CHAPTER 7
Vectors
Chapter 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
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Ch7_2
7.1 Vectors in 2-Space
Review of Vectors
Please refer to Fig 7.1.1 through Fig 7.1.6.
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Ch7_3
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Ch7_4
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Ch7_5
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Ch7_6
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Ch7_7
Example 1 Position Vector
Please refer to Fig 7.1.7.
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Ch7_8
Definition 7.1.1 Addition, Scalar Multiplication, Equality
Let a = <a1, a2>, b = <b1, b2> be vectors in R2
(i) Addition: a + b = <a1 + a2, b1 + b2>
(1)
(ii) Scalar multiplication: ka = <ka1, ka2 >
(2)
(iii)Equality: a = b if and only if a1 = b1, a2 = b2 (3)
a – b = <a1− b1, a2 − b2>
(4)
PP
1 2  OP2  OP1   x2  x1 , y2  y1 
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Ch7_9
Graph Solution
Fig 7.1.8 shows the graph solutions of the addition
and subtraction of two vectors.
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Ch7_10
Example 2 Addition and Subtraction of Two
Vectors
If a = <1, 4>, b = <−6, 3>, find a + b, a − b, 2a + 3b.
Solution:
Using (1), (2), (4), we have
a  b   1  (6), 4  3  5, 7 
a  b   1  (6), 4  3  7, 1 
2a  3b   2, 8    18, 9  16, 17 
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Ch7_11
Theorem 7.1.1 Properties of Vectors
← commutative law
(i)
a+b=b+a
← associative law
(ii) a + (b + c) = (a + b) + c
← additive identity
(iii) a + 0 = a
(iv) a + (−a) = 0
← additive inverse
(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
← (Zero Vector)
(ix) 0a = 0 = <0, 0>
0 = <0, 0>
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Ch7_12
Magnitude, Length, Norm
a = <a1 , a2>, then || a ||  a12  a22
Clearly, we have ||a||  0, ||0|| = 0
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Ch7_13
Unit Vector
A vector that ha magnitude 1 is called a unit vector.
u = (1/||a||)a is a unit vector, since
1
1
|| u || 
a 
|| a || 1
|| a ||
|| a ||
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Ch7_14
Example 3 Unit Vectors
Given a = <2, −1>, then the unit vector in the same
direction u is
1
1
u
a
 2,  1 
5
5
2 1
,
5 5
and
2 1
u   ,
5 5
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Ch7_15
The i, j vectors
If a = <a1, a2>, then
 a1 , a2 
  a1 , 0    0, a2   a1  1, 0   a2  0, 1 
(5)
Let i = <1, 0>, j = <0, 1>, then (5) becomes
a = a1i + a2j
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(6)
Ch7_16
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Ch7_17
Example 4 Vector Operations Using i and j
(i)
<4, 7> = 4i + 7j
(ii)
(2i – 5j) + (8i + 13j) = 10i + 8j
(iii)
|| i  j ||  2
(iv)
10(3i – j) = 30i – 10j
(v)
a = 6i + 4j, b = 9i + 6j are parallel and b = (3/2)a
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Ch7_18
Example 5 Graphs of Vector Sum/Vector
Difference
Let a = 4i + 2j, b = –2i + 5j. Graph a + b, a – b
Solution:
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Ch7_19
7.2 Vectors in 3-Space
Simple Review
Please refer to Fig 7.2.1 through Fig 7.2.3.
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Ch7_20
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Ch7_21
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Ch7_22
Example 1 Graphs of Three Points
Graph the points (4, 5, 6), (3, −3, −1) and (−2, −2, 0).
Solution:
See Fig 7.2.4.
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Ch7_23
Distance Formula

d ( P1 , P2 )  ( x2  x1 ) 2  ( y2  y1 ) 2  ( z2  z1 ) 2
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(1)
Ch7_24
Example 2 Distance Between Two Points
Find the distance between (2, −3, 6) and (−1, −7, 4)
Solution:
d  (2  (1))2  (3  (7))2  (6  4) 2  29
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Ch7_25
Midpoint Formula

 x1  x2 y1  y2 z1  z2 
,
,


2
2 
 2
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(2)
Ch7_26
Example 2 Coordinates of a Midpoint
Find the midpoint of (2, −3, 6) and (−1, −7, 4)
Solution:
From (2), we have
 2  (1) ,  3  (7) , 6  4    1 ,  5, 5 

