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Vectors
CHAPTER 7
Copyright © Jones and Bartlett;滄海書局 Ch7_2
Chapter Contents
7.1 Vectors in 2-Space7.2 Vectors in 3-Space7.3 Dot Product7.4 Cross Product7.5 Lines and Planes in 3-Space7.6 Vector Spaces7.7 Gram-Schmidt Orthogonalization Process
Copyright © Jones and Bartlett;滄海書局 Ch7_3
7.1 Vectors in 2-Space
Review of VectorsPlease refer to Fig 7.1.1 through Fig 7.1.6.
Copyright © Jones and Bartlett;滄海書局 Ch7_4
Copyright © Jones and Bartlett;滄海書局 Ch7_5
Copyright © Jones and Bartlett;滄海書局 Ch7_6
Copyright © Jones and Bartlett;滄海書局 Ch7_7
Copyright © Jones and Bartlett;滄海書局 Ch7_8
Example 1 Position Vector
Please refer to Fig 7.1.7.
Copyright © Jones and Bartlett;滄海書局 Ch7_9
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)
Definition 7.1.1 Addition, Scalar Multiplication, Equality
a – b = <a1− b1, a2 − b2> (4)
1 2 2 1 2 1 2 1,PP OP OP x x y y ������������������������������������������
Copyright © Jones and Bartlett;滄海書局 Ch7_10
Graph Solution
Fig 7.1.8 shows the graph solutions of the addition and subtraction of two vectors.
Copyright © Jones and Bartlett;滄海書局 Ch7_11
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
17,169,188,232
1,734),6(1
7,534),6(1
ba
ba
ba
Copyright © Jones and Bartlett;滄海書局 Ch7_12
(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 = <0, 0>
Theorem 7.1.1 Properties of Vectors← commutative law
← associative law
← additive identity
← additive inverse
← (Zero Vector)
0 = <0, 0>
Copyright © Jones and Bartlett;滄海書局 Ch7_13
Magnitude, Length, Norm
a = <a1 , a2>, then
Clearly, we have ||a|| 0, ||0|| = 0
22
21|||| aa a
Copyright © Jones and Bartlett;滄海書局 Ch7_14
Unit Vector
A vector that ha magnitude 1 is called a unit vector. u = (1/||a||)a is a unit vector, since
1||||||||
1||||
1|||| a
aa
au
Copyright © Jones and Bartlett;滄海書局 Ch7_15
Example 3 Unit Vectors
Given a = <2, −1>, then the unit vector in the same direction u is
and
51
,5
21,2
51
51 au
51
,5
2 u
Copyright © Jones and Bartlett;滄海書局 Ch7_16
The i, j vectors
If a = <a1, a2>, then
(5)
Let i = <1, 0>, j = <0, 1>, then (5) becomes
a = a1i + a2j (6)
1,00,1,00,
,
2121
21
aaaa
aa
Copyright © Jones and Bartlett;滄海書局 Ch7_17
Copyright © Jones and Bartlett;滄海書局 Ch7_18
Example 4 Vector Operations Using i and j
(i) <4, 7> = 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
2|||| ji
Copyright © Jones and Bartlett;滄海書局 Ch7_19
Example 5 Graphs of Vector Sum/Vector Difference
Let a = 4i + 2j, b = –2i + 5j. Graph a + b, a – b
Solution:
Copyright © Jones and Bartlett;滄海書局 Ch7_20
7.2 Vectors in 3-Space
Simple ReviewPlease refer to Fig 7.2.1 through Fig 7.2.3.
Copyright © Jones and Bartlett;滄海書局 Ch7_21
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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.
