In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature.

Transcription

1 The vector product mc-ty-vectorprod Oneofthewaysinwhichtwovectorscanecominedisknownasthevectorproduct.When wecalculatethevectorproductoftwovectorstheresult,asthenamesuggests,isavector. Inthisunityouwilllearnhowtocalculatethevectorproductandmeetsomegeometricalapplications. In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they ecome second nature. Afterreadingthistext,and/orviewingthevideotutorialonthistopic,youshouldealeto: definethevectorproductoftwovectors calculate the vector product when the two vectors are given in cartesian form use the vector product in some geometrical applications Contents 1. Introduction 2 2. Definition of the vector product 2 3. Some properties of the vector product 4 4. The vector product of two vectors given in cartesian form 5 5. Some applications of the vector product c mathcentre 2009

2 1. Introduction Oneofthewaysinwhichtwovectorscanecominedisknownasthevectorproduct.When wecalculatethevectorproductoftwovectorstheresult,asthenamesuggests,isavector. Inthisunityouwilllearnhowtocalculatethevectorproductandmeetsomegeometricalapplications. 2. Definition of the vector product Studythetwovectors aand drawninfigure1.notethatwehavedrawnthetwovectorsso thattheirtailsareatthesamepoint.theangleetweenthetwovectorshaseenlaelled θ. θ a Figure1.Twovectors aand drawnsothattheangleetweenthemis θ. Aswestatedefore, whenwefindavectorproducttheresultisavector. Wedefinethe modulus, or magnitude, of this vector as a sin θ soatthisstage,averysimilardefinitiontothescalarproduct,exceptnowthesineof θappears intheformula.however,thisquantityisnotavector.tootainavectorweneedtospecifya direction. Bydefinitionthedirectionofthevectorproductissuchthatitisatrightanglesto oth aand.thismeansitisatrightanglestotheplaneinwhich aand lie.figure2shows thatwehavetwochoicesforsuchadirection. a Figure2.Therearetwodirectionswhichareperpendiculartooth aand. Theconventionisthatwechoosethedirectionspecifiedytherighthandscrewrule. This meansthatweimagineascrewdriverintherighthand. Thedirectionofthevectorproductis 2 c mathcentre 2009

3 thedirectioninwhichascrewwouldadvanceasthescrewdriverhandleisturnedinthesense from ato.thisisshowninfigure3. a ṋ a Figure3.Thedirectionofthevectorproductisdeterminedytherighthandscrewrule. Weletaunitvectorinthisdirectionelaelled ˆn.Wethendefinethevectorproductof aand as follows: Key Point Thevectorproductof aand isdefinedtoe a = a sin θ ˆn where a isthemodulus,ormagnitudeof a, isthemodulusof, θistheangleetween aand,and ˆnisaunitvector,perpendiculartooth aand inasensedefinedytherighthandscrewrule. Somepeoplefindithelpfultootainthedirectionofthevectorproductusingtherighthand thumrule. Thisisachievedycurlingthefingersoftherighthandinthedirectioninwhich a woulderotatedtomeet.thethumthenpointsinthedirectionof a. Yetanotherviewistoalignthefirstfingeroftherighthandwith a,andthemiddlefingerwith. Ifthesetwofingersandthethumarethenpositiionedatright-angles,thethumpointsin thedirectionof a.trythisforyourself. Notethatthesymolforthevectorproductisthetimessign,orcross,andsowesometimes refertothevectorproductasthecrossproduct.eithernamewilldo.sometextooksandsome teachers and lecturers use the alternative wedge symol. 3 c mathcentre 2009

