Applied Vectors Vectors2 Notes

GeneralTips: ● Drawingadiagramhelpsmassively.Regardlessofwhetherit’sabstractvectorsorapplication.● Practicethestandardmethodsandgetgoodatthem(Footofperpendicular,lineofintersection,findinganglesbetweenvectors,etc.)● Befamiliarwiththedifferentformsoflinesandplanes(Cartesian, Parametric,Vector, Scalar product)● Understandthegeometricsignificanceofthedot/crossproducts○ Dot:Testforperpendicular,lengthofprojection,angle,etc.○ Cross:Areasweptout,normalvector,lengthofoppositeetc. AsummaryofBasicFormulae: Products:𝑎 • 𝑏 = |𝑎||𝑏|𝑐𝑜𝑠(Θ) = 𝑎1𝑏1 + 𝑎2𝑏2 + 𝑎3𝑏3 (𝑖𝑛 3𝐷),whereistheunitvectorperpendiculartobothvectors,whose𝑎 × 𝑏 = |𝑎||𝑏|𝑠𝑖𝑛(Θ)𝑛 𝑛directionisgivenbytherighthandrule.(Component-wiseformulagiveninMF26,althoughifI’mnotmistaken,theyremoveditfromMF27.) Inparticular:If and areperpendicular,𝑎 𝑏 𝑎 • 𝑏 = 0If and areparallel,𝑎 𝑏 𝑎 × 𝑏 = 0givestheareaoftheparallelogramwhosesidesare and.|𝑎 × 𝑏| 𝑎 𝑏 Equation(s)ofaline:,whereisthepositionvectorofapointontheline,and isthedirection𝑟 = 𝑎 + λ𝑑 𝑎 𝑑vectoroftheline.(Alineisdefinedbyapointandadirection.) By:u/A_Strolling_Orca 1

Bywritingtheaboveequationoutcomponent-wiseandequatingineach,weget:λ ,providedthatnoneof arezero.Ifoneis0, 𝑥−𝑎1𝑑1 = 𝑦−𝑎2𝑑2 = 𝑧−𝑎3𝑑3 𝑑1,𝑑2,𝑑3 onethecoordinatesisfixed. E.g.𝑟 = (2,1,0) + λ(1,1,0) ↔ 𝑥 − 2 = 𝑦 − 1, 𝑧 = 0. Equation(s)ofaplane:,whereisthenormalvectortotheplane,and isthepositionvectorof𝑟 • 𝑛 = 𝑎 • 𝑛 𝑛 𝑎agivenpointintheplane.(Aplaneisdefinedbyapointandanormal).WecanalsogettheCartesianequationbywriting andevaluatingtheright𝑟 = (𝑥,𝑦,𝑧), 𝑛 = (𝑛1,𝑛2,𝑛3)handside.Notethatif isaunitvector, representsthelengthofprojectionof𝑛 𝑎 • 𝑛 𝑎onto,whichistheperpendiculardistancefromtheorigintotheplane.𝑛 Aplanecanalsobedefinedintermsof2linearlyindependent(i.e.notparallel)vectorsparalleltoit,orequivalently,threenon-collinearpointsitcontains.Thedifferencevectorsbetweenthe3pointsgiveyouthe2vectorsitcontains. ,whereisapointintheplane,and arevectorsparalleltothe𝑟 = 𝑎 + λ𝑏 + µ𝑐 𝑎 𝑏 𝑐plane,with foranyrealscalar, and, realparameters.𝑏 ≠ 𝑘𝑐 𝑘 λµ E.g. definesaplane.𝑟 = (1,1,1) + λ(1,0,0) + µ(0,1,0) doesnotdefineaplane,asthetwo𝑟 = (1,2,3) + λ(1,0,0) + µ(− 1,0,0)vectorsgivenareparallel. Takingthecrossproductofthetwovectorsgivesyouthenormaltotheplane,therebylettingyouconvertfromthisparametricformtothecartesianform.Convertingfromthecartesiantotheparametricissimplyamatteroffinding3non-collinearpointsontheplane,butIhaveneverseenaquestionaskforthis. Angles:Foranytwovectors ,theangle betweenthemisgivenby:𝑎,𝑏 Θ𝑐𝑜𝑠(Θ) = (𝑎•𝑏)|𝑎||𝑏| By:u/A_Strolling_Orca 2