Copyright © Jones and Bartlett;滄海書局 Ch7_24
Distance Formula
(1)
212
212
21221 )()()(),( zzyyxxPPd
Copyright © Jones and Bartlett;滄海書局 Ch7_25
Example 2 Distance Between Two Points
Find the distance between (2, −3, 6) and (−1, −7, 4)
Solution:
29)46())7(3())1(2( 222 d
Copyright © Jones and Bartlett;滄海書局 Ch7_26
Midpoint Formula
(2)
2,
2,
2212121 zzyyxx
Copyright © Jones and Bartlett;滄海書局 Ch7_27
Example 2 Coordinates of a Midpoint
Find the midpoint of (2, −3, 6) and (−1, −7, 4)
Solution:From (2), we have
5 ,5 ,21
246
,2
)7(3,
2)1(2
Copyright © Jones and Bartlett;滄海書局 Ch7_28
Vectors in 3-Space
321 ,, aaaa
Copyright © Jones and Bartlett;滄海書局 Ch7_29
Let a = <a1, a2 , a3>, b = <b1, b2, b3 > in R3
(i) a + b = <a1 + b1, a2 + b2, a3 + b3>
(ii) ka = <ka1, ka2, ka3>
(iii) a = b if and only if a1 = b1, a2 = b2, a3 = b3
(iv) –b = (−1)b = <− b1, − b2, − b3>
(v) a – b = <a1 − b1, a2 − b2, a3 − b3>
(vi) 0 = <0, 0 , 0>
(vi)
Definition 7.2.1 Component Definitions in 3-Space
23
22
21|||| aaa a
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Copyright © Jones and Bartlett;滄海書局 Ch7_31
Example 4 Vector Between Two Points
Find the vector from P1(4, 6, −2) to P2(1, 8, 3).
Solution:
5,2,3
)2(3,68,411221 OPOPPP
21PP
Copyright © Jones and Bartlett;滄海書局 Ch7_32
Example 5 A Unit Vector
The magnitude of a is
A unit vector in the direction of a is
749632|||| 222 a
76
,73
,72
6 ,3 ,271
aa
Copyright © Jones and Bartlett;滄海書局 Ch7_33
The i, j, k vectors
i = <1, 0, 0>, j = <0, 1, 0>, k = <0, 0, 1>
a = < a1, a2, a3> = a1i + a2j + a3j
1,0,00,1,00,0,1
,0,00,,00,0,
,,
321
321
321
aaa
aaa
aaa
Copyright © Jones and Bartlett;滄海書局 Ch7_34
Copyright © Jones and Bartlett;滄海書局 Ch7_35
Example 6a = <7, −5, 13> = 7i − 5j + 13k
Example 7(a) a = 5i + 3k is in the xz-plane(b)
Example 8If a = 3i − 4j + 8k, b = i − 4k, find 5a − 2b.
Solution:5a − 2b = 13i − 20j + 48k
3435||35|| 22 ki
Copyright © Jones and Bartlett;滄海書局 Ch7_36
7.3 Dot Product
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)
Definition 7.3.1 Dot Product of Two Vectors
Copyright © Jones and Bartlett;滄海書局 Ch7_37
Example 1 Dot Product Using (2)
If a = 10i + 2j – 6k, b = (−1/2)i + 4j – 3k, then
21)3)(6()4)(2(21
)10(
ba.
Copyright © Jones and Bartlett;滄海書局 Ch7_38
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. (3)
Similarly, by (2)i i = 1, j j = 1, k k = 1
(4)
Copyright © Jones and Bartlett;滄海書局 Ch7_39
(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
Theorem 7.3.1 Properties of the Dot Product
← commutative law
← distributive law
23
22
21 aaaaaa
Copyright © Jones and Bartlett;滄海書局 Ch7_40
The dot product of two vectors a and b is
(5)where is the angle between the vectors 0 .
cos|||||||| baba .
Theorem 7.3.2 Alternative Form of the Dot Product
Copyright © Jones and Bartlett;滄海書局 Ch7_41
Component Form of Dot Product
(6)
cos||||||||2|||||||||||| 22 baabc
)||||||||||(||21
cos|||||||| 222 cabba
Copyright © Jones and Bartlett;滄海書局 Ch7_42
Copyright © Jones and Bartlett;滄海書局 Ch7_43
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 Orthogonal Vectors
Copyright © Jones and Bartlett;滄海書局 Ch7_44
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.
Copyright © Jones and Bartlett;滄海書局 Ch7_45
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
(7)||||||||
cos 332211
ba
bababa
Copyright © Jones and Bartlett;滄海書局 Ch7_46
Example 4 Angle Between Two Vectors
Find the angle between a = 2i + 3j + k, b = −i + 5j + k.
Solution:
14,27||||,14|||| baba .
942
271414
cos
44.9
77.0942
cos 1
Copyright © Jones and Bartlett;滄海書局 Ch7_47
Direction Cosines
Referring to Fig 7.3.3, the angles , , are called the direction angles. Now by (7)
We say cos , cos , cos are direction cosines, and
cos2 + cos2 + cos2 = 1
||k||||a||ka
||j||||a||ja
||i||||a||ia ... cos,cos,cos
||a||||a||||a||321 cos,cos,cos
aaa
kjik||a||
j||a||
i||a||
a||a||
)(cos)(cos)(cos1 321 aaa
Copyright © Jones and Bartlett;滄海書局 Ch7_48
Copyright © Jones and Bartlett;滄海書局 Ch7_49
Example 5 Direction Cosines/Angles
Find the direction cosines and the direction angles of a = 2i + 5j + 4k.
