Unit 6.1

6.1
Vectors in the
Plane
Copyright © 2011 Pearson, Inc.

What you’ll learn about
 Two-Dimensional Vectors
 Vector Operations
 Unit Vectors
 Direction Angles
 Applications of Vectors
… and why
These topics are important in many real-world
applications, such as calculating the effect of the wind on
an airplane’s path.
Copyright © 2011 Pearson, Inc. Slide 6.1 - 2

Directed Line Segment

Two-Dimensional Vector
A two - dimensional vector v is an ordered pair of real
numbers, denoted in component form as a,b . The
numbers a and b are the components of the vector v.
The standard representation of the vector a,b is the
arrow from the origin to the point (a,b). The magnitude
of v is the length of the arrow and the direction of v is the
direction in which the arrow is pointing. The vector
0 = 0,0 , called the zero vector, has zero length and
no direction.

Initial Point, Terminal Point,
Equivalent

Head Minus Tail (HMT) Rule
If an arrow has initial point x1, y1   and terminal point
x2 , y2  , it represents the vector x2  x1, y2  y1 .

Magnitude
If v is represented by the arrow from x1, y1   to x2 , y2  ,
then
v  x2  x1  2
 y2  y1  2
.
If v  a,b , then v  a2  b2 .

Example Finding Magnitude of a
Vector
Find the magnitude of v represented by PQ,
where P  (3,4) and Q  (5,2).

Example Finding Magnitude of a
Vector
Find the magnitude of v represented by PQ,
where P  (3,4) and Q  (5,2).
v  x2  x1  2
 y2  y1  2
 5 32
 2  (4)2
 2 10

Vector Addition and Scalar Multiplication
Let u  u1,u2 and v  v1,v2 be vectors and let k be
a real number (scalar). The sum (or resultant) of the
vectors u and v is the vector
u  v  u1  v1,u2  v2 .
The product of the scalar k and the vector u is
ku  k u1,u2  ku1,ku2 .

Example Performing Vector
Operations
Let u  2, 1 and v  5,3 . Find 3u  v.

Example Performing Vector
Operations
Let u  2, 1 and v  5,3 . Find 3u  v.
3u  32, 31 = 6,  3
3u  v = 6,  3  5,3  6  5,3 3  11,0

Unit Vectors
A vector u with | u |  1 is a unit vector. If v is not
the zero vector 0,0 , then the vector u 
v
| v |

1
| v |
v
is a unit vector in the direction of v.

Example Finding a Unit Vector
Find a unit vector in the direction of v  2,  3 .

Example Finding a Unit Vector
Find a unit vector in the direction of v  2,  3 .
v  2,  3  22  32
 13, so
v
v

1
13
2,  3 
2
13
, 
3
13

Standard Unit Vectors
The two vectors i  1,0 and j  0,1 are the standard
unit vectors. Any vector v can be written as an expression
in terms of the standard unit vector:
v  a,b
 a,0  0,b
 a 1,0  b 0,1
 ai  bj

Resolving the Vector
If v has direction angle  , the components of v can
be computed using the formula
v = v cos , v sin .
From the formula above, it follows that the unit vector
in the direction of v is u 
v
v
 cos ,sin .

Example Finding the Components of
a Vector
Find the components of the vector v with
direction angle 120o and magnitude 8.

Example Finding the Components of
a Vector
Find the components of the vector v with
direction angle 120o and magnitude 8.
v  a,b  8cos120o,8sin120o

 8 
1
2
 

 
,8
3
2

 

 
 4,4 3
So a  4 and b  4 3.

Example Finding the Direction Angle
of a Vector
Let u  2,3 and v  4, 1 .
Find the magnitude and direction angle of each vector.

of a Vector
Let u  2,3 and v  4, 1 .
Let  be the direction angle of u, then
| u | 22
 32  13
2  u cos
2  13cos
cos 
2
13

  cos1 2
13
 

 
90¼   180¼
  123.69¼

of a Vector
Let u  2,3 and v  4, 1 .
Let  be the direction angle of v, then
| v | 42
 12
 17
4  v cos
4  17 cos
cos 
4
17

  360¼ cos1 4
17
 

 
180¼   2700¼
  194.04¼

of a Vector
Let u  2,3 and v  4, 1 .
Interpret
The direction angle for u
is   123.69¼.
The direction angle for v
is   194.04¼.

Velocity and Speed
The velocity of a moving object is a vector
because velocity has both magnitude and
direction. The magnitude of velocity is speed.

Quick Review
1. Find the values of x and y.

2. Solve for  in degrees.   sin-1 3
11
 

 

Quick Review
3. A naval ship leaves Port Northfolk and averages
43 knots (nautical mph) traveling for 3 hr on a bearing
of 35o and then 4 hr on a course of 120o.
What is the boat's bearing and distance from
Port Norfolk after 7 hr.
The point P is on the terminal side of the angle  .
Find the measure of  if 0o    360o.
4. P(5,7)
5. P(-5,7)

Quick Review Solutions
1. Find the values of x and y.
x  7.5, y  7.5 3

2. Solve for  in degrees.   sin-1 3
11
 

 
64.8¼

Quick Review Solutions
3. A naval ship leaves Port Northfolk and averages
43 knots (nautical mph) traveling for 3 hr on a bearing
of 35o and then 4 hr on a course of 120o.
What is the boat's bearing and distance from
Port Norfolk after 7 hr.
distance = 224.2; bearing = 84.9¼
The point P is on the terminal side of the angle  .
Find the measure of  if 0o    360o.
4. P(5,7) 54.5¼
5. P(5,7) 125.5¼

Unit 6.1

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Unit 6.1