CALCULUS II Vectors. Paul Dawkins

Transcription

1 CALCULUS II Vectos Paul Dawkins

2 Table of Contents Peface... ii Vectos... 3 Intoduction... 3 Vectos The Basics... 4 Vecto Aithmetic... 8 Dot Poduct Coss Poduct Paul Dawkins i

3 Peface Hee ae my online notes fo my Calculus II couse that I teach hee at Lama Univesity. Despite the fact that these ae my class notes, they should be accessible to anyone wanting to lean Calculus II o needing a efeshe in some of the topics fom the class. These notes do assume that the eade has a good woking knowledge of Calculus I topics including limits, deivatives and basic integation and integation by substitution. Calculus II tends to be a vey difficult couse fo many students. Thee ae many easons fo this. The fist eason is that this couse does equie that you have a vey good woking knowledge of Calculus I. The Calculus I potion of many of the poblems tends to be skipped and left to the student to veify o fill in the details. If you don t have good Calculus I skills, and you ae constantly getting stuck on the Calculus I potion of the poblem, you will find this couse vey difficult to complete. The second, and pobably lage, eason many students have difficulty with Calculus II is that you will be asked to tuly think in this class. That is not meant to insult anyone; it is simply an acknowledgment that you can t just memoize a bunch of fomulas and expect to pass the couse as you can do in many math classes. Thee ae fomulas in this class that you will need to know, but they tend to be faily geneal. You will need to undestand them, how they wok, and moe impotantly whethe they can be used o not. As an example, the fist topic we will look at is Integation by Pats. The integation by pats fomula is vey easy to emembe. Howeve, just because you ve got it memoized doesn t mean that you can use it. You ll need to be able to look at an integal and ealize that integation by pats can be used (which isn t always obvious) and then decide which potions of the integal coespond to the pats in the fomula (again, not always obvious). Finally, many of the poblems in this couse will have multiple solution techniques and so you ll need to be able to identify all the possible techniques and then decide which will be the easiest technique to use. So, with all that out of the way let me also get a couple of wanings out of the way to my students who may be hee to get a copy of what happened on a day that you missed. 1. Because I wanted to make this a faily complete set of notes fo anyone wanting to lean calculus I have included some mateial that I do not usually have time to cove in class and because this changes fom semeste to semeste it is not noted hee. You will need to find one of you fellow class mates to see if thee is something in these notes that wasn t coveed in class. 2. In geneal I ty to wok poblems in class that ae diffeent fom my notes. Howeve, with Calculus II many of the poblems ae difficult to make up on the spu of the moment and so in this class my class wok will follow these notes faily close as fa as woked poblems go. With that being said I will, on occasion, wok poblems off the top of my head when I can to povide moe examples than just those in my notes. Also, I often 2007 Paul Dawkins ii

4 don t have time in class to wok all of the poblems in the notes and so you will find that some sections contain poblems that ween t woked in class due to time estictions. 3. Sometimes questions in class will lead down paths that ae not coveed hee. I ty to anticipate as many of the questions as possible in witing these up, but the eality is that I can t anticipate all the questions. Sometimes a vey good question gets asked in class that leads to insights that I ve not included hee. You should always talk to someone who was in class on the day you missed and compae these notes to thei notes and see what the diffeences ae. 4. This is somewhat elated to the pevious thee items, but is impotant enough to meit its own item. THESE NOTES ARE NOT A SUBSTITUTE FOR ATTENDING CLASS!! Using these notes as a substitute fo class is liable to get you in touble. As aleady noted not eveything in these notes is coveed in class and often mateial o insights not in these notes is coveed in class. Vectos Intoduction This is a faily shot chapte. We will be taking a bief look at vectos and some of thei popeties. We will need some of this mateial in the next chapte and those of you heading on towads Calculus III will use a fai amount of this thee as well. Hee is a list of topics in this chapte. Vectos The Basics In this section we will intoduce some of the basic concepts about vectos. Vecto Aithmetic Hee we will give the basic aithmetic opeations fo vectos. Dot Poduct We will discuss the dot poduct in this section as well as an application o two. Coss Poduct In this section we ll discuss the coss poduct and see a quick application Paul Dawkins iii

