Chapter Linear Systems:The linear system is a combination of 2 or more linear equation (Mathisfun, n.d) where those equations are sharing same points, In graphically when you draw those points on the graph, it will draw straight lines with each of them in PDF

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Linear Systems:The linear system is a combination of 2 or more linear equation (Mathisfun, n.d) where those equations are sharing same points, In graphically when you draw those points on the graph, it will draw straight lines with each of them intersecting on top of other and will give specific poi...

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Applied Mathematics I (Math 231B)

CHAPTER ONE VECTORS AND VECTOR SPACES 1.1. Scalar and vectors in

R2 ∧ R3

Definitions: a) A scalar is a physical quantity that is described by its magnitude only. For example, temperature, length, and speed are scalars because they are completely described by a number that tells "how much"-say a temperature of 20°C, a length of 5 cm, or a speed of 10 m/s. b) A vector is a physical quantity that is described using both magnitude and its direction. For instance, velocity, displacement, and force are some examples of vectors. Vectors in

2

3

R ∧R A vector in the plane

R2 can be described as u=( u1 , u2) , where u1 ,u 2 ∈ R . Similarly, a vector in the space R3 can be described as a triple of numbers v =( v 1 , v 2 , v3 ) where v 1 , v 2 , v 3 ∈ R . EQUAL (OR EQUIVALENT) VECTORS 2 3 u∧v in are said to be equal (or Definition: Two vectors R /R equivalent) if they have the same magnitude and direction, and is denoted by u=v . That is, if u=( u1 , u2) , v =( v 1 , v 2 ) ∈R 2 , u =v iff u 1=v 1∧u 2=v 2 . Definitions: 1. A located vector is a vector

 AB defined as an arrow whose initial point

is at point A and whose terminal point is at B. Y b2

B(b1,b2)

a2 A(a1,a2)

a1

b1

X

2. Position vector is a vector whose initial point is at the origin. Definition (Parallel / collinear Vectors) Page 1

Applied Mathematics I (Math 231B)

Definition Two non – zero vectors

u∧v

of the same dimension are said to be

parallel or, alternatively, collinear if at least one of the vectors is a scalar multiple of the other. If one of the vectors is a positive scalar multiple of the other, then the vectors are said to have the same direction, and if one of them is a negative scalar multiple of the other, then the vectors are said to have opposite directions. In other words, the two vectors are said to be parallel, denoted by

u /¿ v

u∧v

if there exists a scalar c such that

u=c v .

REMARK : 1. The vector 0 is parallel to every vector v in the same dimension , since it can be expressed as the scalar multiple 0 = 0v. 2. The zero vectors has no natural direction, so we will agree that it can be assigned any direction that is convenient for the problem at hand. Definition: For any

u=( u1 , u2) ∧c ∈ R , we define

c u as c u =(cu 1 , cu 2 ) .

1.2. Vector addition and Scalar multiplication Definition: For any two vectors

u=( u1 , u2) ∧v = (v 1 , v 2 )

in R2 , we define their

u + v =( u1 + v1 , u2 +v 2 ) .

sum to be

Geometrically, if we represent the two vectors respectively, then

u + v

u∧v

 AB∧  BC

is represented by  AC , as shown in the diagram

below:

 AC AB+ BC= C U+V V A

by

B U

a) The Triangular Law

Page 2

Applied Mathematics I (Math 231B)

D C U+V = V+U

V

V

A

B U

b) The Parallelogram Law Addition of vectors on the coordinate plane Let

v =( v 1 , v 2) ∧ w = ( w 1 , w2 )

then

→

→

v +w = ( v 1 +w 1 , v 2 +w2 ) .

