The dot product of two vectors A and B, written as A·B, is a scalar equal to the magnitude of A times the magnitude of B times the cosine of the angle between them. The dot product is commutative and distributes over vector addition. It can be used to find the component of a vector that is parallel to a given line and the angle between two vectors. Applications include finding the angle between two lines and resolving a force into components parallel and perpendicular to a structural member.

1. 2.9 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 tails
A·B = AB cosθ where 0°≤ θ ≤180°
Referred to as scalar
product of vectors as
result is a scalar

2. 2.9 Dot Product
Laws of Operation
1. Commutative law
A·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)

3. 2.9 Dot Product
Cartesian Vector Formulation
- Dot product of Cartesian unit vectors
Eg: 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

4. 2.9 Dot Product
Cartesian Vector Formulation
- Dot product of 2 vectors A and B
A·B = (Axi + Ayj + Azk)· (Bxi + Byj + Bzk)
= 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

5. 2.9 Dot Product
Applications
- The angle formed between two vectors or
intersecting lines
θ = cos-1 [(A·B)/(AB)] 0°≤ θ ≤180°
Note: if A·B = 0, cos-10= 90°, A is
perpendicular to B

6. 2.9 Dot Product
Applications
- The components of a vector parallel and
perpendicular to a line
- Component of A parallel or collinear with line aa’ is
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

7. 2.9 Dot Product
Applications
- If A║ is positive, A║ has a directional
sense same as u
- 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

8. 2.9 Dot Product
Applications
For component of A perpendicular to line aa’
1. Since A = A║ + A┴,
then A┴ = A - A║
2. θ = cos-1 [(A·u)/(A)]
then A┴ = Asinθ
3. If A║ is known, by Pythagorean Theorem
A⊥ = A2 + A||2

9. 2.9 Dot Product
For angle θ between the
rope and the beam A,
- Unit vectors along the
beams, uA = rA/rA
- Unit vectors along the
ropes, ur=rr/rr
- Angle θ = cos-1
(rA.rr/rArr)
= cos-1 (uA· ur)

10. 2.9 Dot Product
For projection of the force
along the beam A
- Define direction of the beam
uA = rA/rA
- Force as a Cartesian vector
F = F(rr/rr) = Fur
- Dot product
F║ = F║·uA

11. 2.9 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
member AB.

12. 2.9 Dot Product
Solution
Since r r r
r
r r 2i + 6 j + 3k
u B = rB =
rB (2)2 + (6)2 + (3)2
r r r
= 0.286i + 0.857 j + 0.429k
Then
r r
FAB = F cosθ
rr r r r r
= F .u B = (300 j ) ⋅ (0.286i + 0.857 j + 0.429k )
= (0)(0.286) + (300)(0.857) + (0)(0.429)
= 257.1N

13. 2.9 Dot Product
Solution
Since result is a positive scalar,
FAB has the same sense of
direction as uB. Express in
Cartesian form
r r r
FAB = FAB u AB
r r r
= (257.1N )(0.286i + 0.857 j + 0.429k )
r r r
= {73.5i + 220 j + 110k }N
Perpendicular component
r r r r r r r r r r
F⊥ = F − FAB = 300 j − (73.5i + 220 j + 110k ) = {−73.5i + 80 j − 110k }N

14. 2.9 Dot Product
Solution
Magnitude can be determined
From F┴ or from Pythagorean
Theorem
r r2 r 2
F⊥ = F − FAB
= (300 N )2 − (257.1N )2
= 155 N

15. 2.9 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
parallel and perpendicular to BA.

16. 2.9 Dot Product
Solution
For angle θ
rBA = {-2i - 2j + 1k}m
rBC = {- 3j + 1k}m
Thus,
r r
rBA ⋅ rBC (− 2 )(0 ) + (− 2 )(− 3) + (1)(1)
cos θ = r r =
rBA rBC 3 10
= 0.7379
θ = 42.5o

17. 2.9 Dot Product
Solution
Components ofr F
r r r
r r (−2i − 2 j + 1k )
u AB = rAB =
rAB 3
r
 − 2 i +  − 2  r +  1  k
r
=   j  
 3   3   3
r rr
FAB = F .u B
r r  2 r  2  r  1  r
= (− 758.9 j + 253.0k ) ⋅  − i +  −  j +  k
 3   3   3
= 0 + 506.0 + 84.3
= 590 N

18. 2.9 Dot Product
Solution
Checking from trigonometry,
r r
FAB = F cos θ
= 800 cos 42.5o N
= 540 N
Magnitude can be determined
From F┴
r r
F⊥ = F sin θ = 800 sin 42.5o = 540 N

19. 2.9 Dot Product
Solution
Magnitude can be determined from F┴ or from
Pythagorean Theorem
r r2 r 2
F⊥ = F − FAB
= (800)2 − (590)2
= 540 N