5 Vectos The Basics Let s stat this section off with a quick discussion on what vectos ae used fo. Vectos ae used to epesent quantities that have both a magnitude and a diection. Good examples of quantities that can be epesented by vectos ae foce and velocity. Both of these have a diection and a magnitude. Let s conside foce fo a second. A foce of say 5 Newtons that is applied in a paticula diection can be applied at any point in space. In othe wods, the point whee we apply the foce does not change the foce itself. Foces ae independent of the point of application. To define a foce all we need to know is the magnitude of the foce and the diection that the foce is applied in. The same idea holds moe geneally with vectos. Vectos only impat magnitude and diection. They don t impat any infomation about whee the quantity is applied. This is an impotant idea to always emembe in the study of vectos. In a gaphical sense vectos ae epesented by diected line segments. The length of the line segment is the magnitude of the vecto and the diection of the line segment is the diection of the vecto. Howeve, because vectos don t impat any infomation about whee the quantity is applied any diected line segment with the same length and diection will epesent the same vecto. Conside the sketch below. Each of the diected line segments in the sketch epesents the same vecto. In each case the vecto stats at a specific point then moves 2 units to the left and 5 units up. The notation that we ll use fo this vecto is, v = -2,5 and each of the diected line segments in the sketch ae called epesentations of the vecto. Be caeful to distinguish vecto notation, - 2,5, fom the notation we use to epesent coodinates of points, ( 2,5) -. The vecto denotes a magnitude and a diection of a quantity while the point denotes a location in space. So don t mix the notations up! 2007 Paul Dawkins 4

6 A epesentation of the vecto v = a1, a2 uuu AB, fom the point A= ( xy, ) to the point B ( x a, y a ) the vecto v = a1, a2, a3 in two dimensional space is any diected line segment, = Likewise a epesentation of uuu in thee dimensional space is any diected line segment, AB, fom the point A= ( xyz,, ) to the point B ( x a, y a, z a ) = Note that thee is vey little diffeence between the two dimensional and thee dimensional fomulas above. To get fom the thee dimensional fomula to the two dimensional fomula all we did is take out the thid component/coodinate. Because of this most of the fomulas hee ae given only in thei thee dimensional vesion. If we need them in thei two dimensional fom we can easily modify the thee dimensional fom. Thee is one epesentation of a vecto that is special in some way. The epesentation of the vecto v = a1, a2, a3 that stats at the point ( 0,0,0) B= a, a, a A = and ends at the point ( ) is called the position vecto of the point ( a, a, a ). So, when we talk about position vectos we ae specifying the initial and final point of the vecto. Position vectos ae useful if we eve need to epesent a point as a vecto. As we ll see thee ae times in which we definitely ae going to want to epesent points as vectos. In fact, we e going to un into topics that can only be done if we epesent points as vectos. Next we need to discuss biefly how to geneate a vecto given the initial and final points of the epesentation. Given the two points A= ( a1, a2, a3) and B= ( b1, b2, b3) the vecto with the uuu epesentation AB is, v = b -a, b -a, b -a Note that we have to be vey caeful with diection hee. The vecto above is the vecto that stats uuu at A and ends at B. The vecto that stats at B and ends at A, i.e. with epesentation BA is, w= a -b, a -b, a -b These two vectos ae diffeent and so we do need to always pay attention to what point is the stating point and what point is the ending point. When detemining the vecto between two points we always subtact the initial point fom the teminal point. Example 1 Give the vecto fo each of the following. (a) The vecto fom ( 2, - 7,0) to ( 1, -3,- 5). (b) The vecto fom ( 1, -3,- 5) to( 2, - 7,0). (c) The position vecto fo (- 90,4) (a) Remembe that to constuct this vecto we subtact coodinates of the stating point fom the ending point Paul Dawkins 5

