Euclidean Vector Spaces

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1 CHAPTER 3 Eclidean Vector Spaces CHAPTER CONTENTS 3.1 Vectors in 2-Space, 3-Space, and n-space Norm, Dot Prodct, and Distance in R n Orthogonalit The Geometr of Linear Sstems Cross Prodct 172 INTRODUCTION Engineers and phsicists distingish between two tpes of phsical qantities scalars, which are qantities that can be described b a nmerical ale alone, and ectors, which are qantities that reqire both a nmber and a direction for their complete phsical description. For eample, temperatre, length, and speed are scalars becase the can be fll described b a nmber that tells how mch a temperatre of 20 C, a length of 5 cm, or a speed of 75 km/h. In contrast, elocit and force are ectors becase the reqire a nmber that tells how mch and a direction that tells which wa sa, a boat moing at 10 knots in a direction 45 northeast, or a force of 100 lb acting erticall. Althogh the notions of ectors and scalars that we will std in this tet hae their origins in phsics and engineering, we will be more concerned with sing them to bild mathematical strctres and then appling those strctres to sch dierse fields as genetics, compter science, economics, telecommnications, and enironmental science. 3.1 Vectors in 2-Space, 3-Space, and n-space Linear algebra is primaril concerned with two tpes of mathematical objects, matrices and ectors. In Chapter 1 we discssed the basic properties of matrices, we introdced the idea of iewing n-tples of real nmbers as ectors, and we denoted the set of all sch n-tples as R n. In this section we will reiew the basic properties of ectors in two and three dimensions with the goal of etending these properties to ectors in R n. Geometric Vectors Terminal point Initial point Figre Engineers and phsicists represent ectors in two dimensions (also called 2-space) or in three dimensions (also called 3-space) b arrows. The direction of the arrowhead specifies the direction of the ector and the length of the arrow specifies the magnitde. Mathematicians call these geometric ectors. The tail of the arrow is called the initial point of the ector and the tip the terminal point (Figre 3.1.1). In this tet we will denote ectors in boldface tpe sch as a, b,, w, and, and we will denote scalars in lowercase italic tpe sch as a, k,, w, and. When we want to indicate that a ector has initial point A and terminal point B, then, as shown in Figre 3.1.2, we will write = AB 131

2 132 Chapter 3 Eclidean Vector Spaces B A = AB Figre Vector Addition Vectors with the same length and direction, sch as those in Figre 3.1.3, are said to be eqialent. Since we want a ector to be determined solel b its length and direction, eqialent ectors are regarded as the same ector een thogh the ma be in different positions. Eqialent ectors are also said to be eqal, which we indicate b writing = w The ector whose initial and terminal points coincide has length zero, so we call this the zero ector and denote it b 0. The zero ector has no natral direction, so we will agree that it can be assigned an direction that is conenient for the problem at hand. There are a nmber of important algebraic operations on ectors, all of which hae their origin in laws of phsics. Eqialent ectors Figre Parallelogram Rle for Vector Addition If and w are ectors in 2-space or 3-space that are positioned so their initial points coincide, then the two ectors form adjacent sides of a parallelogram, and the sm + w is the ector represented b the arrow from the common initial point of and w to the opposite erte of the parallelogram (Figre 3.1.4a). Here is another wa to form the sm of two ectors. Triangle Rle for Vector Addition If and w are ectors in 2-space or 3-space that are positioned so the initial point of w is at the terminal point of, then the sm + w is represented b the arrow from the initial point of to the terminal point of w (Figre 3.1.4b). In Figre 3.1.4c we hae constrcted the sms + w and w + b the triangle rle. This constrction makes it eident that + w = w + (1) and that the sm obtained b the triangle rle is the same as the sm obtained b the parallelogram rle. w w + w + w + w w + Figre w (a) (b) w (c) Vector addition can also be iewed as a process of translating points. Vector AdditionViewed astranslation If, w, and + w are positioned so their initial points coincide, then the terminal point of + w can be iewed in two was: 1. The terminal point of + w is the point that reslts when the terminal point of is translated in the direction of w b a distance eqal to the length of w (Figre 3.1.5a). 2. The terminal point of + w is the point that reslts when the terminal point of w is translated in the direction of b a distance eqal to the length of (Figre 3.1.5b). Accordingl, we sa that + w is the translation of b w or, alternatiel, the translation of w b.

3 3.1 Vectors in 2-Space, 3-Space, and n-space w + w Figre w (a) w (b) Vector Sbtraction In ordinar arithmetic we can write a b = a + ( b), which epresses sbtraction in terms of addition. There is an analogos idea in ector arithmetic. Vector Sbtraction The negatie of a ector, denoted b, is the ector that has the same length as bt is oppositel directed (Figre 3.1.6a), and the difference of from w, denoted b w, is taken to be the sm w = w + ( ) (2) The difference of from w can be obtained geometricall b the parallelogram method shown in Figre 3.1.6b, or more directl b positioning w and so their initial points coincide and drawing the ector from the terminal point of to the terminal point of w (Figre 3.1.6c). w w w w Figre (a) (b) (c) Scalar Mltiplication Sometimes there is a need to change the length of a ector or change its length and reerse its direction. This is accomplished b a tpe of mltiplication in which ectors are mltiplied b scalars. As an eample, the prodct 2 denotes the ector that has the same direction as bt twice the length, and the prodct 2 denotes the ector that is oppositel directed to and has twice the length. Here is the general reslt. 1 2 ( 1) Scalar Mltiplication If is a nonzero ector in 2-space or 3-space, and if k is a nonzero scalar, then we define the scalar prodct of b k to be the ector whose length is k times the length of and whose direction is the same as that of if k is positie and opposite to that of if k is negatie. If k = 0or = 0, then we define k to be 0. 2 Figre ( 3) Figre shows the geometric relationship between a ector and some of its scalar mltiples. In particlar, obsere that ( 1) has the same length as bt is oppositel directed; therefore, ( 1) = (3) Parallel and Collinear Vectors Sppose that and w are ectors in 2-space or 3-space with a common initial point. If one of the ectors is a scalar mltiple of the other, then the ectors lie on a common line, so it is reasonable to sa that the are collinear (Figre 3.1.8a). Howeer, if we translate one of the ectors, as indicated in Figre 3.1.8b, then the ectors are parallel bt no longer collinear. This creates a lingistic problem becase translating a ector does not change it. The onl wa to resole this problem is to agree that the terms parallel and

4 134 Chapter 3 Eclidean Vector Spaces collinear mean the same thing when applied to ectors. Althogh the ector 0 has no clearl defined direction, we will regard it as parallel to all ectors when conenient. k k Figre (a) (b) Sms ofthree or More Vectors Vector addition satisfies the associatie law for addition, meaning that when we add three ectors, sa,, and w, it does not matter which two we add first; that is, + ( + w) = ( + ) + w It follows from this that there is no ambigit in the epression + + w becase the same reslt is obtained no matter how the ectors are groped. A simple wa to constrct + + w is to place the ectors tip to tail in sccession and then draw the ector from the initial point of to the terminal point of w (Figre 3.1.9a). The tip-to-tail method also works for for or more ectors (Figre 3.1.9b). The tip-to-tail method makes it eident that if,, and w are ectors in 3-space with a common initial point, then + + w is the diagonal of the parallelepiped that has the three ectors as adjacent sides (Figre 3.1.9c). Figre ( + w) ( + ) + w + + w w + + w + (a) (b) (c) w w + + w Vectors in Coordinate Sstems The component forms of the zero ector are 0 = (0, 0) in 2-space and 0 = (0, 0, 0) in 3- space. Up ntil now we hae discssed ectors withot reference to a coordinate sstem. Howeer, as we will soon see, comptations with ectors are mch simpler to perform if a coordinate sstem is present to work with. If a ector in 2-space or 3-space is positioned with its initial point at the origin of a rectanglar coordinate sstem, then the ector is completel determined b the coordinates of its terminal point (Figre ). We call these coordinates the components of relatie to the coordinate sstem. We will write = ( 1, 2 ) to denote a ector in 2-space with components ( 1, 2 ), and = ( 1, 2, 3 ) to denote a ector in 3-space with components ( 1, 2, 3 ). z ( 1, 2 ) ( 1, 2, 3 ) Figre

5 ( 1, 2 ) 3.1 Vectors in 2-Space, 3-Space, and n-space 135 It shold be eident geometricall that two ectors in 2-space or 3-space are eqialent if and onl if the hae the same terminal point when their initial points are at the origin. Algebraicall, this means that two ectors are eqialent if and onl if their corresponding components are eqal. Ths, for eample, the ectors in 3-space are eqialent if and onl if = ( 1, 2, 3 ) and w = (w 1,w 2,w 3 ) Figre The ordered pair ( 1, 2 ) can represent a point or a ector. Vectors Whose Initial Point Is Not at the Origin P 1 ( 1, 1 ) OP 1 = P 1 P 2 = OP 2 OP 1 Figre OP 2 P 2 ( 2, 2 ) 1 = w 1, 2 = w 2, 3 = w 3 Remark It ma hae occrred to o that an ordered pair ( 1, 2 ) can represent either a ector with components 1 and 2 or a point with coordinates 1 and 2 (and similarl for ordered triples). Both are alid geometric interpretations, so the appropriate choice will depend on the geometric iewpoint that we want to emphasize (Figre ). It is sometimes necessar to consider ectors whose initial points are not at the origin. If P 1 P 2 denotes the ector with initial point P 1 ( 1, 1 ) and terminal point P 2 ( 2, 2 ), then the components of this ector are gien b the formla P 1 P 2 = ( 2 1, 2 1 ) (4) That is, the components of P 1 P 2 are obtained b sbtracting the coordinates of the initial point from the coordinates of the terminal point. For eample, in Figre the ector P 1 P 2 is the difference of ectors OP 2 and OP 1,so P 1 P 2 = OP 2 OP 1 = ( 2, 2 ) ( 1, 1 ) = ( 2 1, 2 1 ) As o might epect, the components of a ector in 3-space that has initial point P 1 ( 1, 1,z 1 ) and terminal point P 2 ( 2, 2,z 2 ) aregienb P 1 P 2 = ( 2 1, 2 1,z 2 z 1 ) (5) EXAMPLE 1 Finding the Components of a Vector The components of the ector = P 1 P 2 with initial point P 1 (2, 1, 4) and terminal point P 2 (7, 5, 8) are = (7 2, 5 ( 1), ( 8) 4) = (5, 6, 12) n-space The idea of sing ordered pairs and triples of real nmbers to represent points in twodimensional space and three-dimensional space was well known in the eighteenth and nineteenth centries. B the dawn of the twentieth centr, mathematicians and phsicists were eploring the se of higher dimensional spaces in mathematics and phsics. Toda, een the laman is familiar with the notion of time as a forth dimension, an idea sed b Albert Einstein in deeloping the general theor of relatiit. Toda, phsicists working in the field of string theor commonl se 11-dimensional space in their qest for a nified theor that will eplain how the fndamental forces of natre work. Mch of the remaining work in this section is concerned with etending the notion of space to n dimensions. To eplore these ideas frther, we start with some terminolog and notation. The set of all real nmbers can be iewed geometricall as a line. It is called the real line and is denoted b R or R 1. The sperscript reinforces the intitie idea that a line is onedimensional. The set of all ordered pairs of real nmbers (called 2-tples) and the set of all ordered triples of real nmbers (called 3-tples) are denoted b R 2 and R 3, respectiel.

