Vectors

1
Vectors

• CHAPTER 7

2
Contents

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

3
7.1 Vectors in 2-Space

• Review of VectorsPlease refer to Fig 7.1 through
  Fig 7.6.

4
Fig 7.1 (Geometric Vectors)
5
Fig 7.2 (Vectors are equal)
6
Fig 7.3 (Parallel vectors)
7
Fig 7.4 (sum)
8
Fig 7.5 (difference)
9
Fig 7.6 (position vectors)

10
Example 1

• Please refer to Fig 7.7.
• Fig 7.7

11

• a b lta1- b1, a2 - b2gt (4)

12
Graph Solution

• Fig 7.8 shows the graph solutions of the addition
  and subtraction of two vectors.

13
Example 2

• If a lt1, 4gt, b lt-6, 3gt, find a b, a - b,
  2a 3b.
• Solution Using (1), (2), (4), we have

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 kb k scalar(vi) (k1 k2)a
  k1a k2a k1, k2 scalars(vii) k1(k2a)
  (k1k2)a k1, k2 scalars(viii) 1a a (ix) 0a
  0 lt0, 0gt
• 0 lt0, 0gt

15
Magnitude, Length, Norm

• a lta1 , a2gt, then Clearly, we have a ? 0,
  0 0

16
Unit Vector

• A vector that ha magnitude 1 is called a unit
  vector. u (1/a)a is a unit vector, since

17
Example 3

• Given a lt2, -1gt, then the unit vector in the
  same direction u is and

18
The i, j vectors

• If a lta1, a2gt, then (5)Let i
  lt1, 0gt, j lt0, 1gt, then (5) becomes a a1i
  a2j (6)

19
Fig 7.10
20
Example 4

• (i) lt4, 7gt 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

21
Example 5

• Let a 4i 2j, b 2i 5j. Graph a b, a b
  
• SolutionFig 7.11

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7.2 Vectors in 3-Space

• Simple ReviewPlease refer to Fig 7.22 through
  Fig 7.24.
• Fig 7.22

23
Fig 7.23
24
Fig 7.24
25
Example 1

• Graph the points (4, 5, 6) and (-2, -2, 0).
• Solution See Fig 7.25.

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Distance Formula

• (1)
• Fig 7.26

27
Example 2

• Find the distance between (2, -3, 6) and (-1, -7,
  4)
• Solution

28
Midpoint Formula

• (2)

29
Example 2

• Find the midpoint of (2, -3, 6) and (-1, -7, 4)
• SolutionFrom (2), we have

30
Vectors in 3-Space

• Fig 7.27.

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DEFINITION 7.2

• Let a lta1, a2 , a3gt, b ltb1, b2, b3 gt in
  R3(i) a b lta1 b1, a2 b2, a3 b3gt(ii)
  ka ltka1, ka2, ka3gt (iii) a b if and only if
  a1 b1, a2 b2, a3 b3 (iv) b (-1)b
  lt- b1, - b2, - b3gt(v) a b lta1 - b1, a2 -
  b2, a3 - b3gt(vi) 0 lt0, 0 , 0gt(vi)

Component Definitions in 3-Spaces
32
Fig 7.28

33
Example 4

• Find the vector from (4, 6, -2) to (1, 8, 3)
• Solution

34
Example 5

• From Definition 7.2, we have

35
The i, j, k vectors

• i lt1, 0, 0gt, j lt0, 1, 0gt, k lt0, 0,
  1gta lt a1, a2, a3gt a1i a2j a3j

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Fig 7.29
37

• Example 6a lt7, -5, 13gt 7i - 5j 13j
• Example 7(a) a 5i 3k is in the
  xz-plance(b)
• Example 8If a 3i - 4j 8k, b i - 4k, find
  5a - 2b
• Solution5a - 2b 13i - 20j 48k

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7.3 Dot Product
DEFINITION 7.3
Dot Product of Two Vectors
The dot product of a and b is the
scalar (1)where ? is the angle
between the vectors 0 ? ? ? ?.
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Fig 7.32
40
Example 1

• From (1) we obtain i ? i 1, j ? j 1, k ?
  k 1 (2)

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Component Form of Dot Product

• (3) (4)
• See Fig 7.33

42
Fig 7.33

43
Example 2

• If a 10i 2j 6k, b (-1/2)i 4j 3k, then

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 a2

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Orthogonal Vectors

• (i) a ? b gt 0 if and only if ? is acute(ii) a ?
  b lt 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.

