The scalar product mc-ty-scalarprod-2009- Oneofthewaysinwhichtwovectorscanbecombinedisknownasthescalarproduct.When wecalculatethescalarproductoftwovectorstheresult,asthenamesuggestsisascalar,rather than a vector. Inthisunityouwilllearnhowtocalculatethescalarproductandmeetsomegeometricalapplications. In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature. Afterreadingthistext,and/orviewingthevideotutorialonthistopic,youshouldbeableto: definethescalarproductoftwovectors state some important properties of the scalar product calculate the scalar product when the two vectors are given in cartesian form use the scalar product in some geometrical applications Contents. Introduction 2 2. Definition of the scalar product 2. Some properties of the scalar product 4. The scalar product of two vectors given in cartesian form 5 5. Some applications of the scalar product 8 www.mathcentre.ac.uk c mathcentre 2009
. Introduction Oneofthewaysinwhichtwovectorscanbecombinedisknownasthescalarproduct.When wecalculatethescalarproductoftwovectorstheresult,asthenamesuggestsisascalar,rather than a vector. Inthisunityouwilllearnhowtocalculatethescalarproductandmeetsomegeometricalapplications. 2. Definition of the scalar product Studythetwovectors aand bdrawninfigure.notethatwehavedrawnthetwovectorsso thattheirtailsareatthesamepoint.theanglebetweenthetwovectorshasbeenlabelled θ. b θ a Figure.Twovectors, aand b,drawnsothattheanglebetweenthemis θ. Wedefinethescalarproductof aand basfollows: Thescalarproductof aand bisdefinedtobe a b = a b cosθ where a isthemodulus,ormagnitudeof a, b isthemodulusof b,and θistheanglebetween aand b. Notethatthesymbolforthescalarproductisthedot,andsowesometimesrefertothescalar productasthedotproduct.eithernamewilldo. www.mathcentre.ac.uk 2 c mathcentre 2009
Example Considerthetwovectors aand bshowninfigure2. Suppose ahasmodulus4units, bhas modulus5units,andtheanglebetweenthemis 60,asshown. b 60 a Figure2. aand bhavelengths4and5unitsrespectively;theanglebetweenthemis 60. Wecanusethedefinitiongivenabovetofindthescalarproductof aand b. a b = a b cosθ = 4 5 cos 60 = 4 5 2 = 0 Sothescalarproductofthesevectorsisthenumber0.Notethattheanswerisascalar,that isanumber,ratherthanavector. So,wehavelearntamethodofcombiningtwovectorsto produce a scalar.. Some properties of the scalar product Commutativity and distributivity Suppose for the two vectors in the previous example we calculate the product in a different order. Thatis,supposewewanttofind b a.thedefinitionof b ais b a = b a cosθ Performing the calculation using the numbers in the Example we find b a = b a cosθ = 5 4 cos 60 = 5 4 2 = 0 So,weseethattheresultisthesamewhicheverwayaroundweperformthecalculation. This is true in general: a b = b a Thispropertyofthescalarproductisknownascommutativity. Wepointitoutbecausein another unit you can learn about another way of combining vectors known as the vector product. Thevectorproductisnotcommutativesotheorderinwhichwewritedownthetwovectorswill be very important. www.mathcentre.ac.uk c mathcentre 2009
The scalar product is commutative. This means a b = b a Another property of the scalar product is that it is distributive over addition. This means that a (b + c) = a b + a c Althoughweshallnotprovethisresulthereweshalluseitlateronwhenwedevelopanalternative formula for finding the scalar product. The scalar product is distributive over addition. This means a (b + c) = a b + a c and also, equivalently (b + c) a = b a + c a The scalar product of two perpendicular vectors Example Considerthetwovectors aand bshowninfigure.theanglebetweenthemis 90,asshown. b a Figure.Theanglebetween aand bis 90. www.mathcentre.ac.uk 4 c mathcentre 2009
