2.9 Dot Product2.9 Dot Product

� Dot product of vectors A and B is written as A·B (Read A dot B)

� Define the magnitudes of A and B and the angle between their tailsangle between their tails

A·B = AB cosθ where 0°≤ θ ≤180°

� Referred to as scalar

product of vectors as

result is a scalar

2.9 Dot Product2.9 Dot Product

� Laws of Operation

1. Commutative law

A·B = B·AA·B = B·A

2. Multiplication by a scalar

a(A·B) = (aA)·B = A·(aB) = (A·B)a

3. Distribution law

A·(B + D) = (A·B) + (A·D)

2.9 Dot Product2.9 Dot Product

� Cartesian Vector Formulation

- Dot product of Cartesian unit vectors

Eg: i·i = (1)(1)cos0° = 1 andEg: i·i = (1)(1)cos0° = 1 and

i·j = (1)(1)cos90° = 0

- Similarly

i·i = 1 j·j = 1 k·k = 1

i·j = 0 i·k = 1 j·k = 1

2.9 Dot Product2.9 Dot Product

� Cartesian Vector Formulation- Dot product of 2 vectors A and B

A·B = (Axi + Ayj + Azk)· (Bxi + Byj + Bzk)

= A B (i·i) + A B (i·j) + A B (i·k)= AxBx(i·i) + AxBy(i·j) + AxBz(i·k)

+ AyBx(j·i) + AyBy(j·j) + AyBz(j·k)

+ AzBx(k·i) + AzBy(k·j) + AzBz(k·k)

= AxBx + AyBy + AzBz

Note: since result is a scalar, be careful of including any unit vectors in the result

2.9 Dot Product2.9 Dot Product

� Applications

- The angle formed between two vectors or intersecting linesintersecting lines

θ = cos-1 [(A·B)/(AB)] 0°≤ θ ≤180°

Note: if A·B = 0, cos-10= 90°, A is

perpendicular to B

2.9 Dot Product2.9 Dot Product

� Applications- The components of a vector parallel and perpendicular to a line

- Component of A parallel or collinear with line aa’ is - Component of A parallel or collinear with line aa’ is defined by A║ (projection of A onto the line)

A║ = A cos θ

- If direction of line is specified by unit vector u (u = 1),

A║ = A cos θ = A·u

2.9 Dot Product2.9 Dot Product

�Applications- If A║ is positive, A║ has a directional sense same as u

- If A is negative, A has a directional - If A║ is negative, A║ has a directional sense opposite to u

- A║ expressed as a vector

A║ = A cos θ u

= (A·u)u

� ApplicationsFor component of A perpendicular to line aa’

1. Since A = A║ + A┴,

then A = A - A

2.9 Dot Product2.9 Dot Product

then A┴ = A - A║2. θ = cos-1 [(A·u)/(A)]

then A┴ = Asinθ

3. If A║ is known, by Pythagorean Theorem

2||

2 AAA +=⊥

2.9 Dot Product2.9 Dot Product

� For angle θ between the rope and the beam A,

- Unit vectors along the beams, uA = rA/rAbeams, uA = rA/rA

- Unit vectors along the ropes, ur=rr/rr

- Angle θ = cos-1

(rA.rr/rArr)

= cos-1 (uA· ur)

2.9 Dot Product2.9 Dot Product

� For projection of the force along the beam A

- Define direction of the beam

u = r /ruA = rA/rA

- Force as a Cartesian vector

F = F(rr/rr) = Fur

- Dot product

F║ = F║·uA

2.9 Dot Product2.9 Dot Product

Example 2.16

The frame is subjected to a horizontal force

F = {300j} N. Determine the components of

this force parallel and perpendicular to the this force parallel and perpendicular to the

member AB.

2.9 Dot Product2.9 Dot Product

Solution

Since

( ) ( ) ( )kji

r

ru

B

BB

362

362222 ++

++==

rrr

r

rr

Then

( ) ( ) ( )

( ) ( )

N

kjijuF

FF

kji

r

B

AB

B

1.257

)429.0)(0()857.0)(300()286.0)(0(

429.0857.0286.0300.

cos

429.0857.0286.0

362 222

=

++=

++⋅==

=

++=

++

rrrrrr

rr

rrr

θ

2.9 Dot Product2.9 Dot Product

Solution

Since result is a positive scalar,

FAB has the same sense of

direction as uB. Express in B

Cartesian form

Perpendicular component

( )( )

NkjikjijFFF

Nkji

kjiN

uFF

AB

ABABAB

}110805.73{)1102205.73(300

}1102205.73{

429.0857.0286.01.257

rrrrrrrrrr

rrr

rrr

rrr

−+−=++−=−=

++=

++=

=

⊥

2.9 Dot Product2.9 Dot Product

Solution

Magnitude can be determined

From F┴ or from Pythagorean TheoremTheorem

( ) ( )N

NN

FFF AB

155

1.257300 22

22

=

−=

−=⊥

rrr

2.9 Dot Product2.9 Dot Product

Example 2.17

The pipe is subjected to F = 800N. Determine the

angle θ between F and pipe segment BA, and the

magnitudes of the components of F, which are magnitudes of the components of F, which are

parallel and perpendicular to BA.

2.9 Dot Product2.9 Dot Product

Solution

For angle θ

rBA = {-2i - 2j + 1k}m

rBC = {- 3j + 1k}mrBC = {- 3j + 1k}m

Thus,

( )( ) ( )( ) ( )( )

o

rr

rr

5.42

7379.0

103

113202cos

=

=

+−−+−=

⋅=

θ

θBCBA

BCBA

rr

rr

2.9 Dot Product2.9 Dot Product

Solution

Components of Fkji

r

ru

AB

ABAB 3

)122( +−−==

rrr

r

rr

( )

N

kjikj

uFF

kji

BAB

AB

590

3.840.5060

31

32

32

0.2539.758

.

31

32

32

=

++=

+

−+

−⋅+−=

=

+

−+

−=

rrrrr

rrr

rrr

2.9 Dot Product2.9 Dot Product

Solution

Checking from trigonometry,

FFAB cos=rr

θ

Magnitude can be determined

From F┴

N

N

540

5.42cos800

=

= o

NFF 5405.42sin800sin ===⊥o

rrθ

2.9 Dot Product2.9 Dot Product

Solution

Magnitude can be determined from F┴ or from Pythagorean Theorem

( ) ( )N

FFF AB

540

590800 22

22

=

−=

−=⊥rrr