 

2
2  2
 2

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Ch7_27
Vectors in 3-Space
a  a1 , a2 , a3 
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Ch7_28
Definition 7.2.1 Component Definitions in 3-Space
Let a
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vi)
= <a1, a2 , a3>, b = <b1, b2, b3 > in R3
a + b = <a1 + b1, a2 + b2, a3 + b3>
ka = <ka1, ka2, ka3>
a = b if and only if a1 = b1, a2 = b2, a3 = b3
–b = (−1)b = <− b1, − b2, − b3>
a – b = <a1 − b1, a2 − b2, a3 − b3>
0 = <0, 0 , 0>
|| a ||  a12  a22  a32
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Ch7_29
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Ch7_30
Example 4 Vector Between Two Points
Find the vector P1P2 from P1(4, 6, −2) to P2(1, 8, 3).
Solution:
P1 P2  OP2  OP1
  1  4, 8  6, 3  (2) 
 3, 2, 5 
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Ch7_31
Example 5 A Unit Vector
The magnitude of a is
|| a || 
 22  32  62  49  7
A unit vector in the direction of a is
a 1
2 3 6
  2, 3, 6   , ,
a 7
7 7 7
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Ch7_32
The i, j, k vectors
i = <1, 0, 0>, j = <0, 1, 0>, k = <0, 0, 1>
 a1 , a2 , a3 
  a1 , 0, 0    0, a2 , 0    0, 0, a3 
 a1  1, 0, 0   a2  0, 1, 0   a3  0, 0, 1 
a = < a1, a2, a3> = a1i + a2j + a3j
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Ch7_33
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Ch7_34
Example 6
a = <7, −5, 13> = 7i − 5j + 13k
Example 7
(a) a = 5i + 3k is in the xz-plane
(b) || 5i  3k ||  52  32  34
Example 8
If a = 3i − 4j + 8k, b = i − 4k, find 5a − 2b.
Solution:
5a − 2b = 13i − 20j + 48k
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Ch7_35
7.3 Dot Product
Definition 7.3.1 Dot Product of Two Vectors
In 2-space the dot product of two vectors a = <a1, a2>
and b = <b1, b2> is the number
a‧b = a1b1 + a2b2
(1)
In 3-space the dot product of two vectors a = <a1, a2 ,
a3> and b = <b1, b2, b3> is the number
a‧b = a1b1 + a2b2 + a3b3
(2)
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Ch7_36
Example 1 Dot Product Using (2)
If a = 10i + 2j – 6k, b = (−1/2)i + 4j – 3k, then
1

a．b  (10)    (2)(4)  (6)(3)  21
 2
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Ch7_37
Example 2 Dot Products of the Basis Vectors
Since i = <1, 0, 0>, j = <0, 1, 0>, and k = <0, 0, 1>,
we see from (2) we that
i‧j = j‧i = 0, j‧k = k‧j = 0,
and k‧i = i‧k = 0.
Similarly, by (2)
i  i = 1, j  j = 1, k  k = 1
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(3)
(4)
Ch7_38
Theorem 7.3.1 Properties of the Dot Product
(i)
(ii)
(iii)
(iv)
(v)
(vi)
a  b = 0 if and only if a = 0 or b = 0
← commutative law
ab=ba
← distributive law
a  (b + c) = a  b + a  c
a  (kb) = (ka)  b = k(a  b)
aa0
a  a = ||a||2
a  a  a  a12  a22  a32
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Ch7_39
Theorem 7.3.2 Alternative Form of the Dot Product
The dot product of two vectors a and b is
a．b  || a || || b || cos 
(5)
where  is the angle between the vectors 0    .
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Ch7_40
Component Form of Dot Product