4 3. Some properties of the vector product Suppose,forthetwovectors aand wecalculatetheproductinadifferentorder. Thatis, supposewewanttofind a.usingthedefinitionof aandusingtheright-handscrewrule to otain the required direction we find a = a sin θ ( ˆn) Weseethatthedirectionof aisoppositetothatof a asshowninfigure4.so a = a Sothevectorproductisnotcommutative. Inpractice,thismeansthattheorderinwhichwe dothecalculationdoesmatter. aisintheoppositedirectionto a. a ˆn a a -n ˆ a Figure4.Thedirectionof aisoppositetothatof a. The vector product is not commutative. Key Point a = a Another property of the vector product is that it is distriutive over addition. This means that a ( + c) = a + a c Althoughweshallnotprovethisresulthereweshalluseitlateronwhenwedevelopanalternative formula for finding the vector product. 4 c mathcentre 2009

5 Key Point The vector product is distriutive over addition. This means a ( + c) = a + a c Equivalently, ( + c) a = a + c a The vector product of two parallel vectors Suppose the two vectors a and are parallel. Strictly speaking the definition of the vector productdoesnotapply,ecausetwoparallelvectorsdonotdefineaplane,andsoitdoesnot makesensetotalkaoutaunitvector ˆnperpendiculartotheplane.Butifweneverthelesswrite downtheformula,wecanseewhattheanswer ought toe: a = a sin θ ˆn = a sin 0 ˆn = 0 ecause sin 0 = 0. So,whentwovectorsareparallelwedefinetheirvectorproducttoethe zero vector, 0. Key Point For two parallel vectors a = 0 4. The vector product of two vectors given in cartesian form Wenowconsiderhowtofindthevectorproductoftwovectorswhenthesevectorsaregivenin cartesian form, for example as a = 3i 2j + 7k and = 5i + 4j 3k where i, jand kareunitvectorsinthedirectionsofthe x, yand zaxesrespectively. Firstofallweneedtodevelopafewresultsinthefollowingexamples. 5 c mathcentre 2009

6 Supposewewanttofind i j. Thevectors iand jareshowninfigure5. Notethatecause these vectors lie along the x and y axes they must e perpendicular. z k i O j y x Figure5.Theunitvectors i, jand k.notethat kisaunitvectorperpendicularto iand j. Theangleetween iand jis 90,and sin 90 = 1.Further,ifweapplytherighthandscrewrule, avectorperpendiculartooth iand jis k.therefore i j = i j sin 90 k = (1)(1)(1)k = k Supposewewanttofind j i. Again,refertoFigure5. Ifweapplytherighthandscrewrule, avectorperpendiculartooth jand i,inthesensedefinedytherighthandscrewrule,is k. Therefore j i = k Supposewewanttofind i i. Becausethesetwovectorsareparalleltheangleetweenthem is 0.WecanusetheKeyPointdevelopedonpage5toshowthat i i = 0. InasimilarmannerwecanderivealltheresultsgiveninthefollowingKeyPoint: Key Point i i = 0 j j = 0 k k = 0 i j = k j k = i k i = j j i = k k j = i i k = j 6 c mathcentre 2009

7 Wecanusetheseresultstodevelopaformulaforfindingthevectorproductoftwovectorsgiven in cartesian form: Suppose a = a 1 i + a 2 j + a 3 kand = 1 i + 2 j + 3 kthen a = (a 1 i + a 2 j + a 3 k) ( 1 i + 2 j + 3 k) = a 1 i ( 1 i + 2 j + 3 k) + a 2 j ( 1 i + 2 j + 3 k) + a 3 k ( 1 i + 2 j + 3 k) = a 1 i 1 i + a 1 i 2 j + a 1 i 3 k + a 2 j 1 i + a 2 j 2 j + a 2 j 3 k + a 3 k 1 i + a 3 k 2 j + a 3 k 3 k = a 1 1 i i + a 1 2 i j + a 1 3 i k + a 2 1 j i + a 2 2 j j + a 2 3 j k + a 3 1 k i + a 3 2 k j + a 3 3 k k Now,fromthepreviousKeyPointthreeofthesetermsarezero.Thosethatarenotzerosimplify to give a = (a 2 3 a 3 2 )i + (a 3 1 a 1 3 )j + (a 1 2 a 2 1 )k Thisistheformulawhichwecanusetocalculateavectorproductwhenwearegiventhecartesian components of the two vectors. If a = a 1 i + a 2 j + a 3 kand = 1 i + 2 j + 3 kthen Key Point a = (a 2 3 a 3 2 )i + (a 3 1 a 1 3 )j + (a 1 2 a 2 1 )k Supposewewishtofindthevectorproductofthetwovectors a = 4i+3j+7kand = 2i+5j+4k. Weusethepreviousresultwith a 1 = 4, a 2 = 3, a 3 = 7and 1 = 2, 2 = 5, 3 = 4.Sustitution into the formula gives a = ((3)(4) (7)(5))i + ((7)(2) (4)(4))j + ((4)(5) (3)(2))k which simplifies to a = 23i 2j + 14k 7 c mathcentre 2009

