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Lecture 9: 3.2 Norm of a Vector Wei-Ta Chu 2008/10/17 Translation of Axes In the figure we have translated the axes of an xy-coordinate system to obtain an x y coordinate system whose O is at point (x, y) = (k,l). A point P in 2-space now has both (x, y) coordinates and (x , y ) coordinates. x = x –k, y = y –l, these formulas are called the translation equations. In 3-space the translation equations are x = x –k, y = y –l, z = z –m, where (k, l, m) are the xyz-coordinates of the x y z -origin. 2008/10/17 Elementary Linear Algebra 2 Example Suppose that an xy-coordinate system is translated to obtain an x ’ y ’ -coordinate system whose origin has xy-coordinate (k, l) = (4, 1) The point with xy-coordinates P(2,0) has x ’= x –4, y ’= y –1. The x ’ y ’ -coordinates of P are (-2,-1) The point with x ’ y ’ -coordinates Q(-1,5) x=x ’ +4, y = y ’ +1 The xy-coordinates of Q are (3, 6) 2008/10/17 Elementary Linear Algebra 3 Theorem 3.2.1 (Properties of Vector Arithmetic) If u, v and w are vectors in 2- or 3-space and k and l are scalars, then the following relationships hold. u+v=v+u (u + v) + w = u + (v + w) u+0=0+u=u u + (-u) = 0 k(lu) = (kl)u k(u + v) = ku + kv (k + l)u = ku + lu 1u = u 2008/10/17 Elementary Linear Algebra 4 Proof of part (b) (geometric) 2008/10/17 Elementary Linear Algebra 5 Norm of a Vector The length of a vector u is often called the norm (範數) of u and is denoted by ||u||. It follows from the Theorem of Pythagoras that the norm of a vector u = (u1,u2,u3) in 3-space is u u12 u22 u32 A vector of norm 1 is called a unit vector. (單位向量) The distance between two points is the norm of the vector. The length of the vector ku = ||ku|| = |k| ||u||. 2008/10/17 Elementary Linear Algebra 6 Distance If P1(x1, y1, z1) and P2(x2, y2, z2) are two points in 3-space, then the distance d between them is the norm of the vector Euclidean distance (歐幾里德距離, 歐式距離) 2008/10/17 Elementary Linear Algebra 7 Lecture 9: 3.3 Dot Product and Projection Wei-Ta Chu 2008/10/17 Definitions Let u and v be two nonzero vectors in 2-space or 3-space, and assume these vectors have been positioned so their initial points coincided. By the angle between u and v, we shall mean the angle determined by u and v that satisfies 0 . If u and v are vectors in 2-space or 3-space and is the angle between u and v, then the dot product (點積) or Euclidean inner product (內積) u · v is defined by u v cos if u 0 and v 0 u v 0 if u 0 or v 0 2008/10/17 Elementary Linear Algebra 9 Example If the angle between the vectors u = (0,0,1) and v = (0,2,2) is 45 , then 1 u v u v cos 0 0 1 0 4 4 2 2 u v (u1 , u2 , u3 ) (v1 , v2 , v3 ) u1v1 u2 v2 u3v3 2 u v 2 1 cos u v 0 0 1 0 4 4 2 2008/10/17 Elementary Linear Algebra 10 Component Form of Dot Product Let u=(u1,u2,u3) and v=(v1,v2,v3) be two nonzero vectors. According to the law of cosine P Q 2008/10/17 Elementary Linear Algebra 11 Theorems Theorem 3.3.1 Let u and v be vectors in 2- or 3-space. v · v = ||v||2; that is, ||v|| = (v · v)½ If the vectors u and v are nonzero and is the angle between them, then u v cos if u 0 and v 0 u v 0 if u 0 or v 0 is acute (銳角) is obtuse (鈍角) = /2 (直角) if and only if u · v > 0 if and only if u · v < 0 if and only if u · v = 0 Theorem 3.3.2 (Properties of the Dot Product) If u, v and w are vectors in 2- or 3-space, and k is a scalar, then 2008/10/17 u·v=v·u u · (v + w) = u · v + u · w k(u · v) = (ku) · v = u · (kv) v · v > 0 if v 0, and v · v = 0 if v = 0 Elementary Linear Algebra 12 Proof of Theorem 3.3.1 (a) Since the angle (b) Since satisfies , it follows that is acute if and only if , that is obtuse if and only if , and that if and only if . But has the same sign as since , . Thus, the result follows. 