7 (b) Same thing hee. 1-2, -3-(-7, ) -5-0 = -1,4, , -7-(-3,0 ) -(- 5) = 1, - 4,5 Notice that the only diffeence between the fist two is the signs ae all opposite. This diffeence is impotant as it is this diffeence that tells us that the two vectos point in opposite diections. (c) Not much to this one othe than acknowledging that the position vecto of a point is nothing moe than a vecto with the point s coodinates as its components. - 90,4 We now need to stat discussing some of the basic concepts that we will un into on occasion. Magnitude The magnitude, o length, of the vecto v = a1, a2, a3 is given by, v = a + a + a Example 2 Detemine the magnitude of each of the following vectos. (a) a = 3, -5, (b) u =, (c) w = 0,0 (d) i = 1,0,0 Thee isn t too much to these othe than plug into the fomula. (a) a = = 134 (c) w = 0+ 0 = 0 (b) 1 4 u = + = 1= (d) i = = 1 We also have the following fact about the magnitude. If a = 0 then a = 0 This should make sense. Because we squae all the components the only way we can get zeo out of the fomula was fo the components to be zeo in the fist place. Unit Vecto Any vecto with magnitude of 1, i.e. u = 1, is called a unit vecto Paul Dawkins 6

8 Example 3 Which of the vectos fom Example 2 ae unit vectos? Both the second and fouth vectos had a length of 1 and so they ae the only unit vectos fom the fist example. Zeo Vecto The vecto w = 0,0 that we saw in the fist example is called a zeo vecto since its components ae all zeo. Zeo vectos ae often denoted by 0. Be caeful to distinguish 0 (the numbe) fom 0 (the vecto). The numbe 0 denotes the oigin in space, while the vecto 0 denotes a vecto that has no magnitude o diection. Standad Basis Vectos The fouth vecto fom the second example, i = 1,0,0, is called a standad basis vecto. In thee dimensional space thee ae thee standad basis vectos, i = 1,0,0 j = 0,1,0 k = 0,0,1 In two dimensional space thee ae two standad basis vectos, i = 1,0 j = 0,1 Note that standad basis vectos ae also unit vectos. Waning We ae petty much done with this section howeve, befoe poceeding to the next section we should point out that vectos ae not esticted to two dimensional o thee dimensional space. Vectos can exist in geneal n-dimensional space. The geneal notation fo a n-dimensional vecto is, v = a1, a2, a3, K, an and each of the a i s ae called components of the vecto. Because we will be woking almost exclusively with two and thee dimensional vectos in this couse most of the fomulas will be given fo the two and/o thee dimensional cases. Howeve, most of the concepts/fomulas will wok with geneal vectos and the fomulas ae easily (and natually) modified fo geneal n-dimensional vectos. Also, because it is easie to visualize things in two dimensions most of the figues elated to vectos will be two dimensional figues. So, we need to be caeful to not get too locked into the two o thee dimensional cases fom ou discussions in this chapte. We will be woking in these dimensions eithe because it s easie to visualize the situation o because physical estictions of the poblems will enfoce a dimension upon us Paul Dawkins 7

9 Vecto Aithmetic In this section we need to have a bief discussion of vecto aithmetic. We ll stat with addition of two vectos. So, given the vectos a = a1, a2, a3 and b = b, b, b the addition of the two vectos is given by the following fomula. a+ b = a + b, a + b, a + b The following figue gives the geometic intepetation of the addition of two vectos. This is sometimes called the paallelogam law o tiangle law. Computationally, subtaction is vey simila. Given the vectos a = a1, a2, a3 b = b, b, b the diffeence of the two vectos is given by, a- b = a -b, a -b, a -b and Hee is the geometic intepetation of the diffeence of two vectos Paul Dawkins 8