Foranytwoplaneswithnormalvectors theacuteanglebetweentheplanesis𝑛1,𝑛2,givenby: 𝑐𝑜𝑠(Θ) = |(𝑛1•𝑛2)||𝑛1||𝑛2| Foralinewithdirectionvector intersectingaplanewithnormal,theangleit𝑑1 𝑛1makeswiththeplaneisgivenby: 𝑠𝑖𝑛(Θ) = |(𝑛1•𝑑1)||𝑛1||𝑑1| Distances/Lengths:Fortwoparallelplanes, and ,thedistancebetweenthemisgiven𝑟 • 𝑛 = 𝑑1 𝑟 • 𝑛 = 𝑑2by: .𝐷 = |𝑑1 − 𝑑2|/|𝑛|(Thisformulacanbederivedbyconsideringtheprojectionofpositionvectorsofpointsonthetwoplanesontotheircommonnormal.) Lengthofprojectionof onto isgivenby:𝑎 𝑏 𝑎 • 𝑏|𝑏| = 𝑎 • 𝑏 GiventwopointsAandBwithpositionvectorsand,thepositionvectorofthepoint𝑎 𝑏C,thathasthepropertyAC:CB= ,isgivenby:λ:µ 𝑂𝐶= (λ𝑂𝐵+ µ𝑂𝐴)/(λ + µ) By:u/A_Strolling_Orca 3

Findingtheshortestdistance/footofperpendicularbetweenXandY. MostofthetimeinanAppliedVectorsquestion,youwillbeaskedtofindtheshortestdistancebetweentwoobjects,XandY.Mostofthetime,theyfallintooneofthefollowingcategories: ShortestDistancebetween:1)Apoint,X,andaline,l.2)ApointX,andaplane,P. Inadditiontothis,theymayaskyoutofindthefootofperpendicular/pointsateitherobjectwheretheyareclosesttoeachother.Solvingforonetendstomakesolvingfortheotherquitesimple,ifyouhaven’talreadyfoundit.Iwilloutlinesomemethodstosolvethesetypesofquestionsbelow. Case1:SupposewehaveapointX,withpositionvectorOX,andalinel.Letdbethedirectionvectoroftheline.WewishtofindtheshortestdistancebetweenXandl. Method1:Wewillsolveforthefootofperpendicular,F,fromXtol,thenfindthelengthofXF.NotethatthevectorXFsatisfies: 𝑋𝐹• 𝑑 = 0 Thatis,XFisperpendiculartothelinel.SinceFliesonl,wemaysubinthegeneralequationforthepositionvectorofapointonl,andsolvefortheparameterthatcorrespondstothepositionvectorOF.WecanthensolveforXFandhencetheshortestdistance. Example:FindtheshortestdistancebetweenthepointA,givenbycoordinates(2,2,-6)andthelinegivenby:𝑥 − 1 = 2 − 𝑦 = 𝑧 + 6 By:u/A_Strolling_Orca 4

WefirstconvertthelineequationfromCartesiantoParametric.Ifyou’renotconfidentindoingthis,pleasereviewyournotes.Weobtain: 𝑂𝑅 = (1 + λ , 2 − λ , − 6 + λ )Thus: 𝑋𝑅 = (− 1 + λ , − λ , λ )Andhence: , forsome .𝑋𝐹 = (− 1 + λ , − λ , λ ) λ Takingthedotproduct: 𝑋𝐹 • 𝑑 = (− 1 + λ , − λ , λ ) • (1,− 1, 1) = 0 Wherethelastequalityfollowsfromthetwovectorsbeingperpendicular. Weget: andthus,3λ − 1 = 0 λ = 1/3 Subbingin ,weget:λ = 1/3 𝑋𝐹 = (− 2/3 , − 1/3 , 1/3 ) Andsotheshortestdistanceis|𝑋𝐹| = √6 / 3 Method2:LetORbethepositionvectorofanarbitrarypointonl.WewillfindXRandhenceobtainanexpressionfor intermsoftheparameter.Thiswillalwaysbethesquare|𝑋𝑅| λrootofaquadraticfunction.Sincethesquarerootisanincreasingfunction,itisminimisedwhenthequadraticinsideitisminimised.Thiswillgiveustheminimisingvalueof ,andhencetheminimalvalueof .λ |𝑋𝑅| Example:FindtheshortestdistancebetweenthepointA,givenbycoordinates(2,2,-6)andthelinegivenby:𝑥 − 1 = 2 − 𝑦 = 𝑧 + 6 Fromabove: 𝑋𝑅 = (− 1 + λ , − λ , λ ) Andthus: |𝑋𝑅| = (λ − 1) 2 +λ 2 +λ 2 = 3λ 2 − 2λ + 1 By:u/A_Strolling_Orca 5

