Vectors CHAPTER 7

Ch7_2 Contents  7.1 Vectors in 2-Space 7.1 Vectors in 2-Space  7.2 Vectors in 3-Space 7.2 Vectors in 3-Space  7.3 Dot Product 7.3 Dot Product  7.4 Cross Product 7.4 Cross Product  7.5 Lines and Planes in 3-Space 7.5 Lines and Planes in 3-Space  7.6 Vector Spaces 7.6 Vector Spaces  7.7 Gram-Schmidt Orthogonalization Process 7.7 Gram-Schmidt Orthogonalization Process

Ch7_3 7.1 Vectors in 2-Space  Review of Vectors Please refer to Fig 7.1 through Fig 7.6.

Ch7_4 Fig 7.1 (Geometric Vectors)

Ch7_5 Fig 7.2 (Vectors are equal)

Ch7_6 Fig 7.3 (Parallel vectors)

Ch7_7 Fig 7.4 (sum)

Ch7_8 Fig 7.5 (difference)

Ch7_9 Fig 7.6 (position vectors)

Ch7_10 Example 1  Please refer to Fig 7.7.  Fig 7.7

Ch7_11 Let a =, b = be vectors in R 2 (i) Addition: a + b = (1) (ii) Scalar multiplication: ka =, k is a scalar(2) (iii)Equality: a = b if and only if a 1 = b 1, a 2 = b 2 (3) DEFINITION 7.1 Addition, Scalar Multiplication, Equality a – b = (4)

Ch7_12 Graph Solution  Fig 7.8 shows the graph solutions of the addition and subtraction of two vectors.

Ch7_13 Example 2 If a =, b =, find a + b, a − b, 2a + 3b. Solution Using (1), (2), (4), we have

Ch7_14 Properties  (i) a + b = b + a (ii) a + (b + c) = (a + b) + c (iii) a + 0 = a (iv) a + (−a) = 0 (v) k(a + b) = ka + kbk scalar (vi) (k 1 + k 2 )a = k 1 a + k 2 ak 1, k 2 scalars (vii) k 1 (k 2 a) = (k 1 k 2 )ak 1, k 2 scalars (viii) 1a = a (ix) 0a = 0 =  0 =

Ch7_15 Magnitude, Length, Norm  a =, then Clearly, we have ||a||  0, ||0|| = 0

Ch7_16 Unit Vector  A vector that ha magnitude 1 is called a unit vector. u = (1/||a||)a is a unit vector, since

Ch7_17 Example 3  Given a =, then the unit vector in the same direction u is and

Ch7_18 The i, j vectors  If a =, then (5) Let i =, j =, then (5) becomes a = a 1 i + a 2 j(6)

Ch7_19 Fig 7.10

Ch7_20 Example 4  (i) = 4i + 7j (ii) (2i – 5j) + (8i + 13j) = 10i + 8j (iii) (iv) 10(3i – j) = 30i – 10j (v) a = 6i + 4j, b = 9i + 6j are parallel and b = (3/2)a

Ch7_21 Example 5 Let a = 4i + 2j, b = –2i + 5j. Graph a + b, a – b Solution Fig 7.11

Ch7_ Vectors in 3-Space  Simple Review Please refer to Fig 7.22 through Fig  Fig 7.22

Ch7_23 Fig 7.23

Ch7_24 Fig 7.24

Ch7_25 Example 1 Graph the points (4, 5, 6) and (−2, −2, 0). Solution See Fig 7.25.

Ch7_26 Distance Formula  (1)  Fig 7.26

Ch7_27 Example 2 Find the distance between (2, −3, 6) and (−1, −7, 4) Solution

Ch7_28 Midpoint Formula  (2)

Ch7_29 Example 2 Find the midpoint of (2, −3, 6) and (−1, −7, 4) Solution From (2), we have

Ch7_30 Vectors in 3-Space  Fig 7.27.

