Vector Algebra and Calculus

Vector Algebra and Calculus 1. Revision of vector algebra, scalar product, vector product 2. Triple products, multiple products, applications to geometry 3. Differentiation of vector functions, applications to mechanics 4. Scalar and vector fields. Line, surface and volume integrals, curvilinear co-ordinates 5.Vectoroperators grad,divandcurl 6. Vector Identities, curvilinear co-ordinate systems 7. Gauss and Stokes Theorems and extensions 8. Engineering Applications

2. More Algebra& Geometry using Vectors Inwhichwediscuss... Vector products: Scalar Triple Product, Vector Triple Product, Vector Quadruple Product GeometryofLinesandPlanes Solving vector equations Angular velocity and moments

Triple and multiple products 2.2 Using mixtures of scalar products and vector products, it is possible to derive triple products between three vectors n-products between n vectors. Nothing new about these but some have nice geometric interpretations... Wewilllookatthe Scalar triple product Vector triple product Vector quadruple product

Scalartripleproducta (b c) 2.3 Scalar triple product given by the true determinant a (b c)= a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 Yourknowledgeofdeterminantstellsyouthatifyou swaponepairofrowsofadeterminant,signchanges; swaptwopairsofrows,itssignstaysthesame. Hence (i)a (b c)=c (a b)=b (c a)(cyclicpermutation.) (ii)a (b c)= b (a c)andsoon.(anti-cyclicpermutation) (iii)thefactthata (b c)=(a b) callowsthescalartripleproducttobewrittenas[a,b,c]. Thisnotationisnotveryhelpful,andwewilltrytoavoiditbelow.

Geometrical interpretation of scalar triple product 2.4 Thescalartripleproductgivesthevolumeoftheparallelopipedwhosesidesarerepresentedbythevectorsa,b,and c. c cos β β c b Vectorproduct(a b)has magnitudeequaltotheareaofthebase direction perpendicular to the base. a Thecomponentofcinthisdirectionisequaltotheheightoftheparallelopiped Hence (a b) c =volumeofparallelopied

Linearly dependent vectors 2.5 Ifthescalartripleproductofthreevectors a (b c)=0 n b c then the vectors are linearly dependent. a=λb+µc a You can see this immediately either using the determinant Thedeterminantwouldhaveonerowthatwasalinearcombinationoftheothers or geometrically for a 3-dimensional vector. the parallelopiped would have zero volume if squashed flat.

Vectortripleproducta (b c) 2.6 a (b c)isperpendicularto(b c) but(b c)isperpendiculartobandc. Soa (b c)mustbecoplanarwithbandc. a (b c)=λb+µc b x c a In arbitrary direction c b a x ( b x c ) Similarlyforcomponents2and3:so (a (b c)) 1 = a 2 (b c) 3 a 3 (b c) 2 = a 2 (b 1 c 2 b 2 c 1 )+a 3 (b 1 c 3 b 3 c 1 ) = (a 2 c 2 +a 3 c 3 )b 1 (a 2 b 2 +a 3 b 3 )c 1 = (a 1 c 1 +a 2 c 2 +a 3 c 3 )b 1 (a 1 b 1 +a 2 b 2 +a 3 b 3 )c 1 = (a c)b 1 (a b)c 1 [a (b c)] =(a c)b (a b)c

Projection using vector triple product 2.7 v Bookssaythatthevectorprojectionofanyoldvectorvintoaplanewith normalˆnis v INPLANE =ˆn (v ˆn). n Thecomponentofvintheˆndirectionisv ˆn soiwouldwritethevectorprojectionas v INPLANE =v (v ˆn)ˆn v INPLANE Canwereconcilethetwoexpressions?Subst.ˆn a,v b,ˆn c,intoourearlierformula a (b c) = (a c)b (a b)c ˆn (v ˆn) = (ˆn ˆn)v (ˆn v)ˆn = v (v ˆn)ˆn Fantastico!Butv (v ˆn)ˆnismucheasiertounderstand,cheapertocompute!

