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CHAPTER 3 Euclidean Vector Spaces CHAPTER CONTENTS 3.1 Vectors in 2-Space, 3-Space, and n-Space 131 n...

CHAPTER 3 Euclidean Vector Spaces CHAPTER CONTENTS 3.1 Vectors in 2-Space, 3-Space, and n-Space 131 n 3.2 Norm, Dot Product, and Distance in R 142 3.3 Orthogonality 155 3.4 The Geometry of Linear Systems 164 3.5 Cross Product 172 INTRODUCTION Engineers and physicists distinguish between two types of physical quantities— scalars, which are quantities that can be described by a numerical value alone, and vectors, which are quantities that require both a number and a direction for their complete physical description. For example, temperature, length, and speed are scalars because they can be fully described by a number that tells “how much”—a temperature of 20◦ C, a length of 5 cm, or a speed of 75 km/h. In contrast, velocity and force are vectors because they require a number that tells “how much” and a direction that tells “which way”—say, a boat moving at 10 knots in a direction 45◦ northeast, or a force of 100 lb acting vertically. Although the notions of vectors and scalars that we will study in this text have their origins in physics and engineering, we will be more concerned with using them to build mathematical structures and then applying those structures to such diverse fields as genetics, computer science, economics, telecommunications, and environmental science. 3.1 Vectors in 2-Space, 3-Space, and n-Space Linear algebra is primarily concerned with two types of mathematical objects, “matrices” and “vectors.” In Chapter 1 we discussed the basic properties of matrices, we introduced the idea of viewing n-tuples of real numbers as vectors, and we denoted the set of all such n-tuples as R n. In this section we will review the basic properties of vectors in two and three dimensions with the goal of extending these properties to vectors in R n. Geometric Vectors Engineers and physicists represent vectors in two dimensions (also called 2-space) or in three dimensions (also called 3-space) by arrows. The direction of the arrowhead specifies the direction of the vector and the length of the arrow specifies the magnitude. Terminal point Mathematicians call these geometric vectors. The tail of the arrow is called the initial point of the vector and the tip the terminal point (Figure 3.1.1). In this text we will denote vectors in boldface type such as a, b, v, w, and x, and we Initial point will denote scalars in lowercase italic type such as a , k , v , w , and x. When we want to indicate that a vector v has initial point A and terminal point B , then, as shown in Figure 3.1.1 Figure 3.1.2, we will write −→ v = AB 131 132 Chapter 3 Euclidean Vector Spaces B Vectors with the same length and direction, such as those in Figure 3.1.3, are said to be equivalent. Since we want a vector to be determined solely by its length and direction, v equivalent vectors are regarded as the same vector even though they may be in different positions. Equivalent vectors are also said to be equal, which we indicate by writing A v=w v = AB The vector whose initial and terminal points coincide has length zero, so we call this the zero vector and denote it by 0. The zero vector has no natural direction, so we will Figure 3.1.2 agree that it can be assigned any direction that is convenient for the problem at hand. Vector Addition There are a number of important algebraic operations on vectors, all of which have their origin in laws of physics. Parallelogram Rule for Vector Addition If v and w are vectors in 2-space or 3-space that are positioned so their initial points coincide, then the two vectors form adjacent sides of a parallelogram, and the sum v + w is the vector represented by the arrow from the common initial point of v and w to the opposite vertex of the parallelogram Equivalent vectors (Figure 3.1.4a). Figure 3.1.3 Here is another way to form the sum of two vectors. Triangle Rule for Vector Addition If v and w are vectors in 2-space or 3-space that are positioned so the initial point of w is at the terminal point of v, then the sum v + w is represented by the arrow from the initial point of v to the terminal point of w (Figure 3.1.4b). In Figure 3.1.4c we have constructed the sums v + w and w + v by the triangle rule. This construction makes it evident that v+w=w+v (1) and that the sum obtained by the triangle rule is the same as the sum obtained by the parallelogram rule. w w v v+w v v v+w v v+w w+v w w Figure 3.1.4 (a) (b) (c) Vector addition can also be viewed as a process of translating points. Vector Addition Viewed asTranslation If v, w, and v + w are positioned so their initial points coincide, then the terminal point of v + w can be viewed in two ways: 1. The terminal point of v + w is the point that results when the terminal point of v is translated in the direction of w by a distance equal to the length of w (Figure 3.1.5a). 2. The terminal point of v + w is the point that results when the terminal point of w is translated in the direction of v by a distance equal to the length of v (Figure 3.1.5b). Accordingly, we say that v + w is the translation of v by w or, alternatively, the translation of w by v.  3.1 Vectors in 2-Space, 3-Space, and n-Space 133 v v+w v v+w w w Figure 3.1.5 (a) (b) Vector Subtraction In ordinary arithmetic we can write a − b = a + (−b), which expresses subtraction in terms of addition. There is an analogous idea in vector arithmetic. Vector Subtraction The negative of a vector v, denoted by −v, is the vector that has the same length as v but is oppositely directed (Figure 3.1.6a), and the difference of v from w, denoted by w − v, is taken to be the sum w − v = w + (−v) (2) The difference of v from w can be obtained geometrically by the parallelogram method shown in Figure 3.1.6b, or more directly by positioning w and v so their ini- tial