|| c ||  || b || 2  || a || 2 2 || a || || b || cos
1
|| a || || b || cos  (|| b || 2  || a || 2  || c || 2 )
2
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(6)
Ch7_41
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Ch7_42
Orthogonal Vectors
(i) a  b > 0
(ii) a  b < 0
(iii) a  b = 0
if and only if  is acute
if and only if  is obtuse
if and only if cos  = 0,  = /2
Theorem 7.1 Criterion for Orthogonal Vectors
Two nonzero vectors a and b are orthogonal if and
only if a  b = 0.
Note: Since 0  b = 0, we say the zero vector is
orthogonal to every vector.
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Ch7_43
Example 3 Orthogonal Vectors
If a = –3i – j + 4k and b = 2i + 14j + 5k, then
a‧b = (–3)(2) + (–1)(14) + (4)(5) = 0.
From Theorem 7.3.3, we conclude that a and b are
orthogonal.
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Ch7_44
Angle between Two Vectors
By equating the two forms of the dot product, (2) and
(5), we can determine the angle between two
vectors from
a1b1  a2b2  a3b3
cos 
|| a || || b ||
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(7)
Ch7_45
Example 4 Angle Between Two Vectors
Find the angle between a = 2i + 3j + k, b = −i + 5j + k.
Solution:
|| a ||  14, || b ||  27 , a．b  14
14
42
cos 

14 27
9
 42 
  0.77
 9 
  cos 1 
  44.9
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Ch7_46
Direction Cosines
Referring to Fig 7.3.3, the angles , ,  are called the
direction angles. Now by (7)
cos  
a．i
a．j
a．k
, cos  
, cos  
|| a || || i ||
|| a || || j ||
|| a || || k ||
cos  
a
a1
a
, cos   2 , cos   3
|| a ||
|| a ||
|| a ||
We say cos , cos , cos  are direction cosines, and
a3
1
a1
a2
a
i
j
k  (cos )i  (cos  ) j  (cos  )k
|| a ||
|| a || || a || || a ||
cos2 + cos2 + cos2 = 1
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Ch7_47
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Ch7_48
Example 5 Direction Cosines/Angles
Find the direction cosines and the direction angles of
a = 2i + 5j + 4k.
Solution:
|| a ||  2 2  52  4 2  45  3 5
2
5
4
cos  
, cos 
, cos  
3 5
3 5
3 5
The direction angles are
 2 
  cos 
  1.27 radians or   72.7
3 5
1  5 
  cos 
  0.73 radians or   41.8
3 5
1  4 
  cos 
  0.93 radians or   53.4
3 5
1
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Ch7_49
Component of a on b
Since a = a1i + a2j + a3k, then
a1  a．i, a2  a．j, a3  a．k
(8)
We write the components of a as
comp ja  a．j,
comp i a  a．i,
comp k a  a．k (9)
See Fig 7.3.4. The component of a on any vector b is
compba = ||a|| cos 
(10)
Rewrite (10) as
comp b a 
|| a || || b || cos  a．b

|| b ||
|| b ||
 1  a b
 a．
b 
 || b ||  b
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(11)
Ch7_50
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Ch7_51
Example 6 Component of a Vector on
Another Vector
Let a = 2i + 3j – 4k, b = i + j + 2k. Find compba and
compab.
Solution:
Form (10), a  b = −3
1
1
b
(i  j  2k )
|| b ||
6
1
3
comp b a  (2i  3 j  4k )． (i  j  2k )  
6
6
1
1
|| a ||  29,
a
(2i  3j  4k )
|| a ||
29
1
3
comp b b  (i  j  2k )．
(2i  3 j  4k )  
29
29
|| b ||  6,
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Ch7_52
Projection of a onto b
See Fig 7.3.5, the projection of a onto i is
proji a  (comp i a)i  (a  i)i  a1i
See Fig 7.3.6, the projection of a onto b is
 1   a b 
projb a  (comp b a) b   
b
 b   b b 
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(12)
Ch7_53
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Ch7_54
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Ch7_55
Example 7 Projection of a Vector on Another
Vector
Find the projection of a = 4i + j onto the vector b = 2i +
3j. Graph.
Solution:
1
11
comp b a  (4i  j) 
(2i  3j) 
13
13
11  1 
22 33