Wecanusethedefinitiontofindthescalarproductof aand b. a b = a b cosθ = a b cos 90 = 0 because cos 90 = 0. Thisistruewhateverthelengthsof aand b. Sothescalarproductof two vectors which are at right-angles is always 0. We say that such vectors are perpendicular or orthogonal. For two perpendicular vectors a b = 0 Theconverseofthisstatementisalsotrue: ifwehavetwonon-zerovectors aand bandwe find that their scalar product is zero, it follows that these vectors must be perpendicular. We canusethisfacttotestwhethertwovectorsareperpendicular,asweshallseeshortly. 4. The scalar product of two vectors given in cartesian form Wenowconsiderhowtofindthescalarproductoftwovectorswhenthesevectorsaregivenin cartesian form, for example as a = i 2j + 7k and b = 5i + 4j k where i, jand kareunitvectorsinthedirectionsofthe x, yand zaxesrespectively. Firstofallweneedtodevelopafewresultsinthefollowingexamples. Example Supposewewanttofind i j.thesevectorsareshowninfigure4. Notethatbecause iand jliealongthe xand yaxestheymustbeperpendicular. So,fromthe resultwehavejustestablished,thescalarproduct i jmustbezero. Forthesamereason,we www.mathcentre.ac.uk 5 c mathcentre 2009
obtainthesameresultifwecalculate i kand j k. z k i O j y x Figure4.Theunitvectors iand jareperpendicular. Example Supposewewanttofind i i.referagaintofigure4.thevector iisaunitvector,soitslength isunit.theanglebetweenavectoranditselfmustbezero.so since cos 0 =. Forthesamereason j j = and k k =. i i = i i cos 0 = = If i, jand kareunitvectorsinthedirectionsofthe x, yand zaxes,then i j = 0 i k = 0 j k = 0 i i = j j = k k = Wecanusetheseresultstodevelopaformulaforfindingthescalarproductoftwovectorsgiven in cartesian form: www.mathcentre.ac.uk 6 c mathcentre 2009
Suppose a = a i + a 2 j + a kand b = b i + b 2 j + b kthen a b = (a i + a 2 j + a k) (b i + b 2 j + b k) = a i (b i + b 2 j + b k) + a 2 j (b i + b 2 j + b k) + a k (b i + b 2 j + b k) = a i b i + a i b 2 j + a i b k + a 2 j b i + a 2 j b 2 j + a 2 j b k + a k b i + a k b 2 j + a k b k = a b i i + a b 2 i j + a b i k + a 2 b j i + a 2 b 2 j j + a 2 b j k + a b k i + a b 2 k j + a b k k NowfromthepreviousKeyPointmostofthesetermsarezero. Thosethatarenotsimplify because i i = j j = k k =.Weobtain a b = a b + a 2 b 2 + a b Thisistheformulawhichwecanusetocalculateascalarproductwhenwearegiventhecartesian components of the two vectors. If a = a i + a 2 j + a kand b = b i + b 2 j + b kthen a b = a b + a 2 b 2 + a b Note that a useful way to remember this is: multiply the i components together, multiply the j components together, multiply the k components together, and finally, add the results. On occasionsyoumayseethisformreferredtoastheinnerproductofthevectorsaand b. In the context of vectors this simply means the sum of the products of the corresponding vector components. Example Supposewewishtofindthescalarproductofthetwovectors a = 4i+j+7kand b = 2i+5j+4k. Theresultwehavejustderivedtellsustomultiplythe icomponentstogether,multiplythe j components together, multiply the k components together, and finally add the results. So a b = (4)(2) + ()(5) + (7)(4) = 8 + 5 + 28 = 5 www.mathcentre.ac.uk 7 c mathcentre 2009
Example Supposewewishtofindthescalarproductofthetwovectors a = 6i+j kand b = 2i+4k. Notethatthe jcomponentof biszero.so a b = ( 6)(2) + ()(0) + ( )(4) = 72 + 0 44 = 6 Itisoftenusefultomakeuseofcolumnvectornotation. Consideragainthelastexample. Writing aand bascolumnvectors 6 2 a = b = 0 4 the scalar product becomes 6 a b = Exercises. 2 0 4 = ( 6)(2) + ()(0) + ( )(4) = 6.If a = 4i + 9jand b = i + 2jfind(a) a b,(b) b a,(c) a a,(d) b b. 2.Findthescalarproductofthevectors 5iand 8j..If p = 4i + j + 2kand q = 2i j + kfind(a) p q,(b) q p,(c) p p,(d) q q. 4.If r = i + 2j + 8kshowthat r r = r 2. 2 2 5.If a = and b = 4 find(a) a b,(b) b a,(c) a a,(d) b b. 4 6. Points A, B, and C have coordinates(,2,),(5,4,2), and ( 4, 2, ) respectively. Find the scalarproductof ABand AC. 5. Some applications of the scalar product Inthissectionwewilllookatsomewaysinwhichthescalarproductcanbeused. Using the scalar product to test whether two vectors are perpendicular A common application of the scalar product is to test whether two vectors are perpendicular. From the definition a b = a b cosθ Ifboththevectors aand barenon-zeroandwefindthat a b = 0thenwecandeduce cosθ mustbezero,sothat θ = 90,i.e. aand bareperpendicular. Considerthefollowingexample. www.mathcentre.ac.uk 8 c mathcentre 2009