points coincide and drawing the vector from the terminal point of v to the terminal point of w (Figure 3.1.6c). v w w–v w w–v –v –v v v Figure 3.1.6 (a) (b) (c) Scalar Multiplication Sometimes there is a need to change the length of a vector or change its length and reverse its direction. This is accomplished by a type of multiplication in which vectors are multiplied by scalars. As an example, the product 2v denotes the vector that has the same direction as v but twice the length, and the product −2v denotes the vector that is oppositely directed to v and has twice the length. Here is the general result. Scalar Multiplication If v is a nonzero vector in 2-space or 3-space, and if k is a nonzero scalar, then we define the scalar product of v by k to be the vector whose v 1 (–1)v length is |k| times the length of v and whose direction is the same as that of v if k is v 2 positive and opposite to that of v if k is negative. If k = 0 or v = 0, then we define k v to be 0. (–3)v Figure 3.1.7 shows the geometric relationship between a vector v and some of its scalar 2v multiples. In particular, observe that (−1)v has the same length as v but is oppositely directed; therefore, Figure 3.1.7 (−1)v = −v (3) Parallel and Collinear Suppose that v and w are vectors in 2-space or 3-space with a common initial point. If Vectors one of the vectors is a scalar multiple of the other, then the vectors lie on a common line, so it is reasonable to say that they are collinear (Figure 3.1.8a). However, if we trans- late one of the vectors, as indicated in Figure 3.1.8b, then the vectors are parallel but no longer collinear. This creates a linguistic problem because translating a vector does not change it. The only way to resolve this problem is to agree that the terms parallel and 134 Chapter 3 Euclidean Vector Spaces collinear mean the same thing when applied to vectors. Although the vector 0 has no clearly defined direction, we will regard it as parallel to all vectors when convenient. kv kv v v Figure 3.1.8 (a) (b) Sums ofThree or More Vector addition satisfies the associative law for addition, meaning that when we add three Vectors vectors, say u, v, and w, it does not matter which two we add first; that is, u + (v + w) = (u + v) + w It follows from this that there is no ambiguity in the expression u + v + w because the same result is obtained no matter how the vectors are grouped. A simple way to construct u + v + w is to place the vectors “tip to tail” in succession and then draw the vector from the initial point of u to the terminal point of w (Figure 3.1.9a). The tip-to-tail method also works for four or more vectors (Figure 3.1.9b). The tip-to-tail method makes it evident that if u, v, and w are vectors in 3-space with a common initial point, then u + v + w is the diagonal of the parallelepiped that has the three vectors as adjacent sides (Figure 3.1.9c). v w x v+ u u+v u+ x v v+ w u + w w v w u + (v + w) v+ (u + v) + w +w u u Figure 3.1.9 (a) (b) (c) Vectors in Coordinate Up until now we have discussed vectors without reference to a coordinate system. How- Systems ever, as we will soon see, computations with vectors are much simpler to perform if a coordinate system is present to work with. If a vector v in 2-space or 3-space is positioned with its initial point at the origin of a rectangular coordinate system, then the vector is completely determined by the coor- dinates of its terminal point (Figure 3.1.10). We call these coordinates the components The component forms of the of v relative to the coordinate system. We will write v = (v1 , v2 ) to denote a vector v in zero vector are 0 = (0, 0) in 2-space with components (v1 , v2 ), and v = (v1 , v2 , v3 ) to denote a vector v in 3-space 2-space and 0 = (0, 0, 0) in 3- space. with components (v1 , v2 , v3 ). y z (v1, v2) (v1, v2, v3) v v y x x Figure 3.1.10  3.1 Vectors in 2-Space, 3-Space, and n-Space 135 It should be evident geometrically that two vectors in 2-space or 3-space are equiv- alent if and only if they have the same terminal point when their initial points are at the origin. Algebraically, this means that two vectors are equivalent if and only if their y corresponding components are equal. Thus, for example, the vectors (v1, v2) v = (v1 , v2 , v3 ) and w = (w1 , w2 , w3 ) in 3-space are equivalent if and only if v1 = w1 , v2 = w2 , v3 = w3 x Remark It may have occurred to you that an ordered pair (v1 , v2 ) can represent either a vector Figure 3.1.11 The ordered with components v1 and v2 or a point with coordinates v1 and v2 (and similarly for ordered triples). pair (v1 , v2 ) can represent a Both are valid geometric interpretations, so the appropriate choice will depend on the geometric point or a vector. viewpoint that we want to emphasize (Figure 3.1.11). Vectors Whose Initial Point It is sometimes necessary to consider vectors whose initial points are not at the origin. Is Not at the Origin −−→ If P1 P2 denotes the vector with initial point P1 (x1 , y1 ) and terminal point P2 (x2 , y2 ), then the components of this vector are given by the formula y P2(x2, y2) −−→ P1(x1, y1) v P1 P2 = (x2 − x1 , y2 − y1 ) (4) OP1 OP2 −−→ That is, the components of P1 P2 are obtained by subtracting the coordinates of the initial point from the coordinates of the terminal point. For example, in Figure 3.1.12 x −−→ −−→ −−→ the vector P1 P2 is the difference of vectors OP2 and OP1 , so −−→ −−→ −−→ P1 P2 = OP2 − OP1 = (x2 , y2 ) − (x1 , y1 ) = (x2 − x1 , y2 − y1 ) v = P1P2 = OP2 – OP1 As you might expect, the components of a vector in 3-space that has initial point Figure 3.1.12 P1 (x1 , y1 , z1 ) and terminal point P2 (x2 , y2 , z2 ) are given by −−→ P1 P2 = (x2 − x1 , y2 − y1 , z2 − z1 ) (5) E X A M P L E 1 Finding the Components of a Vector −−→ The components of the vector v = P1 P2 with initial point P1 (2, −1, 4) and terminal point P2 (7, 5, −8) are v = (7 − 2, 5 − (−1), (−8) − 