10 It is a little hade to see this geometic intepetation. To help see this let s instead think of subtaction as the addition of a and -b. Fist, as we ll see in a bit -b is the same vecto as b with opposite signs on all the components. In othe wods, a + - b. b. Hee is the vecto set up fo ( ) -b goes in the opposite diection as + - As we can see fom this figue we can move the vecto epesenting a ( b) we ve got in the fist figue showing the diffeence of the two vectos. to the position Note that we can t add o subtact two vectos unless they have the same numbe of components. If they don t have the same numbe of components then addition and subtaction can t be done. The next aithmetic opeation that we want to look at is scala multiplication. Given the vecto a = a1, a2, a3 and any numbe c the scala multiplication is, ca = ca, ca, ca So, we multiply all the components by the constant c. To see the geometic intepetation of scala multiplication let s take a look at an example. Example 1 Fo the vecto a = 2,4 the same axis system. compute 3a, 1 2 a and -2a. Gaph all fou vectos on Hee ae the thee scala multiplications. 1 3a = 6,12 a = 1,2-2a = -4,-8 2 Hee is the gaph fo each of these vectos Paul Dawkins 9

11 In the pevious example we can see that if c is positive all scala multiplication will do is stetch (if c > 1) o shink (if c < 1) the oiginal vecto, but it won t change the diection. Likewise, if c is negative scala multiplication will switch the diection so that the vecto will point in exactly the opposite diection and it will again stetch o shink the magnitude of the vecto depending upon the size of c. Thee ae seveal nice applications of scala multiplication that we should now take a look at. The fist is paallel vectos. This is a concept that we will see quite a bit ove the next couple of sections. Two vectos ae paallel if they have the same diection o ae in exactly opposite diections. Now, ecall again the geometic intepetation of scala multiplication. When we pefomed scala multiplication we geneated new vectos that wee paallel to the oiginal vectos (and each othe fo that matte). So, let s suppose that a and b ae paallel vectos. If they ae paallel then thee must be a numbe c so that, a = cb So, two vectos ae paallel if one is a scala multiple of the othe. Example 2 Detemine if the sets of vectos ae paallel o not. (a) a = 2, - 4,1, b = -6,12,-3 (b) a = 4,10, b = 2, -9 (a) These two vectos ae paallel since b =-3a 1 (b) These two vectos aen t paallel. This can be seen by noticing that ( 2 ) 1 10( ) = 5-9. In othe wods we can t make a be a scala multiple of b. 2 The next application is best seen in an example. 4 = 2 and yet 2007 Paul Dawkins 10

12 Example 3 Find a unit vecto that points in the same diection as w = -5,2,1. Okay, what we e asking fo is a new paallel vecto (points in the same diection) that happens to be a unit vecto. We can do this with a scala multiplication since all scala multiplication does is change the length of the oiginal vecto (along with possibly flipping the diection to the opposite diection). Hee s what we ll do. Fist let s detemine the magnitude of w. w = = 30 Now, let s fom the following new vecto, u = w= - 5,2,1 = -,, w The claim is that this is a unit vecto. That s easy enough to check u = + + = = This vecto also points in the same diection as w since it is only a scala multiple of w and we used a positive multiple. So, in geneal, given a vecto w, u = as w. w will be a unit vecto that points in the same diection w Standad Basis Vectos Revisited In the pevious section we intoduced the idea of standad basis vectos without eally discussing why they wee impotant. We can now do that. Let s stat with the vecto a = a, a, a We can use the addition of vectos to beak this up as follows, a = a, a, a = a,0,0 + 0, a,0 + 0,0, a Using scala multiplication we can futhe ewite the vecto as, a = a,0,0 + 0, a,0 + 0,0, a = a 1,0,0 + a 0,1,0 + a 0,0,1 Finally, notice that these thee new vectos ae simply the thee standad basis vectos fo thee dimensional space. a, a, a = ai + a j + ak 2007 Paul Dawkins 11