projba  

(2i  3j)  i  j
13 13
 13  13 
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Ch7_56
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Ch7_57
Physical Interpretation of the Dot Product
See Fig 7.3.8. If F causes a displacement d of a body,
then the work done is
W=Fd
(13)
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Ch7_58
Example 8 Work Done by a Constant Force
Let F = 2i + 4j. If the block moves from P1(1, 1) to
P2(4, 6), find the work done by F.
Solution:
d = 3i + 5j
W = F  d = 26 N-m
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Ch7_59
7.4 Cross Product
a1 a2
 a1b2  a2b1
b1 b2
a1 a2
b1 b2
c1 c2
a3
b2
b3  a1
c2
c3
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b3
b1 b3
b1 b2
 a2
 a3
c3
c1 c3
c1 c3
(1)
(2)
Ch7_60
Definition 7.4.1 Cross Product of Two Vectors
The cross product of two vectors a = <a1, a2 , a3>
and b = <b1, b2, b3> is the vector
a  b  (a2b3  a3b2 )i  (a1b3  a3b1 ) j  (a1b2  a2b1 )k
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(3)
Ch7_61
We also can write (3) as
a2
ab 
b2
a3
a1 a3
a1 a2
i
j
k
b3
b1 b3
b1 b2
(4)
In turn, (4) becomes
i
j k
a  b  a1 a2 a3
b1 b2 b3
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(5)
Ch7_62
Example 1 Cross Product Using (4) and (5)
Let a = 4i – 2j + 5k, b = 3i + j – k, Find a  b.
Solution:
From (5), we have
i
j k
ab  4  2 5
3
1 1
2 5
4 5
4 2

