Math 8. Notes o Dot Product, Cross Product, Plaes, Area, ad Volumes This lecture focuses primarily o the dot product ad its may applicatios, especially i the measuremet of agles ad scalar projectio ad determiig the equatio of a plae. We also itroduce the cross product i R which ca be used to fid a vector orthogoal (perpedicular) to ay pair of ozero, oparallel vectors i R ad ca also be used i the measuremet of area. The Dot Product (or Scalar Product) I additio to the most basic operatios of scalig ad vector additio (both doe compoet-wise), the measuremet of legths ad agles are facilitated by the dot product of vectors (also kow as the ier product or the scalar product). The dot product ca be defied i R for ay which will allow for the defiitio of orthogoality i ay dimesio. Defiitio: The dot product of two vectors u, v i R is a scalar defied as follows: u v= u, u,, u v, v,, v = uv + uv + + uv Measurig the legth of a vector: Note that for ay vector u i uu = u, u,, u u, u,, u = u + u + + u = u, so R, uu = u or u= uu. There are some easy-to-verify algebraic properties of the dot product that follow from its defiitio: Algebraic Properties of the Dot Product: Suppose u, v, ad w are vectors i R ad that t is ay scalar. ) vu = uv (symmetry, dot product is commutative) u ( v+ w) = u v+ u w ) (left ad right distributive laws) ( u+ v) w = u w+ v w ) ( tu) v = t( u v) = u ( tv ) (how the dot product behaves relative to scalig of vectors) 4) uu = u for all u(ad uu = u = oly for u= ) A corollary of the Pythagorea Theorem is the Law of Cosies. Referrig to the figure, the Law of Cosies states that C = A + B AB cos θ where A ad B are the legths of the sides adjacet to the agle θ ad C is the legth of the side opposite this agle. We ca state a vector versio of this usig a modified figure. I this figure the Law of Cosies ca be expressed as u v = u + v u v cos θ. Usig the facts above, the left-had-side gives u v = ( u v) ( u v) = u u u v v u+ v v = u + v u v. Therefore u + v u v= u + v u v cos θ, ad cacellatio gives that u v= u v cos θ. This restatemet of the Law of Cosies is also kow as the geometric defiitio of the dot product. The great importace of this relatio is that it coects the algebraically-defied dot product (sum of the products of the respective compoets) to the geometric measuremets of legths ad agles. Example: A triagle i R has vertices at the poits with coordiates A (,), B(, ), ad C (4, ). Fid the agles αβγ,, show. Solutio: The dot product requires two vectors, so i order to proceed we have to choose vectors appropriate for each of the give agles. For example, to determie the agle α we Revised October 8, 5
will wat to fid the vectors u = AB =, ad v = AC =, emaatig out from this commo vertex. The relatio u v= u v cos α ca be expressed as cos α= u v u v, so,, cos α= = ad α= cos 74.74 cosβ= ad calculate that. Similarly, we ca use the vectors BA =, ad BC =, 4 β= cos 47.7 ; ad we ca use the vectors CB =, 4 7 cos γ= ad 7 β= cos 57.5 7 7 to calculate that ad CA =,. Note that α+β+γ= 8, as expected. Acute, obtuse, right agles We kow that whe a agle θ is acute, the cos θ> ; whe θ is obtuse, the cos θ< ; ad whe θ is a right agle, the cos θ=. If we couple these observatios with the relatio u v= u v cos θ, we get that if u, v i R are ozero vectors emaatig from a commo vertex to form a agle θ, the u v > if ad oly if the agle θ is acute u v < if ad oly if the agle θ is obtuse u v = if ad oly if the agle θ is a right agle, i.e. u v Orthogoal Projectio Referrig to the sketch show, the scalar projectio of the vector v i the directio of the vector u (also called the compoet of the vector v i the directio of the vector u) is the legth l (which will be positive for a acute agle ad egative for a obtuse agle). Basic trigoometry gives that l = v cos θ ad the relatio u v= u v cos θ gives that uv vu u l = v cos θ= = = v. This latter expressio eables us to u u u express this i words simply as To fid the compoet of a vector i a give directio, calculate the dot product of that vector with a uit vector i the desired directio. This is cosistet with our previous use of the word compoet, e.g. the x-compoet of the vector v = xyz,, is give by vi = xyz,,,, = x. We ca ow, however, fid the compoet of a vector i ay give directio ad ot just i the directios of the coordiate axes. We ca the use this fact to defie the vector projectio of v i the directio of u by costructio it by startig with the vector u, ormalizig it to get a uit vector i the same directio, ad the scalig it by the value of the scalar u u vu projectio to get Proju v= v = u. This ca be useful for u u u expressig a vector as the sum of a tagetial compoet vector ad a ormal compoet vector, especially i geometry ad physics. to Revised October 8, 5