13 So, we can take any vecto and wite it in tems of the standad basis vectos. Fom this point on we will use the two notations intechangeably so make sue that you can deal with both notations. Example 4 If a = 3, -9,1 and w=- i + 8k compute 2a -3w. In ode to do the poblem we ll convet to one notation and then pefom the indicated opeations. 2a- 3w= 23, -9,1-3 -1,0,8 = 6, -18,2 - -3,0,24 = 9, -18,-22 We will leave this section with some basic popeties of vecto aithmetic. Popeties If v, w and u ae vectos (each with the same numbe of components) and a and b ae two numbes then we have the following popeties. v+ w= w+ v u+ ( v+ w) = ( u+ v) + w v+ 0= v 1v = v a v+ w = av+ aw a+ b v = av+ bv ( ) ( ) The poofs of these ae petty much just computation poofs so we ll pove one of them and leave the othes to you to pove. Poof of a( v+ w) = av+ aw We ll stat with the two vectos, v = v1, v2, K, vn and w= w1, w2, K, wn and yes we did mean fo these to each have n components. The theoem woks fo geneal vectos so we may as well do the poof fo geneal vectos. Now, as noted above this is petty much just a computational poof. What that means is that we ll compute the left side and then do some basic aithmetic on the esult to show that we can make the left side look like the ight side. Hee is the wok. a v+ w = a v, v, K, v + w, w, K, w ( ) ( 1 2 n 1 2 n ) = a v + w, v + w, K, v + w ( ), ( ), K, ( ) = a v + w a v + w a v + w = av + aw, av + aw, K, av + aw = av1, av2, K, avn + aw1, aw2, K, awn = a v, v, K, v + a w, w, K, w = av+ aw 1 2 n 1 2 n n n n n n n 2007 Paul Dawkins 12

14 Dot Poduct The next topic fo discussion is that of the dot poduct. Let s jump ight into the definition of the dot poduct. Given the two vectos a = a1, a2, a3 and b = b1, b2, b3 the dot poduct is, ab g = ab + ab + ab (1) Sometimes the dot poduct is called the scala poduct. The dot poduct is also an example of an inne poduct and so on occasion you may hea it called an inne poduct. Example 1 Compute the dot poduct fo each of the following. (a) v = 5i - 8 j, w= i + 2j (b) a = 0,3, - 7, b = 2,3,1 Not much to do with these othe than use the fomula. (a) vw= g 5-16=-11 (b) ab= g = 2 Hee ae some popeties of the dot poduct. Popeties ug v w uv g uw g cv gw vg cw c vw g vw g = wv g vg0= 0 2 vv g = v If vv g = 0 then v = 0 ( + ) = + ( ) = ( ) = ( ) The poofs of these popeties ae mostly computational poofs and so we e only going to do a couple of them and leave the est to you to pove. Poof of u g( v + w ) = uv g + uw g We ll stat with the thee vectos, u = u1, u2, K, un, v = v1, v2, K, vn and w= w1, w2, K, wn and yes we did mean fo these to each have n components. The theoem woks fo geneal vectos so we may as well do the poof fo geneal vectos. Now, as noted above this is petty much just a computational poof. What that means is that we ll compute the left side and then do some basic aithmetic on the esult to show that we can make the left side look like the ight side. Hee is the wok Paul Dawkins 13

15 ug v w u u K u g v v K v w w K w ( + ) = 1, 2,, n ( 1, 2,, n + 1, 2,, n ) = u, u, K, u g v + w, v + w, K, v + w 1 2 n n n ( ), ( ), K, ( ) = u v + w u v + w u v + w = uv + uw, uv + uw, K, uv + uw n n n n n n n = uv, uv, K, uv + uw, uw, K, uw n n n n = u1, u2, K, un g v1, v2, K, vn + u1, u2, K, un g w1, w2, K, wn = uv g + uw g Poof of : If vv= g 0 then v = 0 This is a petty simple poof. Let s stat with v = v1, v2, K, vn and compute the dot poduct. vv g = v, v, K, v g v, v, K, v = n 1 2 = v + v + L + v n 2 Now, since we know vi 0 fo all i then the only way fo this sum to be zeo is to in fact have 2 v = 0. This in tun howeve means that we must have v = 0 and so we must have had v = 0. i i n Thee is also a nice geometic intepetation to the dot poduct. Fist suppose that q is the angle between a and b such that 0 q p as shown in the image below. We can then have the following theoem Paul Dawkins 14

16 Theoem ab g = a b cosq (2) Poof Let s give a modified vesion of the sketch above. The thee vectos above fom the tiangle AOB and note that the length of each side is nothing moe than the magnitude of the vecto foming that side. The Law of Cosines tells us that, a- b = a + b -2 a b cosq Also using the popeties of dot poducts we can wite the left side as, 2 a- b = ( a-b) g( a-b) = aa g -ab g - ba g + bb g 2 2 = a - 2ab g + b Ou oiginal equation is then, a- b = a + b -2 a b cosq a - 2ab g + b = a + b -2 a b cosq - 2ab g =-2 a b cosq ab g = a b cosq 2007 Paul Dawkins 15

