte sstem? ' ' ' c g A h i 32 In Fig. 3-31, cue of edge z length sits with one corner t the origin of n z coordinte sstem. A od digonl is line tht etends from one corner to nother through the center. In unit-vector nottion, wht is the od digonl tht etends from the corner t () coordintes (0, Figure 3-31 Prolem 32. 0, 0), () coordintes (, 0, 0), (c) coordintes (0,, 0), nd (d) coordintes (,, 0)? (e) Determine the v w m j k u Figure 3-29 Prolem 29. O ' Figure 3-30 Prolem 31. l p s t ' r n ' o q B