Equatios for lies i R ad plaes i R We ca use the dot product to costruct a equatio for a lie i R give ay kow poit ( x, y ) o the lie ad a ormal vector = AB, perpedicular to the lie. Note that for ay other poit ( xy, ) to be o this lie it must be the case that the differece vector x x = x x, y y is orthogoal to the ormal vector = AB, ad this will be the case if ad oly if ( x x ) =. If we write this out i compoets we get that AB, x x, y y = or Ax ( x) By ( y) + =. Note that the ormal vector has slope B A (assumig it s either horizotal or vertical) ad we ca re-express the previous equatio i the form ( y y) = B A ( x x) which is just the familat poit-slope form of a lie if we recogize that the slope A B is just the egative reciprocal of the slope B A of the ormal vector. The costructio is essetially the same for a plae i R. I this case if x = x, y, z represets the positio vector of ay kow poit i the plae ad if = ABC,, is a ormal vector for the plae (ote that ay ozero scalar multiple will do just as well), the if x = xyz,, is the positio vector of ay other poit i the plae it must ecessarily be the case that the differece vector x x = x x, y y, z z is orthogoal to the ormal vector = ABC,, ad this will be or the case if ad oly if ( x x ) =. If we express this i compoets, this becomes ABC,, x x, y y, z z = Ax ( x) + By ( y) + Cz ( z) =. If we expad this out ad traspose all the costats to the right-hadside we get a equatio of the form Ax + By + Cz = D where D is a costat. Note that the compoets of the ormal vector appear as the coefficiets i this liear equatio. Had we used a scalar multiple for the ormal vector we would get a equivalet liear equatio with these coefficiets still i the same proportio as the compoets of the ormal vector. I problems, we ofte jump to this form oce we kow the ormal vector ad determie D by pluggig i the coordiates of the give poit. Example: Fid a equatio for the plae with ormal vector =,, 4 that passes through the poit with coordiates (,,5). Solutio: We ca simply substitute ito the costructio above to get that,, 4 x, y, z 5 = or ( x ) + ( y ) 4( z 5) = or x+ y 4z = 5. Note that a plae with the equatio x+ y 4z = would have the same (or a parallel) ormal vector but shares o poits with the previous plae. They are parallel plaes. The Cross Product (or Vector Product) Whe fid the equatio of a plae we will ot ecessarily be provided with a kow poit ad a coveiet ormal vector. We might, for example be give three poits that are ot co-liear, i.e. that do ot all lie o a sigle lie. Problem: Fid a equatio for the plae that cotais the three poits P(,, ), Q (,, ), ad R (,,). Revised October 8, 5
I this case there are several productive approaches we ca take. For example, we ca take poits pairwise to produce vectors parallel to the plae ad the try to fid a vector orthogoal to these two vectors. Such a vector will the serve as a ormal vector for this plae. We would like to fid a vector = ABC,, such that PQ = ad PR =. For the give poits we have PQ =,, ad PR =,, 4, so the above coditios traslate ito two equatios i three ukows, A B+ C = amely A+ B+ 4C =. This has ifiitely may solutios (correspodig to that face that ay scalar multiple of a ormal vector will still be a ormal vector), but addig them gives the relatio A+ 7C = or A= 7C. If we simply choose C =, the we must ecessary have A = 7 ad we ca use either of the two origial equatios to solve for B =. Therefore = 7,, will work as a ormal vector. This method gets us a solutio, but it s aythig but routie. Alteratively, we might simply observe that the equatio of the plae must be of the form Ax + By + Cz = D for appropriate costats ABCD.,,, Sice the three give poits all presumably lie o this plae, they must all satisfy the equatio for the plae, so we get three equatios i the four ukows ABCD,,,, amely A+ B C = D A+ B C D= A+ B = D which also be writte as A+ B D=. For those of you who are familiar with A+ B+ C = D A+ B+ C D= 7 row reductio, this ca be solved as RREF 6 7 to yield that A= D, 4 B= 6 D, ad C = 4 D. If iteger values are desired, we ca choose D = which the gives A = 7, B =, ad C =. So a equatio for this plae is 7x y z =. Oce agai, we get a solutio, but there must be a better way. The most coveiet way to do this is defie the cross product of two vectors (defied oly i R ). Give two vectors u = u, u, u ad v = v, v, v i R, we ca use the orthogoality requiremet to show that the followig cross product will be orthogoal to both vectors: u v = u, u, u v, v, v = uv uv, uv uv, uv uv There are several differet ways to express this usig the defiitio of a determiat, amely a b a b det ad bc c d = =. Examiig the above expressio we see that: c d u u u u u u u u u u u u u v =,,,, v v v v v v = v v v v v v Note the sig switch i the middle compoet. This is doe so that you ca coveietly perform the u u u calculatio by creatig a array from the give two vectors v v v ad the respectively coverig the st, d, ad rd colums ad calculatig the determiat of the resultig determiats (with appropriate sig switch of the middle compoet. For example, if u = PQ =,, ad v = PR =,, 4, we would get 4 Revised October 8, 5