Ch7_31 Let a =, b = in R 3 (i) a + b = (ii) ka = (iii) a = b if and only if a 1 = b 1, a 2 = b 2, a 3 = b 3 (iv) –b = (−1)b = (v) a – b = (vi) 0 = (vi) DEFINITION 7.2 Component Definitions in 3-Spaces

Ch7_32 Fig 7.28

Ch7_33 Example 4 Find the vector from (4, 6, −2) to (1, 8, 3) Solution

Ch7_34 Example 5  From Definition 7.2, we have

Ch7_35 The i, j, k vectors  i =, j =, k = a = = a 1 i + a 2 j + a 3 j

Ch7_36 Fig 7.29

Ch7_37 Example 6 a = = 7i − 5j + 13j Example 7 (a) a = 5i + 3k is in the xz-plance (b) Example 8 If a = 3i − 4j + 8k, b = i − 4k, find 5a − 2b Solution 5a − 2b = 13i − 20j + 48k

Ch7_ Dot Product The dot product of a and b is the scalar (1) where  is the angle between the vectors 0    . DEFINITION 7.3 Dot Product of Two Vectors

Ch7_39 Fig 7.32

Ch7_40 Example 1  From (1) we obtain i  i = 1, j  j = 1, k  k = 1(2)

Ch7_41 Component Form of Dot Product  (3) (4)  See Fig 7.33

Ch7_42 Fig 7.33

Ch7_43 Example 2  If a = 10i + 2j – 6k, b = (−1/2)i + 4j – 3k, then

Ch7_44 Properties  (i) a  b = 0 if and only if a = 0 or b = 0 (ii) a  b = b  a (iii) a  (b + c) = a  b + a  c (iv) a  (kb) = (ka)  b = k(a  b) (v) a  a  0 (vi) a  a = ||a|| 2

Ch7_45 Orthogonal Vectors  (i) a  b > 0 if and only if  is acute (ii) a  b < 0 if and only if  is obtuse (iii) a  b = 0 if and only if cos  = 0,  =  /2  Note: Since 0  b = 0, we say the zero vector is orthogonal to every vector. Two nonzero vectors a and b are orthogonal if and only if a  b = 0. THEOREM 7.1 Criterion for an Orthogonal Vectors

Ch7_46 Example 3 i, j, k are orthogonal vectors. i  j = j  i = 0, j  k = k  j = 0, k  i = i  k = 0(5) Example 4 If a = −3i − j + 4k, b = 2i + 14j + 5k, then a  b = –6 – = 0 They are orthogonal.

Ch7_47 Angle between Two Vectors (6)

Ch7_48 Example 5 Find the angle between a = 2i + 3j + k, b = −i + 5j + k. Solution

Ch7_49 Direction Cosines  Referring to Fig 7.34, the angles , ,  are called the direction angles. Now by (6) We say cos , cos , cos  are direction cosines, and cos 2  + cos 2  + cos 2  = 1

Ch7_50 Fig 7.34

Ch7_51 Example 6 Find the direction cosines and the direction angles of a = 2i + 5j + 4k. Solution

Ch7_52 Component of a on b  Since a = a 1 i + a 2 j + a 3 k, then (7) We write the components of a as (8) See Fig The component of a on any vector b is comp b a = ||a|| cos  (9) Rewrite (9) as (10)

Ch7_53 Fig 7.35

Ch7_54 Example 7 Let a = 2i + 3j – 4k, b = i + j + 2k. Find comp b a and comp a b. Solution Form (10), a  b = −3

Ch7_55 Physical Interpretation  See Fig If F causes a displacement d of a body, then the work fone is W = F  d(11)

Ch7_56 Fig 7.36

Ch7_57 Example 8 Let F = 2i + 4j. If the block moves from (1, 1) to (4, 6), find the work done by F. Solution d = 3i + 5j W = F  d = 26 N-m

Ch7_58 Projection of a onto b  See Fig the projection of a onto i is  See Fig the projection of a onto b is (12)

Ch7_59 Fig 7.37

Ch7_60 Fig 7.38

Ch7_61 Example 9 Find the projection of a = 4i + j onto b = 2i + 3j. Solution

Ch7_62 Fig 7.39

Ch7_ Cross Product The cross product of two vectors a and b is (1) where  is the angle between them, 0    , and n is a unit vector perpendicular to the plane of a and b with direction given by right-hand rule. DEFINITION 7.4 Cross Product of Two Vectors