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• 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 4If a -3i - j 4k, b 2i 14j 5k,
  then a ? b 6 14 20 0They are
  orthogonal.

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Angle between Two Vectors

• (6)

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Example 5

• Find the angle between a 2i 3j k, b -i
  5j k.
• Solution

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 cos2? cos2? cos2? 1

50
Fig 7.34

51
Example 6

• Find the direction cosines and the direction
  angles of a 2i 5j 4k.
• Solution

52
Component of a on b

• Since a a1i a2j a3k, then (7)We
  write the components of a as (8)See Fig
  7.35. The component of a on any vector b is
  compba a cos ? (9)Rewrite (9)
  as (10)

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Fig 7.35
54
Example 7

• Let a 2i 3j 4k, b i j 2k. Find compba
  and compab.
• SolutionForm (10), a ? b -3

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Physical Interpretation

• See Fig 7.36. If F causes a displacement d of a
  body, then the work fone is W F ? d (11)

56
Fig 7.36
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

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Projection of a onto b

• See Fig 7.37. the projection of a onto i is
• See Fig 7.38. the projection of a onto b
  is (12)

59
Fig 7.37
60
Fig 7.38
61
Example 9

• Find the projection of a 4i j onto b 2i
  3j.
• Solution

62
Fig 7.39
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7.4 Cross Product
64
Fig 7.46
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Example 1

• To understand the physical meaning of the cross
  product, please see Fig 7.37 and 7.48. The torque
  ? done by a force F acting at the end of
  position vector r is given by ? r ? F.
• Fig 7.47 Fig 7.48

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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

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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

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Example 3

• See Fig 7.49, we have (4)

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Fig 7.49

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Alternative Definition

• Since (5)we have (6)

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

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Example 4

• Let a 4i 2j 5k, b 3i j k, Find a ? b.
• SolutionFrom (8), we have

73
Special Products

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

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Area and Volume

• Area of a parallelogram A a ?
  b (11)Area of a triangle A ½a ?
  b (12)Volume of the parallelepiped V
  a ? (b ? c) (13)See Fig 7.50 and Fig 7.51

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Fig 7.50
76
Fig 7.51

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Example 5

• Find the area of the triangle determined by the
  points (1, 1, 1), (2, 3, 4), (3, 0, 1).
• SolutionUsing (1, 1, 1) as the base point, we
  have two vectors a lt1, 2, 3gt, b lt2, 1, 2gt

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Coplanar Vectors

• a ? (b ? c) 0 if and only if a, b, c are
  coplanar.

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7.5 Lines and Planes in 3-Space

• Lines Vector EquationSee Fig 7.55. We find r2
  r1 is parallel to r r2, then r r2 t(r2
  r1) (1)If we write a r2 r1 ltx2 x1,
  y2 y1, z2 z1gt lta1, a2,
  a3gt (2)then (1) implies a vector equation
  for the line is r r2 tawhere a is called
  the direction vector.

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Fig 7.55
81
Example 1

• Find a vector equation for the line through (2,
  1, 8) and (5, 6, 3).
• SolutionDefine a lt2 5, 1 6, 8 ( 3)gt
  lt3, 7, 11gt.The following are three possible
  vector equations (3) (4)
  (5)

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Parametric equation

• We can also write (2) as (6)The
  equations (6) are called parametric equations.

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Example 2

• Find the parametric equations for the line in
  Example 1.
• SolutionFrom (3), it follows x 2 3t, y
  1 7t, z 8 11t (7)From (5), x 5
  3t, y 6 7t, z 3 11t (8)

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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

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Symmetric Equations

• From (6) provided ai are nonzero.
  Then (9)are said to be symmetric
  equation.