2008/10/17 between and Elementary Linear Algebra is 0, we have 13 Proof of Theorem 3.3.2 k(u · v) = (ku) · v = u · (kv) Let u=(u1,u2,u3) and v=(v1,v2,v3) 2008/10/17 Elementary Linear Algebra 14 Orthogonal Vectors Definition Perpendicular vectors are also called orthogonal (正交) vectors. Two nonzero vectors are orthogonal if and only if their dot product is zero. To indicate that u and v are orthogonal vectors we write u⊥v. 2008/10/17 Elementary Linear Algebra 15 Example Show that in 2-space the nonzero vector n = (a,b) is perpendicular to the line ax + by + c = 0. Let P1(x1,y1) and P2(x2,y2) be distinct points on the line, so that Since the vector runs along the line, we need only show that n and are perpendicular. But on subtracting the equations, we obtain y ax + by + c = 0 P1 P2 x 2008/10/17 Elementary Linear Algebra 16 An Orthogonal Projection To "decompose" a vector u into a sum of two terms, one parallel to a specified nonzero vector a and the other perpendicular to a. We have w2 = u –w1 and w1 + w2 = w1 + (u –w1) = u The vector w1 is called the orthogonal projection (正交投影) of u on a or sometimes the vector component (分向量) of u along a, and denoted by projau The vector w2 is called the vector component of u orthogonal to a, and denoted by w2 = u –projau 2008/10/17 Elementary Linear Algebra 17 w1 proja u Theorem 3.3.3 w 2 u proja u If u and a are vectors in 2-space or 3-space and if a 0, then u a proja u 2 a a (vector component of u along a) u a u proja u u 2 a a (vector component of u orthogonal to a) u a proja u u cos a 2008/10/17 Elementary Linear Algebra 18 Proof of Theorem 3.3.3 Let and . Since w1 is parallel to a, it must be a scalar multiple of a, so it can be written in the form w1 = ka. Thus Taking the dot product But Since w1 = ka, we obtain 2008/10/17 Elementary Linear Algebra 19 Example u a proja u 2 a a u a u proja u u 2 a a Let u ( 2,1,3) and a ( 4,1,2). Find the vector component of u along a and the vector component of u orthogonal to a. Solution: u a ( 2)( 4) ( 1)( 1) (3)( 2) 15 a 2 4 2 ( 1) 2 2 2 21 Thus, the vector component of u along a is u a proj a u 2 a 15 ( 4,1,2) ( 207 ,75 , 107 ) 21 a and the vector component of u orthogonal to a is u proj a u ( 2, 1,3) ( 207 ,75 , 107 ) ( 67 ,72 , 117 ) Verify tha t the vector u proj a u and a are perpendicu lar by showing that their dot product is zero. 2008/10/17 Elementary Linear Algebra 20 Length of Orthogonal Projection scalar Formula (5) of Section 3.2 Since If denotes the angle between u and a, then 2008/10/17 Elementary Linear Algebra 21 Length of Orthogonal Projection u a u a 2008/10/17 Elementary Linear Algebra 22 Distance between a Point and a Line y ax + by + c = 0 n = (a,b) D Q(x1,y1) P0(x0,y0) x Find a formula for the distance D between point P0 ( x0 , y0 ) and the line ax by c 0. Solution: Let Q( x1 , y1 ) be any point on the line and position the vector n (a, b) so that its initial point is at Q. By virtueof Example5, the vector n is perpendicular to the line (Fig 3.3.8). As indicated in the figure, the distance D is equal to the length of the orthogonal projection of QP0 on n; thus, QP0 n D projn QP0 n But QP0 ( x0 x1 , y0 y1 ), 2008/10/17 QP0 n a( x0 x1 ) b( y0 y1 ), Elementary