17 The fomula fom this theoem is often used not to compute a dot poduct but instead to find the angle between two vectos. Note as well that while the sketch of the two vectos in the poof is fo two dimensional vectos the theoem is valid fo vectos of any dimension (as long as they have the same dimension of couse). Let s see an example of this. Example 2 Detemine the angle between a = 3, -4,-1 and b = 0,5,2. We will need the dot poduct as well as the magnitudes of each vecto. ab g =- 22 a = 26 b = 29 The angle is then, ab g -22 cosq = = = a b ( ) q = - = -1 cos adians= degees The dot poduct gives us a vey nice method fo detemining if two vectos ae pependicula and it will give anothe method fo detemining when two vectos ae paallel. Note as well that often we will use the tem othogonal in place of pependicula. Now, if two vectos ae othogonal then we know that the angle between them is 90 degees. Fom (2) this tells us that if two vectos ae othogonal then, ab= g 0 Likewise, if two vectos ae paallel then the angle between them is eithe 0 degees (pointing in the same diection) o 180 degees (pointing in the opposite diection). Once again using (2) this would mean that one of the following would have to be tue. ab g = a b = ab g =- a b = ( q 0 ) OR ( q 180 ) Example 3 Detemine if the following vectos ae paallel, othogonal, o neithe. (a) a = 6, -2,- 1, b = 2,5,2 1 1 (b) u = 2 i - j, v =- i + j 2 4 (a) Fist get the dot poduct to see if they ae othogonal. ab= g = 0 The two vectos ae othogonal. (b) Again, let s get the dot poduct fist Paul Dawkins 16

18 1 5 uv=- g 1- =- 4 4 So, they aen t othogonal. Let s get the magnitudes and see if they ae paallel. 5 5 u = 5 v = = 16 4 Now, notice that, 5 5 uv 5 Ê ˆ g =- =- =- u v 4 Á 4 Ë So, the two vectos ae paallel. Thee ae seveal nice applications of the dot poduct as well that we should look at. Pojections The best way to undestand pojections is to see a couple of sketches. So, given two vectos a and b we want to detemine the pojection of b onto a. The pojection is denoted by poj a b. Hee ae a couple of sketches illustating the pojection. So, to get the pojection of b onto a we dop staight down fom the end of b until we hit (and fom a ight angle) with the line that is paallel to a. The pojection is then the vecto that is paallel to a, stats at the same point both of the oiginal vectos stated at and ends whee the dashed line hits the line paallel to a. Thee is an nice fomula fo finding the pojection of b onto a. Hee it is, g poj a b = a ab2 a Note that we also need to be vey caeful with notation hee. The pojection of a onto b is given by 2007 Paul Dawkins 17

19 g poj b a = b ab2 b We can see that this will be a totally diffeent vecto. This vecto is paallel to b, while poj a b is paallel to a. So, be caeful with notation and make sue you ae finding the coect pojection. Hee s an example. Example 4 Detemine the pojection of b = 2,1, -1 onto a = 1,0, -2. We need the dot poduct and the magnitude of a. 2 ab g = 4 a = 5 The pojection is then, poj a ab g b = a 2 a 4 = 1,0, =,0,- 5 5 Fo compaison puposes let s do it the othe way aound as well. Example 5 Detemine the pojection of a = 1,0, -2 ontob = 2,1, -1. We need the dot poduct and the magnitude of b. 2 ab g = 4 b = 6 The pojection is then, poj b a = ab g b 2 b 4 = 2,1, =,, As we can see fom the pevious two examples the two pojections ae diffeent so be caeful Paul Dawkins 18