Ch7_64 Fig 7.46

Ch7_65 Example 1  To understand the physical meaning of the cross product, please see Fig 7.37 and The torque  done by a force F acting at the end of position vector r is given by  = r  F.  Fig 7.47Fig 7.48

Ch7_66 Properties  (i) a  b = 0, if a = 0 or b = 0 (ii) a  b = −b  a (iii) a  (b + c) = (a  b) + (a  c) (iv) (a + b)  c = (a  c) + (b  c) (v) a  (kb) = (ka)  b = k(a  b) (vi) a  a = 0 (vii) a  (a  b) = 0 (viii) b  (a  b) = 0 Two nonzero vectors a and b are parallel, if and only if a  b = 0. THEOREM 7.2 Criterion for Parallel Vectors

Ch7_67 Example 2  (a) From properties (iv) i  i = 0, j  j = 0, k  k = 0(2) (b) If a = 2i + 3j – k, b = –6i – 3j + 3k = –3a, then a and b are parallel. Thus a  b = 0  If a = i, b = j, then (3) According to the right-hand rule, n = k. So i  j = k

Ch7_68 Example 3  See Fig 7.49, we have (4)

Ch7_69 Fig 7.49

Ch7_70 Alternative Definition  Since (5) we have (6)

Ch7_71 We also can write (6) as (7) In turn, (7) becomes (8)

Ch7_72 Example 4 Let a = 4i – 2j + 5k, b = 3i + j – k, Find a  b. Solution From (8), we have

Ch7_73 Special Products  We have (9) is called the triple vector product. The following results are left as an exercise. (10)

Ch7_74 Area and Volume  Area of a parallelogram A = || a  b||(11) Area of a triangle A = ½||a  b||(12) Volume of the parallelepiped V = |a  (b  c)|(13) See Fig 7.50 and Fig 7.51

Ch7_75 Fig 7.50

Ch7_76 Fig 7.51

Ch7_77 Example 5 Find the area of the triangle determined by the points (1, 1, 1), (2, 3, 4), (3, 0, –1). Solution Using (1, 1, 1) as the base point, we have two vectors a =, b =

Ch7_78 Coplanar Vectors  a  (b  c) = 0 if and only if a, b, c are coplanar.

Ch7_ Lines and Planes in 3-Space  Lines: Vector Equation See Fig We find r 2 – r 1 is parallel to r – r 2, then r – r 2 = t(r 2 – r 1 )(1) If we write a = r 2 – r 1 = = (2) then (1) implies a vector equation for the line is r = r 2 + ta where a is called the direction vector.

Ch7_80 Fig 7.55

Ch7_81 Example 1 Find a vector equation for the line through (2, –1, 8) and (5, 6, –3). Solution Define a = =. The following are three possible vector equations: (3) (4) (5)

Ch7_82 Parametric equation  We can also write (2) as (6) The equations (6) are called parametric equations.

Ch7_83 Example 2 Find the parametric equations for the line in Example 1. Solution From (3), it follows x = 2 – 3t, y = –1 – 7t, z = t(7) From (5), x = 5 + 3t, y = 6 + 7t, z = –3 – 11t(8)

Ch7_84 Example 3 Find a vector a that is parallel to the line: x = 4 + 9t, y = –14 + 5t, z = 1 – 3t Solution a = 9i + 5j – 3k

Ch7_85 Symmetric Equations  From (6) provided a i are nonzero. Then (9) are said to be symmetric equation.

