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
−+−=++−=−=
++=
++=
=
⊥
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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