the array 4 ad use the procedure to calculate 4, (4 6), ( ) 7,, u v = =. This coicides with the ormal vector we obtaied with greater effort i our first attempt. Some people prefer to express this procedure usig { i,, jk } otatio by formally calculatig the i j k u u u u u u determiat u v= u u u = i j+ k. v v v v v v v v v Usig oly this algebraic defiitio for the cross product, we ca derive the followig properties: Algebraic Properties of the Cross Product: Suppose u, v, ad w are vectors i ) v u= u v (aticommutative) [Corollary: u u= for ay vector u ] u ( v+ w) = u v+ u w ) (left ad right distributive laws) ( u+ v) w = u w+ v w R ad that t is ay scalar. ) ( tu) v = t( u v) = u ( tv ) (how the dot product behaves relative to scalig of vectors) 4) u = 5) u ( v w) = ( u v) w (triple scalar product) 6) u ( v w) = ( uwv ) ( u vw ) (triple vector product) All of the above algebraic properties of the cross product except for the last oe are straightforward. You ca prove the last oe by otig that the first compoet would be: u u = u( vw vw ) u( vw vw ) = uvw uvw uvw + uvw vw vw vw vw = uvw uvw uvw + uvw + uvw uvw = ( uw + u w + uw) v ( uv + uv + uv) w = ( u w) v ( u v) w Similarly, we ca show that the d ad rd compoets are ( u w) v ( u v ) w ad ( u w) v ( u v ) w. Together these give that u ( v w) = ( uwv ) ( u vw. ) Physicists (ad others) ofte refer to this property as the BAC-CAB Rule ad express it as A ( B C) = ( ACB ) ( ABC ) = BAC ( ) CAB. ( ) We ca idepedet defie the cross product i purely geometric terms. Geometric defiitio of the cross product: Suppose u ad v are vectors i R. The the cross product u v is the uique vector i R such that: () u v is orthogoal to both u ad v; () the magitude of the cross product u v is equal to the area of the parallelogram determied by u ad v; () u v is orieted accordig to the Right-Had Rule (as explaied i class ad elsewhere). It is true that these three properties uiquely determie the cross product, ad we ca also easily derive the previous algebraic defiitio from these requiremets. We ca also derive these geometric properties from the algebraic defiitio usig the previously stated algebraic properties. Specifically: 5 Revised October 8, 5
() u ( u v) = ( u u) v= v =, so u v is orthogoal to the vector u; ( u v) v= u ( v v) = u =, so u v is orthogoal to the vector v. () If we cosider the parallelogram determied by u ad v ad let θ be the agle betwee these vectors (drawig a picture is advisable), the the area of the parallelogram will be give by (legth of base)( height) = u v siθ. Squarig both sides gives (Area) = u v si θ= u v ( cos θ ) = u v u v cos θ= u v ( u v ). O the other had, Therefore u v = ( u v) ( u v) = u [ v ( u v)] = u [( v v) u ( v u) v] = u v ( u v ). (Area) = u v, so Area = u v. () You ca easily calculate usig the algebraic defiitio that i j= k which satisfies the Right-Had Rule. The argue usig a cotiuity argumet that if this is true for these two vectors tha by cotiuously varyig these vectors i R to alig them with the give two vectors, the right-had rule must be preserved. Volume ad the Triple Scalar Product: Property (5) of the Algebraic Properties of the Cross Product (the triple scalar product) has a iterestig geometric iterpretatio. Note that ( u v) w = u v w cos θ where θ is the agle betwee the vectors u v ad w. [For simplicity, we re cosiderig the case where θ is a acute agle. If it is obtuse, everythig s the same except for a chage i sig.] Note that u v gives the area of the parallelogram that forms the base of the parallelepiped determied by the three vectors u, v, ad w i R ; ad w cos θ correspods to the perpedicular height of this parallelepiped, i.e. the scalar projectio of the vector w i the directio of the vector u v. ( u v) w = u v w cos θ = (area of base)( height) = Volume of Thus ( ) the parallelepiped (up to sig). So the volume is u ( v w ) or ( u v) w. The triple scalar product may also be calculated as u u u ( u v) w = u ( v w ) = v v v, a determiat. w w w Notes by Robert Witers 6 Revised October 8, 5