6 136 Chapter 3 Eclidean Vector Spaces The sperscript reinforces the idea that the ordered pairs correspond to points in the plane (two-dimensional) and ordered triples to points in space (three-dimensional). The following definition etends this idea. DEFINITION 1 If n is a positie integer, then an ordered n-tple is a seqence of n real nmbers ( 1, 2,..., n ). The set of all ordered n-tples is called n-space and is denoted b R n. Remark Yo can think of the nmbers in an n-tple ( 1, 2,..., n ) as either the coordinates of a generalized point or the components of a generalized ector, depending on the geometric image o want to bring to mind the choice makes no difference mathematicall, since it is the algebraic properties of n-tples that are of concern. Here are some tpical applications that lead to n-tples. Eperimental Data A scientist performs an eperiment and makes n nmerical measrements each time the eperiment is performed. The reslt of each eperiment can be regarded as a ector = ( 1, 2,..., n ) in R n in which 1, 2,..., n are the measred ales. Storage and Warehosing A national trcking compan has 15 depots for storing and sericing its trcks. At each point in time the distribtion of trcks in the serice depots can be described b a 15-tple = ( 1, 2,..., 15 ) in which 1 is the nmber of trcks in the first depot, 2 is the nmber in the second depot, and so forth. Electrical Circits A certain kind of processing chip is designed to receie for inpt oltages and prodce three otpt oltages in response. The inpt oltages can be regarded as ectors in R 4 and the otpt oltages as ectors in R 3. Ths, the chip can be iewed as a deice that transforms an inpt ector = ( 1, 2, 3, 4 ) in R 4 into an otpt ector w = (w 1,w 2,w 3 ) in R 3. Graphical Images One wa in which color images are created on compter screens is b assigning each piel (an addressable point on the screen) three nmbers that describe the he, satration, and brightness of the piel. Ths, a complete color image can be iewed as a set of 5-tples of the form = (,,h,s,b)in which and are the screen coordinates of a piel and h, s, and b are its he, satration, and brightness. Economics One approach to economic analsis is to diide an econom into sectors (manfactring, serices, tilities, and so forth) and measre the otpt of each sector b a dollar ale. Ths, in an econom with 10 sectors the economic otpt of the entire econom can be represented b a 10-tple s = (s 1,s 2,...,s 10 ) in which the nmbers s 1,s 2,...,s 10 are the otpts of the indiidal sectors. Albert Einstein ( ) Historical Note The German-born phsicist Albert Einstein immigrated to the United States in 1935, where he settled at Princeton Uniersit. Einstein spent the last three decades of his life working nsccessfll at prodcing a nified field theor that wold establish an nderling link between the forces of grait and electromagnetism. Recentl, phsicists hae made progress on the problem sing a framework known as string theor. In this theor the smallest, indiisible components of the Unierse are not particles bt loops that behae like ibrating strings. Whereas Einstein s space-time nierse was for-dimensional, strings reside in an 11-dimensional world that is the focs of crrent research. [Image: Bettmann/CORBIS]

7 3.1 Vectors in 2-Space, 3-Space, and n-space 137 Mechanical Sstems Sppose that si particles moe along the same coordinate line so that at time t their coordinates are 1, 2,..., 6 and their elocities are 1, 2,..., 6, respectiel. This information can be represented b the ector = ( 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6,t) in R 13. This ector is called the state of the particle sstem at time t. Operations on Vectors in R n Or net goal is to define sefl operations on ectors in R n. These operations will all be natral etensions of the familiar operations on ectors in R 2 and R 3. We will denote a ector in R n sing the notation = ( 1, 2,..., n ) and we will call 0 = (0, 0,...,0) the zero ector. We noted earlier that in R 2 and R 3 two ectors are eqialent (eqal) if and onl if their corresponding components are the same. Ths, we make the following definition. DEFINITION 2 Vectors = ( 1, 2,..., n ) and w = (w 1,w 2,...,w n ) in R n are said to be eqialent (also called eqal) if 1 = w 1, 2 = w 2,..., n = w n We indicate this b writing = w. EXAMPLE 2 Eqalit of Vectors (a,b,c,d)= (1, 4, 2, 7) if and onl if a = 1,b = 4,c = 2, and d = 7. Or net objectie is to define the operations of addition, sbtraction, and scalar mltiplication for ectors in R n. To motiate these ideas, we will consider how these operations can be performed on ectors in R 2 sing components. B stding Figre o shold be able to dedce that if = ( 1, 2 ) and w = (w 1,w 2 ), then In particlar, it follows from (7) that + w = ( 1 + w 1, 2 + w 2 ) (6) k = (k 1,k 2 ) (7) = ( 1) = ( 1, 2 ) (8) ( 1 + w 1, 2 + w 2 ) 2 (w 1, w 2 ) w 2 w + w k (k 1, k 2 ) ( 1, 2 ) k 2 2 ( 1, 2 ) 1 k 1 Figre w 1

8 138 Chapter 3 Eclidean Vector Spaces and hence that w = w + ( ) = (w 1 1,w 2 2 ) (9) Motiated b Formlas (6) (9), we make the following definition. DEFINITION 3 If = ( 1, 2,..., n ) and w = (w 1,w 2,...,w n ) are ectors in R n, and if k is an scalar, then we define + w = ( 1 + w 1, 2 + w 2,..., n + w n ) (10) k = (k 1,k 2,...,k n ) (11) = ( 1, 2,..., n ) (12) w = w + ( ) = (w 1 1,w 2 2,...,w n n ) (13) In words, ectors are added (or sbtracted) b adding (or sbtracting) their corresponding components, and a ector is mltiplied b a scalar b mltipling each component b that scalar. EXAMPLE 3 Algebraic Operations Using Components If = (1, 3, 2) and w = (4, 2, 1), then + w = (5, 1, 3), 2 = (2, 6, 4) w = ( 4, 2, 1), w = + ( w) = ( 3, 5, 1) The following theorem smmarizes the most important properties of ector operations. THEOREM If,, and w are ectors in R n, and if k and m are scalars, then: (a) + = + (b) ( + ) + w = + ( + w) (c) + 0 = 0 + = (d) + ( ) = 0 (e) k( + ) = k + k ( f) (k + m) = k + m (g) k(m) = (km) (h) 1 = We will proe part (b) and leae some of the other proofs as eercises. Proof (b) Let = ( 1, 2,..., n ), = ( 1, 2,..., n ), and w = (w 1,w 2,...,w n ). Then ( + ) + w = ( ( 1, 2,..., n ) + ( 1, 2,..., n ) ) + (w 1,w 2,...,w n ) = ( 1 + 1, 2 + 2,..., n + n ) + (w 1,w 2,...,w n ) [Vector addition] = ( ) ( ) + w 1,( ) + w 2,...,( n + n ) + w n [Vector addition] = ( 1 + ( 1 + w 1 ), 2 + ( 2 + w 2 ),..., n + ( n + w n ) ) [Regrop] = ( 1, 2,..., n ) + ( 1 + w 1, 2 + w 2,..., n + w n ) [Vector addition] = + ( + w) The following additional properties of ectors in R n can be dedced easil b epressing the ectors in terms of components (erif).

9 3.1 Vectors in 2-Space, 3-Space, and n-space 139 THEOREM If is a ector in R n and k is a scalar, then: (a) 0 = 0 (b) k0 = 0 (c) ( 1) = Calclating Withot Components Linear Combinations One of the powerfl conseqences of Theorems and is that the allow calclations to be performed withot epressing the ectors in terms of components. For eample, sppose that, a, and b are ectors in R n, and we want to sole the ector eqation + a = b for the ector withot sing components. We cold proceed as follows: + a = b [ Gien ] ( + a) + ( a) = b + ( a) [ Add the negatie of a to both sides ] + (a + ( a)) = b a [ Part (b) of Theorem ] + 0 = b a [ Part (d) of Theorem ] = b a [ Part (c) of Theorem ] While this method is obiosl more cmbersome than compting with components in R n, it will become important later in the tet where we will enconter more general kinds of ectors. Addition, sbtraction, and scalar mltiplication are freqentl sed in combination to form new ectors. For eample, if 1, 2, and 3 are ectors in R n, then the ectors = and w = are formed in this wa. In general, we make the following definition. Note that this definition of a linear combination is consistent with that gien in the contet of matrices (see Definition 6 in Section 1.3). DEFINITION 4 If w is a ector in R n, then w is said to be a linear combination of the ectors 1, 2,..., r in R n if it can be epressed in the form w = k k k r r (14) where k 1,k 2,...,k r are scalars. These scalars are called the coefficients of the linear combination. In the case where r = 1, Formla (14) becomes w = k 1 1, so that a linear combination of a single ector is jst a scalar mltiple of that ector. Alternatie Notations for Vectors Up to now we hae been writing ectors in R n sing the notation = ( 1, 2,..., n ) (15) We call this the comma-delimited form. Howeer, since a ector in R n is jst a list of its n components in a specific order, an notation that displas those components in the correct order is a alid wa of representing the ector. For eample, the ector in (15) can be written as =[ 1 2 n ] (16) which is called row-ector form,oras 1 2 =. (17) n

10 140 Chapter 3 Eclidean Vector Spaces which is called colmn-ector form. The choice of notation is often a matter of taste or conenience, bt sometimes the natre of a problem will sggest a preferred notation. Notations (15), (16), and (17) will all be sed at arios places in this tet. Application of Linear Combinations to Color Models Colors on compter monitors are commonl based on what is called the RGB color model. Colors in this sstem are created b adding together percentages of the primar colors red (R), green (G), and ble (B). One wa to do this is to identif the primar colors with the ectors r = (1, 0, 0) (pre red), g = (0, 1, 0) (pre green), b = (0, 0, 1) (pre ble) in R 3 and to create all other colors b forming linear combinations of r, g, and b sing coefficients between 0 and 1, inclsie; these coefficients represent the percentage of each pre color in the mi. The set of all sch color ectors is called RGB space or the RGB color cbe (Figre ). Ths, each color ector c in this cbe is epressible as a linear combination of the form c = k 1 r + k 2 g + k 3 b = k 1 (1, 0, 0) + k 2 (0, 1, 0) + k 3 (0, 0, 1) = (k 1,k 2,k 3 ) where 0 k i 1. As indicated in the figre, the corners of the cbe represent the pre primar colors together with the colors black, white, magenta, can, and ellow. The ectors along the diagonal rnning from black to white correspond to shades of gra. Figre Magenta (1, 0, 1) Black (0, 0, 0) Ble (0, 0, 1) Red (1, 0, 0) Yellow (1, 1, 0) Can (0, 1, 1) White (1, 1, 1) Green (0, 1, 0) Eercise Set 3.1 In Eercises 1 2, find the components of the ector. 1. (a) (b) z (1, 5) (0, 0, 4) (4, 1) 2. (a) (b) z (2, 3, 0) (0, 4, 4) ( 3, 3) (2, 3) (3, 0, 4) In Eercises 3 4, find the components of the ector P 1 P (a) P 1 (3, 5), P 2 (2, 8) (b) P 1 (5, 2, 1), P 2 (2, 4, 2) 4. (a) P 1 ( 6, 2), P 2 ( 4, 1) (b) P 1 (0, 0, 0), P 2 ( 1, 6, 1) 5. (a) Find the terminal point of the ector that is eqialent to = (1, 2) and whose initial point is A(1, 1). (b) Find the initial point of the ector that is eqialent to = (1, 1, 3) and whose terminal point is B( 1, 1, 2). 6. (a) Find the initial point of the ector that is eqialent to = (1, 2) and whose terminal point is B(2, 0). (b) Find the terminal point of the ector that is eqialent to = (1, 1, 3) and whose initial point is A(0, 2, 0). 7. Find an initial point P of a nonzero ector = PQ with terminal point Q(3, 0, 5) and sch that (a) has the same direction as = (4, 2, 1). (b) is oppositel directed to = (4, 2, 1).

11 3.1 Vectors in 2-Space, 3-Space, and n-space Find a terminal point Q of a nonzero ector = PQ with initial point P( 1, 3, 5) and sch that (a) has the same direction as = (6, 7, 3). (b) is oppositel directed to = (6, 7, 3). 9. Let = (4, 1), = (0, 5), and w = ( 3, 3). Find the components of (a) + w (b) 3 (c) 2( 5w) (d) 3 2( + 2w) 10. Let = ( 3, 1, 2), = (4, 0, 8), and w = (6, 1, 4). Find the components of (a) w (b) (c) 3( 8w) (d) (2 7w) (8 + ) 11. Let = ( 3, 2, 1, 0), = (4, 7, 3, 2), and w = (5, 2, 8, 1). Find the components of (a) w (b) + ( 4w) (c) 6( 3) (d) (6 w) (4 + ) 12. Let = (1, 2, 3, 5, 0), = (0, 4, 1, 1, 2), and w = (7, 1, 4, 2, 3). Find the components of (a) + w (b) 3(2 ) (c) (3 ) (2 + 4w) (d) 1 (w 5 + 2) Let,, and w be the ectors in Eercise 11. Find the components of the ector that satisfies the eqation 3 + 2w = 3 + 2w. 14. Let,, and w be the ectors in Eercise 12. Find the components of the ector that satisfies the eqation 2 + = 7 + w. 15. Which of the following ectors in R 6, if an, are parallel to = ( 2, 1, 0, 3, 5, 1)? (a) (4, 2, 0, 6, 10, 2) (b) (4, 2, 0, 6, 10, 2) (c) (0, 0, 0, 0, 0, 0) 16. For what ale(s) of t, if an, is the gien ector parallel to = (4, 1)? (a) (8t, 2) (b) (8t,2t) (c) (1,t 2 ) 17. Let = (1, 1, 3, 5) and = (2, 1, 0, 3). Find scalars a and b so that a + b = (1, 4, 9, 18). 18. Let = (2, 1, 0, 1, 1) and = ( 2, 3, 1, 0, 2). Find scalars a and b so that a + b = ( 8, 8, 3, 1, 7). 22. Show that there do not eist scalars c 1, c 2, and c 3 sch that c 1 (1, 0, 1, 0) + c 2 (1, 0, 2, 1) + c 3 (2, 0, 1, 2) = (1, 2, 2, 3) 23. Let P be the point (2, 3, 2) and Q the point (7, 4, 1). (a) Find the midpoint of the line segment connecting the points P and Q. (b) Find the point on the line segment connecting the points P and Q that is 3 of the wa from P to Q In relation to the points P 1 and P 2 in Figre , what can o sa abot the terminal point of the following ector if its initial point is at the origin? = OP1 + 1 ( OP2 OP1 ) In each part, find the components of the ector + + w. (a) w 26. Referring to the ectors pictred in Eercise 25, find the components of the ector + w. 27. Let P be the point (1, 3, 7). If the point (4, 0, 6) is the midpoint of the line segment connecting P and Q,whatisQ? 28. If the sm of three ectors in R 3 is zero, mst the lie in the same plane? Eplain. 29. Consider the reglar heagon shown in the accompaning figre. (a) What is the sm of the si radial ectors that rn from the center to the ertices? (b) How is the sm affected if each radial ector is mltiplied b 1? 2 (c) What is the sm of the fie radial ectors that remain if a is remoed? (d) Discss some ariations and generalizations of the reslt in part (c). f a b (b) w In Eercises 19 20, find scalars c 1, c 2, and c 3 for which the eqation is satisfied. 19. c 1 (1, 1, 0) + c 2 (3, 2, 1) + c 3 (0, 1, 4) = ( 1, 1, 19) 20. c 1 ( 1, 0, 2) + c 2 (2, 2, 2) + c 3 (1, 2, 1) = ( 6, 12, 4) e d c Figre E Show that there do not eist scalars c 1, c 2, and c 3 sch that c 1 ( 2, 9, 6) + c 2 ( 3, 2, 1) + c 3 (1, 7, 5) = (0, 5, 4) 30. What is the sm of all radial ectors of a reglar n-sided polgon? (See Eercise 29.)