Vector Subtraction The negative of a vector v, denoted by -v, is the vector that has the same length as v but is oppositely directed, and the difference of v from w, denoted by w - v, is taken to be the sum

w - v = w+ (-v)

Page 3

Applied Mathematics I (Math 231B)

Example: Provided that

U =(-1,0,1) and V= (2,-1,5 ) ,find each of the

following vectors a) U + V

b) 2U

c) V -2U

Solution: a) U+V= (-1,0,1) + (2,-1,5 ) = (1,-1,6) b) 2U = 2 (-1,0,1) = (-2 , 0 , 2) c) V -2U = (2,-1,5 )-2(-1,0,1)= (4 ,-1 ,3)

Properties of Vector addition & Scalar Multiplication Let

u , v

and

 w

be vectors in

R

2

a)

u + v ∈ R

b)

u + v =v +u

c)

 0+u =u , where 0=( 0,0 ) ∈ R 2 . u +0=

and

c∧m

are scalars. Then:

2

d) There exists

w∈R

e)

w ) =( u +v ) +  w u +( v +

f)

c ( mu) =( cm ) u

g)

(c +m) u =cu +m u

h)

1. u =u

2

such that

u + w =0

for every u ∈ R2 .

Remark: The properties described above also hold true for vectors in where

0=( 0,0,0 ) ∈ R3 , replaces the zero vector

SUMSOFTHREEORMOREVECTORS

Page 4

0

in R2 .

3

R ,

Applied Mathematics I (Math 231B)

1.3. Dot (Scalar) product, Magnitude of a vector, Angle between two Vectors, Orthogonal Projection, Direction angles and direction cosines. 1.3.1. Dot (Scalar) Product Definition: Let

v =( v 1 , v 2 , v 3 )

(norm) of

v,

be a vector in R3 . Then the magnitude

‖v‖

denoted by

is defined by:

‖v‖ 2=( ¿ ) 2+ ( PR )2=(OQ )2 +( QR)2 +( PR )2

‖v‖= √ v 21+v 22+v23 Similarly, for a vector

2 v =( v 1 , v 2 ) ∈ R ,

its norm is given by

‖v‖ =√ v 12+ v 22 Examples: a) If b) If Remarks:

(i) (ii)

Theorem: If

v =(−1,4,3) , then find ‖v‖ .

‖u‖ =6, find x such that u=( −1, x , 5 ) .

‖v‖≠ 0 if v ≠ 0 ‖v‖ =‖− v‖ .

c ∈ R , then ‖c v‖=|c|‖v‖ .

Proof: suppose that

v ∈ R n then

‖c v‖=√ ( c v 1 )2 + ( c v 2 )2 +…+( c v k )2

( v1 ) ¿ 2 2 v +…+ [¿ 2+( 2 ) ( vk ) ¿] c2 ¿ ¿ √¿ Page 5

Applied Mathematics I (Math 231B)

=

|c |‖v ‖

Definition (Unit Vector) Any vector

satisfying

u

‖u‖ =1 is called a unit vector. (0,1) , (− 1,0 ),

Examples: The vectors

, ( 1,0,0) ( √12 , −1 √ 2)

are examples of unit

vectors. N.B: 1. All unit vectors in

are of the form ( cos θ , sin θ ) , where θ ∈ R .

R2

2. For any non-zero vector v , the unit vector the direction of

can be obtained as:

v

P ( u1 , u2) ∧Q ( v 1 , v 2 )

3. For two points distance

d(Ρ , Q)

^u corresponding to

on the plane

v

in

 V ‖V‖

^u=

2

R ,

we calculate the

between the two points, as :

√

d ( Ρ , Q ) =‖ ΡQ‖= ( v 1−u1 ) +( v 2−u2 ) , ΡQ where 

2

is the vector

2

with initial point P and terminal point Q,

 Ρ Q= ( v 1−u 1 , v 2−u2) . 4. Distance between two vectors can be viewed as the length of U =( u1 , u2 ) ∧V = ( v 1 , v2 )

U -V where

.

1.3.2. The Dot Product (or Scalar Product) v =( v 1 , v 2 ) are vectors in R2 , and Definition: Suppose that u=( u1 , u2) ∧¿ that θ ∈[ 0 , π ] represents the angle between them. We define the dot  denoted by u ∙v by: product of u∧V u ∙ v =u1 v 1 +u2 v 2 Or alternatively, we write

{

u ∙ v = ‖u‖‖v ‖cos θ ,if u ≠ 0∧v ≠ 0 0 if u =0∨ v =0 Page 6

Applied Mathematics I (Math 231B)

θ=¿u 1 v 1+ u 2 v2 , for non – zero vector ‖u‖‖v ‖cos ¿

Thus

Similarly, if

u=( u1 , u2 , u3) ∧v = ( v 1 , v2 , v 3 )

u∧v ∈R 2 .

are vectors in

3

R ,

we define

u ∙ v

as

bellow: u ∙ v =u1 v 1 +u2 v 2 +u3 v 3

Remark: The dot product of two vectors is a scalar quantity, and its value is maximum when

and minimum if

θ=0°

θ=180 °∨π

radians.