Solution:
The direction angles are
5345452|||| 222 a
534
cos,53
5cos,
532
cos
4.53orradians 93.053
4cos
8.41orradians 73.053
5cos
7.72orradians 27.153
2cos
1
1
1
Copyright © Jones and Bartlett;滄海書局 Ch7_50
Component of a on b
Since a = a1i + a2j + a3k, then(8)
We write the components of a as(9)
See Fig 7.3.4. The component of a on any vector b is compba = ||a|| cos (10)
Rewrite (10) as
(11)
kajaia ... 321 ,, aaa
,comp iaai . ,comp jaaj . kaak .comp
bba
bb
a
bba
bba
ab
||||1
||||||||cos||||||||
comp
.
.
Copyright © Jones and Bartlett;滄海書局 Ch7_51
Copyright © Jones and Bartlett;滄海書局 Ch7_52
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
)2(6
1||||
1,6|||| kjib
bb
63
)2(6
1)432(comp kjikjiab .
)432(291
||||1
,29|||| kjiaa
a
293
)432(291
)2(comp kjikjibb .
Copyright © Jones and Bartlett;滄海書局 Ch7_53
Projection of a onto b
See Fig 7.3.5, the projection of a onto i is
See Fig 7.3.6, the projection of a onto b is
(12)b
bbba
bb1
aa bb
)(compproj
iaiiaiaa ii 1)()(compproj
Copyright © Jones and Bartlett;滄海書局 Ch7_54
Copyright © Jones and Bartlett;滄海書局 Ch7_55
Copyright © Jones and Bartlett;滄海書局 Ch7_56
Example 7 Projection of a Vector on Another Vector
Find the projection of a = 4i + j onto the vector b = 2i + 3j. Graph.
Solution:
1311
)(2131
)(4comp 3jijiab
jijiab 1333
1322
)3(2131
1311
proj
Copyright © Jones and Bartlett;滄海書局 Ch7_57
Copyright © Jones and Bartlett;滄海書局 Ch7_58
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)
Copyright © Jones and Bartlett;滄海書局 Ch7_59
Example 8 Work Done by a Constant Force
Let F = 2i + 4j. If the block moves from P1(1, 1) toP2(4, 6), find the work done by F.
Solution: d = 3i + 5j
W = F d = 26 N-m
Copyright © Jones and Bartlett;滄海書局 Ch7_60
7.4 Cross Product
(1)
(2)
122121
21 bababb
aa
31
213
31
312
32
321
321
321
321
cc
bba
cc
bba
cc
bba
ccc
bbb
aaa
Copyright © Jones and Bartlett;滄海書局 Ch7_61
The cross product of two vectors a = <a1, a2 , a3> and b = <b1, b2, b3> is the vector
(3)
Definition 7.4.1 Cross Product of Two Vectors
kjiba )()()( 122113312332 babababababa
Copyright © Jones and Bartlett;滄海書局 Ch7_62
We also can write (3) as
(4)
In turn, (4) becomes
(5)
kjiba21
21
31
31
32
32
bb
aa
bb
aa
bb
aa
321
321
bbb
aaa
kji
ba
Copyright © Jones and Bartlett;滄海書局 Ch7_63
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
kji
kji
kji
ba
10193
13
24
13
54
11
52
113
524
Copyright © Jones and Bartlett;滄海書局 Ch7_64
Example 2 Cross Product of Two Basis Vectors
If i = <1, 0, 0> and j = <0, 1, 0>, then
kkji
kji
ji 10
01
00
01
01
00
010
001
001
01
01
01
00
00
001
001 kji
kji
ii
Copyright © Jones and Bartlett;滄海書局 Ch7_65
Proceeding as in Example 2, it is readily shown that
(6)
(7)
(8)
The results in (6) can be obtained using the circular mnemonic illustrated in Fig 7.4.1.
0kk0jj0ii
jkiijkkij
jikikjkji
,,
,,
,,
Copyright © Jones and Bartlett;滄海書局 Ch7_66
Copyright © Jones and Bartlett;滄海書局 Ch7_67
Properties
a b is orthogonal to the plane containing a and b. (9)
(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), k is scalar(vi) a a = 0(vii) a (a b) = 0(viii) b (a b) = 0
Theorem 7.4.1 Properties of the Cross Product
← distributive law
← distributive law
Copyright © Jones and Bartlett;滄海書局 Ch7_68
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.