4) = (5, 6, −12) n-Space The idea of using ordered pairs and triples of real numbers to represent points in two- dimensional space and three-dimensional space was well known in the eighteenth and nineteenth centuries. By the dawn of the twentieth century, mathematicians and physi- cists were exploring the use of “higher dimensional” spaces in mathematics and physics. Today, even the layman is familiar with the notion of time as a fourth dimension, an idea used by Albert Einstein in developing the general theory of relativity. Today, physicists working in the field of “string theory” commonly use 11-dimensional space in their quest for a unified theory that will explain how the fundamental forces of nature work. Much of the remaining work in this section is concerned with extending the notion of space to n dimensions. To explore these ideas further, we start with some terminology and notation. The set of all real numbers can be viewed geometrically as a line. It is called the real line and is denoted by R or R 1. The superscript reinforces the intuitive idea that a line is one- dimensional. The set of all ordered pairs of real numbers (called 2-tuples) and the set of all ordered triples of real numbers (called 3-tuples) are denoted by R 2 and R 3 , respectively. 136 Chapter 3 Euclidean Vector Spaces The superscript 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 extends this idea. DEFINITION 1 If n is a positive integer, then an ordered n-tuple is a sequence of n real numbers (v1 , v2 ,... , vn ). The set of all ordered n-tuples is called n-space and is denoted by R n. Remark You can think of the numbers in an n-tuple (v1 , v2 ,... , vn ) as either the coordinates of a generalized point or the components of a generalized vector, depending on the geometric image you want to bring to mind—the choice makes no difference mathematically, since it is the algebraic properties of n-tuples that are of concern. Here are some typical applications that lead to n-tuples. Experimental Data—A scientist performs an experiment and makes n numerical measurements each time the experiment is performed. The result of each experiment can be regarded as a vector y = (y1 , y2 ,... , yn ) in R n in which y1 , y2 ,... , yn are the measured values. Storage and Warehousing—A national trucking company has 15 depots for storing and servicing its trucks. At each point in time the distribution of trucks in the service depots can be described by a 15-tuple x = (x1 , x2 ,... , x15 ) in which x1 is the number of trucks in the first depot, x2 is the number in the second depot, and so forth. Electrical Circuits—A certain kind of processing chip is designed to receive four input voltages and produce three output voltages in response. The input voltages can be regarded as vectors in R 4 and the output voltages as vectors in R 3. Thus, the chip can be viewed as a device that transforms an input vector v = (v1 , v2 , v3 , v4 ) in R 4 into an output vector w = (w1 , w2 , w3 ) in R 3. Graphical Images—One way in which color images are created on computer screens is by assigning each pixel (an addressable point on the screen) three numbers that describe the hue, saturation, and brightness of the pixel. Thus, a complete color image can be viewed as a set of 5-tuples of the form v = (x, y, h, s, b) in which x and y are the screen coordinates of a pixel and h, s , and b are its hue, saturation, and brightness. Economics—One approach to economic analysis is to divide an economy into sectors (manufacturing, services, utilities, and so forth) and measure the output of each sector by a dollar value. Thus, in an economy with 10 sectors the economic output of the entire economy can be represented by a 10-tuple s = (s1 , s2 ,... , s10 ) in which the numbers s1 , s2 ,... , s10 are the outputs of the individual sectors. Historical Note The German-born physicist Albert Einstein immigrated to the United States in 1935, where he settled at Princeton University. Einstein spent the last three decades of his life working unsuccessfully at producing a unified field theory that would establish an underlying link between the forces of gravity and electromagnetism. Recently, physi- cists have made progress on the problem using a frame- work known as string theory. In this theory the smallest, indivisible components of the Universe are not particles but loops that behave like vibrating strings. Whereas Einstein’s space-time universe was four-dimensional, strings reside in an 11-dimensional world that is the focus of current re- Albert Einstein search. (1879–1955) [Image: © Bettmann/CORBIS]  3.1 Vectors in 2-Space, 3-Space, and n-Space 137 Mechanical Systems—Suppose that six particles move along the same coordinate line so that at time t their coordinates are x1 , x2 ,... , x6 and their velocities are v1 , v2 ,... , v6 , respectively. This information can be represented by the vector v = (x1 , x2 , x3 , x4 , x5 , x6 , v1 , v2 , v3 , v4 , v5 , v6 , t) in R 13. This vector is called the state of the particle system at time t. Operations on Vectors in Rn Our next goal is to define useful operations on vectors in R n. These operations will all be natural extensions of the familiar operations on vectors in R 2 and R 3. We will denote a vector v in R n using the notation v = (v1 , v2 ,... , vn ) and we will call 0 = (0, 0,... , 0) the zero vector. We noted earlier that in R 2 and R 3 two vectors are equivalent (equal) if and only if their corresponding components are the same. Thus, we make the following definition. DEFINITION 2 Vectors v = (v1 , v2 ,... , vn ) and w = (w1 , w2 ,... , wn ) in R n are said to be equivalent (also called equal) if v1 = w1 , v2 = w2 ,... , vn = wn We indicate this by writing v = w. E X A M P L E 2 Equality of Vectors (a, b, c, d) = (1, −4, 2, 7) if and only if a = 1, b = −4, c = 2, and d = 7. Our next objective is to define the operations of