VectorQuadrupleProduct(a b) (c d) 2.8 Wehavejustlearnedthat Regardinga basasinglevector vqpmustbealinearcombinationofcandd Regardingc dasasinglevector vqpmustbealinearcombinationofaandb. Substituting in carefully(you check...) [p (q r)] = (p r)q (p q)r (a b) (c d) =?? (a b) (c d) = [(a b) d]c [(a b) c]d = [(c d) a]b [(c d) b]a

Vector Quadruple Product/ctd 2.9 UsingjusttheR-Hsidesofwhatwejustwrote... [(a b) c]d=[(b c) d]a+[(c a) d]b+[(a b) d]c So Don trememberby d = [(b c) d]a+[(c a) d]b+[(a b) d]c [(a b) c] = αa+βb+γc. a d Keypointisthattheprojectionofa3Dvector dontoabasissetof3non-coplanarvectorsis UNIQUE. c b

Example 2.10 Question Usethequadruplevectorproducttoexpressthevectord=[3,2,1]intermsofthevectorsa=[1,2,3],b=[2,3,1] andc=[3,1,2]. Answer d= [(b c) d]a+[(c a) d]b+[(a b) d]c [(a b) c] So, grinding away at the determinants, we find (a b) c= 18and(b c) d=6 (c a) d= 12and(a b) d= 12. So d = 1 18 (6a 12b 12c) = 1 3 ( a+2b+2c)

Geometry using vectors: Lines 2.11 Equationoflinepassingthroughpointa 1 andlyingin thedirectionofvectorbis r=a+βb a λ^ b r Point r traces out line. NB!Onlywhenyoumakeaunitvectorinthedirnofbdoestheparametertakeonthelengthunitsdefinedbya: r=a+λˆb Foralinedefinedbytwopointsa 1 anda 2 r=a1+β(a2 a1) ortheunitversion... r=a 1 +λ(a 2 a 1 )/ a 2 a 1

Theshortestdistancefromapointtoaline 2.12 λb VectorpfromctoANYlinepointris p=r c=a+λˆb c=(a c)+λˆb which has length squared a r r c p 2 =(a c) 2 +λ 2 +2λ(a c) ˆb. c Easiertominimizep 2 ratherthanpitself. d dλ p2 =0 when λ= (a c) ˆb. Sotheminimumlengthvectoris p=(a c) ((a c) ˆb)ˆb. Nosurprise!It sthecomponentof(a c)perpendiculartoˆb. We could therefore write using the book formula... p = ˆb [(a c) ˆb] p min = ˆb [(a c) ˆb] = (a c) ˆb.

Shortest distance between two straight lines 2.13 Shortestdistancefrompointtolineisalongtheperpline shortest distance between two straight lines is along mutual perpendicular. Thelinesare: r=a+λˆb r=c+µˆd c µ d λb Theunitvectoralongthemutualperpis ˆp= ˆb ˆd ˆb ˆd. a Q P (Yes!Don tforgetthatˆb ˆdisNOTaunitvector.) Theminimumlengthisthereforethecomponentof(a c)inthisdirection ) (ˆb ˆd p min = (a c) ˆb ˆd.

Example 2.14 Question for civil engineers Two long straight pipes are specified using Cartesian co-ordinates as follows: Pipe A: diameter 0.8; axis through points (2,5,3) and (7,10,8). Pipe B: diameter 1.0; axis through points (0,6,3) and ( 12,0,9). Do the pipes need re-aligning to avoid intersection?

Example continued 2.15 Answer PipesAandBhaveaxes: (Non-unit) perpendicular to both their axes is î ĵ ˆk p= 1 1 1 2 1 1 =[2, 3,1] The length of the mutual perpendicular is mod r A = [2,5,3]+λ [5,5,5]=[2,5,3]+λ[1,1,1]/ 3 r B = [0,6,3]+µ [ 12, 6,6]=[0,6,3]+µ[ 2, 1,1]/ 6 (a b) [2, 3,1] 14 =[2, 1,0] [2, 3,1] 14 =1.87. Sumoftheradiiofthepipesis0.4+0.5=0.9. Hence the pipes do not intersect. [0,6,3] [7,10,8] b a p [2,5,3] [ 12,0,9]

Three ways of describing a plane. Number 1 2.16 1. Point + 2 non-parallel vectors Ifbandcnon-parallel,andaisapointontheplane,then where λ, µ are scalar parameters. r=a+λb+µc r a c O b NB that these are parallel to the plane, not necessarily in the plane

Three ways of describing a plane. Number 2 2.17 r c 2. Three points Pointsa,bandcintheplane. r=a+λ(b a)+µ(c a) a b Vectors(b a)and(c a)aresaidtospantheplane. O

Three ways of describing a plane. Number 3 2.18 n ^ 3.UnitnormalUnitnormaltotheplaneisˆn,andapoint intheplaneisa r ˆn=a ˆn=D a r O Notice that D is the perpendicular distance to the plane from the origin. WhynotjustD?