12 142 Chapter 3 Eclidean Vector Spaces Working with Proofs 31. Proe parts (a), (c), and (d) of Theorem Proe parts (e) (h) of Theorem Proe parts (a) (c) of Theorem Tre-False Eercises TF. In parts (a) (k) determine whether the statement is tre or false, and jstif or answer. (a) Two eqialent ectors mst hae the same initial point. (b) The ectors (a, b) and (a, b, 0) are eqialent. (c) If k is a scalar and is a ector, then and k are parallel if and onl if k 0. (d) The ectors + ( + w) and (w + ) + are the same. (e) If + = + w, then = w. (f ) If a and b are scalars sch that a + b = 0, then and are parallel ectors. (g) Collinear ectors with the same length are eqal. (h) If (a,b,c)+ (,,z)= (,,z), then (a,b,c) mst be the zero ector. (i) If k and m are scalars and and are ectors, then (k + m)( + ) = k + m ( j) If the ectors and w are gien, then the ector eqation can be soled for. 3(2 ) = 5 4w + (k) The linear combinations a a 2 2 and b b 2 2 can onl be eqal if a 1 = b 1 and a 2 = b Norm, Dot Prodct, and Distance in R n In this section we will be concerned with the notions of length and distance as the relate to ectors. We will first discss these ideas in R 2 and R 3 and then etend them algebraicall to R n. Norm of a Vector ( 1, 2 ) 2 1 (a) z In this tet we will denote the length of a ector b the smbol, which is read as the norm of, the length of, or the magnitde of (the term norm being a common mathematical snonm for length). As sggested in Figre 3.2.1a, it follows from the Theorem of Pthagoras that the norm of a ector ( 1, 2 ) in R 2 is = (1) Similarl, for a ector ( 1, 2, 3 ) in R 3, it follows from Figre 3.2.1b and two applications of the Theorem of Pthagoras that 2 = (OR) 2 + (RP ) 2 = (OQ) 2 + (QR) 2 + (RP ) 2 = O Q (b) Figre P( 1, 2, 3 ) S R and hence that = (2) Motiated b the pattern of Formlas (1) and (2), we make the following definition. DEFINITION 1 If = ( 1, 2,..., n ) is a ector in R n, then the norm of (also called the length of or the magnitde of ) is denoted b, and is defined b the formla = n (3)

13 3.2 Norm, Dot Prodct, and Distance in R n 143 EXAMPLE 1 Calclating Norms It follows from Formla (2) that the norm of the ector = ( 3, 2, 1) in R 3 is = ( 3) = 14 and it follows from Formla (3) that the norm of the ector = (2, 1, 3, 5) in R 4 is = ( 1) ( 5) 2 = 39 Or first theorem in this section will generalize to R n the following three familiar facts abot ectors in R 2 and R 3 : Distances are nonnegatie. The zero ector is the onl ector of length zero. Mltipling a ector b a scalar mltiplies its length b the absolte ale of that scalar. It is important to recognize that jst becase these reslts hold in R 2 and R 3 does not garantee that the hold in R n their alidit in R n mst be proed sing algebraic properties of n-tples. THEOREM If is a ector in R n, and if k is an scalar, then: (a) 0 (b) =0 if and onl if = 0 (c) k = k We will proe part (c) and leae (a) and (b) as eercises. Proof (c) If = ( 1, 2,..., n ), then k = (k 1,k 2,...,k n ),so k = (k 1 ) 2 + (k 2 ) 2 + +(k n ) 2 = (k 2 )( n ) = k n = k WARNING Sometimes o will see Formla (4) epressed as = This is jst a more compact wa of writing that formla and is not intended to cone that is being diided b. Unit Vectors A ector of norm 1 is called a nit ector. Sch ectors are sefl for specifing a direction when length is not releant to the problem at hand. Yo can obtain a nit ector in a desired direction b choosing an nonzero ector in that direction and mltipling b the reciprocal of its length. For eample, if is a ector of length 2 in R 2 or R 3, then 1 is a nit ector in the same direction as. More generall, if is an nonzero 2 ector in R n, then = 1 (4) defines a nit ector that is in the same direction as. We can confirm that (4) is a nit ector b appling part (c) of Theorem with k = 1/ to obtain = k = k =k = 1 =1

14 144 Chapter 3 Eclidean Vector Spaces The process of mltipling a nonzero ector b the reciprocal of its length to obtain a nit ector is called normalizing. EXAMPLE 2 Normalizing a Vector Find the nit ector that has the same direction as = (2, 2, 1). Soltion The ector has length = ( 1) 2 = 3 Ths, from (4) = 1 (2, 2, 1) = ( 2, 2, ) As a check, o ma want to confirm that =1. The Standard Unit Vectors (0, 1) j i (1, 0) (a) z (0, 0, 1) k j i (0, 1, 0) (1, 0, 0) (b) Figre When a rectanglar coordinate sstem is introdced in R 2 or R 3, the nit ectors in the positie directions of the coordinate aes are called the standard nit ectors. InR 2 these ectors are denoted b i = (1, 0) and j = (0, 1) and in R 3 b i = (1, 0, 0), j = (0, 1, 0), and k = (0, 0, 1) (Figre 3.2.2). Eer ector = ( 1, 2 ) in R 2 and eer ector = ( 1, 2, 3 ) in R 3 can be epressed as a linear combination of standard nit ectors b writing = ( 1, 2 ) = 1 (1, 0) + 2 (0, 1) = 1 i + 2 j (5) = ( 1, 2, 3 ) = 1 (1, 0, 0) + 2 (0, 1, 0) + 3 (0, 0, 1) = 1 i + 2 j + 3 k (6) Moreoer, we can generalize these formlas to R n b defining the standard nit ectors in R n to be e 1 = (1, 0, 0,...,0), e 2 = (0, 1, 0,...,0),..., e n = (0, 0, 0,...,1) (7) in which case eer ector = ( 1, 2,..., n ) in R n can be epressed as = ( 1, 2,..., n ) = 1 e e n e n (8) EXAMPLE 3 Linear Combinations of Standard Unit Vectors (2, 3, 4) = 2i 3j + 4k (7, 3, 4, 5) = 7e 1 + 3e 2 4e 3 + 5e 4 Distance in R n If P 1 and P 2 are points in R 2 or R 3, then the length of the ector P 1 P 2 is eqal to the distance d between the two points (Figre 3.2.3). Specificall, if P 1 ( 1, 1 ) and P 2 ( 2, 2 ) are points in R 2, then Formla (4) of Section 3.1 implies that d = P 1 P 2 = ( 2 1 ) 2 + ( 2 1 ) 2 (9)

15 3.2 Norm, Dot Prodct, and Distance in R n 145 d P 2 This is the familiar distance formla from analtic geometr. Similarl, the distance between the points P 1 ( 1, 1,z 1 ) and P 2 ( 2, 2,z 2 ) in 3-space is d(, ) = P 1 P 2 = ( 2 1 ) 2 + ( 2 1 ) 2 + (z 2 z 1 ) 2 (10) P 1 d = P 1 P 2 Motiated b Formlas (9) and (10), we make the following definition. Figre We noted in the preios section that n-tples can be iewed either as ectors or points in R n. In Definition 2 we chose to describe them as points, as that seemed the more natral interpretation. DEFINITION 2 If = ( 1, 2,..., n ) and = ( 1, 2,..., n ) are points in R n, then we denote the distance between and b d(, ) and define it to be d(, ) = = ( 1 1 ) 2 + ( 2 2 ) 2 + +( n n ) 2 (11) EXAMPLE 4 Calclating Distance in R n If = (1, 3, 2, 7) and = (0, 7, 2, 2) then the distance between and is d(, ) = (1 0) 2 + (3 7) 2 + ( 2 2) 2 + (7 2) 2 = 58 Dot Prodct Or net objectie is to define a sefl mltiplication operation on ectors in R 2 and R 3 and then etend that operation to R n. To do this we will first need to define eactl what we mean b the angle between two ectors in R 2 or R 3. For this prpose, let and be nonzero ectors in R 2 or R 3 that hae been positioned so that their initial points coincide. We define the angle between and to be the angle θ determined b and that satisfies the ineqalities 0 θ π (Figre 3.2.4). θ θ θ θ Figre The angle θ between and satisfies 0 θ π. DEFINITION 3 If and are nonzero ectors in R 2 or R 3, and if θ is the angle between and, then the dot prodct (also called the Eclidean inner prodct) of and is denoted b and is defined as = cos θ (12) If = 0 or = 0, then we define to be 0. The sign of the dot prodct reeals information abot the angle θ that we can obtain b rewriting Formla (12) as cos θ = (13) Since 0 θ π, it follows from Formla (13) and properties of the cosine fnction stdied in trigonometr that θ is acte if > 0. θ is obtse if < 0. θ = π/2if = 0.

16 146 Chapter 3 Eclidean Vector Spaces z (0, 2, 2) (0, 0, 1) θ = 45 Figre EXAMPLE 5 Dot Prodct Find the dot prodct of the ectors shown in Figre Soltion The lengths of the ectors are =1 and = 8 = 2 2 and the cosine of the angle θ between them is cos(45 ) = 1/ 2 Ths, it follows from Formla (12) that = cos θ = (1)(2 2)(1/ 2) = 2 Component Form of the Dot Prodct z P( 1, 2, 3 ) θ Figre Q( 1, 2, 3 ) Althogh we deried Formla (15) and its 2-space companion nder the assmption that and are nonzero, it trned ot that these formlas are also applicable if = 0 or = 0 (erif). For comptational prposes it is desirable to hae a formla that epresses the dot prodct of two ectors in terms of components. We will derie sch a formla for ectors in 3-space; the deriation for ectors in 2-space is similar. Let = ( 1, 2, 3 ) and = ( 1, 2, 3 ) be two nonzero ectors. If, as shown in Figre 3.2.6, θ is the angle between and, then the law of cosines ields PQ 2 = cos θ (14) Since PQ =, we can rewrite (14) as cos θ = 1 2 ( ) or = 1 2 ( ) Sbstitting 2 = , 2 = and 2 = ( 1 1 ) 2 + ( 2 2 ) 2 + ( 3 3 ) 2 we obtain, after simplifing, The companion formla for ectors in 2-space is = (15) = (16) Motiated b the pattern in Formlas (15) and (16), we make the following definition. Josiah Willard Gibbs ( ) Historical Note The dot prodct notation was first introdced b the American phsicist and mathematician J. Willard Gibbs in a pamphlet distribted to his stdents at Yale Uniersit in the 1880s. The prodct was originall written on the baseline, rather than centered as toda, and was referred to as the direct prodct. Gibbs s pamphlet was eentall incorporated into a book entitledvector Analsis that was pblished in 1901 and coathored with one of his stdents. Gibbs made major contribtions to the fields of thermodnamics and electromagnetic theor and is generall regarded as the greatest American phsicist of the nineteenth centr. [Image: SCIENCE SOURCE/Photo Researchers/ Gett Images]

17 3.2 Norm, Dot Prodct, and Distance in R n 147 In words, to calclate the dot prodct (Eclidean inner prodct) mltipl corresponding components and add the reslting prodcts. DEFINITION 4 If = ( 1, 2,..., n ) and = ( 1, 2,..., n ) are ectors in R n, then the dot prodct (also called the Eclidean inner prodct)of and is denoted b and is defined b = n n (17) EXAMPLE 6 Calclating Dot Prodcts Using Components (a) Use Formla (15) to compte the dot prodct of the ectors and in Eample 5. (b) Calclate for the following ectors in R 4 : = ( 1, 3, 5, 7), = ( 3, 4, 1, 0) 3 (k, k, k) 2 1 θ (0, k, 0) (k, 0, 0) z (0, 0, k) d Figre Note that the angle θ obtained in Eample 7 does not inole k. Wh was this to be epected? Soltion (a) The component forms of the ectors are = (0, 0, 1) and = (0, 2, 2). Ths, = (0)(0) + (0)(2) + (1)(2) = 2 which agrees with the reslt obtained geometricall in Eample 5. Soltion (b) = ( 1)( 3) + (3)( 4) + (5)(1) + (7)(0) = 4 EXAMPLE 7 A Geometr Problem Soled Using Dot Prodct Find the angle between a diagonal of a cbe and one of its edges. Soltion Let k be the length of an edge and introdce a coordinate sstem as shown in Figre If we let 1 = (k, 0, 0), 2 = (0,k,0), and 3 = (0, 0,k), then the ector d = (k,k,k)= is a diagonal of the cbe. It follows from Formla (13) that the angle θ between d and the edge 1 satisfies cos θ = 1 d 1 d = k 2 (k)( 3k 2 ) = 1 3 With the help of a calclator we obtain ( ) 1 θ = cos Algebraic Properties of the Dot Prodct In the special case where = in Definition 4, we obtain the relationship = n = 2 (18) This ields the following formla for epressing the length of a ector in terms of a dot prodct: = (19) Dot prodcts hae man of the same algebraic properties as prodcts of real nmbers. THEOREM If,, and w are ectors in R n, and if k is a scalar, then: (a) = [ Smmetr propert ] (b) ( + w) = + w [ Distribtie propert ] (c) k( ) = (k) [ Homogeneit propert ] (d) 0 and = 0 if and onl if = 0 [ Positiit propert ] We will proe parts (c) and (d) and leae the other proofs as eercises.