Copyright © Jones and Bartlett;滄海書局 Ch7_69
Copyright © Jones and Bartlett;滄海書局 Ch7_70
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)where n is a unit vector given by the right-hand rule that is orthogonal to the plane of a and b.
For nonzero vectors a and b, if is the angle between a and b (0 ≤ ≤ ), then
(11)
Theorem 7.4.2 Magnitude of the Cross Product
sin|||||||| baba
nbaba )sin||||||(||
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Parallel Vectors
Two nonzero vectors a and b are parallel, if and only if a b = 0.
Theorem 7.4.3 Criterion for Parallel Vectors
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Example 3 Parallel Vectors
Determine whether a = 2i + 3j – k and b = –6i – 3j + 3k are parallel vectors.Solution:
0kji
kji
kji
ba
000
36
12
36
12
33
11
336
112
Copyright © Jones and Bartlett;滄海書局 Ch7_73
Special Products
We have(13)
is called the triple vector product.
(14)
The following results are left as an exercise.
(15)
321
321
321
)(
ccc
bbb
aaa
cba.
cbabcacba )()()( ..
cbacba )()(
Copyright © Jones and Bartlett;滄海書局 Ch7_74
Area and Volume
Area of a parallelogram A = || a b|| (16)
Area of a triangleA = 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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Copyright © Jones and Bartlett;滄海書局 Ch7_76
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).SolutionUsing (1, 1, 1) as the base point, we have two vectors a = <1, 2, 3>, b = <1, –3, –5>
kji
kji
kji
58
31
21
51
31
53
32
531
3213221
PPPP
1023
||58||21 kjiA
Copyright © Jones and Bartlett;滄海書局 Ch7_77
Coplanar Vectors
a (b c) = 0 if and only if a, b, c are coplanar.
Copyright © Jones and Bartlett;滄海書局 Ch7_78
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.
Copyright © Jones and Bartlett;滄海書局 Ch7_79
7.5 Lines and Planes in 3-Space
Lines: Vector EquationSee 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 + tawhere a is called the direction vector.
Copyright © Jones and Bartlett;滄海書局 Ch7_80
Copyright © Jones and Bartlett;滄海書局 Ch7_81
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:
(3)
(4)
(5)
11,7,38,1,2,, tzyx
11,7,33,6,5,, tzyx
11,7,33,6,5,, tzyx
Copyright © Jones and Bartlett;滄海書局 Ch7_82
Parametric equation
We can also write (2) as
(6)
The equations (6) are called parametric equations.
tazztayytaxx 322212 ,,
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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 (8)
Copyright © Jones and Bartlett;滄海書局 Ch7_84
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
Copyright © Jones and Bartlett;滄海書局 Ch7_85
Symmetric Equations
From (6)
provided ai are nonzero. Then
(9)
are said to be symmetric equation.
3
2
2
2
1
2
azz
ayy
axx
t
3
2
2
2
1
2
azz
ayy
axx
Copyright © Jones and Bartlett;滄海書局 Ch7_86
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
82
19
37
zyx
Copyright © Jones and Bartlett;滄海書局 Ch7_87
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
1,2
33
5 z
yx
Copyright © Jones and Bartlett;滄海書局 Ch7_88
Copyright © Jones and Bartlett;滄海書局 Ch7_89
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, Symmetric:
23
106
54
zyx
Copyright © Jones and Bartlett;滄海書局 Ch7_90
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)
Copyright © Jones and Bartlett;滄海書局 Ch7_91
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.
Copyright © Jones and Bartlett;滄海書局 Ch7_92
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) = 0or
2x + 8y – 5z + 15 = 0
Copyright © Jones and Bartlett;滄海書局 Ch7_93
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 7.5.1 Plane with Normal Vector
Copyright © Jones and Bartlett;滄海書局 Ch7_94
Example 8 A Vector Normal to a Plane
A vector normal to the plane 3x – 4y + 10z – 8 = 0 is n = 3i – 4j + 10k.
Copyright © Jones and Bartlett;滄海書局 Ch7_95
Given three noncollinear points, P1, P2, P3, we arbitrarily choose P1 as the base point. See Fig 7.5.4, Then we can obtain
(13)
0)()]()[( 11312 rrrrrr .