20 Diection Cosines This application of the dot poduct equies that we be in thee dimensional space unlike all the othe applications we ve looked at to this point. Let s stat with a vecto, a, in thee dimensional space. This vecto will fom angles with the x- axis (a ), the y-axis (b ), and the z-axis (g ). These angles ae called diection angles and the cosines of these angles ae called diection cosines. Hee is a sketch of a vecto and the diection angles. The fomulas fo the diection cosines ae, ai g a agj a ak g a3 a = = b = = g = = a a a a a a 1 2 cos cos cos whee i, j and k ae the standad basis vectos. Let s veify the fist dot poduct above. We ll leave the est to you to veify. ai g = a, a, a g 1,0,0 = a 1 Hee ae a couple of nice facts about the diection cosines. 1. The vecto u = cos a,cos b,cosg is a unit vecto cos a + cos b + cos g = a = a cos a,cos b,cosg Let s do a quick example involving diection cosines Paul Dawkins 19

21 Example 6 Detemine the diection cosines and diection angles fo a = 2,1, -4. We will need the magnitude of the vecto. a = = 21 The diection cosines and angles ae then, cosa = 2 21 a = adians= degees cosb = 1 21 b = adians= degees cosg = g = adians= degees 2007 Paul Dawkins 20

22 Coss Poduct In this final section of this chapte we will look at the coss poduct of two vectos. We should note that the coss poduct equies both of the vectos to be thee dimensional vectos. Also, befoe getting into how to compute these we should point out a majo diffeence between dot poducts and coss poducts. The esult of a dot poduct is a numbe and the esult of a coss poduct is a vecto! Be caeful not to confuse the two. So, let s stat with the two vectos a = a1, a2, a3 given by the fomula, b = b, b, b and a b = ab -ab, ab -ab, ab -ab then the coss poduct is This is not an easy fomula to emembe. Thee ae two ways to deive this fomula. Both of them use the fact that the coss poduct is eally the deteminant of a 3x3 matix. If you don t know what this is that is don t woy about it. You don t need to know anything about matices o deteminants to use eithe of the methods. The notation fo the deteminant is as follows, i j k a b = a a a b b b The fist ow is the standad basis vectos and must appea in the ode given hee. The second ow is the components of a and the thid ow is the components of b. Now, let s take a look at the diffeent methods fo getting the fomula. The fist method uses the Method of Cofactos. If you don t know the method of cofactos that is fine, the esult is all that we need. Hee is the fomula. whee, a a a a a a a b = i - j + k b b b b b b a b ad bc c d = - This fomula is not as difficult to emembe as it might at fist appea to be. Fist, the tems altenate in sign and notice that the 2x2 is missing the column below the standad basis vecto that multiplies it as well as the ow of standad basis vectos. The second method is slightly easie; howeve, many textbooks don t cove this method as it will only wok on 3x3 deteminants. This method says to take the deteminant as listed above and then copy the fist two columns onto the end as shown below Paul Dawkins 21

23 i j k i j a b = a a a a a 1 2 b b b b b 1 2 We now have thee diagonals that move fom left to ight and thee diagonals that move fom ight to left. We multiply along each diagonal and add those that move fom left to ight and subtact those that move fom ight to left. This is best seen in an example. We ll also use this example to illustate a fact about coss poducts. Example 1 If a = 2,1, -1 and b = -3,4,1 (a) a b (b) b a (a) Hee is the computation fo this one. i j k i j a b = compute each of the following = i + j k - j -i - -k - = 5i + j + 11k (b) And hee is the computation fo this one. i j k i j b a = ()() 1 1 ( 1)( 3) ( 2)( 4) ( 2)() 1 ( 1)( 4) ()( 1 3) = i - + j + k - - j - - -i -k =-5i - j -11k ( 4)( 1) ()( 1 2) ( 3)() 1 ( 3)( 1) ()() 1 1 ( 4)( 2) Notice that switching the ode of the vectos in the coss poduct simply changed all the signs in the esult. Note as well that this means that the two coss poducts will point in exactly opposite diections since they only diffe by a sign. We ll fomalize up this fact shotly when we list seveal facts. Thee is also a geometic intepetation of the coss poduct. Fist we will let q be the angle between the two vectos a and b and assume that 0 q p, then we have the following fact, a b = a b sinq (1) and the following figue Paul Dawkins 22

