ENGR 1990 Engineering Mathematics
Application of Two-Dimensional Vectors – ME 2560, ME 2570, ME 2580
Scalars and Vectors
A scalar is a quantity represented by a positive or negative number. It contains a
magnitude (its absolute value) and a sign. They are sometimes called one-dimensional
vectors, because the sign refers to the direction along a single axis. Examples include
length, area, volume, mass, pressure, and temperature.
A two-dimensional vector is represented by a magnitude
Y
and a direction related to two reference axes. Usually, the
reference axes (X and Y) are perpendicular to each other. In
application, vectors can be categorized as fixed or free. A
j
V
fixed vector is defined to be anchored at a specific point,
whereas free vectors can be located anywhere without

changing their meaning. Whether a vector is thought to be
X
i
fixed or free depends on the quantity the vector represents.
For example, if vector V in the diagram represents a force acting on some object, it is
considered to be a fixed vector, because its point of application is important. In contrast,
consider vectors i and j shown in the diagram. These vectors have unit magnitude and
point along the X and Y axes, respectively. They are called unit vectors and are used to
define directions of interest. Since their point of origin is not important, they are free
vectors. The mathematical representation of a vector does not indicate whether it is fixed
or free, so we must be mindful of this as we use them.
Given: The magnitude  V
 and direction ( ) of vector V .
Find: Its X and Y components.
Solution:
The diagram shows the X and Y components of V labeled
as Vx and Vy . The components are also 2D vectors. Their
Y
j
V
Vy
magnitudes are given by right-triangle trigonometry and
their directions are along i and j , respectively. Vector V
 V
x
is the sum of these two components.
Vx  V cos( ) i
V y  V sin( ) j
V  V cos( ) i  V sin( ) j
i
X
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Example 1:
Given: A force F has a magnitude F  100 (lbs) and an angle   30 (deg) .
Find: Express the force F in terms of the unit vectors i and j .
F
Solution:
30
F  100cos(30) i  100sin(30) j  86.6 i  50 j (lbs)
50 j
o
86.6 i
Example 2:
Given: A force F has a magnitude F  100 (lbs) and an angle   120 (deg) .
Find: Express the force F in terms of the unit vectors i and j .
Solution:
86.6 j
F
120o
F  100cos(120) i  100sin(120) j  50 i  86.6 j (lbs)
50 i
Example 3:
?
Given: A force F  70 i  50 j (lbs) .
50 j
?
Find: The magnitude and direction of F .
70 i
Solution:
 35.54 (deg)
F  702  502  86.0 (lbs) and   tan 1 (50 / 70)  
0.6202 (rad)
Example 4:
Given: A force F  50 i  70 j (lbs) .
70 j
Find: The magnitude and direction of F .
Solution:
F  (50)2  702  86.0 (lbs)
?
?
50 i
54.46  180  125.5 (deg)
  tan 1 (70 / 50)  
 0.9505    2.191 (rad)
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Vector Addition
To add two or more vectors, simply express them in terms of the same unit vectors,
and then add like components.
j
F2
Example 5:
Given:
F1  150 (lbs) , 1  20 (deg) ,
F1
2
F2  100 (lbs) ,
1
 2  60 (deg) .
i
Find: a) The total force F acting on the support in terms of the unit vectors shown, and
b) the magnitude and direction of F .
Solution:
a) The total force is the vector sum of the two forces.
F1  150cos(20) i  150sin(20) j  140.95 i  51.3 j (lbs)
F2  100cos(60) i  100sin(60) j  50 i  86.6 j (lbs)
F  F1  F2  (140.95  50) i  (51.3  86.6) j  90.95 i  137.9 j (lbs)
j
b) F  90.952  137.92  165.2 (lbs)
F
 56.59 (deg)
  tan 137.9 / 90.95   
0.9877 (rad)
1
F2

F1
i
As you can see from the figure, F1 and F2 form the sides of a parallelogram, and the
sum F forms the diagonal. The observation that vectors can be added geometrically in
this way is called the parallelogram law of addition. In general, the triangle formed by
F1 , F2 , and F is a non-right triangle. The lengths and angles within this triangle can be
studied using the laws of cosines and sines as discussed in previous notes.
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Example 6:
Given: The lengths and angles of a two link planar
Y
robot are r1  3 (ft) , r2  2 (ft) , 1  20 (deg) , and
2  70 (deg) .
j
Find: a) The position vector R that defines the
position of the end-point of the robot relative to O,
and b) the magnitude and direction of R .
B
R
r2
r1
2
A
1
O
X
i
Solution:
a) The position vector R may be calculated by adding the vectors r1 and r2 .
R  r1  r2   3cos(20) i  3sin(20) j    2cos(70) i  2sin(70) j 
  2.8191 i  1.0261 j    0.684 i  1.8794 j 
 3.503 i  2.906 j (ft)
 39.7 (deg)
b) R  3.5032  2.9062  4.55 (ft) and   tan 1  2.906 / 3.503  
0.6925 (rad)
Scalar (Dot) Product
Geometric Definition
The scalar (or dot) product of two vectors is defines as follows.
A  B  A B cos( A, B )
Here, cos( A, B ) represents the cosine of the angle between the tails of the two vectors. If
one of the vectors is a unit vector, then the scalar product is the projection of the vector in
the direction of the unit vector.
A  n  A n cos( )  A cos( )
The components of A that are parallel and perpendicular
to n are A   A  n  n
and
A  A  A .
A
A