Copyright © Jones and Bartlett;滄海書局 Ch7_96
Copyright © Jones and Bartlett;滄海書局 Ch7_97
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>.
.)2()2() , ,(
)0 ,2 ,2(
,43)0 ,2 ,2(
)4 ,1 ,3(,52
)4 ,1 ,3(
)1 ,0 ,1(
kjiw
kjivkjiu
zyxzyx
Copyright © Jones and Bartlett;滄海書局 Ch7_98
Example 9 (2)
If we choose (2, −2, 0) as the base point, then<x – 2, y + 2, z – 0> <−11, −3, 5> = 0
kji
kji
vu 5311
431
512
05)2(3)2(11 zyx
0165311 zyx
Copyright © Jones and Bartlett;滄海書局 Ch7_99
Graphs
The graph of (12) with one or two variables missing is still a plane.
Copyright © Jones and Bartlett;滄海書局 Ch7_100
Example 10 Graph of a Plane
Graph 2x + 3y + 6z = 18.
Solution:Setting:y = z = 0 gives x = 9x = z = 0 gives y = 6x = y = 0 gives z = 3See Fig 7.5.5.
Copyright © Jones and Bartlett;滄海書局 Ch7_101
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 = 2See Fig 7.5.6.
Copyright © Jones and Bartlett;滄海書局 Ch7_102
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.
Copyright © Jones and Bartlett;滄海書局 Ch7_103
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.
Copyright © Jones and Bartlett;滄海書局 Ch7_104
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 + tthen x = 14 + 7t, y = 9 + 6t, z = t.
Copyright © Jones and Bartlett;滄海書局 Ch7_105
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 = −4Thus, (x0, y0, z0) = (−3, −10, −16)
Copyright © Jones and Bartlett;滄海書局 Ch7_106
7.6 Vector Spaces
n-SpaceSimilar to 3-space
(1)
(2)
nn bababa ,,, 2211 ba
nkakakak ,,, 21 a
nn
nn
bababa
bbbaaa
2211
2121 ,,,,,, ..ba
Copyright © Jones and Bartlett;滄海書局 Ch7_107
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 7.6.1 Vector Space
Copyright © Jones and Bartlett;滄海書局 Ch7_108
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 7.6.1 Vector Space
← commutative law
← associative law
← zero vector
← negative of a vector
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Properties (i) and (vi) are called the closure axioms.
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) (k1+k2)x = k1x+ k2x
(ix) k1(k2x) = (k1k2)x
(x) 1x = x
Definition 7.6.1 Vector Space
← distributive law
← distributive law
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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.
Copyright © Jones and Bartlett;滄海書局 Ch7_111
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 = xyand scalar multiplication is defined by
kx = xk
Determine whether V is a vector space.
Copyright © Jones and Bartlett;滄海書局 Ch7_112
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 Vx + (y + z) = x(yz) = (xy)z = (x + y) + z
(iv) Since 1 + x = 1x = x = x, x + 1 = x1 = x = xThe zero vector 0 is 1 = 1
Copyright © Jones and Bartlett;滄海書局 Ch7_113
Example 2 (3)
(v) If we define −x = 1/x, thenx + (−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,
(ix) For scalars k1 and k2,
(x) 1x = x1 = x = x
xxx 21)(
212121)( kkxxxkk kkkk
xx )()()( 21212112 kkxxkk kkkk
Copyright © Jones and Bartlett;滄海書局 Ch7_114
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 7.6.2 Subspace
Copyright © Jones and Bartlett;滄海書局 Ch7_115
A nonempty subset W is a subspace of V if and only ifW 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 7.6.1 Criteria for a Subspace
Copyright © Jones and Bartlett;滄海書局 Ch7_116
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 (−, ).
Copyright © Jones and Bartlett;滄海書局 Ch7_117
Example 4 A Subspace
The set Pn of polynomials of degree less than or equal to n is a subspace of C(−, ).
Copyright © Jones and Bartlett;滄海書局 Ch7_118
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.
Definition 7.6.3 Linear Independence
Copyright © Jones and Bartlett;滄海書局 Ch7_119
For example: i, j, k are linearly independent.
a = <1, 1, 1>, b = <2, –1, 4> and c = <5, 2, 7> are
linearly dependent, because3<1, 1, 1> + <2, –1, 4> − <5, 2, 7> = <0, 0, 0>3a + b – c = 0
Copyright © Jones and Bartlett;滄海書局 Ch7_120
Basis
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>
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.