The shortest distance from a point to a plane 2.19 Theplaneisr ˆn=a ˆn=D Theshortestdistanced min fromanypointtotheplaneisalongtheperpendicular. So,theshortestdistancefromtheorigintotheplaneis d min = D = a ˆn = a (b c) b c. Now,theshortestdistancefrompointdtotheplane...? d n ^ 1. Must be along the perpendicular 2.d+λˆnmustbeapointonplane 3.(d+λˆn) ˆn=D 4.λ=D d ˆn 5.d min = λ = D d ˆn r O

Solution of vector equations 2.20 Find the most general vector x satisfying a given vector relationship. Eg General Method(assuming 3 dimensions) x=x a+b 1. Set up a system of three basis vectors using two non-parallel vectors appearing in the original vector relationship. For example a,b,(a b) 2. Write whereλ,µ,νarescalarstobefound. x=λa+µb+νa b 3. Substitute expression for x into the vector relationship to determine the set of constraints on λ,µ, and ν.

Example:Solvex=x a+b. 2.21 Step1:Basisvectorsa,bandv.p.a b. Step2:x=λa+µb+νa b. Step3:Bungxbackintotheequation! λa+µb+νa b = (λa+µb+νa b) a+b Equatingcoefficientsofa,banda bintheequationgives = 0+µ(b a)+ν(a b) a+b = ν(a b)a+(νa 2 +1)b µ(a b) λ= ν(a b) µ=νa 2 +1 ν= µ so that µ= 1 1+a 2 ν= 1 1+a 2 λ= a b 1+a 2. So finally the solution is the single point: x= 1 1+a 2[(a b)a+b (a b)]

Another example 2.22 Oftennotalltheparametersaredetermined:µandνmightdependonanarbitrarychoiceofλ(see2A1Asheet). Andwhathappensiftherearenottwofixedvectorsintheexpression? Question.Findxwhenx a=k. Answer. Step1Usea,introduceanarbitraryvectorb,anda b Step2:x=λa+µb+νa b. Step3:Bungxbackintotheequation! So,hereλ,νANDbarearbitary... λa 2 +µb a = K µ = K λa2 b a x=λa+ K λa2 b a b+νa b

A random comment about solving vector identities 2.23 Supposeyouarefacedwith µa+λb=c andyouwantµ. Whatisthefastwayofgettingridofb? Useb b=0... µa b = c b µ(a b) (a b) = (c b) (a b) µ = (c b) (a b) (a b) (a b)

A random comment about solving vector identities 2.24 µa+λb=c An alternative is to construct two simultaneous equations µa b+λb 2 = c b µa 2 +λa b = a c and eliminate λ Compare with previous µ= (a b)(b c) (a c)b2 (a b) 2 a 2 b 2 µ= (c b) (a b) (a b) (a b)

Rotation, angular velocity and acceleration 2.25 Arotationcanrepresentedbyavectorwhose directionisalongtheaxisofrotationinthesense of a right-handed screw, magnitudeisproportionaltothesizeoftherotation. Thesameideacanbeextendedtothederivatives angular velocity ω angular acceleration ω. ω r v Theinstantaneousvelocityv(r)ofanypointPatronarigidbodyundergoingpurerotationcanbedefinedbya vector product v=ω r.

Vector Moments 2.26 Angular accelerations arise because of moments. ThevectorequationforthemomentMofaforceF aboutapointqis M=r F whererisavectorfromqtoanypointonthelineof actionlofforcef. Theresultingangularacceleration ωisinthesamedirectionasthemomentvectorm.(howaretheyrelated?) Q M α F r

Alwaysrunaconsistencycheckthatequationsarevectororscalaronbothsides. Summary 2.27 Today we ve discussed... Vector products Geometry of Lines and Planes Solving vector equations Angular velocity and moments(briefly!!!) Key point from this week: Use vectors and their algebra constructively to solve problems.(the elastic collision was a good example.) Don t be afraid to produce solutions that involve vector operations Eg:µ=a b/ c a Workingoutdetailcouldbelefttoacomputerprogram If you are constantly breaking vectors into their components, you are not using their power.