Review: Analysis vector. VECTOR ANALYSIS 1.1SCALARS AND VECTORS 1.2VECTOR COMPONENTS AND UNIT VECTOR...

Review: Analysis vector

VECTOR ANALYSIS

1.1 SCALARS AND VECTORS

1.2 VECTOR COMPONENTS AND UNIT VECTOR

1.3 VECTOR ALGEBRA

1.4 POSITION AND DISTANCE VECTOR

1.5 SCALAR AND VECTOR PRODUCT OF

VECTORS

• A scalar quantity – has only magnitude• A vector quantity – has both magnitude

and direction

1.1 SCALARS & VECTORS

electric field intensity

• A vector in Cartesian Coordinates maybe represented as

zyx RRR ,,R

1.2 VECTOR COMPONENTS & UNIT VECTOR

R

Or

zzyyxx RRR aaaR

R

yRThe vector has three component vectors, which are , and

zRxR

VECTOR COMPONENTS & UNIT VECTOR (Cont’d)

• Each component vectors have magnitude

which depend on the given vector and they

have a known and constant direction.

• A unit vector along is defined as a

vector whose magnitude is unity and

directed along the coordinate axes in the

direction of the increasing coordinate

values

R

VECTOR COMPONENTS & UNIT VECTOR (Cont’d)

Any vector maybe described asR

zzyyxx RRR aaaR

The magnitude of written or simply given by

R R

R

222zyx RRR R

VECTOR COMPONENTS & UNIT VECTOR (Cont’d)

Unit vector in the direction of the vector is:

R

R

RRa

222zyx

RRRR

VECTOR COMPONENTS & UNIT VECTOR (Cont’d)

EXAMPLE 1

Specify the unit vector extending from

the origin toward the point

1,2,2 G

SOLUTION TO EXAMPLE 1

Construct the vector extending from origin to point G

Find the magnitude of

zyx aaaG

22G

3122 222 G

So, unit vector is:

zyx

zyxG

aaa

aaaG

Ga

333.0667.0667.0

3

1

3

2

3

2

SOLUTION TO EXAMPLE 1 (Cont’d)

1.3 VECTOR ALGEBRA

• Two vectors, and can be added together to give another vector

A

B

C

BAC

• Let zyx AAA ,,A zyx BBB ,,

B

zzzyyyxxx BABABA aaaC

VECTOR ALGEBRA (Cont’d)

Vectors in 2 components

• Vector subtraction is similarly carried out as:

)B(ABAD

zzzyyyxxx BABABA aaaD

VECTOR ALGEBRA (Cont’d)

VECTOR ALGEBRA (Cont’d)

• Laws of Vectors:

Associative Law

Commutative Law

Distributive Law

Multiplication by Scalar

CB)(AC)(BA

ABBA

BAB)(A aaa

BABAB)(AB)(AB)(A ssrrsrsr )(

EXAMPLE 2

If

Find: (a) The component of along

(b) The magnitude of

(c) A unit vector along

zyx aaaA 6410

yx aaB

2

A ya

BA3

BA 2

(a) The component of along is

A ya

4yA

(b)

zyx aaa

BA

181328

18,13,28

0,1,218,12,30

0,1,26,4,1033

SOLUTION TO EXAMPLE 2

Hence, the magnitude of is:

BA3

74.351813283 222 BA

(c) Let

zyx aaa

B2AC

6214

6,2,14

0,2,46,4,10

SOLUTION TO EXAMPLE 2 (Cont’d)

zyx

zyx

C

aaa

aaa

C

Ca

391.0130.0911.036.15

6

36.15

2

36.15

14

6214

6,2,14222

So, the unit vector along is:C

SOLUTION TO EXAMPLE 2 (Cont’d)

• A point P in Cartesian coordinate maybe represented as

• The position vector (radius vector) of point P is as the directed distance from the origin O to point P is

zyxP zyxOP aaar

zyx ,,P

Pr

1.4 POSITION AND DISTANCE VECTOR

zyxP aaar 543

POSITION AND DISTANCE VECTOR (Cont’d)

POSITION AND DISTANCE VECTOR (Cont’d)

• If we have two

position vectors,

and , the third

vector or “distance

vector” can be defined

as:

Pr

Qr

PQPQ rrr

Point P and Q are located at

and . Calculate:

4,2,0 5,1,3

(a) The position vector P

(b) The distance vector from P to Q

(c) The distance between P and Q

(d) A vector parallel to with magnitude

of 10

PQ

EXAMPLE 3

(a)

(b)

(c)

zyzyxP aaaaar 42420

zyx

PQPQ

aaa

rrr

3

4,2,05,1,3

Since is the distance vector, the distance between P and Q is the magnitude of this distance vector.

PQr

SOLUTION TO EXAMPLE 3

SOLUTION TO EXAMPLE 3 (Cont’d)

Distance, d

317.3113 222 PQd r

(d) Let the required vector be then

Where is the magnitude of

AAaA

A

10AA

Since is parallel to , it must

have same unit vector as or

A

PQ

SOLUTION TO EXAMPLE 3 (Cont’d)

PQr

QPr

317.3

1,1,3

PQ

PQAa

r

r

So, 317.3

1,1,310

A

cos ( , )

cosAB

ABAB

A B A B A B

cos ABB

AB

A

B

cos ABA

ABEnclosed Angle

cos

cosBA

AB

BA

AB

A B B A

cos cosAB AB

cos

arccos

AB

AB

A B

A B

A B

A B

SCALAR PRODUCT OF VECTORS

1.5 SCALAR AND VECTOR PRODUCT OF VECTORS

sin ( , )

sin

AB

AB

AB

C

AB

S

C A×B

A B A B

AB

A

B

C

ABS

and /

und C A C B

Surface

and

VECTOR PRODUCT OF VECTORS

Add the first two Columns

( )

+ ( )

( )

x y z

x y z

x y z

x y z x y

x y z x y

x y z x y

y z z y x

z x x z y

x y y x z

A A A

B B B

A A A A A

B B B B B

A B A B

A B A B

A B A B

e e e

A×B

e e e e e

e

e

e

Sarrus Law[Pierre Frédéric Sarrus, 1831]

http://de.wikipedia.org/wiki/Regel_von_Sarrus

VECTOR PRODUCT OF VECTORS (Cont’d)

Properties of cross product of unit vectors:

yxzxzyzyx aaaaaaaaa ,,

Or by using cyclic permutation:

VECTOR PRODUCT OF VECTORS (Cont’d)

Determine the dot product and cross product of the following vectors:

zyx

zyx

aaaB

aaaA

65

432

EXAMPLE 4

The dot product is:

41

)6)(4()5)(3()1)(2(

BA zzyyxx BABABA

SOLUTION TO EXAMPLE 4

zyx

z

y

x

zyx

zyx

zyx

zyx

BBB

AAA

aaa

a

a

a

aaaaaa

BA

782

)1)(3()5)(2(

)1)(4()6)(2(

)5)(4()6)(3(

651

432

The cross product is:

SOLUTION TO EXAMPLE 4 (Cont’d)