addition, subtraction, and scalar multiplication for vectors in R n. To motivate these ideas, we will consider how these op- erations can be performed on vectors in R 2 using components. By studying Figure 3.1.13 you should be able to deduce that if v = (v1 , v2 ) and w = (w1 , w2 ), then v + w = (v1 + w1 , v2 + w2 ) (6) k v = (kv1 , kv2 ) (7) In particular, it follows from (7) that −v = (−1)v = (−v1 , −v2 ) (8) y (v1 + w1, v2 + w2) v2 (w1, w2) w y v+ w (kv1, kv2) w2 kv (v1, v2) kv2 (v1, v2) v v2 v x x v1 Figure 3.1.13 v1 w1 kv1 138 Chapter 3 Euclidean Vector Spaces and hence that w − v = w + (−v) = (w1 − v1 , w2 − v2 ) (9) Motivated by Formulas (6)–(9), we make the following definition. DEFINITION 3 If v = (v1 , v2 ,... , vn ) and w = (w1 , w2 ,... , wn ) are vectors in R n , and if k is any scalar, then we define v + w = (v1 + w1 , v2 + w2 ,... , vn + wn ) (10) k v = (kv1 , kv2 ,... , kvn ) (11) −v = (−v1 , −v2 ,... , −vn ) (12) w − v = w + (−v) = (w1 − v1 , w2 − v2 ,... , wn − vn ) (13) E X A M P L E 3 Algebraic Operations Using Components If v = (1, −3, 2) and w = (4, 2, 1), then In words, vectors are added (or subtracted) by adding (or sub- v + w = (5, −1, 3), 2v = (2, −6, 4) tracting) their corresponding −w = (−4, −2, −1), v − w = v + (−w) = (−3, −5, 1) components, and a vector is multiplied by a scalar by multi- plying each component by that The following theorem summarizes the most important properties of vector opera- scalar. tions. THEOREM 3.1.1 If u, v, and w are vectors in R n , and if k and m are scalars, then: (a) u+v=v+u (b) (u + v) + w = u + (v + w) (c) u+0=0+u=u (d ) u + (−u) = 0 (e) k(u + v) = k u + k v ( f) (k + m)u = k u + mu ( g) k(mu) = (km)u (h) 1u = u We will prove part (b) and leave some of the other proofs as exercises. Proof (b) Let u = (u1 , u2 ,... , un ), v = (v1 , v2 ,... , vn ), and w = (w1 , w2 ,... , wn ). Then   (u + v) + w = (u1 , u2 ,... , un ) + (v1 , v2 ,... , vn ) + (w1 , w2 ,... , wn ) = (u1 + v1 , u2 + v2 ,... , un + vn ) + (w1 , w2 ,... , wn ) [Vector addition]   = (u1 + v1 ) + w1 , (u2 + v2 ) + w2 ,... , (un + vn ) + wn [Vector addition]   = u1 + (v1 + w1 ), u2 + (v2 + w2 ),... , un + (vn + wn ) [Regroup] = (u1 , u2 ,... , un ) + (v1 + w1 , v2 + w2 ,... , vn + wn ) [Vector addition] = u + (v + w) The following additional properties of vectors in R n can be deduced easily by ex- pressing the vectors in terms of components (verify).  3.1 Vectors in 2-Space, 3-Space, and n-Space 139 THEOREM 3.1.2 If v is a vector in R n and k is a scalar, then: (a) 0v = 0 (b) k 0 = 0 (c) (−1)v = −v Calculating Without One of the powerful consequences of Theorems 3.1.1 and 3.1.2 is that they allow cal- Components culations to be performed without expressing the vectors in terms of components. For example, suppose that x, a, and b are vectors in R n , and we want to solve the vector equation x + a = b for the vector x without using components. We could proceed as follows: x+a=b [ Given ] (x + a) + (−a) = b + (−a) [ Add the negative of a to both sides ] x + (a + (−a)) = b − a [ Part (b) of Theorem 3.1.1 ] x+0=b−a [ Part (d ) of Theorem 3.1.1 ] x=b−a [ Part (c) of Theorem 3.1.1 ] While this method is obviously more cumbersome than computing with components in R n , it will become important later in the text where we will encounter more general kinds of vectors. Linear Combinations Addition, subtraction, and scalar multiplication are frequently used in combination to form new vectors. For example, if v1 , v2 , and v3 are vectors in R n , then the vectors u = 2v1 + 3v2 + v3 and w = 7v1 − 6v2 + 8v3 are formed in this way. In general, we make the following definition. Note that this definition of a DEFINITION 4 If w is a vector in R n , then w is said to be a linear combination of the linear combination is consis- vectors v1 , v2 ,... , vr in R n if it can be expressed in the form tent with that given in the con- text of matrices (see Definition w = k1 v1 + k2 v2 + · · · + kr vr (14) 6 in Section 1.3). where k1 , k2 ,... , kr are scalars. These scalars are called the coefficients of the linear combination. In the case where r = 1, Formula (14) becomes w = k1 v1 , so that a linear combination of a single vector is just a scalar multiple of that vector. Alternative Notations for Up to now we have been writing vectors in R n using the notation Vectors v = (v1 , v2 ,... , vn ) (15) n We call this the comma-delimited form. However, since a vector in R is just a list of its n components in a specific order, any notation that displays those components in the correct order is a valid way of representing the vector. For example, the vector in (15) can be written as v = [v1 v2 · · · vn ] (16) which is called row-vector form, or as ⎡ ⎤ v1 ⎢v ⎥ ⎢ 2⎥ v=⎢ ⎥ ⎢.. ⎥ (17) ⎣.⎦ vn 140 Chapter 3 Euclidean Vector Spaces which is called column-vector form. The choice of notation is often a matter of taste or convenience, but sometimes the nature of a problem will suggest a preferred notation. Notations (15), (16), and (17) will all be used at various places in this text. Application of Linear Combinations to Color Models Colors on computer monitors are commonly based on what is called The set of all such color vectors is called RGB space or the RGB the RGB color model. Colors in this system are created by adding color cube (Figure 3.1.14). Thus, each color vector c in this cube is together percentages of the primary colors red (R), green (G), and expressible as a linear combination of the form blue (B). One way to do this is to identify the primary colors with c = k1 r + k2 g + k3 b the vectors r = (1, 0, 0) (pure red), = k1 (1, 0, 0) + k2 (0, 1, 0) + k3 (0, 0, 1) g = (0, 1, 0) (pure green), = (k1 , k2 , k3 ) b = (0, 0, 1) (pure blue) where 0 ≤ ki ≤ 1. As indicated in the figure, the corners of the cube in R 3 and to create all other colors by forming linear combinations represent the pure primary colors together with the colors black, of r, g, and b using coefficients between 0 and 1, inclusive; these white, magenta, cyan, and yellow. The vectors along the diagonal coefficients represent the percentage of each pure color in the mix. running from black to white correspond to shades of gray. Blue Cyan (0, 0, 1) (0, 1, 1) Magenta White (1, 0, 1) (1, 1, 1) Black Green (0, 0, 0) (0, 1, 0) Red Yellow Figure 3.1.14 (1, 0, 0) (1, 1, 0) Exercise Set 3.1 −−→ In Exercises 1–2, find the components of the vector. In Exercises 3–4, find the components of the vector P1 P2. 