i
j
k
1 1
3 1
3
1
 3i  19 j  10k
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Ch7_63
Example 2 Cross Product of Two Basis
Vectors
If i = <1, 0, 0> and j = <0, 1, 0>, then
i j k
0 0 1 0 1 0
i j  1 0 0 
i
j
k k
1 0 0 0 0 1
0 1 0
i j k
0 0 1 0 1 0
ii  1 0 0 
i
j
k 0
0 0 1 0 1 0
1 0 0
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Ch7_64
Proceeding as in Example 2, it is readily shown that
i  j  k,
j  k  i,
k i  j
j  i  k , k  j  i, i  k   j
i  i  0,
j  j  0, k  k  0
(6)
(7)
(8)
The results in (6) can be obtained using the circular
mnemonic illustrated in Fig 7.4.1.
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Ch7_65
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Ch7_66
Properties
Theorem 7.4.1 Properties of the Cross Product
(i) a  b = 0, if a = 0 or b = 0
(ii) a  b = −b  a
(iii) a  (b + c) = (a  b) + (a  c) ← distributive law
(iv) (a + b)  c = (a  c) + (b  c)
← distributive law
(v) a  (kb) = (ka)  b = k(a  b), k is scalar
(vi) a  a = 0
(vii) a  (a  b) = 0
(viii) b  (a  b) = 0
a  b is orthogonal to the plane containing a and b.
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(9)
Ch7_67
Right-Hand Rule
The vectors a, b, and a  b form a right-handed
system or a right-handed triple. This means that a 
b points in the direction given by the right-hand rule:
If the fingers of the right hand point along
the vector a and then curl toward the vector b,
the thumb will give the direction of a  b.
(10)
See Fig 7.4.2(a). In Figure 7.4.2(b), the right-hand
rule shows the direction of b  a.
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Ch7_68
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Ch7_69
Theorem 7.4.2 Magnitude of the Cross Product
For nonzero vectors a and b, if  is the angle between a
and b (0 ≤  ≤ ), then
(11)
a  b || a || || b || sin 
Combining (9), (10), and Theorem 7.4.2 we see for
any pair of vectors a and b in R3 that the cross
product has the alternative form
(12)
a  b  (|| a || || b || sin  )n
where n is a unit vector given by the right-hand rule
that is orthogonal to the plane of a and b.
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Ch7_70
Parallel Vectors
Theorem 7.4.3 Criterion for Parallel Vectors
Two nonzero vectors a and b are parallel, if and only
if a  b = 0.
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Ch7_71
Example 3 Parallel Vectors
Determine whether a = 2i + 3j – k and b = –6i – 3j + 3k
are parallel vectors.
Solution:
i
j k
1 1
2 1
2
1
ab  2
1 1 
i
j
k
3 3
6 3
6 3
6 3 3
 0i  0 j  0k  0
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Ch7_72
Special Products
We have
a1 a2
a．(b  c)  b1 b2
c1 c2
a3
b3
c3
(13)
is called the triple vector product.
a  (b  c)  (a  b)  c
(14)
The following results are left as an exercise.
a  (b  c)  (a．c)b  (a．b)c
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(15)
Ch7_73
Area and Volume
Area of a parallelogram
A = || a  b||
(16)
Area of a triangle
A = ½||a  b||
(17)
Volume of the parallelepiped
V = |a  (b  c)|
(18)
See Fig 7.4.3 and Fig 7.4.4.
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Ch7_74
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Ch7_75
Example 4 Area of a Triangle
Find the area of the triangle determined by the points
P1(1, 1, 1), P2(2, 3, 4) and P3(3, 0, –1).
Solution
Using (1, 1, 1) as the base point, we have two vectors a
= <1, 2, 3>, b = <1, –3, –5>
i
j k
2
3
1
3
1
2
P1 P2  P2 P3  1
2
3 
i
j
k
3 5
1 5
1 3
1 3 5
 i  8 j  5k
1
3
A  || i  8 j  5k ||  10
2
2
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Ch7_76
Coplanar Vectors
a  (b  c) = 0 if and only if a, b, c are coplanar.
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Ch7_77
Physical Interpretation of the Cross Product
To understand the physical meaning of the cross
product, please see Fig 7.4.5 and 7.4.6. The torque 
done by a force F acting at the end of position vector
r is given by  = r  F.
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Ch7_78
7.5 Lines and Planes in 3-Space
Lines: Vector Equation
See Fig 7.5.1. We find r2 – r1 is parallel to r – r2, then
r – r2 = t(r2 – r1)
(1)
If we write
a = r2 – r1 = <x2 – x1, y2 – y1, z2 – z1>
= <a1, a2, a3>
(2)
then (1) implies a vector equation for the line L a is
r = r2 + ta
where a is called the direction vector.
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Ch7_79
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Ch7_80
Example 1 Vector Equation of a Line
Find a vector equation for the line through (2, –1, 8) and
(5, 6, –3).
Solution:
Define a = <2 – 5, –1 – 6, 8 – (– 3)> = <–3, –7, 11>.
The following are three possible vector equations:
 x, y, z    2,  1, 8   t  3,  7, 11 
(3)
 x, y, z    5, 6,  3   t  3,  7, 11 
(4)
 x, y, z    5, 6,  3   t  3, 7,  11 
(5)
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Ch7_81
Parametric equation
We can also write (2) as
x  x2  a1t ,
y  y2  a2t ,
z  z2  a3t
(6)
The equations (6) are called parametric equations.
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Ch7_82
Example 2 Parametric Equations of a Line
Find the parametric equations for the line in Example 1.
Solution:
From (3), it follows
x = 2 – 3t, y = –1 – 7t, z = 8 + 11t
(7)
From (5),
x = 5 + 3t, y = 6 + 7t, z = –3 – 11t
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(8)
Ch7_83
Example 3 Vector Parallel to a Line
Find a vector a that is parallel to the line L a :
x = 4 + 9t, y = –14 + 5t, z = 1 – 3t
Solution:
a = 9i + 5j – 3k
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Ch7_84
Symmetric Equations
From (6)
x  x2 y  y 2 z  z 2
t


a1
a2
a3
provided ai are nonzero. Then
x  x2 y  y 2 z  z 2


a1
a2
a3
(9)
are said to be symmetric equation.
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Ch7_85
Example 4 Symmetric Equations of a Line
Find the symmetric equations for the line through
(4, 10, −6) and (7, 9, 2).
Solution:
Define a1 = 7 – 4 = 3, a2 = 9 – 10 = –1, a3 = 2 – (–6) =
8, then
x 7 y 9 z 2


3
1
8
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Ch7_86
Example 5 Symmetric Equations of a Line
Find the symmetric equations for the line through
(5, 3, 1) and (2, 1, 1).
Solution:
Define a1 = 5 – 2 = 3, a2 = 3 – 1 = 2, a3 = 1 – 1 = 0,
then
x 5 y 3