RecallfromE-maththatasmileyparabolaisminimisedat :− 𝑏/2𝑎λ =− (− 2)/2(3) = 1/3 Whichisthesamevalueof weobtainedviaMethod1,andhencewearriveattheλsameanswer. Strictlyspeaking,youcandifferentiate withthesquarerootandsolveforstationary|𝑋𝑅|points/check2ndderivativeetc,butIbelievetheargumentoutlinedaboveshouldsuffice.Thatsaid,Iamnotateacher,andIdonotknowhowindividualJCsallotmarksforworking. Method3:Letdbetheunitdirectionvectorofl.TakeanypointR,onl.Then,theperpendiculardistancefromAtolisgivenby: 𝐷 = |𝐴𝑅× 𝑑| Thismethodworkswellifyouarenotaskedtofindthecoordinatesofthefootoftheperpendicular,onlythedistance. Example:FindtheshortestdistancebetweenthepointA,givenbycoordinates(2,2,-6)andthelinegivenby:𝑥 − 1 = 2 − 𝑦 = 𝑧 + 6 𝑑 = (1/ 3,− 1 3,1 3)liesonl.(1,2,− 6) Thus:𝐴𝑅= (− 1,0,0)Hence:𝐷 = |(− 1,0,0) × (1/ 3,− 1/ 3,1/ 3)| = 6/3 Case2: SupposeAisapoint,andPaplane.WewanttofindtheshortestdistancefromAtoP. Method1:Wewillsolveforthefootofperpendicular,F,fromAtoP,andthentakethelengthAF. By:u/A_Strolling_Orca 6

SinceFisintheplane,OFsatisfiestheplaneequationforP.Additionally,sinceFisthefootofperpendicularfromA,AFissomemultipleofthenormalvectorforP.ThiswillbeenoughtosolveforthecoordinatesofF. Example:FindtheshortestdistancefromthepointA,withcoordinates totheplane(5,− 6,5),givenby .2𝑥 − 3𝑦 + 𝑧 = 5 Method1:LetFbethefootofperpendicularfromAtoP.,forsome.𝑂𝐹= 𝑂𝐴+ 𝐴𝐹= (5,− 6,5) + λ(2,− 3,1) λ SinceFliesinP: 𝑂𝐹• (2,− 3,1) = 5Thus: ((5,− 6,5) + λ(2,− 3,1)) • (2, − 3,1) = 513+ 14λ = 5λ =− 4/7 Andhence: |𝐴𝐹| = | − 4/7* (2,− 3,1)| = 4 14/7 NotethatifthequestionhadaskedforthecoordinatesofF,wesimplyremovetheabsolutevaluesignfromthelastline.Thus,thismethodworksthesamewayregardlessofwhichtheyaskfor. Method2:WewillconstructaplaneparalleltoPatA,thenfindthedistancebetweenthetwoplanes. Example:FindtheshortestdistancefromthepointA,withcoordinates totheplane(5,− 6,5),givenby .2𝑥 − 3𝑦 + 𝑧 = 5 Ourconstructedplanehasequation:2𝑥 − 3𝑦 + 𝑧 = (5,− 6,5) • (2,− 3,1) = 13 Thus: 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒= |13− 5|/|(2,− 3,1)| = 8/ 14= 4 14/7 Note:Thismethodisfasteratgettingtheshortestdistance(inmyopinion),butwhenitcomestofindingthefootofperpendicular,alittlebitofcareisrequired.Inthiscase,we By:u/A_Strolling_Orca 7

knowthedistanceAF,andthatAFisparalleltothenormal, ,butinwhich(2,− 3,1)direction? Herearetwowaysofreasoningthisout: Thebrainlessapproachistosimplytrybothof: , —𝑂𝐹= (5,− 6,5)±4 14/7* (2/ 14,− 3/ 14,1/ 14) (†) AndseewhichvectorsatisfiestheplaneequationofP. Amoremethodicalapproachistoconsiderthetwoplaneequations: P: 2𝑥 − 3𝑦 + 𝑧 = 5ConstructedPlane: 2𝑥 − 3𝑦 + 𝑧 = 13 Notethattherighthandsideofbothequationsbeing