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Higher Level Mathematics - Vectors

1:General• 𝑟 = 𝑥! + 𝑦! + 𝑧!

• Themagnitudeis𝑟 = 𝑏𝑎𝑠𝑒! + 𝑧!.

𝐵𝑎𝑠𝑒 = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑜𝑓 𝑥 − 𝑦 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 = 𝑥! + 𝑦!.Subbingitin𝑟 = 𝑥! + 𝑦! + 𝑧!

• • 𝐴𝐵 = 𝑂𝐵 − 𝑂𝐴

• 𝑂𝑃 = !!"!!!"

!!!

• 𝑎 = !!

• Theparallelsidesofaparallelogramhavethesamevectors.• Usethecarmethodtofindnewvectors.

2:ScalarandVectorProducts

• Thescalarproductisdefinedas𝑎 ⋅ 𝑏 = 𝑎 𝑏 cos𝜃,0 ≤ 𝜃 ≤ 180• Thevectorsmustconvergeordivergeatapoint• 𝑎 ⋅ 𝑎 = 𝑎 !• 𝑎 ⋅ 𝑎 = !

!𝑎 ⋅ 𝑎 = ! !

!= 𝑎

• Wecanusebasicalgebrahere,butnodividingorcancellation.E.g.𝑎 − 𝑏 ⋅ 𝑎 + 𝑏 = 𝑎 ! − 𝑏 !

• If2vectorsareperpendicular,thentheirdot(scalar)productisequalto0• Wecanusethedotproducttofindthelengthofprojection.

• Fromhere,wecanworkout𝑎!,whichisequaltotheunitvectorofb

dottedtothevector𝑎.Hence,𝑎! = 𝑎 ⋅ !!= 𝑎 ⋅ 𝑏.Ifaskedtofindthe

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vectorparalleltoit,wesimplymultiplythisvalueby𝑏,asthelengthoftheprojectionisthemagnitudeoftheprojectionvector.

• Wecanfind𝑎! byoneof2ways.Thedistance𝑎! = 𝑎 − 𝑎! .• However,tofindthevectorperpendicularto𝑏,thevector(cross)product

mustbeused,asthisiswhatthecrossproductgivesus,avectorperpendiculartothe2linesinquestion.Notethatthecrossproductisdefinedas𝑎 × 𝑏 = 𝑎 𝑏 sin𝜃 𝑛,0 ≤ 𝜃 ≤ 180andwhere𝑛isaunitvectorperpendicularto 𝑎and𝑏

• 𝑎!𝑏!𝑐!

×𝑎!𝑏!𝑐!

=𝑏!𝑐! − 𝑐!𝑏!

−(𝑎!𝑐! − 𝑐!𝑎!)𝑎!𝑏! − 𝑏!𝑎!

,covertopfirstanddodet,thencover

middle,thenbottom• 𝑎×𝑏 = −(𝑏×𝑎)• 𝑎×𝑎 = 0• 𝑎×𝑏 ! = 𝑎 ! 𝑏 ! − 𝑎 ! 𝑏 ! cos! 𝜃 = 𝑎 ! 𝑏 ! − 𝑎 ⋅ 𝑏 !• Tofindthevector𝑎!,wedo𝑎 × 𝑏.Thedistancecanbefoundthrough

𝑎 × 𝑏 • Thecrossproductcanalsobeusedtofindvectorsperpendiculartoa

plane,wetake2linesontheplane(notfromtheorigin,whichwillnotbeonthesameplane),andcrossthemtogethertofindavectorperpendiculartotheplane(maybeintheformof“findtheresolvedcomponentofavectorperpendiculartothelineON.”)

• Youcantakefactors,e.g.the !!"#$%&'()

foraunitvectoroutforacrossandmultiplythevectorproductbyit.