3. (a) P1 (3, 5), P2 (2, 8) (b) P1 (5, −2, 1), P2 (2, 4, 2) 1. (a) y (b) z (1, 5) (0, 0, 4) 4. (a) P1 (−6, 2), P2 (−4, −1) (b) P1 (0, 0, 0), P2 (−1, 6, 1) 5. (a) Find the terminal point of the vector that is equivalent to u = (1, 2) and whose initial point is A(1, 1). y (b) Find the initial point of the vector that is equivalent to (4, 1) x u = (1, 1, 3) and whose terminal point is B(−1, −1, 2). x (2, 3, 0) 6. (a) Find the initial point of the vector that is equivalent to z u = (1, 2) and whose terminal point is B(2, 0). 2. (a) y (b) (0, 4, 4) (b) Find the terminal point of the vector that is equivalent to u = (1, 1, 3) and whose initial point is A(0, 2, 0). (–3, 3) (2, 3) (3, 0, 4) −→ 7. Find an initial point P of a nonzero vector u = PQ with ter- y minal point Q(3, 0, −5) and such that x (a) u has the same direction as v = (4, −2, −1). x (b) u is oppositely directed to v = (4, −2, −1).  3.1 Vectors in 2-Space, 3-Space, and n-Space 141 −→ 8. Find a terminal point Q of a nonzero vector u = PQ with 22. Show that there do not exist scalars c1 , c2 , and c3 such that initial point P (−1, 3, −5) and such that c1 (1, 0, 1, 0) + c2 (1, 0, −2, 1) + c3 (2, 0, 1, 2) = (1, −2, 2, 3) (a) u has the same direction as v = (6, 7, −3). 23. Let P be the point (2, 3, −2) and Q the point (7, −4, 1). (b) u is oppositely directed to v = (6, 7, −3). (a) Find the midpoint of the line segment connecting the 9. Let u = (4, −1), v = (0, 5), and w = (−3, −3). Find the points P and Q. components of (b) Find the point on the line segment connecting the points (a) u + w (b) v − 3u P and Q that is 43 of the way from P to Q. (c) 2(u − 5w) (d) 3v − 2(u + 2w) 24. In relation to the points P1 and P2 in Figure 3.1.12, what can 10. Let u = (−3, 1, 2), v = (4, 0, −8), and w = (6, −1, −4). you say about the terminal point of the following vector if its Find the components of initial point is at the origin? −−→ −−→ −−→ (a) v − w (b) 6u + 2v u = OP1 + 21 (OP2 − OP1 ) (c) −3(v − 8w) (d) (2u − 7w) − (8v + u) 25. In each part, find the components of the vector u + v + w. 11. Let u = (−3, 2, 1, 0), v = (4, 7, −3, 2), and (a) y (b) y w = (5, −2, 8, 1). Find the components of (a) v − w (b) −u + (v − 4w) w v (c) 6(u − 3v) (d) (6v − w) − (4u + v) x x 12. Let u = (1, 2, −3, 5, 0), v = (0, 4, −1, 1, 2), and u w u w = (7, 1, −4, −2, 3). Find the components of (a) v + w (b) 3(2u − v) v (c) (3u − v) − (2u + 4w) (d) 1 2 (w − 5v + 2u) + v 26. Referring to the vectors pictured in Exercise 25, find the com- 13. Let u, v, and w be the vectors in Exercise 11. Find the com- ponents of the vector u − v + w. ponents of the vector x that satisfies the equation 3u + v − 2w = 3x + 2w. 27. Let P be the point (1, 3, 7). If the point (4, 0, −6) is the mid- point of the line segment connecting P and Q, what is Q? 14. Let u, v, and w be the vectors in Exercise 12. Find the com- ponents of the vector x that satisfies the equation 28. If the sum of three vectors in R 3 is zero, must they lie in the 2u − v + x = 7x + w. same plane? Explain. 15. Which of the following vectors in R 6 , if any, are parallel to 29. Consider the regular hexagon shown in the accompanying fig- u = (−2, 1, 0, 3, 5, 1)? ure. (a) (4, 2, 0, 6, 10, 2) (a) What is the sum of the six radial vectors that run from the center to the vertices? (b) (4, −2, 0, −6, −10, −2) (b) How is the sum affected if each radial vector is multiplied (c) (0, 0, 0, 0, 0, 0) by 21 ? 16. For what value(s) of t, if any, is the given vector parallel to (c) What is the sum of the five radial vectors that remain if a u = (4, −1)? is removed? (a) (8t, −2) (b) (8t, 2t) (c) (1, t 2 ) (d) Discuss some variations and generalizations of the result in part (c). 17. Let u = (1, −1, 3, 5) and v = (2, 1, 0, −3). Find scalars a and b so that a u + bv = (1, −4, 9, 18). a 18. Let u = (2, 1, 0, 1, −1) and v = (−2, 3, 1, 0, 2). Find scalars f b a and b so that a u + bv = (−8, 8, 3, −1, 7). In Exercises 19–20, find scalars c1 , c2 , and c3 for which the equation is satisfied. e c 19. c1 (1, −1, 0) + c2 (3, 2, 1) + c3 (0, 1, 4) = (−1, 1, 19) 20. c1 (−1, 0, 2) + c2 (2, 2, −2) + c3 (1, −2, 1) = (−6, 12, 4) d Figure Ex-29 21. Show that there do not exist scalars c1 , c2 , and c3 such that 30. What is the sum of all radial vectors of a regular n-sided poly- c1 (−2, 9, 6) + c2 (−3, 2, 1) + c3 (1, 7, 5) = (0, 5, 4) gon? (See Exercise 29.) 142 Chapter 3 Euclidean Vector Spaces Working with Proofs (f ) If a and b are scalars such that a u + bv = 0, then u and v are 31. Prove parts (a), (c), and (d) of Theorem 3.1.1. parallel vectors. 32. Prove parts (e)–(h) of Theorem 3.1.1. (g) Collinear vectors with the same length are equal. 33. Prove parts (a)–(c) of Theorem 3.1.2. (h) If (a, b, c) + (x, y, z) = (x, y, z), then (a, b, c) must be the zero vector. True-False Exercises (i) If k and m are scalars and u and v are vectors, then TF. In parts (a)–(k) determine whether the statement is true or false, and justify your answer. (k + m)(u + v) = k u + mv (a) Two equivalent vectors must have the same initial point. ( j) If the vectors v and w are given, then the vector equation (b) The vectors (a, b) and (a, b, 0) are equivalent. 3(2v − x) = 5x − 4w + v (c) If k is a scalar and v is a vector, then v and k v are parallel if and only if k ≥ 0. can be solved for x. (d) The vectors v + (u + w) and (w + v) + u are the same. (k) The linear combinations a1 v1 + a2 v2 and b1 v1 + b2 v2 can only (e) If u + v = u + w, then v = w. be equal if a1 = b1 and a2 = b2. 