Vectors
Vector is used by scientists to indicate a quantity (such as velocity or force) that has both magnitude and
direction. A vector is often represented by an arrow or a directed line segment. The length of the arrow
represents the magnitude of the vector and the arrow points in the direction of the vector. Figure 1
shows a particle moving along a path in the plane and its velocity vector v at a specific location of the
particle. Here the length of the arrow represents the speed of the particle and it points in the direction
that the particle is moving.
Vector has two characteristics: magnitude (length) and direction. As long as the vectors have same
direction and same length, the vectors are the same. In Figure 2 notice that all of the arrows are
equivalent in the sense that they have the same length and point in the same direction even though
they are in different locations. All vectors in Figure 2 are the same as vector OP which starts at origin
(0, 0) and ends at P (3, 2). We use <3, 2> to represents OP .
Definition: A two-dimensional vector is an ordered pair a  a1 , a2  of real numbers. The numbers
a1 , a2 are called the components of a
A representation of the vector a  a1 , a2  is a directed line
segment AB from any point A( x, y ) to the point
B( x  a1, y  a2 ) . A particular representation of vector a is the
directed line segment OP from the origin to the point P(a1 , a2 ) ,
and  a1 , a2  is called the position vector of the point P(a1 , a2 )
(see Figure 3).
Theorem 1: Given the points A( x1 , y1 ) and B( x2 , y2 ) , the vector a with representation AB is
a  AB  x2  x1, y2  y1 
Example 1: Find the vector represented by the directed line segment with initial point A(2,4) and
terminal point B(3,7) .
Solution: AB  x2  x1 , y2  y1  3  2,7  4  1,3 
The magnitude or length of the vector a  a1 , a2  is denoted by | a | or || a || , by using the distance
formula, we have
| a | a12  a2 2
The length of the vector AB from A( x1 , y1 ) to B( x2 , y2 ) is
| AB | ( x2  x1 )2  ( y2  y1 )2
The only vector with length 0 is the zero vector 0  0,0  . This vector is also the only vector with no
specific direction.
A unit vector is a vector whose length is 1. In general, if a  0 , then the unit vector that has the same
direction as a is
u
1
a
a
|a |
|a |
Example 2: Find the unit vector in the direction of the vector <-4, 3>.
Solution: the length of vector <-4, 3>, | 4,3 | ( 4)2  32  5
The unit vector in the direction of <-4, 3> is
1
4 3
 4,3   ,  .
5
5 5
Vector Addition (Triangle Law and Parallelogram Law)
By the subtraction a  b of two vectors, we mean a  b  a  (b) .
Vector Addition (subtraction). If a  a1 , a2 , b  b1, b2  , then
a  b  a1  b1 , a2  b2 
Multiplication of a Vector by a Scalar. If c is a scalar (a real number), and a  a1 , a2 , then
ca  ca1 , ca2 
Properties of Vectors. If a, b and c are vectors, c and d are scalars , then
ab  ba
a  (b  c)  (a  b)  c
c(a  b)  ca  cb
(c  d ) a  c a  d a
a0  a
a  ( a )  0
(cd) a  c(d a)
Example 3: If a  4, 2 , b  3,5  . Find vectors 2a  3b and 5b  a .
Solution:
2a  3b  2  4,-2  3  3,5  8,-4    9,15  17,11 
5b  a  5  3,5  -  4,-2  15, 25  -  4,-2  11, 27 
There are two special unit vectors: i  1,0 , j  0,1  (See Figure 9). Any vector a  a1 , a2  can
write as a  a1 , a2  a1 i  a2 j
t
Example 4. If a  2i  3 j, b  3i  j , find 2a  b
Solution: 2a  b  2(2i  3 j )  (3i  j )  4i  6 j  3i  j  i  7 j
We conclude this section by considering one of the many applications of vectors in physice and
engineering. A force is represented by a vector because it has both a magnitude (measured in pounds or
newtons, N) and a direction. If several forces are acting on an object, the resultant force experienced by
the object is the vector sum of thes forces.
Example 5. A 100-lb weight hangs from two wires as shown in Figure 11. Find the tensions (forces) T1
and T2 in both wires and their magnitudes.
Solution: The resultant force T1  T2 counterbalances the weight w and so must have
T1  T2   w  100 j
(1)
We can express T1 and T2 in terms of their horizontal and vertical components. From Figure 12, and
the Triangle Law we have
(2)
T1   T1 cos500 i  T1 sin500 j
T2  T2 cos320 i  T2 sin 320 j
Plugging them into equation (1), we have
 T1 cos500 i  T1 sin 500 j+ T2 cos320 i  T2 sin 320 j  100j
(  T1 cos500  T2 cos320 )i + ( T1 sin 500  T2 sin 320 ) j  100 j
Equating the components, we have
 T1 cos500  T2 cos320  0
T1 sin500  T2 sin 320  100
Solving the first of these equations for T2 and substituting into the second, we get
T1 sin 50 
0
T1 cos500
cos32
0
sin 320  100 
T1 
100
 85.64 lb
cos500
0
0
sin 50 
sin 32
cos320
and
T2 
T1 cos500
cos320
 64.91 lb
Thus, substituting these values in (2), we have the forces
T1   T1 cos500 i  T1 sin 500 j  85.64 cos500 i  85.64sin 500 j  55.05 i  65.6 j
T2  T2 cos320 i  T2 sin 320 j  64.91cos320 i  64.91sin 32 0 j  55.05 i  34.4 j