• ForatriangleABC,wecanusetherule𝐴𝑟𝑒𝑎 = !!( 𝐴𝐵×𝐴𝐶 ),seenotebook

forproof.• Similarly,fortheparallelogramABCD,therule𝐴𝑟𝑒𝑎 = 𝐴𝐵×𝐴𝐶 ,the2

linesusedjustneedtomeet,suchthatthereisananglebetweenthem.• Wecanfindthecosineoftheanglesavectormakeswiththex,yandz

axisbythefollowingformulaecos𝛼 = !!,cos𝛽 = !

!,cos 𝛾 = !

!

• Provingthesinerule

𝑎×𝑎 = 0 = 𝑎× 𝑏 − 𝑐 = 0𝑎×𝑏 = 𝑎×𝑐

𝑎 𝑏 sin𝐶 = 𝑎 𝑐 sin𝐵DividingonLHSandRHSby 𝑎 sin𝐵 sin𝐶

𝑏sin𝐵 =

𝑐sin𝐶

QED

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3:StraightLines• Thevectorequationisdefinedas𝑟 = 𝑎 + 𝜆𝑑,where𝑎isanypointonthe

line,𝜆 ∈ ℝ,and𝑑isadirectionvectorthatisparallelto𝑟. • Recallthat𝑟 isapositionvectorforapoint𝑁ontheline.• Itmustpassthroughthefixedpoint𝑎andmustpassthrough2known

points• Thereare3formstothisequation

1. Vector:𝑟 =𝑎!𝑏!𝑐!

+ 𝜆𝑑!𝑑!𝑑!

2. Parametric:𝑥 = 𝑎! + 𝜆𝑑!𝑦 = 𝑎! + 𝜆𝑑!𝑧 = 𝑎! + 𝜆𝑑!

3. Cartesian:𝜆 = !!!!!!

= !!!!!!

= !!!!!!

• ToconvertCartesiantoparametric,justmake𝑧/𝑦/𝑧thesubject,ensuretheyaredevoidofminussigns

• Vectorlinescanbeoneof3typesoflines1. Parallel

a) Theyareparalleliftheyhavethesamesimplifieddirectionvector.

2. Intersectinga) Theyintersectif,whenyouequatethelines,thereare

consistentvaluesof𝜆 and𝜇,i.e.whenequating𝑥,𝑦and𝑧,thesamevaluesof𝜆 and𝜇producethesame𝑥,𝑦and𝑧values.

b) Iffindingtheintersectionwithaplane(e.g.𝑥 − 𝑦plane),lookforthetermyouknowhasasetvalue,e.g.𝑥 − 𝑦planewillhave𝑧 = 0,sosettheparametricequationfor𝑧 = 0,andfindthevalueof𝜆

3. Skew(notparallelorintersecting)a) Ifthereareinconsistentsolutionsfor𝜇 and𝜆andtheir

directionvectorsarenotparallel,thelinesareskewlines.• Ifyouwanttheanglebetween2vectorlines,dottheirdirectionvectors,

dividebytheirindividualmagnitudesandequatethisvaluetocos𝜃• Wecanbeaskedforthepositionvectoroftheperpendicularfootofa

vectorequationwithanothervector.Todothis,wefindthevectorofthelinethatexitsthefootperpendiculartothevectorequation.Thiswillbeintermsof𝜆.Wethendotthistotheunitvectorofthedirectionvectorandequateittozero.

• Example:Findthepositionvectorofthefootoftheperpendicularfromthe

point 2,1,4 totheline𝑟 =105

+ 𝜆−203

.