Definition 7.6.4 Basis for a Vector Space
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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
(5)where these scalars ci, i = 1, 2, .., n, are called the coordinates of v related to the basis B.
cnccc xxxv 2211
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The number of vectors in a basis B for vector space V is said to be the dimension of the space.
Definition 7.6.5 Dimension of a Vector Space
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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.
Copyright © Jones and Bartlett;滄海書局 Ch7_124
Linear Differential Equations
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.
0)()()()( 011
1
1
yxadxdy
xadx
ydxa
dxyd
xa n
n
nn
n
n
Copyright © Jones and Bartlett;滄海書局 Ch7_125
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.
Copyright © Jones and Bartlett;滄海書局 Ch7_126
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).
Copyright © Jones and Bartlett;滄海書局 Ch7_127
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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7.7 Gram-Schmidt Orthogonalization Process
Orthonormal Basis All the vectors of a basis are mutually orthogonal and have unit length .
Copyright © Jones and Bartlett;滄海書局 Ch7_129
Example 1 Orthonormal Basis for R3
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.
21
,2
1 0,
,6
1 ,
61
,6
2
,3
1 ,
31
,3
1
3
2
1
w
w
w
Copyright © Jones and Bartlett;滄海書局 Ch7_130
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)
Suppose B = {w1, w2, …, wn} is an orthonormal basisfor Rn, If u is any vector in Rn, then
u = (u w1)w1 + (u w2)w2 + … + (u wn)wn
Theorem 7.7.1 Coordinates Relative to an Orthonormal Basis
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Example 2
Find the coordinate of u = <3, – 2, 9> relative to the orthonormal basis in Example 1.
Solution:
321
321
211
61
310
211
,6
1 ,
310
wwwu
wuwuwu
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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.
(3)1
11
1222
11
vvvvu
uv
uv
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Copyright © Jones and Bartlett;滄海書局 Ch7_134
Copyright © Jones and Bartlett;滄海書局 Ch7_135
Example 3 Gram–Schmidt Process in R2
Let u1 = <3, 1>, u2 = <1, 1>. Transform them into an orthonormal basis.
Solution: From (3)
Normalizing:
See Fig 7.7.2.
53
,51
1 3,104
1 1,
1 3,
2
11
v
uv
103
,1011
101
,1031
22
2
11
1
vv
w
vv
w
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Copyright © Jones and Bartlett;滄海書局 Ch7_137
Constructing an Orthogonal Basis for R 3
For R3:
(4)2
22
231
11
1333
111
1222
11
vvvvu
vvvvu
uv
vvvvu
uv
uv
Copyright © Jones and Bartlett;滄海書局 Ch7_138
See Fig 7.7.3. Suppose W2 = Span{v1, v2}, then
is in W2 and is called the orthogonal projection of u3 onto W2, denoted by
(5)
(6)
3
222
23
3
111
133
21
2
projproj
proj
u
vvv
vu
u
vvv
vuux
vv
w
2
111
122
1
1
proj
proj
u
vvvvu
ux
v
w
222
231
11
13 vvv
vuv
vv
vux
.proj 32ux w
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Copyright © Jones and Bartlett;滄海書局 Ch7_140
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)
21
,21
0,
31 ,
31 ,
32
1 1, 1,35
2 2, 1,
1 1, 1,
222
231
11
1333
111
1222
11
vvv
vuv
vv
vuuv
vvvvu
uv
uv
Copyright © Jones and Bartlett;滄海書局 Ch7_141
Example 4 (2)
2
1 ,
21
0,
,6
1 ,
61
,6
2 ,
31
,3
1 ,
31
, ,
3, 2, 1, ,1
and 22
,36
,3
21
,21
,0 ,31 ,
31 ,
32
,1 1, 1, , ,
3
21
321
321
321
w
ww
www
vv
wvvv
vvv
B
i
B
ii
i
Copyright © Jones and Bartlett;滄海書局 Ch7_142
Let B = {u1, u2, …, um}, m n, be a basis for a Subspace Wm of
Rn. Then B’ = {v1, v2, …, vm}, where
is an orthogonal basis for Wm. An orthonormal basis for Wm is
Theorem 7.7.2 Gram–Schmidt Orthogonalization Process
111
12
22
21
11
1
222
231
11
1333
111
1222
11
mmm
mmmmmm v
vvvu
vvvvu
vvvvu
uv
vvvvu
vvvvu
uv
vvvvu
uv
uv
mm
mB vv
vv
vv
www1
, ,1
,1
, , , 22
11
21