8 For those familiar with evaluation of determinants there is a convenient way of rememering and representing this formula which is given in the following Key Point and which is explained in the accompanying video and in the elow. Key Point If a = a 1 i + a 2 j + a 3 kand = 1 i + 2 j + 3 kthen i j k a = a 1 a 2 a = a 2 a i a 1 a j + a 1 a k = (a 2 3 a 3 2 )i (a 1 3 a 3 1 )j + (a 1 2 a 2 1 ) Supposewewishtofindthevectorproductofthetwovectors a = 4i+3j+7kand = 2i+5j+4k. Wewritedownadeterminant,whichisanarrayofnumers:inthefirstrowwewritethethree unitvectors i, jand k.inthesecondandthirdrowswewritethethreecomponentsof aand respectively: a = i j k Wethenconsiderthefirstelementinthefirstrow, i. Imaginecoveringuptheelementsinits rowandcolumn,togivethearray Thisisaso-called 2 2determinantandisevaluated yfindingtheproductoftheelementsontheleadingdiagonal(toplefttoottomright)and sutractingtheproductoftheelementsontheotherdiagonal( = 23). The resulting numer gives the i component of the final answer. Wethenconsiderthesecondelementinthefirstrow, j. Imaginecoveringuptheelementsin itsrowandcolumn,togivethearray This 2 2determinantisevaluated,asefore, yfindingtheproductoftheelementsontheleadingdiagonal(toplefttoottomright)and sutractingtheproductoftheelementsontheotherdiagonal, ( = 2).Theresult isthenmultipliedy 1andthisgivesthe jcomponentofthefinalanswer,thatis 2. Finally,weconsiderthethirdelementinthefirstrow, k.imaginecoveringuptheelementsinits rowandcolumn,togivethearray Thisdeterminantisevaluated,asefore,yfinding 8 c mathcentre 2009

9 the product of the elements on the leading diagonal(top left to ottom right) and sutracting theproductoftheelementsontheotherdiagonal( = 14).Theresultingnumer givesthe kcomponentofthefinalanswer. Wewriteallthisasfollows: Exercises 1 a = i j k = i j k = ( )i ( )j + ( )k = 23i 2j + 14k 1. Usetheformula a = (a 2 3 a 3 2 )i + (a 3 1 a 1 3 )j + (a 1 2 a 2 1 )ktofindthevector product a ineachofthefollowingcases. (a) a = 2i + 3j, = 2i + 9j. () a = 4i 2j, = 5i 7j. Comment upon your solutions. 2.UsetheformulainQ1tofindthevectorproduct a ineachofthefollowingcases. (a) a = 5i + 3j + 4k, = 2i 8j + 9k. () a = i + j 12k, = 2i + j + k. 3.Usedeterminantstofindthevectorproduct p qineachofthefollowingcases. (a) p = i + 4j + 9k, q = 2i k. () p = 3i + j + k, q = i 2j 3k. 4.Forthevectors p = i + j + k, q = i j kshowthat,inthisspecialcase, p q = q p. 5.Forthevectors a = i + 2j + 3k, = 2i + 3j + k, c = 7i + 2j + k,showthat a ( + c) = (a ) + (a c) 5. Some applications of the vector product Inthissectionwewilllookatsomewaysinwhichthevectorproductcaneused. Using the vector product to find a vector perpendicular to two given vectors. One of the common applications of the vector product is to finding a vector which is perpendicular totwogivenvectors.thetwovectorsshouldenon-zeroandmustnoteparallel. Supposewewishtofindavectorwhichisperpendiculartoothofthevectors a = i + 3j 2k and = 5i 3k. Weknowfromthedefinitionofthevectorproductthatthevector a willeperpendicular tooth aand.sofirstofallwecalculate a. 9 c mathcentre 2009