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𝐿𝑒𝑡 𝑁 = 𝑡ℎ𝑒 𝑓𝑜𝑜𝑡,𝑎𝑛𝑑 𝐴 = 2,1,4

𝑁𝐴 =214

−1− 2𝜆0

5+ 3𝜆=

1+ 2𝜆1

−1− 3𝜆

𝑁𝐴 ⋅ 𝑑 = 0 =113

1+ 2𝜆1

−1− 3𝜆⋅−203

0 = −2− 4𝜆 − 3− 9𝜆 = −13𝜆 − 5

∴ 𝜆 = −513

𝑂𝑁 =105

−513

−203

=

13+ 1013013

65− 1513

=𝟏𝟏𝟑

𝟐𝟑𝟎𝟓𝟎

• Thisisthepositionvector.Ifwewantedthevectorintheparalleldirectiontotheline,wecouldfindthemagnitudeoftheparallelbitbydotting𝐴𝑃withtheunitvectorof𝑑,assuming𝐴𝑃isthehypotenuse.Wethenmultiplythisbytheunitvectorof𝑑.

• Ifwewantedthevector𝑁𝑃,weonlyneedtocross𝐴𝑃with𝑑.Wecanthenfindthemagnitudeifnecessary.

• Wecouldalsobeaskedtofindthepositionvectorofareflectionoftheperpendicularfoot,whichissimple;wejustusetheratiotheorem.Thedistancebetweenthepoint𝑃and𝑁isthesameas𝑃!and𝑁

• Theline𝑦 = 4𝑥 + 6canbeexpressedas𝑟 = 06 + 𝜆 1

4 • Displacementcanbe𝑟 − 𝑎 = 𝑣𝑡,hencewecanrearrangeitasfollows

𝑟 = 𝑎 +𝑣𝑡,where𝑟isanypointtheparticletravelsthrough,tistimeand𝑎isthestartpoint.

• Themagnitudeofthedirectionvector, 𝑣,isthevelocityoftheparticle.• Thevelocityvectorisdifferentfromthedirectionvector.Wecanrelate

themusingthefollowingequation. 𝑣 = 𝜆 𝑑 • TheCartesianformofavectorforanapplicationsquestionsiswhenwe

findtwoequationsfor𝑡andthenequatethem,tofindanequationwith𝑥,𝑦andarealnumber

• TheCartesianformisusedtofindtheintersectionbetweenthepathsof2objects;thevectorformisusedtofindiftheywillcollide.

• Ifwewanttheminimumdistancebetween2skewlines,wefirstwriteavectorforalinebetweenapointAononelineandthepointBontheother(sothisis𝐴𝐵).Next,crossthedirectionvectorsofthelinetofindthevector𝑛.Wethenequatethefollowing-𝐴𝐵 = 𝑎 𝑛,where𝑎 ∈ ℝ.UseaGDCormanuallyfindthevaluesof𝑎, 𝜇 𝑎𝑛𝑑 𝜆. 𝐴𝐵isthenourdirectionvectorandAorBbecomesourstartpoint.

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4:Planes• Asseenonthisdiagram,𝐴𝑃liesontheplaneand𝑛isperpendiculartothe

plane.Let𝑟bethepositionvectorofanypointoftheplaneandlet𝑃bethatparticularpoint.𝐴𝑃 ⋅ 𝑛 = 0 = (𝑟 − 𝑎) ⋅ 𝑛,leadingusto𝑟 ⋅ 𝑛 = 𝑎 ⋅ 𝑛,where𝑎isthepositionvectorofapointontheplane.

• Aplanecanbeexpressedinoneof3ways1. Parametricform(absolutelyuseless,neverdoitunlessasked)a. Require2non-paralleldirectionvectorsparalleltotheplaneand

onepointontheplane.b. 𝑟 = 𝑎 + 𝜆𝑑! + 𝜇𝑑!c. Wecanfindthenormalbycrossing𝑑!and𝑑!2. VectorEquation–Scalar

a. Weneedanormalandapointonthelineb. 𝑟 ⋅ 𝑛 = 𝐷,Where𝐷isaconstantc. IfwearetoldthataplaneisperpendiculartoplaneAof

equation𝑟 ⋅ 𝑛! = 𝐷!andhas2points(AandB),thenormaltothefirstplaneissimplythecrossproductof𝐴𝐵and𝑛!,becauseiftheplaneisperpendiculartoplaneA,planeA’snormalisonthefirstplane.