Ch7_86 Example 4 Find the symmetric equations for the line through (4, 10, −6) and (7, 9, 2) Solution Define a 1 = 7 – 4 = 3, a 2 = 9 – 10 = –1, a 3 = 2 – (–6) = 8, then

Ch7_87 Example 5 Find the symmetric equations for the line through (5, 3, 1) and (2, 1, 1) Solution Define a 1 = 5 – 2 = 3, a 2 = 3 – 1 = 2, a 3 = 1 – 1 = 0, then

Ch7_88 Fig 7.56

Ch7_89 Example 6 Write vector, parametric and symmetric equations for the line through (4, 6, –3) and parallel to a = 5i – 10j + 2k. Solution Vector: = + t(5, –10, 2) Parametric: x = 4 + 5t, y = 6 – 10t, z = –3 + 2t, Symmetric:

Ch7_90 Planes: Vector Equations  Fig 7.57(a) shows the concept of the normal vector to a plane. Any vector in the plane should be perpendicular to the normal vector, that is n  (r – r 1 ) = 0 (10)

Ch7_91 Fig 7.57

Ch7_92 Cartesian Equations  If the normal vector is ai + bj + ck, then the Cartesian equation of the plane containing P 1 (x 1, y 1, z 1 ) is a(x – x 1 ) + a(y – y 1 ) + c(z – z 1 ) = 0(11)

Ch7_93 Example 7 Find the plane contains (4, −1, 3) and is perpendicular to n = 2i + 8j − 5k Solution From (11): 2(x – 4) + 8(y + 1) – 5(z – 3) = 0 or 2x + 8y – 5z + 15 = 0

Ch7_94  Equation (11) can always be written as ax + by + cz + d = 0(12) The graph of any ax + by + cz + d = 0, a, b, c not all zero, is a plane with the normal vector n = ai + bj + ck THEOREM 7.3 Plane with Normal Vector

Ch7_95 Example 8  A vector normal to the plane 3x – 4y + 10z – 8 = 0 is n = 3i – 4j + 10k.

Ch7_96  Given three noncollinear points, P 1, P 2, P 3, we arbitrarily choose P 1 as the base point. See Fig 7.58, Then we can obtain (13)

Ch7_97 Fig 7.58

Ch7_98 Example 9 Find an equation of the plane contains (1, 0 −1), (3, 1, 4) and (2, −2, 0). Solution We arbitrarily construct two vectors from these three points, say, u = and v =.

Ch7_99 Example 9 (2) If we choose (2, −2, 0) as the base point, then  = 0

Ch7_100 Graphs  The graph of (12) with one or two variables missing is still a plane.

Ch7_101 Example 10 Graph 2x + 3y + 6z = 18 Solution Setting:y = z = 0 gives x = 9 x = z = 0 gives y = 6 x = y = 0 gives z = 3 See Fig 7.59.

Ch7_102 Fig 7.59

Ch7_103 Example 11 Graph 6x + 4y = 12 Solution This equation misses the variable z, so the plane is parallel to the z-axis. Sincex = 0 gives y = 3 y = 0 gives x = 2 See Fig 7.60.

Ch7_104 Fig 7.60

Ch7_105 Example 12 Graph x + y – z = 0 Solution First we observe that the plane passes through (0, 0, 0). Let y = 0, then z = x; x = 0, then z = y.

Ch7_106 Fig 7.61

Ch7_107  Two planes that are not parallel must intersect in a line. See Fig Fig 7.63 shows the intersection of a line and a plane.

Ch7_108 Fig 7.62

Ch7_109 Fig 7.63

Ch7_110 Example 13 Find the parametric equation of the line of the intersection of 2x – 3y + 4z = 1 x – y – z = 5 Solution First we let z = t, 2x – 3y = 1 – 4t x – y = 5 + t then x = t, y = 9 + 6t, z = t.

Ch7_111 Example 14 Find the point of intersection of the plane 3x – 2y + z = −5 and the line x = 1 + t, y = −2 + 2t, z = 4t. Solution Assume (x 0, y 0, z 0 ) is the intersection point. 3x 0 – 2y 0 + z 0 = −5 and x 0 = 1 + t 0, y 0 = −2 + 2t 0, z 0 = 4t 0 then 3(1 + t 0 ) – 2(−2 + 2t 0 ) + 4t 0 = −5, t 0 = −4 Thus, (x 0, y 0, z 0 ) = (−3, −10, −16)

Ch7_ Vector Spaces  n-Space Similar to 3-space (1) (2)

Ch7_113 Let V be a set of elements on which two operations, vector addition and scalar multiplication, are defined. Then V is said to be a vector spaces if the following are satisfied. DEFINITION 7.5 Vector Space