n
A
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Calculation
Given two vectors A and B expressed in terms of a pair of mutually perpendicular
unit vectors i and j , we calculate the dot product as
A  B   ax i  a y j    bx i  by j   axbx  a yby
The dot product of two vectors is zero if they are perpendicular to each other.
Example 7:
Given: Two vectors, A  10 i  2 j and B  3 i  7 j
Find: The angle between the two vectors,  .
Solution:
We can calculate the angle using the inverse cosine function.
 A B 
1  (10  3)  (2  7)
  cos 
2
2
2
2
 10  2 3  7
 A B
  cos 1 

44   55.49 (deg)
1 
  cos 

77.666

 0.9685 (rad)

Example 8:
Given: A vector, A  2 i  8 j , and a unit vector n  53 i  54 j .
Find: a) the angle between the two vectors (θ ), b the component of A parallel to n , and
c) the component of A perpendicular to n .
Solution:
a) We can calculate the angle using the inverse cosine function as before.
3
4
 An 
7.6   22.83 (deg)
1  (2  5 )  (8  5 ) 
1 
cos
cos


  cos 





2
2
8.2462
A

 0.3985 (rad)
2
8





1
b) The component of A parallel to n : A  ( A  n)n  7.6  53 i  54 j   4.56 i  6.08 j
c) The component of A perpendicular to n :
A  A  A   2 i  8 j    4.56 i  6.08 j   2.56 i  1.92 j
Check: A  A   2.56 i  1.92 j    4.56 i  6.08 j    2.56  4.56   1.92  6.08   0 
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Vector (Cross) Product
Geometric Definition
The cross product of two vectors is defines as follows.
A  B   A B sin( A, B )  n
A B
Here, sin( A, B ) is the cosine of the angle between the tails of
n
the two vectors, and n is a unit vector perpendicular to the
plane formed by A and B . The sense of n is defined by the
right-hand-rule, that is, the right thumb points in the direction
of n when the fingers of the right hand point from A to B .

A
Calculation
Given two vectors A and B expressed in terms of
mutually perpendicular unit vectors i and j , we calculate the
cross product as
B
k i j
j
A  B   ax i  a y j    bx i  by j    axby  a ybx  k
i
The cross product of two vectors is zero if they are parallel to each other.
This calculation may be calculated from the determinant form. Since the cross product
of two dimensional vectors has no i or j components, we write
i
j
A  B  ax
bx
ay
by
k
0   axby  a ybx  k
0
Moment of a Force – Torque
The moment (or torque) of a force about
a point O is defined as the magnitude of the
force  F  multiplied by the perpendicular
distance from the point to the line of action
of the force  d  r sin( )  . The direction is
defined by the right-hand-rule. So, it can be
calculated using the cross product.
F
MO
r
A
O
θ
d  r sin( )
Line of
Action
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MO  r  F
Here, r is a position vector from O to the line of action of F .
Example 9:
A force F  50 i  100 j (lbs) is applied at a point A whose coordinates are
Given:
 3,2  (ft) .
Find: a) M O the moment of the force about the origin O at (0,0) , and b) d the
perpendicular distance from O to the line of action of the force.
Solution:
a) M O  r  F   3 i  2 j    50 i  100 j    (3  100)  (2  50)  k  200 k (ft-lbs)
b) d 
| MO |
200
200


 1.79 (ft)
2
2
|F |
111.80
50  100
Example 10:
Given:
A force F  50 i  100 j (lbs) is applied at a point A whose coordinates are
 3, 2  (ft) .
Find: a) M B the moment of the force about point B whose coordinates are (5,10) (ft) , and
b) d the perpendicular distance from B to the line of action of the force.
Solution:
To calculate the moment using the cross product, we must first calculate the position
vector that defines the position of A relative to B.
a) r  r A/ B  r A  rB   3 i  2 j    5 i  10 j   2 i  12 j (ft)
M B  r  F   2 i  12 j    50 i  100 j    (2  100)  (12  50)  k  400 k (ft-lb)
b) d 
| MB |
400
400


 3.58 (ft)
|F |
502  1002 111.80
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