3. VectorEquation–Cartesiana. Werequirethescalarform

b. Simply𝑥𝑦𝑧

⋅𝑎𝑏𝑐

= 𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 = 𝐷

• Theshortestdistancebetweentheplaneandthenormalisequalto𝑟 ⋅ 𝑛 = !⋅!

!= !⋅!

!

• LinesandPlanes1. Parallelandontheplane-If𝑑 ⋅ 𝑛 = 0,thenparallel.If𝑎’s𝑥,𝑦

and𝑧coordinatesaresubbedintotheCartesianformandstillequal𝐷,henceitsatisfiestheequation.

2. Parallelbutnotontheplane.If𝑑 ⋅ 𝑛 = 0,thenparallel.Ifthevectorwhensubbedindoesnotequal𝐷,soitdoesnotsatisfytheequation.

3. Intersectionbetweenlineandplane.DotlinetothenormalandseewhenDisreachedforeach𝑥,𝑦and𝑧value.Musthaveconsistentvaluesfor𝜆.Solvetofindthecoordinates

• Inthecontextoffreevectors,anyvectorparalleltotheplaneistechnicallyonit,ifitisn’t,it’snotontheplane.

• Wecanfindtheangleofthepointbetweenthenormalandthelineifwewant;wefindtheanglebetween𝑙and𝜋.

• sin𝜃 = cos(90− 𝜃) = !⋅!! !

,thensolvefor𝜃.

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• Fortheanglebetweenplanes,cos(𝜃) = !!⋅!!! !

• Tofindif2planesareperpendicular,wedottheirnormals,ifitis0,then

theyareperpendicular.Wecanalsofindtheanglebetweenthemusingthenormal.

• 2planescanbeparallel,perpendicularorintersectinginaline• Tofindanormalvectortoanewplanegivensomeinformation,create

CartesianequationsandusetheGDCtosolve,thecoefficientsof𝑥! arethe𝑥,𝑦and𝑧coefficientsofthecartesianequationoftheplane,allowingustofindthenormal.

• NonGDCmethod(for2planes)

• Non-GDCguidetosolvingrelationshipsbetween3planeso LineofIntersection:Iffinalrowisall0s.Solveusingfirst2

1 −1 2 40 3 4 90 3 4 90 0 0 0

o Intersectonce:numericalsolutionpossibleforthefinalrow.1 −1 2 40 3 4 90 3 5 100 0 1 1

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𝑧 = 1o Nointersection:all𝑥,𝑦 and 𝑧sare0sbutthenumberontheright

isnotzero.Inconsistentsolutions,hencenointersection.1 −1 2 40 3 4 90 3 4 160 0 0 7

• ExampleQuestion

3𝑥 − 𝑦 + 𝑧 = 2𝑥 + 2𝑦 − 𝑧 = −15𝑥 − 4𝑦 + 𝑑𝑧 = 3

Findthevalueof𝑑suchthatthese3planesdonotintersect.Westartbywritingeachequationinmatrixform

3 −1 1 21 2 −1 −15 −4 𝑑 30 0 0 0

Ourtargetistohaveanexpressionwith𝑥,𝑦 = 0

𝑎 = 𝑅! − 3𝑅!𝑏 = 𝑅! − 5𝑅!

𝑏 − 2𝑎

0 −7 4 50 −14 𝑑 + 5 80 0 𝑑 − 3 −20 0 0 0

Wewantasituationwherewehave0,0,0,−2.Hence,𝑑 = 3

• GDCmethod-Tofindtherelationshipbetween3planes,wewritethem

outasCartesianequationsandusepolysimultaneoustosolve.Ifthereareany𝑥! terms,thentheplanesintersectinalineandthedirectionvectorofthelineissimplythe𝑥! coefficients.Iftherearesimplenumbersolutions,theyintersectonce.Iftherearenosolutions,theydonotintersect.