Example: If u=i−2 j+3 k∧v = ⟨ 0,1 ,−5 ⟩ , then find: a)

b)

u ∙v

c)

u ∙u

( u +v ) ∙ v

Properties of the dot product If

are vectors in the same dimension, and

w u , v ,∧

c ∈ R , then:

1.

u ∙u =‖u‖

2

4.

0 ∙u =0

2.

u ∙ v =v ∙ u

5.

(c u )∙ v =c (u ∙ v) =v ∙ ( c u )

u ∙( v +  w )= u ∙ v +u ∙  w

3.

6.

u ∙u ≥ 0∧u ∙ u=0 iff u=0  .

1.3.3. Angle between two vectors If

θ

is the angle between two non – zero vectors

u∧v ,

then, the angle

between the two vectors can be obtained by: cos θ=

u ∙ v

⇒

‖u‖‖v‖

Exercise: Find the angle between the vectors Definition: Two non – zero iff

u ∙v =0,

i.e, if θ=

u∧v

( ‖uu‖‖∙ vv‖ ),∧θ ∈ [ 0 , π ] .

θ=cos−1

u=⟨ 1,0 ,−1⟩∧ v =( 1,1,0 ) .

are said to be orthogonal (perpendicular)

π . 2

Example: Find the value (s) of

x

such that the vectors

A= (1 , 4, 3 ) and Β=( x ,−1, 2 )

are

orthogonal.

Definition: If P and Q are points in 2 or 3 spaces, the distance between P and Q, denoted by

‖Ρ−Q‖ is given by ‖Ρ−Q‖= √ ( Ρ−Q ). ( Ρ−Q ) Page 7

Applied Mathematics I (Math 231B)

Theorem: Given two vectors orthogonal vectors. Pythagoras

u∧v

‖u + v‖=‖u− v‖

in space,

iff

u∧v

are

Proof : Exercise.

Theorem:

If

A

and

B

are

orthogonal

vectors,

then

‖A+ B‖2=‖A ‖2 +‖ B‖2 Proof: Exercise . Note: If A is perpendicular to B, then it is also perpendicular to any scalar multiple of B.

1.3.4. Orthogonal Projection Definition: Suppose S is the foot of the perpendicular from R to the line containing

 PQ ,

then the vector with representation

 PS

is called the

vector projection of B on to A, and is denoted by project A B . R

R B

B A

S

A

Q P

P Proj AB

PS=tA ⟹ proj = B A

S

Q

projAB

B− proj =  SR⟹ B A

B− proj ¿ ).A=0 ¿ ¿

⟹

(B- t A).A =0

⟹

B.A –t A.A =0 ⟹

A∙ B 2 ‖A‖

t=

A∙ B proj BA = PS=t . A= A ‖ A‖2

⟹

The scalar projection of B onto A (also called the component of B a long A) is defined to be the length of project A B , which is equal to is denoted by comp A B . Thus:

( A‖ ∙AB‖ ) ‖AA‖ =‖AA∙‖B A

proj AB =

2

Page 8

‖B ‖cos θ and

Applied Mathematics I (Math 231B)

comp A =‖ proj A‖= B

and Example: Let

B

A= (−1,3,1)=−i+ 3 j + k

projBA [ans .

Then find (i)

‖ ‖‖ ‖ A∙B

2

‖A‖

and

13 ( 2,4,3 ) ] 29

,

(ii)

A =

A∙B

‖A ‖= A ∙ B ‖A‖ ‖A ‖ 2

B=( 2,4,3) =2i+ 4 j+3 k B

proj A

[ ans .