Ch7_114 Axioms for Vector Addition (i) If x and y are in V, then x + y is in V. (ii) For all x, y in V, x + y = y + x (iii) For all x, y, z in V, x + (y + z) = (x + y) + z (iv) There is a unique vector 0 in V, such that 0 + x = x + 0 = x (v) For each x in V, there exists a vector −x in V, such that x + (−x) = (−x) + x = 0 DEFINITION 7.5 Vector Space

Ch7_115 Axioms for Scalars Multiplication (vi) If k is any scalar and x is in V, then kx is in V. (vii) k(x + y) = kx + ky (viii) (k 1 +k 2 )x = k 1 x+ k 2 x (ix) k 1 (k 2 x) = (k 1 k 2 )x (x) 1x = x Properties (i) and (vi) are called the closure axioms. DEFINITION 7.5 Vector Space

Ch7_116 Example 1 Determine whether the sets (a) V = {1} and (b) V = {0} under ordinary addition and multiplication by real numbers are vectors spaces. Solution (a) V = {1}, violates many of the axioms. (b) V = {0}, it is easy to check this is a vector space. Moreover, it is called the trivial or zero vector space.

Ch7_117 Example 2 Consider the set V of all positive real numbers. If x and y denote positive real numbers, then we write vectors as x = x, y = y. Now addition of vectors is defined by x + y = xy and scalar multiplication is defined by kx = x k Determine whether the set is a vector space.

Ch7_118 Example 2 (2) Solution We go through all 10 axioms. (i) For x = x > 0, y = y > 0 in V, x + y = x + y > 0 (ii) For all x = x, y = y in V, x + y = x + y = y + x = y + x (iii) For all x = x, y = y, z = z in V x + (y + z) = x(yz) = (xy) = (x + y) + z (iv) Since 1 + x = 1x = x = x, x + 1 = x1 = x = x The zero vector 0 is 1 = 1

Ch7_119 Example 2 (3) (v) If we define −x = 1/x, then x + (−x) = x(1/x) = 1 = 1 = 0 −x + x = (1/x)x = 1 = 1 = 0 (vi) If k is any scalar and x = x > 0 is in V, then kx = x k > 0 (vii) If k is any scalar, k(x + y) = (xy) k = x k y k = kx + ky (viii) (k 1 +k 2 )x = x k1+k2 = x k1 x k2 = k 1 x+ k 2 x (ix) k 1 (k 2 x) = (x k2 ) k1 = x k1k2 = (k 1 k 2 )x (x) 1x = x 1 = x = x

Ch7_120 If a subset W of a vector space V is itself a vector space under the operations of vector addition and scalar multiplication defined on V, then W is called a subspace of V. DEFINITION 7.6 Subspace

Ch7_121 A nonempty subset W is a subspace of V if and only if W is closed under vector addition and scalar multiplication defined on V: (i) If x and y are in W, then x + y is in W. (ii) If x is in W and k is any scalar, then kx is in W. THEOREM 7.4 Criteria for a Subspace

Ch7_122 Example 3  Suppose f and g are continuous real-valued functions defined on (− ,  ). We know f + g and kf, for any real number k, are continuous real-valued. From this, we conclude that C(− ,  ) is a subspace of the vector space of real-valued function defined on (− ,  ).

Ch7_123 Example 4  The set P n of polynomials of degree less than or equal to n is a subspace of C(− ,  ).

Ch7_124 A set of vectors {x 1, x 2, …, x n } is said to be linearly independent, if the only constants satisfying k 1 x 1 + k 2 x 2 + …+ k n x n = 0(3) are k 1 = k 2 = … = k n = 0. If the set of vectors is not linearly independent, it is linearly dependent. DEFINITION 7.7 Linear Independence

Ch7_125  For example: i, j, k are linearly independent., and are linearly dependent, because 3 + − = 3a + b – c = 0

Ch7_126  It can be shown that any set of three linearly independent vectors is a basis for R 3. For example,, Consider a set of vectors B = {x 1, x 2, …, x n } in a vector space V. If the set is linearly independent and if every vector in V can be expressed as a linear combination of these vectors, then B is said to be a basis for V. DEFINITION 7.8 Basis for a Vector Space

Ch7_127  Standard Basis: {i, j, k} For R n : e 1 =, e 2 = ….. e n = (4) If B is a basis, then there exists scalars such that (5) where these scalars c i, i = 1, 2,.., n, are called the coordinates of v related to the basis B.