18 148 Chapter 3 Eclidean Vector Spaces Proof (c) Proof (d) Let = ( 1, 2,..., n ) and = ( 1, 2,..., n ). Then k( ) = k( n n ) = (k 1 ) 1 + (k 2 ) 2 + +(k n ) n = (k) The reslt follows from parts (a) and (b) of Theorem and the fact that = n n = n = 2 The net theorem gies additional properties of dot prodcts. The proofs can be obtained either b epressing the ectors in terms of components or b sing the algebraic properties established in Theorem THEOREM If,, and w are ectors in R n, and if k is a scalar, then: (a) 0 = 0 = 0 (b) ( + ) w = w + w (c) ( w) = w (d) ( ) w = w w (e) k( ) = (k) We will show how Theorem can be sed to proe part (b) withot breaking the ectors into components. The other proofs are left as eercises. Proof (b) ( + ) w = w ( + ) [B smmetr] = w + w [B distribtiit] = w + w [B smmetr] Formlas (18) and (19) together with Theorems and make it possible to maniplate epressions inoling dot prodcts sing familiar algebraic techniqes. EXAMPLE 8 Calclating with Dot Prodcts ( 2) (3 + 4) = (3 + 4) 2 (3 + 4) = 3( ) + 4( ) 6( ) 8( ) = 3 2 2( ) 8 2 Cach Schwarz Ineqalit and Angles in R n Or net objectie is to etend to R n the notion of angle between nonzero ectors and. We will do this b starting with the formla ( ) θ = cos 1 (20) which we preiosl deried for nonzero ectors in R 2 and R 3. Since dot prodcts and norms hae been defined for ectors in R n, it wold seem that this formla has all the ingredients to sere as a definition of the angle θ between two ectors, and, inr n. Howeer, there is a fl in the ointment, the problem being that the inerse cosine in Formla (20) is not defined nless its argment satisfies the ineqalities 1 1 (21) Fortnatel, these ineqalities do hold for all nonzero ectors in R n as a reslt of the following fndamental reslt known as the Cach Schwarz ineqalit.

19 3.2 Norm, Dot Prodct, and Distance in R n 149 THEOREM Cach Schwarz Ineqalit If = ( 1, 2,..., n ) and = ( 1, 2,..., n ) are ectors in R n, then (22) or in terms of components n n ( n )1/2 ( n )1/2 (23) We will omit the proof of this theorem becase later in the tet we will proe a more general ersion of which this will be a special case. Or goal for now will be to se this theorem to proe that the ineqalities in (21) hold for all nonzero ectors in R n. Once that is done we will hae established all the reslts reqired to se Formla (20) as or definition of the angle between nonzero ectors and in R n. To proe that the ineqalities in (21) hold for all nonzero ectors in R n, diide both sides of Formla (22) b the prodct to obtain 1 or eqialentl 1 from which (21) follows. Geometr in R n Figre Earlier in this section we etended arios concepts to R n with the idea that familiar reslts that we can isalize in R 2 and R 3 might be alid in R n as well. Here are two fndamental theorems from plane geometr whose alidit etends to R n : The sm of the lengths of two side of a triangle is at least as large as the third (Figre 3.2.8). The shortest distance between two points is a straight line (Figre 3.2.9). The following theorem generalizes these theorems to R n. THEOREM If,, and w are ectors in R n, then: (a) + + [ Triangle ineqalit for ectors ] (b) d(, ) d(, w) + d(w, ) [ Triangle ineqalit for distances ] w d(, ) d(, w) + d(w, ) Figre Hermann Amands Schwarz ( ) Viktor Yakoleich Bnakosk ( ) Historical Note The Cach Schwarz ineqalit is named in honor of the French mathematician Agstin Cach (see p. 121) and the German mathematician Hermann Schwarz. Variations of this ineqalit occr in man different settings and nder arios names. Depending on the contet in which the ineqalit occrs, o ma find it called Cach s ineqalit, the Schwarz ineqalit, or sometimes een the Bnakosk ineqalit, in recognition of the Rssian mathematician who pblished his ersion of the ineqalit in 1859, abot 25 ears before Schwarz. [Images: Rdolph Dehrkoop/ llstein bild/the Image Works (Schwarz); Biographies/Bnakosk.html (Bnakosk)]

20 150 Chapter 3 Eclidean Vector Spaces Proof (a) + 2 = ( + ) ( + ) = ( ) + 2( ) + ( ) = 2 + 2( ) = ( + ) 2 Propert of absolte ale Cach Schwarz ineqalit This completes the proof since both sides of the ineqalit in part (a) are nonnegatie. Proof (b) It follows from part (a) and Formla (11) that d(, ) = = ( w) + (w ) w + w =d(, w) + d(w, ) + Figre It is proed in plane geometr that for an parallelogram the sm of the sqares of the diagonals is eqal to the sm of the sqares of the for sides (Figre ). The following theorem generalizes that reslt to R n. THEOREM Parallelogram Eqation for Vectors If and are ectors in R n, then = 2 ( 2 + 2) (24) Proof = ( + ) ( + ) + ( ) ( ) = 2( ) + 2( ) = 2 ( 2 + 2) We cold state and proe man more theorems from plane geometr that generalize to R n, bt the ones alread gien shold sffice to conince o that R n is not so different from R 2 and R 3 een thogh we cannot isalize it directl. The net theorem establishes a fndamental relationship between the dot prodct and norm in R n. THEOREM If and are ectors in R n with the Eclidean inner prodct, then = (25) Note that Formla (25) epresses the dot prodct in terms of norms. Proof + 2 = ( + ) ( + ) = 2 + 2( ) = ( ) ( ) = 2 2( ) + 2 from which (25) follows b simple algebra. Dot Prodcts as Matri Mltiplication There are arios was to epress the dot prodct of ectors sing matri notation. The formlas depend on whether the ectors are epressed as row matrices or colmn matrices. Table 1 shows the possibilities.

21 3.2 Norm, Dot Prodct, and Distance in R n 151 Table 1 Form Dot Prodct Eample 1 5 = 3 T =[1 3 5] 4 = 7 a colmn 5 0 matri and a = T = T colmn matri 5 1 = 4 T =[5 4 0] 3 = =[1 3 5] =[ ] 4 = 7 a row matri 0 and a colmn = = T T 5 matri = T T =[5 4 0] 3 = 7 5 a colmn matri and a row matri a row matri and arow matri 1 = 3 = = T T 5 =[5 4 0] =[1 3 5] = T = T =[5 4 0] 1 =[5 4 0] 3 = T T =[1 3 5] 4 = 7 0 T =[ ] 4 = T =[5 4 0] 3 = 7 5 Application of Dot Prodcts to ISBN Nmbers Althogh the sstem has recentl changed, most older books hae been assigned a niqe 10-digit nmber called an International Standard Book Nmber or ISBN. The first nine digits of this nmber are split into three grops the first grop representing the contr or grop of contries in which the book originates, the second identifing the pblisher, and the third assigned to the book title itself. The tenth and final digit, called a check digit, is compted from the first nine digits and is sed to ensre that an electronic transmission of the ISBN, sa oer the Internet, occrs withot error. To eplain how this is done, regard the first nine digits of the ISBN as a ector b in R 9, and let a be the ector a = (1, 2, 3, 4, 5, 6, 7, 8, 9) For eample, the ISBN of the brief edition of Calcls, sith edition, b Howard Anton is which has a check digit of 9. This is consistent with the first nine digits of the ISBN, since a b = (1, 2, 3, 4, 5, 6, 7, 8, 9) (0, 4, 7, 1, 1, 5, 3, 0, 7) = 152 Diiding 152 b 11 prodces a qotient of 13 and a remainder of 9, so the check digit is c = 9. If an electronic order is placed for a book with a certain ISBN, then the warehose can se the aboe procedre to erif that the check digit is consistent with the first nine digits, thereb redcing the possibilit of a costl shipping error. Then the check digit c is compted sing the following procedre: 1. Form the dot prodct a b. 2. Diide a b b 11, thereb prodcing a remainder c thatisan integer between 0 and 10, inclsie. The check digit is taken to be c, with the proiso that c = 10 is written as X to aoid doble digits.

22 152 Chapter 3 Eclidean Vector Spaces If A is an n n matri and and are n 1 matrices, then it follows from the first row in Table 1 and properties of the transpose that The reslting formlas A = T (A) = ( T A) = (A T ) T = A T A = (A) T = ( T A T ) = T (A T ) = A T A = A T (26) A = A T (27) proide an important link between mltiplication b an n n matri A and mltiplication b A T. EXAMPLE 9 Verifing that A = A T Sppose that Then from which we obtain A = 2 4 1, = 2, = A = = A T = = A = 7( 2) + 10(0) + 5(5) = 11 A T = ( 1)( 7) + 2(4) + 4( 1) = 11 Ths, A = A T as garanteed b Formla (26). We leae it for o to erif that Formla (27) also holds. A Dot Prodct View of Matri Mltiplication Dot prodcts proide another wa of thinking abot matri mltiplication. Recall that if A =[a ij ] is an m r matri and B =[b ij ] is an r n matri, then the ij th entr of AB is a i1 b 1j + a i2 b 2j + +a ir b rj which is the dot prodct of the ith row ector of A [a i1 a i2 a ir ] and the jth colmn ector of B b 1j b 2j. b rj

