lec5

Lecture no.5 of 4

Lecture no.5

Vector Products

• Vectors are not ordinary numbers so normal multiplication has no direct

analog. None the less, there are two way to "multiply" two vectors which are

useful in physics.

• The scalar or dot product "multiplication" of two vectors produces a single

number plus units. The form of the dot product is:

• The vector or cross product "multiplication" of two vectors produces a new

single vector plus units. The form of the cross product is:

Dot Product(or Scalar Product):

• The dot product takes one of the vectors and projects its length onto the other

vector. This projected length is then multiplied by length of the other vector to

produce the value of the dot product. You get the same results no matter which

vector you project onto the other.

• In the most trivial case the dot product of two vectors that are in the same

direction is just the product of the magnitudes of the two vectors. On the other

hand, if the vectors are at right angles to each other their dot product is zero.

Vector (or Cross) Product:

• The magnitude of the cross product is equal to the area of a parallelogram

formed using the vectors as the sides of a parallelogram.

Lecture no.5 of 4

• The direction of the cross product is perpendicular to the plane formed by the

two vectors and follows the right hand rule.

∧

=× nBABA ϑsin

where

∧

n is a unit vector perpendicular to the plane formed by the two vectors .

According to the above rules we can easily conclude that,

jkiijkkij

jikikjkjioiikokjoji

kkjjiikkkjjjiii

−=×−=×−=×

=×=×=×===

=×=×=×======

...

0001.1.1. 222

.

That is if kBjBiBBkAjAiAA zyxzyx ++=++= & we can write:

zzyyxx BABABABA ++=•

And

zyx

zyx

BBB

AAA

kji

BA =×

Lecture no.5 of 4

Example 1

Find the angel between BandA if

kjiBkjiA 4123&22 +−=+−=

Answer

BA

BABABA

•=⇒=• ϑϑ coscos

zzyyxx BABABABA ++=• ⇒ =•BA 2 (3)-2 (-12) +1 (4) = 34

134123

&3122

222222

222222

=++=++=

=++=++=

zyx

zyx

BBBB

AAAA

o

BA

BA3.29

133

34coscos =⇒

×=⇒

•= θϑϑ

Example 2

Find the angel θ between EB and EC shown in figure.

θ is the angel between the two vectors ECEB &

Lecture no.5 of 4

i.e ECEB

ECEB•=ϑcos

then we start by expressing the vectors

ECEB & in terms of i ,j and k.

kjiikjEBABOAEOEB 346634 +−=++−=⇒++=

kjiikjECACOAEOEC 34121234 +−=++−=⇒++=

133412

61346

97)3(3)4(4)12(6

222

222

=++=

=++=

=+−−=•

EC

EB

ECEB

o

ECEB

ECEB

ECEB

ECEB

2.17

6113

97cos

=⇒

=•

=•

=⇒

ϑ

ϑ

(Continued)

lec5

Documents