13 (−1,3,1) ] 11

(iii)

compBA 1.3.5. Directional angles and cosines Let

be a vector positioned at the origin in

A=a1 i+a 2 j+ a3 k

angle of angles

α , β∧γ α , β∧ γ

with the positive

3

R ,

making an

x , y ∧ z axes respectively. Then the

are called the directional angles of A, and the quantities

cos α , cos β∧cos γ

are called the directional cosines of A, which can be

computed as follows: a A .i = 1 ‖ A‖‖i ‖ ‖A‖

cos α=

a A.j = 2 ‖A ‖‖j ‖ ‖A‖

,

and

cos β=

a A.k = 3 ‖A ‖‖k‖ ‖A‖

cos γ=

From this relation we can deduce a unit vector A=‖ A‖(cosα ,cosβ , cosγ ) A

‖A‖

=( cosα , cosβ , cosγ )

a1 , α ∈[ 0, π ] ‖ A‖

cos α=

is a unit vector. ,

a2 , β ∈[ 0, π ] ‖A ‖

cos β=

Remark:

a3 ,γ ∈ ‖A ‖

cos γ=

A = a1 i + a2 j +a3k

� � �

,

�

cos 2 α + cos2 β+ cos2 γ =1 . (Verify!)

Exercise: Let A= (−1,2,2 ) . Then find the directional cosines of A. 1.4. The Cross (or Vector) Product and Triple Products

Page 9

[ 0, π ]

Applied Mathematics I (Math 231B)

Given two nonzero vectors

A=( a1 , a2 , a3 ) =a1 i+a 2 j+a3 k

B=( b 1 , b2 , b3 )=b1 i+ b2 j+ b3 k

and

, it is very useful to be able to find a nonzero vector C that is C=( c 1 , c2 , c 3 ) is such a vector, then A.C = 0 and B.C =

perpendicular to both A and B . If 0

A .C=( a1 c 1+ a2 c2 +a 3 c 3) =0 … ..(1)

B . C=( b 1 c 1 +b2 c 2+b3 c 3 )=0 … ..(2) To eliminate c3 multiply (1) by b3 and multiply (2) by a3 , and subtract b3 ( a1 c 1 +a2 c 2+ a3 c 3 )=0 a3 ( b1 c 1 + b2 c 2+ b3 c 3) =0

b b 3 a −a b ¿ (¿ 2 3 2 )c 2 =0 … … .(3) (¿ ¿ 3 a1 −a3 b1 )c 1 +¿ ¿ Equation (3) has the form pc1 + qc2 =0 .Where the solution for c1 = q and c2 = -p So the solution of equation (3) is b (¿ ¿ 3 a2 −a3 b2 ) c 1=¿

and

c 2=(a3 b1− b3 a1)

Substituting equations (1) and (2) we get c 3=a 1 b 2−a2 b1

That means a vector perpendicular to both A and B is ¿ C=( c 1 , c2 , c 3 )=¿

a2 b3 −a3 b2 , a3 b1−b 3 a 1 , a1 b2 −a2 b1 )

The resulting vector is called the cross product of A and B denoted by A x B Definition: Suppose that B=( b 1 , b2 , b3 )=b1 i+ b2 j+ b3 k

A ×B

A=( a1 , a2 , a3) =a1 i+a 2 j+ a3 k and

be two vectors in R3 . Then the cross product

of the two vectors is defined as: A × B= ( a2 b3−a3 b 2 ) i+ ( a3 b1−a 1 b3 ) j+( a1 b2−a2 b 1 ) k

Page 10

Applied Mathematics I (Math 231B)

Or

i j k AXB= Det a1 a2 a3 b1 b2 b3

( )

Example: Let

A=4 i −3 j + 2k

B=2 i−5 j−k Then find a)

b)

A ×B

B× A

Remarks: For two non – zero vectors A & B, 1.

A ×B

is a vector which is orthogonal to both A and B.

2.