23 3.2 Norm, Dot Prodct, and Distance in R n 153 Ths, if the row ectors of A are r 1, r 2,...,r m and the colmn ectors of B are c 1, c 2,...,c n, then the matri prodct AB can be epressed as r 1 c 1 r 1 c 2 r 1 c n r 2 c 1 r 2 c 2 r 2 c n AB =.. (28). r m c 1 r m c 2 r m c n Eercise Set 3.2 In Eercises 1 2, find the norm of, and a nit ector that is oppositel directed to. 1. (a) = (2, 2, 2) (b) = (1, 0, 2, 1, 3) 2. (a) = (1, 1, 2) (b) = ( 2, 3, 3, 1) In Eercises 3 4, ealate the gien epression with = (2, 2, 3), = (1, 3, 4), and w = (3, 6, 4). 3. (a) + (b) + (c) (d) w 4. (a) + + w (b) (c) 3 3 (d) In Eercises 5 6, ealate the gien epression with = ( 2, 1,4,5), = (3,1, 5,7), and w = ( 6,2,1,1). 5. (a) w (b) w (c) 6. (a) w (b) w 7. Let = ( 2, 3, 0, 6). Find all scalars k sch that k = Let = (1, 1, 2, 3, 1). Find all scalars k sch that k =4. In Eercises 9 10, find,, and. 9. (a) = (3, 1, 4), = (2, 2, 4) (b) = (1, 1, 4, 6), = (2, 2, 3, 2) 10. (a) = (1, 1, 2, 3), = ( 1, 0, 5, 1) (b) = (2, 1, 1, 0, 2), = (1, 2, 2, 2, 1) In Eercises 11 12, find the Eclidean distance between and and the cosine of the angle between those ectors. State whether that angle is acte, obtse, or (a) = (3, 3, 3), = (1, 0, 4) (b) = (0, 2, 1, 1), = ( 3, 2, 4, 4) 12. (a) = (1, 2, 3, 0), = (5, 1, 2, 2) (b) = (0, 1, 1, 1, 2), = (2, 1, 0, 1, 3) 13. Sppose that a ector a in the -plane has a length of 9 nits and points in a direction that is 120 conterclockwise from the positie -ais, and a ector b in that plane has a length of 5 nits and points in the positie -direction. Find a b. 14. Sppose that a ector a in the -plane points in a direction thatis47 conterclockwise from the positie -ais, and a ector b in that plane points in a direction that is 43 clockwise from the positie -ais. What can o sa abot the ale of a b? In Eercises 15 16, determine whether the epression makes sense mathematicall. If not, eplain wh. 15. (a) ( w) (b) ( + w) (c) (d) ( ) 16. (a) (b) ( ) w (c) ( ) k (d) k In Eercises 17 18, erif that the Cach Schwarz ineqalit holds. 17. (a) = ( 3, 1, 0), = (2, 1, 3) (b) = (0, 2, 2, 1), = (1, 1, 1, 1) 18. (a) = (4, 1, 1), = (1, 2, 3) (b) = (1, 2, 1, 2, 3), = (0, 1, 1, 5, 2) 19. Let r 0 = ( 0, 0 ) be a fied ector in R 2. In each part, describe in words the set of all ectors r = (, ) that satisf the stated condition. (a) r r 0 = 1 (b) r r 0 1 (c) r r 0 > Repeat the directions of Eercise 19 for ectors r = (,,z) and r 0 = ( 0, 0,z 0 ) in R 3. Eercises The direction of a nonzero ector in an zcoordinate sstem is completel determined b the angles α, β, and γ between and the standard nit ectors i, j, and k (Figre E-21). These are called the direction angles of, and their cosines are called the direction cosines of. 21. Use Formla (13) to show that the direction cosines of a ector = ( 1, 2, 3 ) in R 3 are cos α = 1, cos β = 2, cos γ = 3

24 154 Chapter 3 Eclidean Vector Spaces k z 30. Under what conditions will the triangle ineqalit (Theorem 3.2.5a) be an eqalit? Eplain or answer geometricall. i γ α β 22. Use the reslt in Eercise 21 to show that j cos 2 α + cos 2 β + cos 2 γ = 1 Figre E Show that two nonzero ectors 1 and 2 in R 3 are orthogonal if and onl if their direction cosines satisf cos α 1 cos α 2 + cos β 1 cos β 2 + cos γ 1 cos γ 2 = The accompaning figre shows a cbe. (a) Find the angle between the ectors d and to the nearest degree. (b) Make a conjectre abot the angle between the ectors d and, and confirm or conjectre b compting the angle. Eercises The effect that a force has on an object depends on the magnitde of the force and the direction in which it is applied. Ths, forces can be regarded as ectors and represented as arrows in which the length of the arrow specifies the magnitde of the force, and the direction of the arrow specifies the direction in which the force is applied. It is a fact of phsics that force ectors obe the parallelogram law in the sense that if two force ectors F 1 and F 2 are applied at a point on an object, then the effect is the same as if the single force F 1 + F 2 (called the resltant) were applied at that point (see accompaning figre). Forces are commonl measred in nits called ponds-force (abbreiated lbf) or Newtons (abbreiated N). F 2 F 1 F 1 + F 2 The single force F 1 + F 2 has the same effect as the two forces F 1 and F 2. z d 31. A particle is said to be in static eqilibrim if the resltant of all forces applied to it is zero. For the forces in the accompaning figre, find the resltant F that mst be applied to the indicated point to prodce static eqilibrim. Describe F b giing its magnitde and the angle in degrees that it makes with the positie -ais. Figre E Estimate, to the nearest degree, the angles that a diagonal of a bo with dimensions 10 cm 15 cm 25 cm makes with the edges of the bo. 26. If =2 and w =3, what are the largest and smallest ales possible for w? Gie a geometric eplanation of or reslts. 27. What can o sa abot two nonzero ectors, and, that satisf the eqation + = +? 28. (a) What relationship mst hold for the point p = (a,b,c) to be eqidistant from the origin and the z-plane? Make sre that the relationship o state is alid for positie and negatie ales of a, b, and c. (b) What relationship mst hold for the point p = (a,b,c)to be farther from the origin than from the z-plane? Make sre that the relationship o state is alid for positie and negatie ales of a, b, and c. 29. State a procedre for finding a ector of a specified length m that points in the same direction as a gien ector. 32. Follow the directions of Eercise lb Figre E lb 120 N Figre E-32 Working with Proofs 33. Proe parts (a) and (b) of Theorem Proe parts (a) and (c) of Theorem Proe parts (d) and (e) of Theorem N 100 N Tre-False Eercises TF. In parts (a) (j) determine whether the statement is tre or false, and jstif or answer.

25 3.3 Orthogonalit 155 (a) If each component of a ector in R 3 is dobled, the norm of that ector is dobled. (b) In R 2, the ectors of norm 5 whose initial points are at the origin hae terminal points ling on a circle of radis 5 centered at the origin. (c) Eer ector in R n has a positie norm. (d) If is a nonzero ector in R n, there are eactl two nit ectors that are parallel to. (e) If =2, =1, and = 1, then the angle between and is π/3 radians. (f ) The epressions ( ) + w and ( + w) are both meaningfl and eqal to each other. (g) If = w, then = w. (h) If = 0, then either = 0 or = 0. (i) In R 2,if lies in the first qadrant and lies in the third qadrant, then cannot be positie. ( j) For all ectors,, and w in R n,wehae + + w + + w Working withtechnolog T1. Let be a ector in R 100 whose ith component is i, and let be the ector in R 100 whose ith component is 1/(i + 1). Find the dot prodct of and. T2. Find, to the nearest degree, the angles that a diagonal of a bo with dimensions 10 cm 11 cm 25 cm makes with the edges of the bo. 3.3 Orthogonalit In the last section we defined the notion of angle between ectors in R n. In this section we will focs on the notion of perpendiclarit. Perpendiclar ectors in R n pla an important role in a wide ariet of applications. Orthogonal Vectors Recall from Formla (20) in the preios section that the angle θ between two nonzero ectors and in R n is defined b the formla ( ) θ = cos 1 It follows from this that θ = π/2 if and onl if = 0. Ths, we make the following definition. DEFINITION 1 Two nonzero ectors and in R n are said to be orthogonal (or perpendiclar)if = 0. We will also agree that the zero ector in R n is orthogonal to eer ector in R n. EXAMPLE 1 Orthogonal Vectors (a) Show that = ( 2, 3, 1, 4) and = (1, 2, 0, 1) are orthogonal ectors in R 4. (b) Let S ={i, j, k} be the set of standard nit ectors in R 3. Show that each ordered pair of ectors in S is orthogonal. Soltion (a) The ectors are orthogonal since = ( 2)(1) + (3)(2) + (1)(0) + (4)( 1) = 0 Soltion (b) It sffices to show that i j = i k = j k = 0

26 156 Chapter 3 Eclidean Vector Spaces Using the comptations in R 3 as a model, o shold be able to see that each ordered pair of standard nit ectors in R n is orthogonal. becase it will follow atomaticall from the smmetr propert of the dot prodct that j i = k i = k j = 0 Althogh the orthogonalit of the ectors in S is eident geometricall from Figre 3.2.2, it is confirmed algebraicall b the comptations i j = (1, 0, 0) (0, 1, 0) = 0 i k = (1, 0, 0) (0, 0, 1) = 0 j k = (0, 1, 0) (0, 0, 1) = 0 Lines and Planes Determined b Points and Normals Formla (1) is called the pointnormal form of a line or plane and Formlas (2) and (3) the component forms. One learns in analtic geometr that a line in R 2 is determined niqel b its slope and one of its points, and that a plane in R 3 is determined niqel b its inclination and one of its points. One wa of specifing slope and inclination is to se a nonzero ector n, called a normal, that is orthogonal to the line or plane in qestion. For eample, Figre shows the line throgh the point P 0 ( 0, 0 ) that has normal n = (a, b) and the plane throgh the point P 0 ( 0, 0,z 0 ) that has normal n = (a,b,c). Both the line and the plane are represented b the ector eqation n P 0 P = 0 (1) where P is either an arbitrar point (, ) on the line or an arbitrar point (,,z)in the plane. The ector P 0 P can be epressed in terms of components as P 0 P = ( 0, 0 ) [ line ] P 0 P = ( 0, 0,z z 0 ) [ plane ] Ths, Eqation (1) can be written as a( 0 ) + b( 0 ) = 0 [ line ] (2) a( 0 ) + b( 0 ) + c(z z 0 ) = 0 [ plane ] (3) These are called the point-normal eqations of the line and plane. z P(, ) (a, b) P(,, z) (a, b, c) n n P 0 ( 0, 0 ) P 0 ( 0, 0, z 0 ) Figre EXAMPLE 2 Point-Normal Eqations It follows from (2) that in R 2 the eqation 6( 3) + ( + 7) = 0 represents the line throgh the point (3, 7) with normal n = (6, 1); and it follows from (3) that in R 3 the eqation 4( 3) + 2 5(z 7) = 0 represents the plane throgh the point (3, 0, 7) with normal n = (4, 2, 5).

27 3.3 Orthogonalit 157 When conenient, the terms in Eqations (2) and (3) can be mltiplied ot and the constants combined. This leads to the following theorem. THEOREM (a) If a and b are constants that are not both zero, then an eqation of the form a + b + c = 0 (4) (b) represents a line in R 2 with normal n = (a, b). If a, b, and c are constants that are not all zero, then an eqation of the form represents a plane in R 3 with normal n = (a,b,c). a + b + cz + d = 0 (5) (a) (b) E X A M P L E 3 Vectors Orthogonal to Lines and Planes Throgh the Origin The eqation a + b = 0 represents a line throgh the origin in R 2. Show that the ector n 1 = (a, b) formed from the coefficients of the eqation is orthogonal to the line, that is, orthogonal to eer ector along the line. The eqation a + b + cz = 0 represents a plane throgh the origin in R 3. Show that the ector n 2 = (a,b,c)formed from the coefficients of the eqation is orthogonal to the plane, that is, orthogonal to eer ector that lies in the plane. Soltion We will sole both problems together. The two eqations can be written as (a, b) (, ) = 0 and (a,b,c) (,,z)= 0 or, alternatiel, as n 1 (, ) = 0 and n 2 (,,z)= 0 These eqations show that n 1 is orthogonal to eer ector (, ) on the line and that n 2 is orthogonal to eer ector (,,z)in the plane (Figre 3.3.1). Referring to Table 1 of Section 3.2, in what other was can o write (6) if n and are epressed in matri form? Recall that a + b = 0 and a + b + cz = 0 are called homogeneos eqations. Eample 3 illstrates that homogeneos eqations in two or three nknowns can be written in the ector form n = 0 (6) where n is the ector of coefficients and is the ector of nknowns. In R 2 this is called the ector form of a line throgh the origin, and in R 3 it is called the ector form of a plane throgh the origin. Orthogonal Projections In man applications it is necessar to decompose a ector into a sm of two terms, one term being a scalar mltiple of a specified nonzero ector a and the other term being orthogonal to a. For eample, if and a are ectors in R 2 that are positioned so their initial points coincide at a point Q, then we can create sch a decomposition as follows (Figre 3.3.2):

28 158 Chapter 3 Eclidean Vector Spaces Drop a perpendiclar from the tip of to the line throgh a. Constrct the ector w 1 from Q to the foot of the perpendiclar. Constrct the ector w 2 = w 1. Since w 1 + w 2 = w 1 + ( w 1 ) = we hae decomposed into a sm of two orthogonal ectors, the first term being a scalar mltiple of a and the second being orthogonal to a. w 2 w 2 w 2 Q w 1 a Q a w 1 w 1 Q (a) (b) (c) Figre Three possible cases. a The following theorem shows that the foregoing reslts, which we illstrated sing ectors in R 2, appl as well in R n. THEOREM Projection Theorem If and a are ectors in R n, and if a = 0, then can be epressed in eactl one wa in the form = w 1 + w 2, where w 1 is a scalar mltiple of a and w 2 is orthogonal to a. Proof Since the ector w 1 is to be a scalar mltiple of a, it mst hae the form w 1 = ka (7) Or goal is to find a ale of the scalar k and a ector w 2 that is orthogonal to a sch that = w 1 + w 2 (8) We can determine k b sing (7) to rewrite (8) as = w 1 + w 2 = ka + w 2 and then appling Theorems and to obtain a = (ka + w 2 ) a = k a 2 + (w 2 a) (9) Since w 2 is to be orthogonal to a, the last term in (9) mst be 0, and hence k mst satisf the eqation a = k a 2 from which we obtain k = a a 2 as the onl possible ale for k. The proof can be completed b rewriting (8) as w 2 = w 1 = ka = a a 2 a and then confirming that w 2 is orthogonal to a b showing that w 2 a = 0 (we leae the details for o).