3.2 Norm, Dot Product, and Distance in R n In this section we will be concerned with the notions of length and distance as they relate to vectors. We will first discuss these ideas in R 2 and R 3 and then extend them algebraically to R n. Norm of a Vector In this text we will denote the length of a vector v by the symbol v , which is read as the norm of v, the length of v, or the magnitude of v (the term “norm” being a common mathematical synonym for length). As suggested in Figure 3.2.1a, it follows from the y (v1, v2) Theorem of Pythagoras that the norm of a vector (v1 , v2 ) in R 2 is ||v|| √ v2 v = v12 + v22 (1) x v1 Similarly, for a vector (v1 , v2 , v3 ) in R 3 , it follows from Figure 3.2.1b and two applica- (a) tions of the Theorem of Pythagoras that z v 2 = (OR)2 + (RP )2 = (OQ)2 + (QR)2 + (RP )2 = v12 + v22 + v32 P(v1, v2, v3) and hence that √ ||v|| v = v12 + v22 + v32 (2) y O Motivated by the pattern of Formulas (1) and (2), we make the following definition. S Q x R (b) DEFINITION 1 If v = (v1 , v2 ,... , vn ) is a vector in R n , then the norm of v (also called the length of v or the magnitude of v) is denoted by v , and is defined by the formula Figure 3.2.1 " v = v12 + v22 + · · · + vn2 (3)  3.2 Norm, Dot Product, and Distance in R n 143 E X A M P L E 1 Calculating Norms It follows from Formula (2) that the norm of the vector v = (−3, 2, 1) in R 3 is " √ v = (−3)2 + 22 + 12 = 14 and it follows from Formula (3) that the norm of the vector v = (2, −1, 3, −5) in R 4 is " √ v = 22 + (−1)2 + 32 + (−5)2 = 39 Our first theorem in this section will generalize to R n the following three familiar facts about vectors in R 2 and R 3 : Distances are nonnegative. The zero vector is the only vector of length zero. Multiplying a vector by a scalar multiplies its length by the absolute value of that scalar. It is important to recognize that just because these results hold in R 2 and R 3 does not guarantee that they hold in R n —their validity in R n must be proved using algebraic properties of n-tuples. THEOREM 3.2.1 If v is a vector in R n , and if k is any scalar, then: (a) v ≥0 (b) v = 0 if and only if v = 0 (c) k v = |k| v We will prove part (c) and leave (a) and (b) as exercises. Proof (c) If v = (v1 , v2 ,... , vn ), then k v = (kv1 , kv2 ,... , kvn ), so " k v = (kv1 )2 + (kv2 )2 + · · · + (kvn )2 # = (k 2 )(v12 + v22 + · · · + vn2 ) # = |k| v12 + v22 + · · · + vn2 = |k| v Unit Vectors A vector of norm 1 is called a unit vector. Such vectors are useful for specifying a direction when length is not relevant to the problem at hand. You can obtain a unit vector in a desired direction by choosing any nonzero vector v in that direction and multiplying v by the reciprocal of its length. For example, if v is a vector of length 2 in R 2 or R 3 , then 21 v is a unit vector in the same direction as v. More generally, if v is any nonzero WARNING Sometimes you will vector in R n , then see Formula (4) expressed as 1 v u= v (4) u= v v This is just a more compact defines a unit vector that is in the same direction as v. We can confirm that (4) is a unit way of writing that formula vector by applying part (c) of Theorem 3.2.1 with k = 1/ v to obtain and is not intended to convey that v is being divided by v. 1 u = k v = |k| v = k v = v =1 v 144 Chapter 3 Euclidean Vector Spaces The process of multiplying a nonzero vector by the reciprocal of its length to obtain a unit vector is called normalizing v. E X A M P L E 2 Normalizing a Vector Find the unit vector u that has the same direction as v = (2, 2, −1). Solution The vector v has length " v = 22 + 22 + (−1)2 = 3 Thus, from (4) 2  u = 13 (2, 2, −1) = 3 , 23 , − 13 As a check, you may want to confirm that u = 1. The Standard Unit Vectors When a rectangular coordinate system is introduced in R 2 or R 3 , the unit vectors in the positive directions of the coordinate axes are called the standard unit vectors. In R 2 these y vectors are denoted by (0, 1) i = (1, 0) and j = (0, 1) and in R 3 by j x i = (1, 0, 0), j = (0, 1, 0), and k = (0, 0, 1) i (1, 0) (Figure 3.2.2). Every vector v = (v1 , v2 ) in R 2 and every vector v = (v1 , v2 , v3 ) in R 3 (a) can be expressed as a linear combination of standard unit vectors by writing z v = (v1 , v2 ) = v1 (1, 0) + v2 (0, 1) = v1 i + v2 j (5) (0, 0, 1) v = (v1 , v2 , v3 ) = v1 (1, 0, 0) + v2 (0, 1, 0) + v3 (0, 0, 1) = v1 i + v2 j + v3 k (6) k j y Moreover, we can generalize these formulas to R n by defining the standard unit vectors i in R n to be (0, 1, 0) (1, 0, 0) x e1 = (1, 0, 0,... , 0), e2 = (0, 1, 0,... , 0),... , en = (0, 0, 0,... , 1) (7) (b) in which case every vector v = (v1 , v2 ,... , vn ) in R n can be expressed as Figure 3.2.2 v = (v1 , v2 ,... , vn ) = v1 e1 + v2 e2 + · · · + vn en (8) E X A M P L E 3 Linear Combinations of Standard Unit Vectors (2, −3, 4) = 2i − 3j + 4k (7, 3, −4, 5) = 7e1 + 3e2 − 4e3 + 5e4 −−→ Distance in Rn If P1 and P2 are points in R 2 or R 3 , then the length of the vector P1 P2 is equal to the distance d between the two points (Figure 3.2.3). Specifically, if P1 (x1 , y1 ) and P2 (x2 , y2 ) are points in R 2 , then Formula (4) of Section 3.1 implies that −−→ " d = P1 P2 = (x2 − x1 )2 + (y2 − y1 )2 (9)  3.2 Norm, Dot Product, and Distance in R n 145 P2 This is the familiar distance formula from analytic geometry. Similarly, the distance d between the points P1 (x1 , y1 , z1 ) and P2 (x2 , y2 , z2 ) in 3-space is −−→ " P1 d(u, v) = P1 P2 = (x2 − x1 )2 + (y2 − y1 )2 + (z2 − z1 )2 (10) d = ||P1P2|| Motivated by Formulas (9) and (10), we make the following definition. Figure 3.2.3 DEFINITION 2 If u = (u1 , u2 ,... , un ) and v = (v1 , v2 ,... , vn ) are points in R n , then we denote the distance between u and v by d(u, v) and define it to be We noted in the previous " section that n-tuples