A ×B

is not defined for

3.

i× j=− j× i=k

A , B ∈ R2 .

j ×k =− ( k × j )=i k ×i=− ( i× k )= j A XB B A

-A XB = B XA

Properties of Cross Product Let A, B and C be vectors in

R3

and α

(1) A ×0=0 × A=0 , where 0=⟨ 0,0,0 ⟩ (2) A × B=−B × A (3) A × ( B× C )≠ ( A × B ) × C (4) (αA ) × B= A × (αB )=α ( A × B ) (5) A × ( B+C )= A × B+ A ×C (6) A ∙( A × B ) =Β ∙ ( A × B )= 0 (7)If A and B are parallel, then

A × B=0 Page 11

be any scalar. Then:

Applied Mathematics I (Math 231B)

(8) ‖A × Β‖ 2=‖A‖2 ‖B ‖2− ( A ∙ Β ) 2 (9) ‖A × Β‖ =‖ A‖‖ Β‖sin θ , θ ∈[ 0 , π] . (10)

‖A‖‖B‖sin θ , where n^

A × B= n^

of

is the unit vector in the direction

A × B∧θ ∈ [ 0 , π ] is the angle between A and B.

‖A‖=2 ,‖Β‖=4∧θ= π

Example: If

4

for two vectors A and B then find ‖A × Β‖

. Note: The angle

θ

between A and B can be obtained by

‖ A × Β‖ , ‖ A‖‖Β‖

sin θ=

for two non – zero vectors A and B. Definition (Scalar Triple Product) Let A, B and C is vectors in R3 . Their scalar triple product is given by A ∙ ( B ×C ) ,

which is a scalar.

Applications of cross Product: (i)Area: The area of a parallelogram whose adjacent sides coincides with

‖A × Β‖=‖ A‖‖ Β‖|sin θ|

the vectors A and B is given by d

b

B h O A

So,

a

a ( oabc ) =Base x height ¿‖A ‖‖Β ‖|sin θ|

N.B: The area of the triangle formed by A and B as its adjacent sides is given by 1 area= ‖A × Β‖ . 2 ii.Volume The volume V of a parallelepiped with the three vectors A, B and C in three of its adjacent edges is given by: Page 12

R3

as

Applied Mathematics I (Math 231B)

a 1 a 2 a3 V =|A ∙ ( B× C )|= det b1 b 2 b3 c 1 c2 c 3

| ( )|

BXC

A

B C

‖‖

‖

|A ∙ ( B ×C )| A ∙ (B ×C ) A ( ) B ×C h= proj = = ‖B× C‖ B×C ‖B ×C‖2

‖

Hence, V = B h=|A ∙ (B × C )| a Examples: 1. Find the area of a triangle whose vertices are A ( 1,−1,0) , B (2,1 ,−1 ) ¿ C (−1,1,2 ) . Solution:

The

vectors on

the sides

of the

triangle

∆ A ΒC

are

 AB=B− A= ( 1,2 ,−1 )∧ AC=(−2,2,2 ) =C−A . So,

a(∆ A Β C)=

1  ‖AB × AC ‖=3 √2 square units . 2

2. Find the volume of the parallelepiped with edges u=i+k ,v =2 i+ j+4 k ∧  w= j+k . 3 Solution: V= |u ∙ (v ×  w )|=1 u

N.B: Three vectors

A , B ∧C are coplanar iff A ∙ ( B ×C ) =0 .

1.5. Lines and Planes in Definition: A vector parallel to

 Ρ 0 Ρ1

R3

v =⟨ a ,b , c ⟩

is said to be parallel to a line

for any two distinct points

Page 13

Ρ0∧Ρ1 on l .

lif v

is

Applied Mathematics I (Math 231B)

A line

l∈R

3

is determined by a given point

parallel vector

v (directional vector)

Ρ0( x0 , y0 , z 0 )

on

and a

l

v =⟨ a ,b , c ⟩¿ l .

Equations of a line in space: Let

Ρ0 ( x0 , y 0 , z 0 )

be a given point on a line

arbitrary point on

l . If

 V = ⟨ a , b , c⟩

y = y 0 +bt

,

P( x , y , z )

is the parallel vector to

(1) The parametric equation of x = x 0 +at

and

l

l , then

l is given by z=z 0 +ct ,t ∈ R ,

,

be any

where t is called the

parameter. (2) The symmetric form of equation of

l

is given by:

x − x 0 y− y 0 z − z 0 = = , for a , b , c ≠ 0. b a c (3) The vector equation of

l

is written as

r −r 0=t v , wheret ∈ R .

Remarks: 1. The above equations ...