29 3.3 Orthogonalit 159 The ectors w 1 and w 2 in the Projection Theorem hae associated names the ector w 1 is called the orthogonal projection of on a or sometimes the ector component of along a, and the ector w 2 is called the ector component of orthogonal to a. The ector w 1 is commonl denoted b the smbol proj a, in which case it follows from (8) that w 2 = proj a. In smmar, proj a = a a (ector component of along a) (10) a 2 proj a = a a (ector component of orthogonal to a) (11) a 2 e 2 = (0, 1) L EXAMPLE 4 Orthogonal Projection on a Line Find the orthogonal projections of the ectors e 1 = (1, 0) and e 2 = (0, 1) on the line L that makes an angle θ with the positie -ais in R 2. Soltion As illstrated in Figre 3.3.3, a = (cos θ,sin θ) is a nit ector along the line L, so or first problem is to find the orthogonal projection of e 1 along a. Since a = sin 2 θ + cos 2 θ = 1 and e 1 a = (1, 0) (cos θ,sin θ) = cos θ (cos θ, sin θ) 1 sin θ θ cos θ e 1 = (1, 0) Figre it follows from Formla (10) that this projection is proj a e 1 = e 1 a a 2 a = (cos θ)(cos θ,sin θ) = (cos2 θ,sin θ cos θ) Similarl, since e 2 a = (0, 1) (cos θ,sin θ) = sin θ, it follows from Formla (10) that proj a e 2 = e 2 a a 2 a = (sin θ)(cos θ,sin θ) = (sin θ cos θ,sin2 θ) EXAMPLE 5 Vector Component of Along a Let = (2, 1, 3) and a = (4, 1, 2). Find the ector component of along a and the ector component of orthogonal to a. Soltion a = (2)(4) + ( 1)( 1) + (3)(2) = 15 a 2 = ( 1) = 21 Ths the ector component of along a is proj a = a a a = 15 (4, 1, 2) = ( 20, 5, ) and the ector component of orthogonal to a is proj a = (2, 1, 3) ( 20 7, 5 7, 10 7 ) ( = 6, 2, ) As a check, o ma wish to erif that the ectors proj a and a are perpendiclar b showing that their dot prodct is zero.

30 160 Chapter 3 Eclidean Vector Spaces θ cos θ (a) 0 θ < π 2 a Sometimes we will be more interested in the norm of the ector component of along a than in the ector component itself. A formla for this norm can be deried as follows: proj a = a a a 2 = a a a 2 a = a a 2 where the second eqalit follows from part (c) of Theorem and the third from the fact that a 2 > 0. Ths, proj a = a a (12) cos θ π (b) < θ π 2 Figre θ a If θ denotes the angle between and a, then a = a cos θ, so (12) can also be written as proj a = cos θ (13) (Verif.) A geometric interpretation of this reslt is gien in Figre TheTheorem of Pthagoras In Section 3.2 we fond that man theorems abot ectors in R 2 and R 3 also hold in R n. Another eample of this is the following generalization of the Theorem of Pthagoras (Figre 3.3.5). + THEOREM Theorem of Pthagoras in R n If and are orthogonal ectors in R n with the Eclidean inner prodct, then + 2 = (14) Figre Proof Since and are orthogonal, we hae = 0, from which it follows that + 2 = ( + ) ( + ) = 2 + 2( ) + 2 = EXAMPLE 6 Theorem of Pthagoras in R 4 We showed in Eample 1 that the ectors = ( 2, 3, 1, 4) and = (1, 2, 0, 1) are orthogonal. Verif the Theorem of Pthagoras for these ectors. Soltion We leae it for o to confirm that Ths, + 2 = = ( 1, 5, 1, 3) + 2 = = OPTIONAL Distance Problems We will now show how orthogonal projections can be sed to sole the following three distance problems: Problem 1. Find the distance between a point and a line in R 2. Problem 2. Find the distance between a point and a plane in R 3. Problem 3. Find the distance between two parallel planes in R 3.

31 3.3 Orthogonalit 161 A method for soling the first two problems is proided b the net theorem. Since the proofs of the two parts are similar, we will proe part (b) and leae part (a) as an eercise. THEOREM (a) In R 2 the distance D between the point P 0 ( 0, 0 ) and the line a + b + c = 0 is D = a 0 + b 0 + c (15) a2 + b 2 (b) In R 3 the distance D between the point P 0 ( 0, 0,z 0 ) and the plane a + b + cz + d = 0 is D = a 0 + b 0 + cz 0 + d a2 + b 2 + c 2 (16) n = (a, b, c) P 0 ( 0, 0, z 0 ) proj n QP 0 D D Proof (b) The nderling idea of the proof is illstrated in Figre As shown in that figre, let Q( 1, 1,z 1 ) be an point in the plane, and let n = (a,b,c)be a normal ector to the plane that is positioned with its initial point at Q. It is now eident that the distance D between P 0 and the plane is simpl the length (or norm) of the orthogonal projection of the ector QP 0 on n, which b Formla (12) is Q( 1, 1, z 1 ) Distance from P 0 to plane. Figre Bt Ths D = proj nqp0 = QP 0 n n QP 0 = ( 0 1, 0 1,z 0 z 1 ) QP 0 n = a( 0 1 ) + b( 0 1 ) + c(z 0 z 1 ) n = a 2 + b 2 + c 2 D = a( 0 1 ) + b( 0 1 ) + c(z 0 z 1 ) a2 + b 2 + c 2 (17) Since the point Q( 1, 1, z 1 ) lies in the gien plane, its coordinates satisf the eqation of that plane; ths a 1 + b 1 + cz 1 + d = 0 or d = a 1 b 1 cz 1 Sbstitting this epression in (17) ields (16). EXAMPLE 7 Distance Between a Point and a Plane Find the distance D between the point (1, 4, 3) and the plane z = 1. Soltion Since the distance formlas in Theorem reqire that the eqations of the line and plane be written with zero on the right side, we first need to rewrite the eqation of the plane as z + 1 = 0 from which we obtain D = 2(1) + ( 3)( 4) + 6( 3) ( 3) = 3 7 = 3 7

32 162 Chapter 3 Eclidean Vector Spaces V P 0 The third distance problem posed aboe is to find the distance between two parallel planes in R 3. As sggested in Figre 3.3.7, the distance between a plane V and a plane W can be obtained b finding an point P 0 in one of the planes, and compting the distance between that point and the other plane. Here is an eample. W Figre The distance between the parallel planes V and W is eqal to the distance between P 0 and W. EXAMPLE 8 Distance Between Parallel Planes The planes + 2 2z = 3 and z = 7 are parallel since their normals, (1, 2, 2) and (2, 4, 4), are parallel ectors. Find the distance between these planes. Soltion To find the distance D between the planes, we can select an arbitrar point in one of the planes and compte its distance to the other plane. B setting = z = 0in the eqation + 2 2z = 3, we obtain the point P 0 (3, 0, 0) in this plane. From (16), the distance between P 0 and the plane z = 7is D = 2(3) + 4(0) + ( 4)(0) ( 4) 2 = 1 6 Eercise Set 3.3 In Eercises 1 2, determine whether and are orthogonal ectors. 1. (a) = (6, 1, 4), = (2, 0, 3) (b) = (0, 0, 1), = (1, 1, 1) (c) = (3, 2, 1, 3), = ( 4, 1, 3, 7) (d) = (5, 4, 0, 3), = ( 4, 1, 3, 7) 2. (a) = (2, 3), = (5, 7) (b) = (1, 1, 1), = (0, 0, 0) (c) = (1, 5, 4), = (3, 3, 3) (d) = (4, 1, 2, 5), = ( 1, 5, 3, 1) In Eercises 3 6, find a point-normal form of the eqation of the plane passing throgh P and haing n as a normal. 3. P( 1, 3, 2); n = ( 2, 1, 1) 4. P(1, 1, 4); n = (1, 9, 8) 5. P(2, 0, 0); n = (0, 0, 2) 6. P(0, 0, 0); n = (1, 2, 3) In Eercises 7 10, determine whether the gien planes are parallel z = 5 and z = z 2 = 0 and z 7 = = 8 4z + 5 and = 1 2 z ( 4, 1, 2) (,,z)= 0 and (8, 2, 4) (,,z)= 0 In Eercises 11 12, determine whether the gien planes are perpendiclar z 4 = 0, + 2z = z = 4, z = 1 In Eercises 13 14, find proj a. 13. (a) = (1, 2), a = ( 4, 3) (b) = (3, 0, 4), a = (2, 3, 3) 14. (a) = (5, 6), a = (2, 1) (b) = (3, 2, 6), a = (1, 2, 7) In Eercises 15 20, find the ector component of along a and the ector component of orthogonal to a. 15. = (6, 2), a = (3, 9) 16. = ( 1, 2), a = ( 2, 3) 17. = (3, 1, 7), a = (1, 0, 5) 18. = (2, 0, 1), a = (1, 2, 3) 19. = (2, 1, 1, 2), a = (4, 4, 2, 2) 20. = (5, 0, 3, 7), a = (2, 1, 1, 1) In Eercises 21 24, find the distance between the point and the line. 21. ( 3, 1); = ( 1, 4); = (2, 5); = (1, 8); 3 + = 5 In Eercises 25 26, find the distance between the point and the plane. 25. (3, 1, 2); + 2 2z = 4

33 3.3 Orthogonalit ( 1, 1, 2); z = 4 In Eercises 27 28, find the distance between the gien parallel planes z = 5 and z = z = 1 and 2 + z = Find a nit ector that is orthogonal to both = (1, 0, 1) and = (0, 1, 1). 30. (a) Show that = (a, b) and w = ( b, a) are orthogonal ectors. (b) Use the reslt in part (a) to find two ectors that are orthogonal to = (2, 3). (c) Find two nit ectors that are orthogonal to = ( 3, 4). 31. Do the points A(1, 1, 1), B( 2, 0, 3), and C( 3, 1, 1) form the ertices of a right triangle? Eplain. 32. Repeat Eercise 31 for the points A(3, 0, 2), B(4, 3, 0), and C(8, 1, 1). 33. Show that if is orthogonal to both w 1 and w 2, then is orthogonal to k 1 w 1 + k 2 w 2 for all scalars k 1 and k Is it possible to hae proj a = proj a? Eplain. Eercises In phsics and engineering the work W performed b a constant force F applied in the direction of motion to an object moing a distance d on a straight line is defined to be W = F d (force magnitde times distance) In the case where the applied force is constant bt makes an angle θ with the direction of motion, and where the object moes along a line from a point P to a point Q, we call PQ the displacement and define the work performed b the force to be W = F PQ = F PQ cos θ (see accompaning figre). Common nits of work are ft-lb (foot ponds) or Nm (Newton meters). F F θ F cos θ PQ Work = ( F cos θ) PQ 35. Show that the work performed b a constant force (not necessaril in the direction of motion) can be epressed as W =± PQ proj F PQ and eplain when the + sign shold be sed and when the sign shold be sed. 36. As illstrated in the accompaning figre, a wagon is plled horizontall b eerting a force of 10 lb on the handle at an angle of 60 with the horizontal. How mch work is done in moing the wagon 50 ft? F 10 lb ft 37. A sailboat traels 100 m de north while the wind eerts a force of 500 N toward the northeast. How mch work does the wind do? Working with Proofs 38. Let and be nonzero ectors in 2- or 3-space, and let k = and l =. Proe that the ector w = l + k bisects the angle between and. 39. Proe part (a) of Theorem Tre-False Eercises TF. In parts (a) (g) determine whether the statement is tre or false, and jstif or answer. (a) The ectors (3, 1, 2) and (0, 0, 0) are orthogonal. (b) If and are orthogonal ectors, then for all nonzero scalars k and m, k and m are orthogonal ectors. (c) The orthogonal projection of on a is perpendiclar to the ector component of orthogonal to a. (d) If a and b are orthogonal ectors, then for eer nonzero ector,wehae proj a (proj b ()) = 0 (e) If a and are nonzero ectors, then proj a (proj a ()) = proj a () (f ) If the relationship proj a = proj a holds for some nonzero ector a, then =. (g) For all ectors and, it is tre that + = + Working withtechnolog T1. Find the lengths of the sides and the interior angles of the triangle in R 4 whose ertices are P(2, 4, 2, 4, 2), Q(6, 4, 4, 4, 6), R(5, 7, 5, 7, 2) T2. Epress the ector = (2, 3, 1, 2) in the form = w 1 + w 2, where w 1 is a scalar mltiple of a = ( 1, 0, 2, 1) and w 2 is orthogonal to a.