can be d(u, v) = u − v = (u1 − v1 )2 + (u2 − v2 )2 + · · · + (un − vn )2 (11) viewed either as vectors or points in R n. In Definition 2 we chose to describe them E X A M P L E 4 Calculating Distance in R n as points, as that seemed the more natural interpretation. If u = (1, 3, −2, 7) and v = (0, 7, 2, 2) then the distance between u and v is " √ d(u, v) = (1 − 0)2 + (3 − 7)2 + (−2 − 2)2 + (7 − 2)2 = 58 Dot Product Our next objective is to define a useful multiplication operation on vectors in R 2 and R 3 and then extend that operation to R n. To do this we will first need to define exactly what we mean by the “angle” between two vectors in R 2 or R 3. For this purpose, let u and v be nonzero vectors in R 2 or R 3 that have been positioned so that their initial points coincide. We define the angle between u and v to be the angle θ determined by u and v that satisfies the inequalities 0 ≤ θ ≤ π (Figure 3.2.4). u u θ θ θ v v v u v u θ The angle θ between u and v satisfies 0 ≤ θ ≤ π. Figure 3.2.4 DEFINITION 3 If u and v are nonzero vectors in R 2 or R 3 , and if θ is the angle between u and v, then the dot product (also called the Euclidean inner product) of u and v is denoted by u · v and is defined as u·v= u v cos θ (12) If u = 0 or v = 0, then we define u · v to be 0. The sign of the dot product reveals information about the angle θ that we can obtain by rewriting Formula (12) as u·v cos θ = (13) u v Since 0 ≤ θ ≤ π , it follows from Formula (13) and properties of the cosine function studied in trigonometry that θ is acute if u · v > 0. θ is obtuse if u · v < 0. θ = π/2 if u · v = 0. 146 Chapter 3 Euclidean Vector Spaces z E X A M P L E 5 Dot Product Find the dot product of the vectors shown in Figure 3.2.5. (0, 2, 2) v Solution The lengths of the vectors are √ √ (0, 0, 1) u = 1 and v = 8=2 2 u θ = 45° y and the cosine of the angle θ between them is √ cos(45◦ ) = 1/ 2 x Thus, it follows from Formula (12) that Figure 3.2.5 √ √ u·v= u v cos θ = (1)(2 2)(1/ 2) = 2 Component Form of the For computational purposes it is desirable to have a formula that expresses the dot Dot Product product of two vectors in terms of components. We will derive such a formula for vectors in 3-space; the derivation for vectors in 2-space is similar. z Let u = (u1 , u2 , u3 ) and v = (v1 , v2 , v3 ) be two nonzero vectors. If, as shown in Figure 3.2.6, θ is the angle between u and v, then the law of cosines yields P(u1, u2, u3) −→ PQ 2 = u 2 + v 2 − 2 u v cos θ (14) u −→ v Q(v1, v2, v3) Since PQ = v − u, we can rewrite (14) as θ y u v cos θ = 21 ( u 2 + v 2 − v − u 2) x or Figure 3.2.6 u · v = 21 ( u 2 + v 2 − v − u 2) Substituting u 2 = u21 + u22 + u23 , v 2 = v12 + v22 + v32 and v−u 2 = (v1 − u1 )2 + (v2 − u2 )2 + (v3 − u3 )2 we obtain, after simplifying, Although we derived Formula (15) and its 2-space compan- u · v = u1 v1 + u2 v2 + u3 v3 (15) ion under the assumption that u and v are nonzero, it turned The companion formula for vectors in 2-space is out that these formulas are also applicable if u = 0 or u · v = u1 v1 + u2 v2 (16) v = 0 (verify). Motivated by the pattern in Formulas (15) and (16), we make the following definition. Historical Note The dot product notation was first in- troduced by the American physicist and mathemati- cian J. Willard Gibbs in a pamphlet distributed to his students at Yale University in the 1880s. The prod- uct was originally written on the baseline, rather than centered as today, and was referred to as the direct product. Gibbs’s pamphlet was eventually incorpo- rated into a book entitled Vector Analysis that was pub- lished in 1901 and coauthored with one of his students. Gibbs made major contributions to the fields of ther- modynamics and electromagnetic theory and is gen- erally regarded as the greatest American physicist of Josiah Willard Gibbs the nineteenth century. (1839–1903) [Image: SCIENCE SOURCE/Photo Researchers/ Getty Images]  3.2 Norm, Dot Product, and Distance in R n 147 In words, to calculate the DEFINITION 4 If u = (u1 , u2 ,... , un ) and v = (v1 , v2 ,... , vn ) are vectors in R n , then dot product (Euclidean inner the dot product (also called the Euclidean inner product) of u and v is denoted by u · v product) multiply correspond- and is defined by ing components and add the u · v = u1 v1 + u2 v2 + · · · + un vn (17) resulting products. E X A M P L E 6 Calculating Dot Products Using Components (a) Use Formula (15) to compute the dot product of the vectors u and v in Example 5. (b) Calculate u · v for the following vectors in R 4 : u = (−1, 3, 5, 7), v = (−3, −4, 1, 0) Solution (a) The component forms of the vectors are u = (0, 0, 1) and v = (0, 2, 2). Thus, u · v = (0)(0) + (0)(2) + (1)(2) = 2 which agrees with the result obtained geometrically in Example 5. z (0, 0, k) Solution (b) u · v = (−1)(−3) + (3)(−4) + (5)(1) + (7)(0) = −4 u3 (k, k, k) d E X A M P L E 7 A Geometry Problem Solved Using Dot Product u2 y Find the angle between a diagonal of a cube and one of its edges. u1 θ (0, k, 0) Solution Let k be the length of an edge and introduce a coordinate system as shown in x Figure 3.2.7. If we let u1 = (k, 0, 0), u2 = (0, k, 0), and u3 = (0, 0, k), then the vector (k, 0, 0) d = (k, k, k) = u1 + u2 + u3 Figure 3.2.7 is a diagonal of the cube. It follows from Formula (13) that the angle θ between d and the edge u1 satisfies u1 · d k2 1 Note that the angle θ obtained cos θ = = √ =√ in Example 7 does not involve u1 d (k)( 3k 2 ) 3 With the help of a calculator we obtain k. Why was this to be ex- ! pected? −1 1 θ = cos √ ≈ 54.74◦ 3 Algebraic Properties of the In the special case where u = v in Definition 4, we obtain the relationship Dot Product v · v = v12 + v22 + · · · + vn2 = v 2 (18) This yields the following formula for expressing the length of a vector in terms of a dot product: √ v = v·v (19) Dot products have many of the same algebraic properties as products of real numbers. THEOREM 3.2.2 If u, v, and w are vectors in R n , and if k is a scalar, then: (a) u · v = v · u [ Symmetry property ] (b) u · (v + w) = u · v + u · w [ Distributive property ] (c) k(u · v) = (k u) · v [ Homogeneity property ] (d ) v · v ≥ 0 and v · v = 0 if and only if v = 0 [ Positivity property ] We will prove parts (c) and (d) and leave the other proofs as exercises. 