Linear Algebra n a 2 b 2 23 Distance between a Point and a Line Since the point Q(x1,y1) lies on the lines, it coordinates satisfy the equation of the line, so ax1 + by1 + c = 0 or c = -ax1 –by1 Example: the distance D from the point (1,-2) to the line 3x+4y-6 = 0 is 2008/10/17 Elementary Linear Algebra 24 Cross Product (叉積, 外積) Definition If u = (u1, u2, u3) and v = (v1, v2, v3) are vectors in 3-space, then the cross product u v is the vector defined by u v (u2 v3 u3v2 , u3v1 u1v3 , u1v2 u2 v1 ) or in determinant notation u2 u v v 2 2008/10/17 u3 u1 u3 u1 u2 , , v3 v1 v3 v1 v2 Elementary Linear Algebra 25 Example Find u v, where u = (1, 2, -2) and v = (3, 0, 1). 2008/10/17 Elementary Linear Algebra 26 Cross Product Form the matrix whose first row contains the components of u and whose second row contains the components of v. To find the first component of , delete the first column and take the determinant; to find the second component, delete the second column and take the negative of the determinant; and to find the third component, delete the third column and take the determinant. 2008/10/17 Elementary Linear Algebra 27 Theorems Theorem 3.4.1 (Relationships Involving Cross Product and Dot Product) If u, v and w are vectors in 3-space, then 2008/10/17 u · (u v) = 0 v · (u v) = 0 || u v ||2 = ||u||2||v||2 –(u · v)2 u (v w) = (u · w) v –(u · v) w product) (u v) w = (u · w) v –(v · w) u product) (u v is orthogonal to u) (u v is orthogonal to v) ( La gr a nge ’ si de nt i t y) (relationship between cross & dot (relationship between cross & dot Elementary Linear Algebra 28 Proof of Theorem 3.4.1(a) Let u=(u1,u2,u3) and v=(v1,v2,v3) Example: u=(1, 2, -2) and v=(3, 0, 1) 2008/10/17 Elementary Linear Algebra 29 Proof of Theorem 3.4.1(c) u v (u2 v3 u3v2 , u3v1 u1v3 , u1v2 u2 v1 ) 2008/10/17 Elementary Linear Algebra 30 Theorems Theorem 3.4.2 (Properties of Cross Product) If u, v and w are any vectors in 3-space and k is any scalar, then u v = - (v u) u (v + w) = u v + u w (u + v) w = u w + v w k(u v) = (ku) v = u (kv) u 0 = 0 u = 0 u u = 0 Proof of (a) Interchanging u and v interchanges the rows of the three determinants and hence changes the sign of each component in the cross product. Thus u v = - (v u). 2008/10/17 Elementary Linear Algebra 31 Standard Unit Vectors z (0,0,1) x The vectors (0,1,0) y (1,0,0) i = (1,0,0), j = (0,1,0), k = (0,0,1) have length 1 and lie along the coordinate axes. They are called the standard unit vectors in 3-space. Every vector v = (v1, v2, v3) in 3-space is expressible in terms of i, j, k since we can write v = (v1, v2, v3) = v1(1,0,0) + v2 (0,1,0) + v3 (0,0,1) = v1i + v2j + v3k For example, (2, -3, 4) = 2i –3j +4k Note that i i = 0, j j = 0, k k = 0 i j = k, j k = i, k i = j j i = -k, k j = -i, i k = -j 2008/10/17 Elementary Linear Algebra 32 Cross Product A cross product can be represented symbolically in the form of 33 determinant: i j u v u1 u2 v1 v2 k u2 u3 v2 v3 u3 u1 u3 u1 u2 i j k v3 v1 v3 v1 v2 Example: if u=(1,2,-2) and v=(3,0,1) 2008/10/17 Elementary Linear Algebra 33 Cross Product I t ’ snott r uei nge ne r a lt ha t For example: Right-hand rule If the fingers of the right hand are cupped so they point in the direction of rotation, then the thumb indicates the direction of 2008/10/17 Elementary Linear Algebra u v 34 Exercises Sec. 3.1: 6, 12, 21 Sec. 3.2: 2, 9, 12 Sec. 3.3: 3, 4, 8, 16, 24 Sec. 3.4: 2, 8, 13, 17, 36 Sec. 3.5: 3, 6, 9, 15, 20, 27 2008/10/17 Elementary Linear Algebra 35