34 164 Chapter 3 Eclidean Vector Spaces 3.4 The Geometr of Linear Sstems In this section we will se parametric and ector methods to std general sstems of linear eqations. This work will enable s to interpret soltion sets of linear sstems with n nknowns as geometric objects in R n jst as we interpreted soltion sets of linear sstems with two and three nknowns as points, lines, and planes in R 2 and R 3. Vector and Parametric Eqations of Lines in R 2 and R 3 In the last section we deried eqations of lines and planes that are determined b a point and a normal ector. Howeer, there are other sefl was of specifing lines and planes. For eample, a niqe line in R 2 or R 3 is determined b a point 0 on the line and a nonzero ector parallel to the line, and a niqe plane in R 3 is determined b a point 0 in the plane and two noncollinear ectors 1 and 2 parallel to the plane. The best wa to isalize this is to translate the ectors so their initial points are at 0 (Figre 3.4.1). 0 0 z 1 2 Figre L Let s begin b deriing an eqation for the line L that contains the point 0 and is parallel to. If is a general point on sch a line, then, as illstrated in Figre 3.4.2, the ector 0 will be some scalar mltiple of, sa 0 = t or eqialentl = 0 + t Figre As the ariable t (called a parameter) aries from to, the point traces ot the line L. Accordingl, we hae the following reslt. Althogh it is not stated eplicitl, it is nderstood in Formlas (1) and (2) that the parameter t aries from to. This applies to all ector and parametric eqations in this tet ecept where stated otherwise. THEOREM Let L be the line in R 2 or R 3 that contains the point 0 and is parallel to the nonzero ector. Then the eqation of the line throgh 0 that is parallel to is = 0 + t (1) If 0 = 0, then the line passes throgh the origin and the eqation has the form = t (2) Vector and Parametric Eqations of Planes in R 3 Net we will derie an eqation for the plane W that contains the point 0 and is parallel to the noncollinear ectors 1 and 2. As shown in Figre 3.4.3, if is an point in the plane, then b forming sitable scalar mltiples of 1 and 2,sat 1 1 and t 2 2, we can create a parallelogram with diagonal 0 and adjacent sides t 1 1 and t 2 2. Ths, we hae 0 = t t 2 2 or eqialentl = 0 + t t 2 2 As the ariables t 1 and t 2 (called parameters) ar independentl from to, the point aries oer the entire plane W. In smmar, we hae the following reslt.

35 3.4 The Geometr of Linear Sstems 165 z W t t1 1 Figre THEOREM Let W be the plane in R 3 that contains the point 0 and is parallel to the noncollinear ectors 1 and 2. Then an eqation of the plane throgh 0 that is parallel to 1 and 2 is gien b = 0 + t t 2 2 (3) If 0 = 0, then the plane passes throgh the origin and the eqation has the form = t t 2 2 (4) Remark Obsere that the line throgh 0 represented b Eqation (1) is the translation b 0 of the line throgh the origin represented b Eqation (2) and that the plane throgh 0 represented b Eqation (3) is the translation b 0 of the plane throgh the origin represented b Eqation (4) (Figre 3.4.4). z = 0 + t 0 = 0 + t t = t 2 1 = t t 2 2 Figre Motiated b the forms of Formlas (1) to (4), we can etend the notions of line and plane to R n b making the following definitions. DEFINITION 1 If 0 and are ectors in R n, and if is nonzero, then the eqation = 0 + t (5) defines the line throgh 0 that is parallel to. In the special case where 0 = 0, the line is said to pass throgh the origin. DEFINITION 2 If 0, 1, and 2 are ectors in R n, and if 1 and 2 are not collinear, then the eqation = 0 + t t 2 2 (6) defines the plane throgh 0 that is parallel to 1 and 2. In the special case where 0 = 0, the plane is said to pass throgh the origin. Eqations (5) and (6) are called ector forms of a line and plane in R n. If the ectors in these eqations are epressed in terms of their components and the corresponding components on each side are eqated, then the reslting eqations are called parametric eqations of the line and plane. Here are some eamples. EXAMPLE 1 Vector and Parametric Eqations of Lines in R 2 and R 3 (a) Find a ector eqation and parametric eqations of the line in R 2 that passes throgh the origin and is parallel to the ector = ( 2, 3). (b) Find a ector eqation and parametric eqations of the line in R 3 that passes throgh the point P 0 (1, 2, 3) and is parallel to the ector = (4, 5, 1). (c) Use the ector eqation obtained in part (b) to find two points on the line that are different from P 0.

36 166 Chapter 3 Eclidean Vector Spaces Soltion (a) It follows from (5) with 0 = 0 that a ector eqation of the line is = t. If we let = (, ), then this eqation can be epressed in ector form as (, ) = t( 2, 3) Eqating corresponding components on the two sides of this eqation ields the parametric eqations = 2t, = 3t Soltion (b) It follows from (5) that a ector eqation of the line is = 0 + t. Ifwe let = (,,z), and if we take 0 = (1, 2, 3), then this eqation can be epressed in ector form as (,,z)= (1, 2, 3) + t(4, 5, 1) (7) Eqating corresponding components on the two sides of this eqation ields the parametric eqations = 1 + 4t, = 2 5t, z = 3 + t Soltion (c) A point on the line represented b Eqation (7) can be obtained b sbstitting a specific nmerical ale for the parameter t. Howeer, since t = 0 prodces (,,z)= (1, 2, 3), which is the point P 0, this ale of t does not sere or prpose. Taking t = 1 prodces the point (5, 3, 2) and taking t = 1 prodces the point ( 3, 7, 4). An other distinct ales for t (ecept t = 0) wold work jst as well. We wold hae obtained different parametric and ector eqations in Eample 2 had we soled (8) for or z rather than. Howeer, one can show the same plane reslts in all three cases as the parameters ar from to. EXAMPLE 2 Vector and Parametric Eqations of a Plane in R 3 Find ector and parametric eqations of the plane + 2z = 5. Soltion We will find the parametric eqations first. We can do this b soling the eqation for an one of the ariables in terms of the other two and then sing those two ariables as parameters. For eample, soling for in terms of and z ields = 5 + 2z (8) and then sing and z as parameters t 1 and t 2, respectiel, ields the parametric eqations = 5 + t 1 2t 2, = t 1, z = t 2 To obtain a ector eqation of the plane we rewrite these parametric eqations as or, eqialentl, as (,,z)= (5 + t 1 2t 2,t 1,t 2 ) (,,z)= (5, 0, 0) + t 1 (1, 1, 0) + t 2 ( 2, 0, 1) EXAMPLE 3 Vector and Parametric Eqations of Lines and Planes in R 4 (a) Find ector and parametric eqations of the line throgh the origin of R 4 that is parallel to the ector = (5, 3, 6, 1). (b) Find ector and parametric eqations of the plane in R 4 that passes throgh the point 0 = (2, 1, 0, 3) and is parallel to both 1 = (1, 5, 2, 4) and 2 = (0, 7, 8, 6). Soltion (a) If we let = ( 1, 2, 3, 4 ), then the ector eqation = t can be epressed as ( 1, 2, 3, 4 ) = t(5, 3, 6, 1) Eqating corresponding components ields the parametric eqations 1 = 5t, 2 = 3t, 3 = 6t, 4 = t

37 Soltion (b) 3.4 The Geometr of Linear Sstems 167 The ector eqation = 0 + t t 2 2 can be epressed as ( 1, 2, 3, 4 ) = (2, 1, 0, 3) + t 1 (1, 5, 2, 4) + t 2 (0, 7, 8, 6) which ields the parametric eqations 1 = 2 + t 1 2 = 1 + 5t 1 + 7t 2 3 = 2t 1 8t 2 4 = 3 4t 1 + 6t 2 LinesThroghTwo Points in R n 1 0 Figre If 0 and 1 are distinct points in R n, then the line determined b these points is parallel to the ector = 1 0 (Figre 3.4.5), so it follows from (5) that the line can be epressed in ector form as = 0 + t( 1 0 ) (9) or, eqialentl, as = (1 t) 0 + t 1 (10) These are called the two-point ector eqations of a line in R n = Figre EXAMPLE 4 A Line Throgh Two Points in R 2 Find ector and parametric eqations for the line in R 2 that passes throgh the points P(0, 7) and Q(5, 0). Soltion We will see below that it does not matter which point we take to be 0 and which we take to be 1, so let s choose 0 = (0, 7) and 1 = (5, 0). It follows that 1 0 = (5, 7) and hence that which we can rewrite in parametric form as (, ) = (0, 7) + t(5, 7) (11) = 5t, = 7 7t Had we reersed or choices and taken 0 = (5, 0) and 1 = (0, 7), then the reslting ector eqation wold hae been and the parametric eqations wold hae been (, ) = (5, 0) + t( 5, 7) (12) = 5 5t, = 7t (erif). Althogh (11) and (12) look different, the both represent the line whose eqation in rectanglar coordinates is = 35 (Figre 3.4.6). This can be seen b eliminating the parameter t from the parametric eqations (erif). The point = (, ) in Eqations (9) and (10) traces an entire line in R 2 as the parameter t aries oer the interal (, ). If, howeer, we restrict the parameter to ar from t = 0tot = 1, then will not trace the entire line bt rather jst the line segment joining the points 0 and 1. The point will start at 0 when t = 0 and end at 1 when t = 1. Accordingl, we make the following definition.

38 168 Chapter 3 Eclidean Vector Spaces DEFINITION 3 If 0 and 1 are ectors in R n, then the eqation = 0 + t( 1 0 ) (0 t 1) (13) defines the line segment from 0 to 1. When conenient, Eqation (13) can be written as = (1 t) 0 + t 1 (0 t 1) (14) EXAMPLE 5 A Line Segment from One Point to Another in R 2 It follows from (13) and (14) that the line segment in R 2 from 0 = (1, 3) to 1 = (5, 6) can be represented either b the eqation or b the eqation = (1, 3) + t(4, 9) (0 t 1) = (1 t)(1, 3) + t(5, 6) (0 t 1) Dot Prodct Form of a Linear Sstem Or net objectie is to show how to epress linear eqations and linear sstems in dot prodct notation. This will lead s to some important reslts abot orthogonalit and linear sstems. Recall that a linear eqation in the ariables 1, 2,..., n has the form a a a n n = b (a 1,a 2,...,a n not all zero) (15) and that the corresponding homogeneos eqation is a a a n n = 0 (a 1,a 2,...,a n not all zero) (16) These eqations can be rewritten in ector form b letting a = (a 1,a 2,...,a n ) and = ( 1, 2,..., n ) in which case Formla (15) can be written as a = b (17) and Formla (16) as a = 0 (18) Ecept for a notational change from n to a, Formla (18) is the etension to R n of Formla (6) in Section 3.3. This eqation reeals that each soltion ector of a homogeneos eqation is orthogonal to the coefficient ector a. To take this geometric obseration a step frther, consider the homogeneos sstem a a a 1n n = 0 a a a 2n n = a m1 1 + a m a mn n = 0 If we denote the sccessie row ectors of the coefficient matri b r 1, r 2,...,r m, then we can rewrite this sstem in dot prodct form as r 1 = 0 r 2 = 0.. r m = 0 from which we see that eer soltion ector is orthogonal to eer row ector of the coefficient matri. In smmar, we hae the following reslt. (19)

39 3.4 The Geometr of Linear Sstems 169 THEOREM If A is an m n matri, then the soltion set of the homogeneos linear sstem A = 0 consists of all ectors in R n that are orthogonal to eer row ector of A. EXAMPLE 6 Orthogonalit of Row Vectors and Soltion Vectors We showed in Eample 6 of Section 1.2 that the general soltion of the homogeneos linear sstem = is 1 = 3r 4s 2t, 2 = r, 3 = 2s, 4 = s, 5 = t, 6 = 0 which we can rewrite in ector form as = ( 3r 4s 2t,r, 2s, s, t, 0) According to Theorem 3.4.3, the ector mst be orthogonal to each of the row ectors r 1 = (1, 3, 2, 0, 2, 0) r 2 = (2, 6, 5, 2, 4, 3) r 3 = (0, 0, 5, 10, 0, 15) r 4 = (2, 6, 0, 8, 4, 18) We will confirm that is orthogonal to r 1, and leae it for o to erif that is orthogonal to the other three row ectors as well. The dot prodct of r 1 and is r 1 = 1( 3r 4s 2t) + 3(r) + ( 2)( 2s) + 0(s) + 2(t) + 0(0) = 0 which establishes the orthogonalit. The Relationship Between A = 0 and A = b We will conclde this section b eploring the relationship between the soltions of a homogeneos linear sstem A = 0 and the soltions (if an) of a nonhomogeneos linear sstem A = b that has the same coefficient matri. These are called corresponding linear sstems. To motiate the reslt we are seeking, let s compare the soltions of the corresponding linear sstems = 0 0 and = We showed in Eamples 5 and 6 of Section 1.2 that the general soltions of these linear sstems can be written in parametric form as homogeneos 1 = 3r 4s 2t, 2 = r, 3 = 2s, 4 = s, 5 = t, 6 = 0 nonhomogeneos 1 = 3r 4s 2t, 2 = r, 3 = 2s, 4 = s, 5 = t, 6 =