148 Chapter 3 Euclidean Vector Spaces Proof (c) Let u = (u1 , u2 ,... , un ) and v = (v1 , v2 ,... , vn ). Then k(u · v) = k(u1 v1 + u2 v2 + · · · + un vn ) = (ku1 )v1 + (ku2 )v2 + · · · + (kun )vn = (k u) · v Proof (d) The result follows from parts (a) and (b) of Theorem 3.2.1 and the fact that v · v = v1 v1 + v2 v2 + · · · + vn vn = v12 + v22 + · · · + vn2 = v 2 The next theorem gives additional properties of dot products. The proofs can be obtained either by expressing the vectors in terms of components or by using the algebraic properties established in Theorem 3.2.2. THEOREM 3.2.3 If u, v, and w are vectors in R n , and if k is a scalar, then: (a) 0 · v = v · 0 = 0 (b) (u + v) · w = u · w + v · w (c) u · (v − w) = u · v − u · w (d ) (u − v) · w = u · w − v · w (e) k(u · v) = u · (k v) We will show how Theorem 3.2.2 can be used to prove part (b) without breaking the vectors into components. The other proofs are left as exercises. Proof (b) (u + v) · w = w · (u + v) [By symmetry] =w·u+w·v [By distributivity] =u·w+v·w [By symmetry] Formulas (18) and (19) together with Theorems 3.2.2 and 3.2.3 make it possible to manipulate expressions involving dot products using familiar algebraic techniques. E X A M P L E 8 Calculating with Dot Products (u − 2v) · (3u + 4v) = u · (3u + 4v) − 2v · (3u + 4v) = 3(u · u) + 4(u · v) − 6(v · u) − 8(v · v) =3 u 2 − 2(u · v) − 8 v 2 Cauchy–Schwarz Inequality Our next objective is to extend to R n the notion of “angle” between nonzero vectors u and Angles in Rn and v. We will do this by starting with the formula ! −1 u·v θ = cos (20) u v which we previously derived for nonzero vectors in R 2 and R 3. Since dot products and norms have been defined for vectors in R n , it would seem that this formula has all the ingredients to serve as a definition of the angle θ between two vectors, u and v, in R n. However, there is a fly in the ointment, the problem being that the inverse cosine in Formula (20) is not defined unless its argument satisfies the inequalities u·v −1 ≤ ≤1 (21) u v Fortunately, these inequalities do hold for all nonzero vectors in R n as a result of the following fundamental result known as the Cauchy–Schwarz inequality.  3.2 Norm, Dot Product, and Distance in R n 149 THEOREM 3.2.4 Cauchy–Schwarz Inequality If u = (u1 , u2 ,... , un ) and v = (v1 , v2 ,... , vn ) are vectors in R n , then |u · v| ≤ u v (22) or in terms of components |u1 v1 + u2 v2 + · · · + un vn | ≤ (u21 + u22 + · · · + u2n )1/2 (v12 + v22 + · · · + vn2 )1/2 (23) We will omit the proof of this theorem because later in the text we will prove a more general version of which this will be a special case. Our goal for now will be to use this theorem to prove that the inequalities in (21) hold for all nonzero vectors in R n. Once that is done we will have established all the results required to use Formula (20) as our definition of the angle between nonzero vectors u and v in R n. To prove that the inequalities in (21) hold for all nonzero vectors in R n , divide both sides of Formula (22) by the product u v to obtain   |u · v|  u·v  ≤ 1 or equivalently  ≤1 u v u v  from which (21) follows. Geometry in Rn Earlier in this section we extended various concepts to R n with the idea that familiar results that we can visualize in R 2 and R 3 might be valid in R n as well. Here are two fundamental theorems from plane geometry whose validity extends to R n : The sum of the lengths of two side of a triangle is at least as large as the third (Figure u+v 3.2.8). v The shortest distance between two points is a straight line (Figure 3.2.9). The following theorem generalizes these theorems to R n. u ||u + v|| ≤ ||u|| + ||v|| THEOREM 3.2.5 If u, v, and w are vectors in R n , then: (a) u+v ≤ u + v [ Triangle inequality for vectors ] Figure 3.2.8 (b) d(u, v) ≤ d(u, w) + d(w, v) [ Triangle inequality for distances ] v Historical Note The Cauchy–Schwarz in- equality is named in honor of the French mathematician Augustin Cauchy (see p. 121) and the German mathemati- cian Hermann Schwarz. Variations of this inequality occur in many different settings w and under various names. Depending on the context in which the inequality occurs, u you may find it called Cauchy’s inequal- d(u, v) ≤ d(u, w) + d(w, v) ity, the Schwarz inequality, or sometimes even the Bunyakovsky inequality, in recog- Figure 3.2.9 nition of the Russian mathematician who Hermann Amandus Viktor Yakovlevich published his version of the inequality in Schwarz Bunyakovsky 1859, about 25 years before Schwarz. (1843–1921) (1804–1889) [Images: © Rudolph Duehrkoop/ ullstein bild/The Image Works (Schwarz); http://www-history.mcs.st-and.ac.uk/ Biographies/Bunyakovsky.html (Bunyakovsky)] 150 Chapter 3 Euclidean Vector Spaces Proof (a) u+v 2 = (u + v) · (u + v) = (u · u) + 2(u · v) + (v · v) = u 2 + 2(u · v) + v 2 ≤ u 2 + 2|u · v| + v 2 Property of absolute value ≤ u 2+2 u v + v 2 Cauchy–Schwarz inequality =( u + v ) 2 This completes the proof since both sides of the inequality in part (a) are nonnegative. Proof (b) It follows from part (a) and Formula (11) that d(u, v) = u − v = (u − w) + (w − v) ≤ u − w + w − v = d(u, w) + d(w, v) It is proved in plane geometry that for any parallelogram the sum of the squares of u+v v the diagonals is equal to the sum of the squares of the four sides (Figure 3.2.10). The u–v following theorem generalizes that result to R n. u Figure 3.2.10 THEOREM 3.2.6 Parallelogram Equation for Vectors If u and v are vectors in R n , then   u+v 2 + u−v 2 =2 u 2 + v 2 (24) Proof u+v 2 + u−v 2