Ch7_128 The number of vectors in a basis B for vector space V is said to be the dimension of the space. DEFINITION 7.8 Dimension for a Vector Space

Ch7_129 Example 5 (a)The dimensions of R, R 2, R 3, R n are in turn 1, 2, 3, n. (b)There are n + 1 vectors in B = {1, x, x 2, …, x n }. The dimension is n + 1 (c)The dimension of the zero space {0} is zero.

Ch7_130 Linear DEs  The general solution of following DE (6) can be written as y = c 1 y 1 + c 1 y 1 + … c n y n and it is said to be the solution space. Thus {y 1, y 2, …, y n } is a basis.

Ch7_131 Example 6 The general solution of y” + 25y = 0 is y = c 1 cos 5x + c 2 sin 5x then {cos 5x, sin 5x} is a basis.

Ch7_132 Span  If S denotes any set of vectors {x 1, x 2, …, x n } then the linear combination k 1 x 1 + k 2 x 2 + … + k n x n is called a span of the vectors and written as Span(S) or Span{x 1, x 2, …, x n }.

Ch7_133 Rephrase Definition 7.8 and 7.9  A set S of vectors {x 1, x 2, …, x n } in a vector space V is a basis, if S is linearly independent and is a spanning set for V. The number of vectors in this spanning set S is the dimension of the space V.

Ch7_ Gram-Schmidt Orthogonalization Process  Orthonormal Basis All the vectors of a basis are mutually orthogonal and have unit length.

Ch7_135 Example 1  The set of vectors (1) is linearly independent in R 3. Hence B = {w 1, w 2, w 3 } is a basis. Since ||w i || = 1, i = 1, 2, 3, w i  w j = 0, i  j, B is an orthonormal basis.

Ch7_136  Proof Since B = {w 1, w 2, …, w n } is an orthonormal basis, then any vector can be expressed as u = k 1 w 1 + k 2 w 2 + … + k n w n (2) (u  w i ) = (k 1 w 1 + k 2 w 2 + … + k n w n )  w i = k i (w i  w i ) = k i Suppose B = {w 1, w 2, …, w n } is an orthonormal basis for R n, If u is any vector in R n, then u = (u  w 1 )w 1 + (u  w 2 )w 2 + … + (u  w n )w n THEOREM 7.5 Coordinates Relative ti an Orthonormal Basis

Ch7_137 Example 2 Find the coordinate of u = relative to the orthonormal basis in Example 1. Solution

Ch7_138 Gram-Schmidt Orthogonalization Process  The transformation of a basis B = {u 1, u 2 } into an orthogonal basis B’= {v 1, v 2 } consists of two steps. See Fig (3)

Ch7_139 Fig 7.64(a)

Ch7_140 Fig 7.64(b)

Ch7_141 Fig 7.64(c)

Ch7_142 Example 3 Let u 1 =, u 2 =. Transform them into an orthonormal basis. Solution From (3) Normalizing: See Fig 7.65

Ch7_143 Fig 7.65

Ch7_144 Gram-Schmidt Orthogonalization Process  For R 3 : (4)

Ch7_145  See Fig Suppose W 2 = Span{v 1, v 2 }, then is in W 2 and is called the orthogonal projection of u 3 onto W 2, denoted by x = proj w2 u 3. (5) (6)

Ch7_146 Fig 7.66

Ch7_147 Example 4 Let u 1 =, u 2 =, u 3 =. Transform them into an orthonormal basis. Solution From (4)

Ch7_148 Example 4 (2)

Ch7_149 Let B = {u 1, u 2, …, u m }, m  n, be a basis for a Subspace W m of R n. Then {v 1, v 2, …, v m }, where is an orthogonal basis for W m. An orthonormal basis for W m is THEOREM 7.6 Orthogonalization Process