,
3
2
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z 1
Ch7_87
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Ch7_88
Example 6 Line Parallel to a Vector
Write vector, parametric and symmetric equations for
the line through (4, 6, –3) and parallel to a = 5i – 10j +
2k.
Solution:
Vector: <x, y, z> = < 4, 6, –3> + t(5, –10, 2)
Parametric: x = 4 + 5t, y = 6 – 10t, z = –3 + 2t,
x4 y6 z 3
Symmetric:


5
 10
2
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Ch7_89
Planes: Vector Equations
Fig 7.5.3(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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Ch7_90
Cartesian Equations
If the normal vector is n = 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)
Equation (11) is sometimes called the point-normal
form of the equation of a plane.
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Ch7_91
Example 7 Equation of a Plane
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
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Ch7_92
Equation (11) can always be written as
ax + by + cz + d = 0
(12)
Theorem 7.5.1 Plane with Normal Vector
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
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Ch7_93
Example 8 A Vector Normal to a Plane
A vector normal to the plane 3x – 4y + 10z – 8 = 0 is
n = 3i – 4j + 10k.
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Ch7_94
Given three noncollinear points, P1, P2, P3, we
arbitrarily choose P1 as the base point. See Fig 7.5.4,
Then we can obtain
[(r2  r1 )  (r3  r1 )]．(r  r1 )  0
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(13)
Ch7_95
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Ch7_96
Example 9 Three Points That Determine a
Plane
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 = <2, 1, 5> and v = <1, 3, 4>.
(1, 0,  1)
u  2i  j  5k ,
(3, 1, 4)
(3, 1, 4)
 v  i  3 j  4k ,
(2,  2, 0)
(2,  2, 0)
 w  ( x  2) i  ( y  2) j  z k .
( x, y , z ) 
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Ch7_97
Example 9 (2)
i j k
u  v  2 1 5  11i  3j  5k
1 3 4
If we choose (2, −2, 0) as the base point, then
<x – 2, y + 2, z – 0>  <−11, −3, 5> = 0
 11( x  2)  3( y  2)  5 z  0
 11x  3 y  5 z  16  0
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Ch7_98
Graphs
The graph of (12) with one or two variables missing
is still a plane.
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Ch7_99
Example 10 Graph of a Plane
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 7.5.5.
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Ch7_100
Example 11 Graph of a Plane
Graph 6x + 4y = 12.
Solution:
This equation misses the variable z, so the plane is
parallel to the z-axis.
Since x = 0 gives y = 3
y = 0 gives x = 2
See Fig 7.5.6.
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Ch7_101
Example 12 Graph of a Plane
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. See Fig 7.5.7.
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Ch7_102
Two planes P1 and P2 that are not parallel must
intersect in a line L. See Fig 7.5.8. Fig 7.5.9 shows
the intersection of a line and a plane.
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Ch7_103
Example 13 Line of Intersection of Two
Planes
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 = 14 + 7t, y = 9 + 6t, z = t.
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Ch7_104
Example 14 Point of Intersection of a Line
and a Plane
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 (x0, y0, z0) is the intersection point.
3x0 – 2y0 + z0 = −5
and x0 = 1 + t0, y0 = −2 + 2t0, z0 = 4t0
then 3(1 + t0) – 2(−2 + 2t0) + 4t0 = −5, t0 = −4
Thus, (x0, y0, z0) = (−3, −10, −16)
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Ch7_105
7.6 Vector Spaces
n-Space
Similar to 3-space
a  b   a1  b1 , a2  b2 , , an  bn 
ka   ka1 , ka2 , , kan 
a．b   a1 , a2 , , an ． b1 , b2 , , bn 
 a1b1  a2b2    anbn
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(1)
(2)
Ch7_106
Definition 7.6.1 Vector Space
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.
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Ch7_107
Definition 7.6.1 Vector Space
Axioms for Vector Addition
(i) If x and y are in V, then x + y is in V.
← commutative law
(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 ← associative law
(iv) There is a unique vector 0 in V, such that
← zero vector
0+x=x+0=x
(v) For each x in V, there exists a vector −x in V,
← negative of a vector
such that x + (−x) = (−x) + x = 0
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Ch7_108
Definition 7.6.1 Vector Space
Axioms for Scalars Multiplication
(vi) If k is any scalar and x is in V, then kx is in V.
← distributive law
(vii) k(x + y) = kx + ky
← distributive law
(viii) (k1+k2)x = k1x+ k2x
(ix) k1(k2x) = (k1k2)x
(x) 1x = x
Properties (i) and (vi) are called the closure axioms.
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Ch7_109
Example 1 Checking the Closure Axioms
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.
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Ch7_110
Example 2 An Example of a Vector S
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 = xk
Determine whether V is a vector space.
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Ch7_111
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 = xy = yx = y + x
(iii) For all x = x , y = y, z = z in V
x + (y + z) = x(yz) = (xy)z = (x + y) + z
(iv) Since 1 + x = 1x = x = x, x + 1 = x1 = x = x
The zero vector 0 is 1 = 1
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Ch7_112
