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Chapter One: Vector Analysis The use of vectors and vector analysis ...

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Electromagnetic <strong>The</strong>orem<br />

(Dr. Omed Ghareb Abdullah) University <strong>of</strong> Sulaimani –College <strong>of</strong> Science – Physics Department<br />

OP (1 0) aˆ<br />

x<br />

( 3<br />

0) aˆ<br />

y<br />

(5 0) aˆ<br />

z<br />

aˆ<br />

(a).<br />

OR (0 0) aˆ<br />

(3 0) aˆ<br />

(8 0) aˆ<br />

3aˆ<br />

x<br />

y<br />

z<br />

x<br />

y<br />

3aˆ<br />

8aˆ<br />

z<br />

y<br />

5aˆ<br />

z<br />

(b). r <br />

QP<br />

( 1<br />

2) aˆ<br />

( 3<br />

4) aˆ<br />

(5 6) aˆ<br />

aˆ<br />

7 aˆ<br />

aˆ<br />

x<br />

y<br />

z<br />

x<br />

y<br />

z<br />

(c).<br />

2<br />

2<br />

2<br />

QR (0 2) (3 4) (8 6) 4 1<br />

4 3unit<br />

Example(12):<br />

Derive the cosine formula<br />

c<br />

2<br />

2 2<br />

a b 2abcos<br />

<strong>and</strong> the sine formula<br />

sin A sin B sin C<br />

using dot <strong>and</strong> cross product respectively.<br />

a b c<br />

b<br />

C<br />

Cosine formula:<br />

<br />

c a<br />

b<br />

<br />

c c (<br />

a b)<br />

(<br />

a b)<br />

a<br />

c<br />

2<br />

a<br />

2<br />

b<br />

2<br />

2<br />

b<br />

2<br />

A<br />

c<br />

2a<br />

b<br />

2 2<br />

2ab<br />

cos c a b 2abcos<br />

a<br />

B<br />

Sine formula:<br />

1 1 <br />

a b b c<br />

2 2<br />

<br />

1<br />

2<br />

a bsin<br />

c <br />

1<br />

2<br />

1 <br />

c a<br />

2<br />

b c sin a <br />

1<br />

2<br />

c a<br />

sin b<br />

dividing<br />

the above equation<br />

by<br />

( a b c ) we get :<br />

sin c<br />

c<br />

<br />

sin a<br />

a<br />

<br />

sin b<br />

b<br />

Example(13):<br />

<br />

<br />

Let E 3aˆ<br />

4aˆ<br />

<strong>and</strong> F 4aˆ<br />

10aˆ<br />

5aˆ<br />

, then find:<br />

x<br />

z<br />

x<br />

y<br />

z<br />

(a). the component <strong>of</strong><br />

E along<br />

F<br />

(b). A unit <strong>vector</strong> perpendicular to both<br />

E <strong>and</strong><br />

F<br />

Solution:<br />

(a). the <strong>vector</strong> component <strong>of</strong><br />

E along F is given by:<br />

63

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