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Vectors

CHAPTER 7

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

7.1 Vectors in 2-Space

Review of VectorsPlease refer to Fig 7.1.1 through Fig 7.1.6.

Example 1 Position Vector

Please refer to Fig 7.1.7.

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 ��

Graph Solution

Fig 7.1.8 shows the graph solutions of the addition and subtraction of two vectors.

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

(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>

Magnitude, Length, Norm

a = <a1 , a2>, then

Clearly, we have ||a|| 0, ||0|| = 0

21|||| aa a

Unit Vector

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

1||||||||

1|||| a

Example 3 Unit Vectors

Given a = <2, −1>, then the unit vector in the same direction u is

The i, j vectors

If a = <a1, a2>, then

Let i = <1, 0>, j = <0, 1>, then (5) becomes

a = a1i + a2j (6)

1,00,1,00,

Example 4 Vector Operations Using i and j

(i) <4, 7> = 4i + 7j

(ii) (2i – 5j) + (8i + 13j) = 10i + 8j

(iv) 10(3i – j) = 30i – 10j

(v) a = 6i + 4j, b = 9i + 6j are parallel and b = (3/2)a

2|||| ji

Example 5 Graphs of Vector Sum/Vector Difference

Let a = 4i + 2j, b = –2i + 5j. Graph a + b, a – b

Solution:

7.2 Vectors in 3-Space

Simple ReviewPlease refer to Fig 7.2.1 through Fig 7.2.3.

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.

Distance Formula

21221 )()()(),( zzyyxxPPd

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

Midpoint Formula

2212121 zzyyxx

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

Vectors in 3-Space

321 ,, aaaa

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>

Definition 7.2.1 Component Definitions in 3-Space

21|||| aaa a

Example 4 Vector Between Two Points

Find the vector from P1(4, 6, −2) to P2(1, 8, 3).

Solution:

)2(3,68,411221 OPOPPP

Example 5 A Unit Vector

The magnitude of a is

A unit vector in the direction of a is

749632|||| 222 a

6 ,3 ,271

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,

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

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

Example 1 Dot Product Using (2)

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

21)3)(6()4)(2(21

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

(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

21 aaaaaa

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

Component Form of Dot Product

cos||||||||2|||||||||||| 22 baabc

)||||||||||(||21

cos|||||||| 222 cabba

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

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.

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

bababa

Example 4 Angle Between Two Vectors

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

Solution:

14,27||||,14|||| baba ．

271414

77.0942

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

kjik||a||

j||a||

i||a||

a||a||

)(cos)(cos)(cos1 321 aaa

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

cos,53

4.53orradians 93.053

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

kajaia ．．． 321 ,, aaa

,comp iaai ． ,comp jaaj ． kaak ．comp

||||||||cos||||||||

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

1,6|||| kjib

1)432(comp kjikjiab ．

)432(291

,29|||| kjiaa

)432(291

)2(comp kjikjibb ．

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

)(compproj

iaiiaiaa ii 1)()(compproj

Example 7 Projection of a Vector on Another Vector

Find the projection of a = 4i + j onto the vector b = 2i + 3j. Graph.

Solution:

)(2131

)(4comp 3jijiab

jijiab 1333

)3(2131

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)

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

7.4 Cross Product

122121

21 bababb

The cross product of two vectors a = <a1, a2 , a3> and b = <b1, b2, b3> is the vector

Definition 7.4.1 Cross Product of Two Vectors

kjiba )()()( 122113312332 babababababa

We also can write (3) as

In turn, (4) becomes

kjiba21

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

Example 2 Cross Product of Two Basis Vectors

If i = <1, 0, 0> and j = <0, 1, 0>, then

001 kji

Proceeding as in Example 2, it is readily shown that

The results in (6) can be obtained using the circular mnemonic illustrated in Fig 7.4.1.

0kk0jj0ii

jkiijkkij

jikikjkji

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

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.

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

Theorem 7.4.2 Magnitude of the Cross Product

sin|||||||| baba

nbaba )sin||||||(||

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

Example 3 Parallel Vectors

Determine whether a = 2i + 3j – k and b = –6i – 3j + 3k are parallel vectors.Solution:

Special Products

We have(13)

is called the triple vector product.

The following results are left as an exercise.

cba．

cbabcacba )()()( ．．

cbacba )()(

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.

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>

3213221

||58||21 kjiA

Coplanar Vectors

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

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.

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.

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:

11,7,38,1,2,, tzyx

11,7,33,6,5,, tzyx

Parametric equation

We can also write (2) as

The equations (6) are called parametric equations.

tazztayytaxx 322212 ,,

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)

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

Symmetric Equations

From (6)

provided ai are nonzero. Then

are said to be symmetric equation.

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

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

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:

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)

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.

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

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

Example 8 A Vector Normal to a Plane

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

Given three noncollinear points, P1, P2, P3, we arbitrarily choose P1 as the base point. See Fig 7.5.4, Then we can obtain

0)()]()[( 11312 rrrrrr ．

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(

kjivkjiu

zyxzyx

Example 9 (2)

If we choose (2, −2, 0) as the base point, then<x – 2, y + 2, z – 0> <−11, −3, 5> = 0

vu 5311

05)2(3)2(11 zyx

0165311 zyx

Graphs

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

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.

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.

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.

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.

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.

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)

7.6 Vector Spaces

n-SpaceSimilar to 3-space

nn bababa ,,, 2211 ba

nkakakak ,,, 21 a

bababa

bbbaaa

2121 ,,,,,, ．．ba

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

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

← commutative law

← associative law

← zero vector

← negative of a vector

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

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.

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.

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

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

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

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

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 (−, ).

Example 4 A Subspace

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

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

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

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

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

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

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.

Linear Differential Equations

The general solution of following DE

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

yxadxdy

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.

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).

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.

7.7 Gram-Schmidt Orthogonalization Process

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

Example 1 Orthonormal Basis for R3

The set of vectors

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.

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

Example 2

Find the coordinate of u = <3, – 2, 9> relative to the orthonormal basis in Example 1.

Solution:

wuwuwu

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.

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.

1 3,104

Constructing an Orthogonal Basis for R 3

For R3:

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

projproj

13 vvv

.proj 32ux w

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)

1 1, 1,35

2 2, 1,

1 1, 1,

Example 4 (2)

3, 2, 1, ,1

and 22

,0 ,31 ,

,1 1, 1, , ,

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

mmmmmm v

, , , 22

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