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 = xk > 0
(vii) If k is any scalar,
k(x + y) = (xy)k = xkyk = kx + ky
(viii) For scalars k1 and k2,
(k1  k2 )x  x ( k1k2 )  x k1  x k2  k1x  k2 x
(ix) For scalars k1 and k2, k1 (k2 x)  ( x k ) k  x k k  (k1k2 )x
(x) 1x = x1 = x = x
2
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1
1 2
Ch7_113
Definition 7.6.2 Subspace
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.
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Ch7_114
Theorem 7.6.1 Criteria for a Subspace
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.
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Ch7_115
Example 3 A Subspace
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 (−, ).
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Ch7_116
Example 4 A Subspace
The set Pn of polynomials of degree less than or equal
to n is a subspace of C(−, ).
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Ch7_117
Definition 7.6.3 Linear Independence
A set of vectors B = {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.
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Ch7_118
For example:
i, j, k are linearly independent.
a = <1, 1, 1>, b = <2, –1, 4> and c = <5, 2, 7> are
linearly dependent, because
3<1, 1, 1> + <2, –1, 4> − <5, 2, 7> = <0, 0, 0>
3a + b – c = 0
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Ch7_119
Basis
Definition 7.6.4 Basis for a Vector Space
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.
It can be shown that any set of three linearly
independent vectors is a basis for R3. For example
<1, 0, 0>, <1, 1, 0>, <1, 1, 1>
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Ch7_120
Standard Basis
Standard Basis: {i, j, k}
For Rn:
e1 = <1, 0, …, 0>, e2 = <0, 2, …, 0> …..
en = <0, 0, …, 1>
(4)
If B is a basis for a vector space, then there exists
scalars such that
v  c1x1  c2 x 2    cn x c
(5)
where these scalars ci, i = 1, 2, .., n, are called the
coordinates of v related to the basis B.
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Ch7_121
Definition 7.6.5 Dimension of a Vector Space
The number of vectors in a basis B for vector space V
is said to be the dimension of the space.
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Ch7_122
Example 5 Dimensions of Some Vector
Spaces
(a) The dimensions of R, R2, R3, and Rn are in turn 1, 2,
3, and n.
(b) There are n + 1 vectors in B = {1, x, x2, …, xn}.
The dimension is n + 1
(c) The dimension of the zero space {0} is zero.
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Ch7_123
Linear Differential Equations
The general solution of following DE
dny
d n1 y
dy
an ( x) n  an1 ( x) n1    a1 ( x)  a0 ( x) y  0
dx
dx
dx
(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.
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Ch7_124
Example 6 Dimension of a Solution Space
The general solution of y” + 25y = 0 is
y = c1 cos 5x + c2 sin 5x
then {cos 5x , sin 5x} is a basis.
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Ch7_125
Span
If S denotes any set of vectors {x1, x2, …, xn} then the
linear combination of the vector x1, x2, …, xn in S,
{k1x1 + k2x2 + … + knxn}
where the ki, i = 1, 2, …, n are scalars, is called a
span of the vectors and written as Span(S) or Span(x1,
x2, …, xn).
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Ch7_126
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.
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Ch7_127
7.7 Gram-Schmidt Orthogonalization
Process
Orthonormal Basis
All the vectors of a basis are mutually orthogonal and
have unit length .
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Ch7_128
Example 1 Orthonormal Basis for R3
The set of vectors
1 1 1
w1 
,
,
,
3 3 3
2 1 1
w2  
,
,
,
6 6 6
1
1
w 3  0,
,
2
2
(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.
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Ch7_129
Theorem 7.7.1 Coordinates Relative to an Orthonormal Basis
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
Proof:
Since B = {w1, w2, …, wn} is an orthonormal basis,
then any vector can be expressed as
u = k1w1 + k2w2 + … + knwn
(u  wi) = (k1w1 + k2w2 + … + knwn)  wi
= ki(wi  wi) = ki
(2)
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Ch7_130
Example 2
Find the coordinate of u = <3, – 2, 9> relative to the
orthonormal basis in Example 1.
Solution:
10
, u  w2 
3
10
1
u
w1 
w2 
3
6
u  w1 
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1
11
, u  w3  
6
2
11
w3
2
Ch7_131
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 7.7.1.
v1  u1
 u 2  v1 
v2  u2  
 v1
 v1  v1 
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(3)
Ch7_132
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Ch7_133
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Ch7_134
Example 3 Gram–Schmidt Process in R2
Let u1 = <3, 1>, u2 = <1, 1>. Transform them into an
orthonormal basis.
Solution:
v1  u1  3, 1 
From (3)
4
1 3
v 2  1, 1    3, 1   ,
10
5 5
Normalizing:
See Fig 7.7.2.
w1 
1
v1 
v1
w2 
1
1
3
v2  
,
v2
10 10
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3
1
,
10 10
Ch7_135
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Ch7_136
Constructing an Orthogonal Basis for R
3
For R3:
v1  u1
 u 2  v1 
v2  u2  
 v1
 v1  v1 
u v 
u v 
v 3  u 3   3 1  v1   3 2  v 2
 v1  v1 
 v2  v2 
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(4)
Ch7_137
See Fig 7.7.3. Suppose W2 = Span{v1, v2}, then
u v 
u v 
x   3 1  v1   3 2  v 2
 v1  v1 
 v2  v2 
is in W2 and is called the orthogonal projection of u3
onto W2, denoted by x  projw u3 .
2
projv1u3
projv 2u3
 
