Vector Products and Scalar Products: Fundamental Concepts, Cheat Sheet of Statics

Download Vector Products and Scalar Products: Fundamental Concepts and more Cheat Sheet Statics in PDF only on Docsity!

VECTORS: FUNDAMENTAL CONCEPTS

The vector product of two vectors is defined as

V = P x Q.

The vector product of P and Q forms a vector which is perpendicular

to both P and Q , of magnitude

V = PQ sin θ.

This vector is directed in such a way that a person located at the

tip of V observes as counterclockwise the rotation through θ

which brings vector P in line with vector Q. The three vectors P ,

Q , and V - taken in that order - form a right-hand triad. It

follows that

Q x P = - ( P x Q )

θ

V = P x Q

P

Q

j

i k x

z

y

O

j x k produces a vector i perpendicular

to both vectors.

j x k = i when viewed from the tip of this

vector it appears as if it rotates

counterclockwise into k.

j

i k x

z

y

O

k x i produces a vector j perpendicular

to both vectors.

k x i = j when viewed from the tip of

this vector it appears as if it rotates

counterclockwise into i.

x ;

x ;

x ;

k k 0

j j 0

i i 0

= x ;

x ;

x ;

j k i

k i j

i j k

x ;

x ;

x ;

k j i

i k j

j i k

i

j

k

i

j

k

Cross Product Combinations

P = P x i + P y j + P z k ; Q = Q x i + Q y j + Q z k

where its scalar components are given by:

V x = P y Q z - P z Qy

V y = P z Q x - P x Q z

V z = P x Q y - P y Q x

i j k ,

i j k

V P Q x y z

x y z

x y z V V V

Q Q Q

= × = P P P = + +

The determinant containing each component of P and Q is expanded

to define the vector V , as well as its scalar components

Scalar Product of Two Vectors

It follows from the definition that the scalar product of two

vectors is commutative, i. e., that

P

Q

θ

C = P • Q = Q • P

The scalar product of two vectors P and Q

is defined as: P

.

Q = PQ cosθ, where θ

is the angle formed by both vectors.

Using the Scalar Product to Find the

Angle formed by Two Given Vectors.

Example. Determine the angle θ formed by the vectors OA and BC.

A O

C

B

(0,0,6)^6

(2,3,0)

(2,3,6)

z

x

y

(0,0,0)

Solution: From Figure., we notice that vectors OA and BC are given by: OA = 2 i + 3 j + 6 k ; BC = 2 i + 3 j – 6 k. Hence, by applying the scalar product definition, we have:

49

23

49

4 9 36

2 3 6 2 3 ( 6 )

( 2 3 6 ) ( 2 3 6 )

cos

2 3 2 2 2 2

=− = −

• − =

• • • • −
• • • + − =

=

i j k i j k

OA BC

OA BC

θ

Calculating its arc cosine,

yields θ = 118 o^.

The projection of a vector P on an axis OL can

be obtained by forming the scalar product of P

and the unit vector λOL along OL.

x

y

z

O

L

A

θx

P

λOL

θz

θy

POL = P • λ OL

( ) (cos cos cos ) cos cos cos ,

( ) ( ) or

,

OL x y z x y z x x y y z z

x y z OL x y z

OL OL

P P P P P P P

OL

OL OL

OL

OL

OL P P P P

OL

P

= + + • θ + θ + θ = θ + θ + θ

= + + • + +

= • = •

i j k i j k

i j k i j k

OL P λ P

. 2 2 2 where OL = OLx + OLy + OLz

2. Projection of a vector P on an arbitrary axis.