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Example 4

• Find the symmetric equations for the line through
  (4, 10, -6) and (7, 9, 2)
• SolutionDefine a1 7 4 3, a2 9 10 1,
  a3 2 (6) 8, then

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Example 5

• Find the symmetric equations for the line through
  (5, 3, 1) and (2, 1, 1)
• SolutionDefine a1 5 2 3, a2 3 1 2,
  a3 1 1 0,then

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Fig 7.56
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Example 6

• Write vector, parametric and symmetric equations
  for the line through (4, 6, 3) and parallel to a
  5i 10j 2k.
• SolutionVector ltx, y, zgt lt 4, 6, 3gt t(5,
  10, 2) Parametric x 4 5t, y 6 10t, z
  3 2t, Symmetric

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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 r1) 0 (10)

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Fig 7.57
92
Cartesian Equations

• If the normal vector is ai bj ck , then the
  Cartesian equation of the plane containing
  P1(x1, y1, z1) is a(x x1) a(y y1) c(z
  z1) 0 (11)

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Example 7

• Find the plane contains (4, -1, 3) and is
  perpendicular to n 2i 8j - 5k
• SolutionFrom (11) 2(x 4) 8(y 1) 5(z
  3) 0or 2x 8y 5z 15 0

94

• Equation (11) can always be written as ax by
  cz d 0 (12)

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Example 8

• A vector normal to the plane 3x 4y 10z 8
  0 is n 3i 4j 10k.

96

• Given three noncollinear points, P1, P2, P3, we
  arbitrarily choose P1 as the base point. See Fig
  7.58, Then we can obtain (13)

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Fig 7.58
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Example 9

• Find an equation of the plane contains (1, 0 -1),
  (3, 1, 4) and (2, -2, 0).
• SolutionWe arbitrarily construct two vectors
  from these three points, say, u lt2, 1, 5gt and v
  lt1, 3, 4gt.

99
Example 9 (2)

• If we choose (2, -2, 0) as the base point,
  thenltx 2, y 2, z 0gt ? lt-11, -3, 5gt 0

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Graphs

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

101
Example 10

• Graph 2x 3y 6z 18
• SolutionSetting y z 0 gives x 9 x z
  0 gives y 6 x y 0 gives z 3See Fig
  7.59.

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Fig 7.59

103
Example 11

• Graph 6x 4y 12
• SolutionThis equation misses the variable z, so
  the plane is parallel to the z-axis. Since x
  0 gives y 3 y 0 gives x 2See Fig 7.60.

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Fig 7.60

105
Example 12

• Graph x y z 0
• SolutionFirst we observe that the plane passes
  through (0, 0, 0). Let y 0, then z x x 0,
  then z y.

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Fig 7.61
107

• Two planes that are not parallel must intersect
  in a line. See Fig 7.62. Fig 7.63 shows the
  intersection of a line and a plane.

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Fig 7.62
109
Fig 7.63
110
Example 13

• Find the parametric equation of the line of the
  intersection of 2x 3y 4z 1 x y
  z 5
• SolutionFirst we let z t, 2x 3y 1 4t
  x y 5 tthen x
  14 7t, y 9 6t, z t.

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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.
• SolutionAssume (x0, y0, z0) is the intersection
  point. 3x0 2y0 z0 -5 and x0 1 t0,
  y0 -2 2t0, z0 4t0then 3(1 t0) 2(-2
  2t0) 4t0 -5, t0 -4Thus, (x0, y0, z0)
  (-3, -10, -16)

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7.6 Vector Spaces

• n-SpaceSimilar to 3-space (1)
  (2)

113
DEFINITION 7.5

• 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.

Vector Space
114
DEFINITION 7.5

• 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

Vector Space
115
DEFINITION 7.5

• 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) (k1k2)x k1x k2x
• (ix) k1(k2x) (k1k2)x
• (x) 1x x
• Properties (i) and (vi) are called the closure
  axioms.

Vector Space
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.

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 xyand scalar
  multiplication is defined by kx
  xkDetermine whether the set is a vector space.

118
Example 2 (2)

• Solution We go through all 10 axioms.(i) For x
  x gt 0, y y gt 0 in V, x y x y gt 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

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 gt 0 is in V,
  then kx xk gt 0
• (vii) If k is any scalar, k(x y) (xy)k
  xkyk kx ky
• (viii) (k1k2)x xk1k2 xk1xk2 k1x k2x
• (ix) k1(k2x) (xk2 )k1 xk1k2 (k1k2)x
• (x) 1x x1 x x

120
DEFINITION 7.6
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.
Subspace
121
THEOREM 7.4
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.
Criteria for a Subspace
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 (-?, ?).

123
Example 4

• The set Pn of polynomials of degree less than or
  equal to n is a subspace of C(-?, ?).

124
DEFINITION 7.7

• A set of vectors x1, x2, , xn is said to be
  linearly
• independent, if the only constants satisfying
  k1x1 k2x2 knxn 0 (3)
• are k1 k2 kn 0. If the set of vectors is
  not
• linearly independent, it is linearly dependent.