10 a = i j k = (3 3 ( 2) 0)i (1 3 ( 2) 5)j + ( )k = 9i 7j 15k Thisvectorisperpendicularto aand. Onoccasionsyoumayeaskedtofindaunitvectorwhichisperpendiculartotwogivenvectors. Toconvertavectorintoaunitvectorinthesamedirectionwemustdivideityitsmodulus. Themodulusof 9i 7j 15kis So, finally, the required unit vector is a = ( 9) 2 + ( 7) 2 + ( 15) 2 = ( 9i 7j 15k). Usingthevectorproducttofindtheareaofaparallelogram. ConsidertheparallelogramshowninFigure6whichhassidesgivenyvectors and c. c θ h Figure6.Aparallelogramwithtwosidesgiveny and c. The area of the parallelogram is the length of the ase multiplied y the perpendicular height, h. Now sin θ = h andso h = c sinθ.therefore c area = c sinθ whichissimplythemodulusofthevectorproductof and c.wededucethattheareaofthe parallelogram is given y area = c Using the vector product to find the volume of a parallelepiped. Consider Figure 7 which illustrates a parallelepiped. This is a six sided solid, the sides of which are parallelograms. Opposite parallelograms are identical. The volume, V, of a parallelepiped withedges a, and cisgiveny V = a ( c) Thisformulacaneotainedyunderstandingthatthevolumeistheproductoftheareaofthe aseandtheperpendicularheight. Becausetheaseisaparallelogramitsareais c.the perpendicular height is the component of a in the direction perpendicular to the plane containing and a,andthisis h = a c.sothevolumeisgiveny V = (height)( area of ase) = a c c c = a c c = a ( c) 10 c mathcentre 2009

11 Thiscouldturnouttoenegative,soinfact,forthevolumewetakeitsmodulus: V = a ( c). h a c Figure7.Aparallelepipedwithedgesgiveny a, and c. Supposewewishtofindthevolumeoftheparallelepipedwithedges a = 3i+2j+k, = 2i+j+k and c = i + 2j + 4k. Wefirstevaluatethevectorproduct c. c = i j k = ( )i ( )j + ( )k = 2i 7j + 3k Thenweneedtofindthescalarproductof awith c. a ( c) = (3i + 2j + k) (2i 7j + 3k) = = 5 Finally, we want the modulus, or asolute value, of this result. We conclude the parallelepiped has volume 5(units cued). Exercises 2. 1.Findaunitvectorwhichisperpendiculartooth a = i + 2j 3kand = 2i + 3j + k. 2. Findtheareaoftheparallelogramwithedgesrepresentedythevectors 2i j + 3kand 7i + j + k. 3. Find the volume of the parallelepiped with edges represented y the vectors i+j+k, 2i+3j+4k and 3i 2j + k. 4. Calculatethetriple scalar product (a ) cwhen a = 2i 2j + k, = 2i + jand c = 3i + 2j + k. Answers to Exercises Exercises 1. 1.(a) 24k,() 18k.Bothanswersarevectorsinthe zdirection.thegivenvectors, aand, lieinthe xyplane. 2.(a) 59i 37j 46k,() 13i 25j k. 3.(a) 4i + 19j 8k,() i + 10j 7k. 4. Bothcrossproductsequalzero,andso,inthisspecialcase p q = q p. Thetwogiven vectors are anti-parallel. 5.Bothequal 11i + 25j 13k c mathcentre 2009

12 Exercises (11i 7j k) squareunits. 3.8unitscued c mathcentre 2009