u v 
u v 
x  projw2 u 3   3 1  v1   3 2  v 2
 v1  v1 
 v2  v2 
projv1u2

 u 2  v1 
x  projw1 u 2  
 v1
 v1  v 1 
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(5)
(6)
Ch7_138
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Ch7_139
Example 4 Gram–Schmidt Process in R3
Let u1 = <1, 1, 1>, u2 = <1, 2, 2>, u3 = <1, 1, 0>.
Transform them into an orthonormal basis.
Solution:
From (4)
v1  u1  1, 1, 1 
 u 2  v1 
5
2 1 1
v2  u2  
 v1  1, 2, 2    1, 1, 1   , ,
3
3 3 3
 v1  v1 
u v 
u v 
1 1
v 3  u 3   3 1  v1   3 2  v 2  0, , 
2 2
 v1  v1 
 v2  v2 
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Ch7_140
Example 4 (2)
2 1 1
1 1 


B  v1 , v 2 , v 3    1, 1, 1 ,  , , , 0, ,  
3 3 3
2 2 

6
2
1
v1  3, v 2 
, v3 
and w i 
v i , i  1, 2, 3,
3
2
vi
B  w1 , w 2 , w 3 
1 1
w1 
,
,
3 3
1
w 3  0,
,
2
1
2 1 1
, w2  
,
,
,
3
6 6 6
1
2
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Ch7_141
Theorem 7.7.2 Gram–Schmidt Orthogonalization Process
Let B = {u1, u2, …, um}, m  n, be a basis for a Subspace Wm of
Rn. Then B’ = {v1, v2, …, vm}, where
v1  u1
 u 2  v1 
v2  u2  
 v1
 v 1  v1 
 u 3  v1 
 u3  v 2 
v3  u3  
 v1  
v2
 v1  v1 
 v2  v2 

 u m  v m1 
 u m  v1 
 um  v2 
 v m1
vm  um  
 v1  
 v 2    
 v1  v1 
 v2  v2 
 v m1  v m1 
is an orthogonal basis for Wm. An orthonormal basis for Wm is
 1

1
1


B  w1 , w 2 ,  , w m   
v1 ,
v 2 , ,
vm 
v2
vm
 v1

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Ch7_142