Linear Independence
125

• For example i, j, k are linearly
  independent.lt1, 1, 1gt , lt2, 1, 4gt and lt5, 2,
  7gt are linearly dependent, because 3lt1, 1, 1gt
  lt2, 1, 4gt - lt5, 2, 7gt lt0, 0, 0gt 3a b c 0

126
DEFINITION 7.8
Consider a set of vectors B x1, x2, , xn
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.
Basis for a Vector Space

• It can be shown that any set of three linearly
  independent vectors is a basis for R3. For
  example lt1, 0, 0gt, lt1, 1, 0gt, lt1, 1, 1gt

127

• Standard Basis i, j, k For Rn e1 lt1, 0,
  , 0gt, e2 lt0, 2, , 0gt .. en lt0, 0, ,
  1gt (4)If B is a basis, then there exists
  scalars such that (5)where these
  scalars ci, i 1, 2, .., n, are called the
  coordinates of v related to the basis B.

128
DEFINITION 7.8

• The number of vectors in a basis B for vector
  space V
• is said to be the dimension of the space.

Dimension for a Vector Space
129
Example 5

• The dimensions of R, R2, R3, Rn are in turn 1, 2,
  3, n.
• There are n 1 vectors in B 1, x, x2, , xn.
  The dimension is n 1
• The dimension of the zero space 0 is zero.

130
Linear DEs

• The general solution of following
  DE (6)can be written as y c1y1
  c1y1 cnyn and it is said to be the solution
  space. Thus y1, y2, , yn is a basis.

131
Example 6

• The general solution of y 25y 0 is y c1
  cos 5x c2 sin 5xthen cos 5x , sin 5x is a
  basis.

132
Span

• If S denotes any set of vectors x1, x2, , xn
  then the linear combination k1x1 k2x2
  knxnis called a span of the vectors and written
  as Span(S) or Spanx1, x2, , xn.

133
Rephrase Definition 7.8 and 7.9

• A set S of vectors x1, x2, , xn 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.

134
7.7 Gram-Schmidt Orthogonalization Process

• Orthonormal Basis All the vectors of a basis are
  mutually orthogonal and have unit length .

135
Example 1

• The set of vectors (1) is
  linearly independent in R3. Hence B w1, w2,
  w3 is a basis. Since wi 1, i 1, 2, 3,
  wi ? wj 0, i ? j, B is an orthonormal basis.

136
THEOREM 7.5
Suppose B w1, w2, , wn is an orthonormal
basis for Rn, If u is any vector in Rn, then u
(u ? w1)w1 (u ? w2)w2 (u ? wn)wn
Coordinates Relative ti an Orthonormal Basis

• Proof Since B w1, w2, , wn is an
  orthonormal basis, then any vector can be
  expressed as u k1w1 k2w2
  knwn (2) (u ? wi) (k1w1 k2w2 knwn) ?
  wi ki(wi ? wi) ki

137
Example 2

• Find the coordinate of u lt3, 2, 9gt relative
  to the orthonormal basis in Example 1.
• Solution

138
Gram-Schmidt Orthogonalization Process

• The transformation of a basis B u1, u2 into
  an orthogonal basis B v1, v2 consists of two
  steps. See Fig 7.64. (3)

139
Fig 7.64(a)

140
Fig 7.64(b)
141
Fig 7.64(c)
142
Example 3

• Let u1 lt3, 1gt, u2 lt1, 1gt. Transform them into
  an orthonormal basis.
• Solution From (3) Normalizing See Fig
  7.65

143
Fig 7.65

144
Gram-Schmidt Orthogonalization Process

• For R3 (4)

145

• See Fig 7.66. Suppose W2 Spanv1, v2, then
  is in W2 and is called the orthogonal
  projection of u3 onto W2, denoted by x
  projw2u3. (5) (6)

146
Fig 7.66
147
Example 4

• Let u1 lt1, 1, 1gt, u2 lt1, 2, 2gt, u3 lt1, 1,
  0gt. Transform them into an orthonormal basis.
• Solution From (4)

148
Example 4 (2)
149
THEOREM 7.6
Let B u1, u2, , um, m ? n, be a basis for a
Subspace Wm of Rn. Then v1, v2, , vm,
whereis an orthogonal basis for Wm. An
orthonormal basis for Wm is
Orthogonalization Process
150
Thank You !