24 Thee should be a natual question at this point. How did we know that the coss poduct pointed in the diection that we ve given it hee? Fist, as this figue, implies the coss poduct is othogonal to both of the oiginal vectos. This will always be the case with one exception that we ll get to in a second. Second, we knew that it pointed in the upwad diection (in this case) by the ight hand ule. This says that if we take ou ight hand, stat at a and otate ou finges towads b ou thumb will point in the diection of the coss poduct. Theefoe, if we d sketched in b a above we would have gotten a vecto in the downwad diection. Example 2 A plane is defined by any thee points that ae in the plane. If a plane contains the 1,0,0 1,1,1 R = 2, - 1,3 find a vecto that is othogonal to the plane. points P = ( ), Q = ( ) and ( ) The one way that we know to get an othogonal vecto is to take a coss poduct. So, if we could find two vectos that we knew wee in the plane and took the coss poduct of these two vectos we know that the coss poduct would be othogonal to both the vectos. Howeve, since both the vectos ae in the plane the coss poduct would then also be othogonal to the plane. So, we need two vectos that ae in the plane. This is whee the points come into the poblem. Since all thee points lie in the plane any vecto between them must also be in the plane. Thee ae many ways to get two vectos between these points. We will use the following two, uuu PQ = 1-1,1-0,1-0 = 0,1,1 uuu PR = 2-1, -1-0,3-0 = 1, -1,3 The coss poduct of these two vectos will be othogonal to the plane. So, let s find the coss poduct Paul Dawkins 23

25 uuu uuu PQ PR = i j k i j = 4i + j -k So, the vecto 4i + j -k will be othogonal to the plane containing the thee points. Now, let s addess the one time whee the coss poduct will not be othogonal to the oiginal vectos. If the two vectos, a and b, ae paallel then the angle between them is eithe 0 o 180 degees. Fom (1) this implies that, a b = 0 Fom a fact about the magnitude we saw in the fist section we know that this implies a b = 0 In othe wods, it won t be othogonal to the oiginal vectos since we have the zeo vecto. This does give us anothe test fo paallel vectos howeve. Fact If a b = 0 then a and b will be paallel vectos. Let s also fomalize up the fact about the coss poduct being othogonal to the oiginal vectos. Fact Povided a b 0 then a b is othogonal to both a and b. Hee ae some nice popeties about the coss poduct. Popeties If u, v and w ae vectos and c is a numbe then, u v =- v u cu v = u cv = cu v u v+ w = u v+ u w ug v w = u v gw ( ) ( ) ( ) ( ) ( ) ( ) u u u ug v w = v v v ( ) w w w The deteminant in the last fact is computed in the same way that the coss poduct is computed. We will see an example of this computation shotly. Thee ae a couple of geometic applications to the coss poduct as well. Suppose we have thee vectos a, b and c and we fom the thee dimensional figue shown below Paul Dawkins 24

26 The aea of the paallelogam (two dimensional font of this object) is given by, Aea = a b and the volume of the paallelepiped (the whole thee dimensional object) is given by, Volume = ag b c ( ) Note that the absolute value bas ae equied since the quantity could be negative and volume isn t negative. We can use this volume fact to detemine if thee vectos lie in the same plane o not. If thee vectos lie in the same plane then the volume of the paallelepiped will be zeo. Example 3 Detemine if the thee vectos a = 1,4, -7 in the same plane o not., b = 2, -1,4 and c = 0, -9,18 lie So, as we noted pio to this example all we need to do is compute the volume of the paallelepiped fomed by these thee vectos. If the volume is zeo they lie in the same plane and if the volume isn t zeo they don t lie in the same plane Paul Dawkins 25

27 ag b = ( c) ()( 1 1)( 18) ( 4)( 4)( 0) ( 7)( 2)( 9) ( 4)( 2)( 18) -()( 1 4)( -9)-(-7)(-1)( 0) = = = 0 So, the volume is zeo and so they lie in the same plane Paul Dawkins 26

28 2007 Paul Dawkins 27