40 170 Chapter 3 Eclidean Vector Spaces which we can then rewrite in ector form as homogeneos ( 1, 2, 3, 4, 5, 6 ) = ( 3r 4s 2t,r, 2s, s, t, 0) nonhomogeneos ( 1, 2, 3, 4, 5, 6 ) = ( 3r 4s 2t,r, 2s, s, t, 1 3 B splitting the ectors on the right apart and collecting terms with like parameters, we can rewrite these eqations as homogeneos ( 1, 2, 3, 4, 5 ) = r( 3, 1, 0, 0, 0) + s( 4, 0, 2, 1, 0, 0) + t( 2, 0, 0, 0, 1, 0) (20) nonhomogeneos ( 1, 2, 3, 4, 5 ) = r( 3, 1, 0, 0, 0) + s( 4, 0, 2, 1, 0, 0) + t( 2, 0, 0, 0, 1, 0) + ( ) 0, 0, 0, 0, 0, 1 (21) 3 Formlas (20) and (21) reeal that each soltion of the nonhomogeneos sstem can be obtained b adding the fied ector ( ) 0, 0, 0, 0, 0, 1 3 to the corresponding soltion of the homogeneos sstem. This is a special case of the following general reslt. ) THEOREM The general soltion of a consistent linear sstem A = b can be obtained b adding an specific soltion of A = b to the general soltion of A = 0. Proof Let 0 be an specific soltion of A = b, let W denote the soltion set of A = 0, and let 0 + W denote the set of all ectors that reslt b adding 0 to each ector in W. We mst show that if is a ector in 0 + W, then is a soltion of A = b, and conersel that eer soltion of A = b is in the set 0 + W. Assme first that is a ector in 0 + W. This implies that is epressible in the form = 0 + w, where A 0 = b and Aw = 0. Ths, 0 A = b A = 0 A = A( 0 + w) = A 0 + Aw = b + 0 = b which shows that is a soltion of A = b. Conersel, let be an soltion of A = b. To show that is in the set 0 + W we mst show that is epressible in the form = 0 + w (22) where w is in W (i.e., Aw = 0). We can do this b taking w = 0. This ector obiosl satisfies (22), and it is in W since Aw = A( 0 ) = A A 0 = b b = 0 0 Figre The soltion set of A = b is a translation of the soltion space of A = 0. Remark Theorem has a sefl geometric interpretation that is illstrated in Figre If, as discssed in Section 3.1, we interpret ector addition as translation, then the theorem states that if 0 is an specific soltion of A = b, then the entire soltion set of A = b can be obtained b translating the soltion space of A = 0 b the ector 0. Eercise Set 3.4 In Eercises 1 4, find ector and parametric eqations of the line containing the point and parallel to the ector. 1. Point: ( 4, 1); ector: = (0, 8) 2. Point: (2, 1); ector: = ( 4, 2) 3. Point: (0, 0, 0); ector: = ( 3, 0, 1) 4. Point: ( 9, 3, 4); ector: = ( 1, 6, 0) In Eercises 5 8, se the gien eqation of a line to find a point on the line and a ector parallel to the line. 5. = (3 5t, 6 t) 6. (,,z)= (4t,7, 4 + 3t) 7. = (1 t)(4, 6) + t( 2, 0) 8. = (1 t)(0, 5, 1)

41 3.4 The Geometr of Linear Sstems 171 In Eercises 9 12, find ector and parametric eqations of the plane that contains the gien point and is parallel to the two ectors. 9. Point: ( 3, 1, 0); ectors: 1 = (0, 3, 6) and 2 = ( 5, 1, 2) 10. Point: (0, 6, 2); ectors: 1 = (0, 9, 1) and 2 = (0, 3, 0) 11. Point: ( 1, 1, 4); ectors: 1 = (6, 1, 0) and 2 = ( 1, 3, 1) 12. Point: (0, 5, 4); ectors: 1 = (0, 0, 5) and 2 = (1, 3, 2) In Eercises 13 14, find ector and parametric eqations of the line in R 2 that passes throgh the origin and is orthogonal to. 13. = ( 2, 3) 14. = (1, 4) In Eercises 15 16, find ector and parametric eqations of the plane in R 3 that passes throgh the origin and is orthogonal to. 15. = (4, 0, 5) [Hint: Constrct two nonparallel ectors orthogonal to in R 3 ]. 16. = (3, 1, 6) In Eercises 17 20, find the general soltion to the linear sstem and confirm that the row ectors of the coefficient matri are orthogonal to the soltion ectors = = = = = = = = = (a) The eqation + + z = 1 can be iewed as a linear sstem of one eqation in three nknowns. Epress a general soltion of this eqation as a particlar soltion pls a general soltion of the associated homogeneos eqation. (b) Gie a geometric interpretation of the reslt in part (a). 22. (a) The eqation + = 1 can be iewed as a linear sstem of one eqation in two nknowns. Epress a general soltion of this eqation as a particlar soltion pls a general soltion of the associated homogeneos sstem. (b) Gie a geometric interpretation of the reslt in part (a). 23. (a) Find a homogeneos linear sstem of two eqations in three nknowns whose soltion space consists of those ectors in R 3 that are orthogonal to a = (1, 1, 1) and b = ( 2, 3, 0). (b) What kind of geometric object is the soltion space? (c) Find a general soltion of the sstem obtained in part (a), and confirm that Theorem holds. 24. (a) Find a homogeneos linear sstem of two eqations in three nknowns whose soltion space consists of those ectors in R 3 that are orthogonal to a = ( 3, 2, 1) and b = (0, 2, 2). (b) What kind of geometric object is the soltion space? (c) Find a general soltion of the sstem obtained in part (a), and confirm that Theorem holds. 25. Consider the linear sstems = and = (a) Find a general soltion of the homogeneos sstem. (b) Confirm that 1 = 1, 2 = 0, 3 = 1 is a soltion of the nonhomogeneos sstem. (c) Use the reslts in parts (a) and (b) to find a general soltion of the nonhomogeneos sstem. (d) Check or reslt in part (c) b soling the nonhomogeneos sstem directl. 26. Consider the linear sstems = and = (a) Find a general soltion of the homogeneos sstem. (b) Confirm that 1 = 1, 2 = 1, 3 = 1 is a soltion of the nonhomogeneos sstem. (c) Use the reslts in parts (a) and (b) to find a general soltion of the nonhomogeneos sstem. (d) Check or reslt in part (c) b soling the nonhomogeneos sstem directl. In Eercises 27 28, find a general soltion of the sstem, and se that soltion to find a general soltion of the associated homogeneos sstem and a particlar soltion of the gien sstem =

42 172 Chapter 3 Eclidean Vector Spaces = Let = 0 + t be a line in R n, and let T : R n R n be a matri operator on R n. What kind of geometric object is the image of this line nder the operator T? Eplain or reasoning. Tre-False Eercises TF. In parts (a) (f ) determine whether the statement is tre or false, and jstif or answer. (a) The ector eqation of a line can be determined from an point ling on the line and a nonzero ector parallel to the line. (b) The ector eqation of a plane can be determined from an point ling in the plane and a nonzero ector parallel to the plane. (c) The points ling on a line throgh the origin in R 2 or R 3 are all scalar mltiples of an nonzero ector on the line. (d) All soltion ectors of the linear sstem A = b are orthogonal to the row ectors of the matri A if and onl if b = 0. (e) The general soltion of the nonhomogeneos linear sstem A = b can be obtained b adding b to the general soltion of the homogeneos linear sstem A = 0. (f ) If 1 and 2 are two soltions of the nonhomogeneos linear sstem A = b, then 1 2 is a soltion of the corresponding homogeneos linear sstem. Working withtechnolog T1. Find the general soltion of the homogeneos linear sstem = 0 0 and confirm that each soltion ector is orthogonal to eer row ector of the coefficient matri in accordance with Theorem Cross Prodct This optional section is concerned with properties of ectors in 3-space that are important to phsicists and engineers. It can be omitted, if desired, since sbseqent sections do not depend on its content. Among other things, we define an operation that proides a wa of constrcting a ector in 3-space that is perpendiclar to two gien ectors, and we gie a geometric interpretation of 3 3 determinants. Cross Prodct of Vectors In Section 3.2 we defined the dot prodct of two ectors and in n-space. That operation prodced a scalar as its reslt. We will now define a tpe of ector mltiplication that prodces a ector as the reslt bt which is applicable onl to ectors in 3-space. DEFINITION 1 If = ( 1, 2, 3 ) and = ( 1, 2, 3 ) are ectors in 3-space, then the cross prodct is the ector defined b = ( , , ) or, in determinant notation, ( ) 2 3 = 2 3, , (1) Remark Instead of memorizing (1), o can obtain the components of as follows: [ ] 1 Form the 2 3 matri 2 3 whose first row contains the components of and whose second row contains the components of.

43 3.5 Cross Prodct 173 To find the first component of, delete the first colmn and take the determinant; to find the second component, delete the second colmn and take the negatie of the determinant; and to find the third component, delete the third colmn and take the determinant. EXAMPLE 1 Calclating a Cross Prodct Find, where = (1, 2, 2) and = (3, 0, 1). Soltion From either (1) or the mnemonic in the preceding remark, we hae ( ) 2 2 = 0 1, , = (2, 7, 6) The following theorem gies some important relationships between the dot prodct and cross prodct and also shows that is orthogonal to both and. The formlas for the ector triple prodcts in parts (d) and (e) of Theorem are sefl becase the allow s to se dot prodcts and scalar mltiplications to perform calclations that wold otherwise reqire determinants to calclate the reqired cross prodcts. THEOREM Relationships Inoling Cross Prodct and Dot Prodct If,, and w are ectors in 3-space, then (a) ( ) = 0 [ is orthogonal to ] (b) ( ) = 0 [ is orthogonal to ] (c) 2 = 2 2 ( ) 2 [ Lagrange s identit ] (d) ( w) = ( w) ( )w [ ector triple prodct ] (e) ( ) w = ( w) ( w) [ ector triple prodct ] Proof (a) Let = ( 1, 2, 3 ) and = ( 1, 2, 3 ). Then ( ) = ( 1, 2, 3 ) ( , , ) = 1 ( ) + 2 ( ) + 3 ( ) = 0 Proof (b) Proof (c) and Similar to (a). Since 2 = ( ) 2 + ( ) 2 + ( ) 2 (2) 2 2 ( ) 2 = ( )( ) ( ) 2 (3) the proof can be completed b mltipling ot the right sides of (2) and (3) and erifing their eqalit. Proof (d) and (e) See Eercises 40 and 41. Historical Note The cross prodct notation A B was introdced b the American phsicist and mathematician J. Willard Gibbs, (see p. 146) in a series of npblished lectre notes for his stdents atyale Uniersit. It appeared in a pblished work for the first time in the second edition of the book Vector Analsis, b Edwin Wilson ( ), a stdent of Gibbs. Gibbs originall referred to A B as the skew prodct.

44 174 Chapter 3 Eclidean Vector Spaces EXAMPLE 2 Is Perpendiclar to and to Consider the ectors = (1, 2, 2) and = (3, 0, 1) In Eample 1 we showed that = (2, 7, 6) Since ( ) = (1)(2) + (2)( 7) + ( 2)( 6) = 0 and ( ) = (3)(2) + (0)( 7) + (1)( 6) = 0 is orthogonal to both and, as garanteed b Theorem The main arithmetic properties of the cross prodct are listed in the net theorem. THEOREM Properties of Cross Prodct If,, and w are an ectors in 3-space and k is an scalar, then: (a) = ( ) (b) ( + w) = ( ) + ( w) (c) ( + ) w = ( w) + ( w) (d) k( ) = (k) = (k) (e) 0 = 0 = 0 ( f) = 0 The proofs follow immediatel from Formla (1) and properties of determinants; for eample, part (a) can be proed as follows. Proof (a) Interchanging and in (1) interchanges the rows of the three determinants on the right side of (1) and hence changes the sign of each component in the cross prodct. Ths = ( ). The proofs of the remaining parts are left as eercises. Joseph Lois Lagrange ( ) Historical Note Joseph Lois Lagrange was a French-Italian mathematician and astronomer. Althogh his father wanted him to become a lawer, Lagrange was attracted to mathematics and astronom after reading a memoir b the astronomer Halle. At age 16 he began to std mathematics on his own and b age 19 was appointed to a professorship at the Roal Artiller School in Trin. The following ear he soled some famos problems sing new methods that eentall blossomed into a branch of mathematics called the calcls of ariations. These methods and Lagrange s applications of them to problems in celestial mechanics were so monmental that b age 25 he was regarded b man of his contemporaries as the greatest liing mathematician. One of Lagrange s most famos works is a memoir, Mécaniqe Analtiqe, in which he redced the theor of mechanics to a few general formlas from which all other necessar eqations cold be deried. Napoleon was a great admirer of Lagrange and showered him with man honors. In spite of his fame, Lagrange was a sh and modest man. On his death, he was bried with honor in the Pantheon. [Image: traeler1116/istockphoto]