Example Supposewewishtotestwhetherornotthevectors aand bareperpendicular,where a = 2 b = 2 Notethatneither anor biszero.calculatingthescalarproductwefind a b = 2 2 andso aand bareindeedperpendicular. = ()() + (2)( 2) + ( )( ) = 4 + = 0 If aand barenon-zerovectorsforwhich a b = 0,then aand bareperpendicular. Using the scalar product to find the angle between two vectors. Oneofthecommonapplicationsofthescalarproductistofindtheanglebetweentwovectors when they are expressed in cartesian form. From the definition of the scalar product a b = a b cosθ Wecanrearrangethistoobtainanexpressionfor cosθ: cosθ = a b a b () Ifwearegiven aand bincartesianformwecanusetheresultobtainedinsection4tocalculate a b.wecanalsocalculatethemodulusofeachof aand bsince a = a 2 + a2 2 + a2 and b = b 2 + b2 2 + b2 WiththisinformationEquation()canbeusedtofindtheanglebetweenthetwovectors. www.mathcentre.ac.uk 9 c mathcentre 2009
Theangle, θ,betweenthetwovectors aand bcanbefoundfrom cosθ = a b a b Example Supposewewishtofindtheanglebetweenthevectors a = 4i + j + 7kand b = 2i + 5j + 4k. Thescalarproduct, a b,hasalreadybeencalculatedonpage7andfoundtobe5. Themodulusofeachvectorisfound: a = 4 2 + 2 + 7 2 = 74 Then, from Equation(), b = 2 2 + 5 2 + 4 2 = 45 cosθ = = a b a b 5 74 45 Finally, using a calculator = 0.888(4 sig fig) θ = cos 0.888 = 27.90 Sotheanglebetweenthevectors aand bis 27.90. Finding the component of a vector in the direction of another vector Anotherapplicationofthescalarproductistofindthecomponentofonevectorinthedirection of another. ConsiderFigure5.Wecanthinkofthevector bbeingmadeupofacomponentinthedirection of a,( OA),togetherwithaperpendicularcomponent,( AB). The component in the direction of aistheprojectionof bonto a. b B O θ l A a Figure5.Theprojectionof bonto ais l. www.mathcentre.ac.uk 0 c mathcentre 2009
From the right angled triangle OAB and using trigonometry we see that and therefore cosθ = l b l = b cos θ Usingtheformulafor cosθobtainedinthepreviousapplicationwehave This can be written in the alternative form l = b cosθ = b a b a b = a b a l = b Sotheprojectionof bonto acanbefoundbytakingthescalarproductof bandaunitvector inthedirectionof a,i.e. l = b â. a a Theprojection, l,of bonto acanbefoundbytakingthescalarproductof bandaunitvector inthedirectionof a: l = b â Example Supposewewishtofindthecomponentof b = i + j + 4kinthedirectionof a = i j + k. Aunitvectorinthedirectionof ais Then b â = 4 â = a a = (i j + k) = ( + 4) = 6 = 2. Sothecomponentof b = i + j + 4kinthedirectionof a = i j + kis 2. www.mathcentre.ac.uk c mathcentre 2009
Exercises 2..PointsA,B,andChavecoordinates(,2,),(5,4,2),and ( 4, 2, )respectively. The scalarproductof ABand AChasbeenfoundinthepreviousExercisesQ6.Findtheangle between ABand AC. 2.Determinewhetherornotthevectors 2i + 4jand i + 2 jareperpendicular..evaluate p iwhere p = 4i + 8j.Hencefindtheanglethat pmakeswiththe xaxis. 4.Findthecomponentofthevector r = i j + kinthedirectionof a = 7i 2j + 2k. Answers to Exercises Exercises.(a)0,(b)0,(c)97,(d). 2.0..(a)27,(b)27,(c)29,(d)26. 4.Bothequal77. 5.(a) 50,(b) 50,(c)29,(d)20. 6. 4. Exercises 2...8. 2. Theirscalarproductiszero. Theyarenon-zerovectors. Wededucethattheymustbe perpendicular.. p i = 4.Therequiredangleis 6.4. 5 4. 57 =.987(d.p.). www.mathcentre.ac.uk 2 c mathcentre 2009