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Unit 6.1
- 2. 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
- 4. 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.
Copyright © 2011 Pearson, Inc. Slide 6.1 - 4
- 6. 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 .
Copyright © 2011 Pearson, Inc. Slide 6.1 - 6
- 7. 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 .
Copyright © 2011 Pearson, Inc. Slide 6.1 - 7
- 8. Example Finding Magnitude of a
Vector
Find the magnitude of v represented by PQ,
where P (3,4) and Q (5,2).
Copyright © 2011 Pearson, Inc. Slide 6.1 - 8
- 9. 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 32
2 (4)2
2 10
Copyright © 2011 Pearson, Inc. Slide 6.1 - 9
- 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 .
Copyright © 2011 Pearson, Inc. Slide 6.1 - 10
- 11. Example Performing Vector
Operations
Let u 2, 1 and v 5,3 . Find 3u v.
Copyright © 2011 Pearson, Inc. Slide 6.1 - 11
- 12. Example Performing Vector
Operations
Let u 2, 1 and v 5,3 . Find 3u v.
3u 32, 31 = 6, 3
3u v = 6, 3 5,3 6 5,3 3 11,0
Copyright © 2011 Pearson, Inc. Slide 6.1 - 12
- 13. 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.
Copyright © 2011 Pearson, Inc. Slide 6.1 - 13
- 14. Example Finding a Unit Vector
Find a unit vector in the direction of v 2, 3 .
Copyright © 2011 Pearson, Inc. Slide 6.1 - 14
- 15. Example Finding a Unit Vector
Find a unit vector in the direction of v 2, 3 .
v 2, 3 22 32
13, so
v
v
1
13
2, 3
2
13
,
3
13
Copyright © 2011 Pearson, Inc. Slide 6.1 - 15
- 16. 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
Copyright © 2011 Pearson, Inc. Slide 6.1 - 16
- 17. 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 .
Copyright © 2011 Pearson, Inc. Slide 6.1 - 17
- 18. Example Finding the Components of
a Vector
Find the components of the vector v with
direction angle 120o and magnitude 8.
Copyright © 2011 Pearson, Inc. Slide 6.1 - 18
- 19. 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.
Copyright © 2011 Pearson, Inc. Slide 6.1 - 19
- 20. 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.
Copyright © 2011 Pearson, Inc. Slide 6.1 - 20
- 21. 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.
Let be the direction angle of u, then
| u | 22
32 13
2 u cos
2 13cos
cos
2
13
cos1 2
13
90¼ 180¼
123.69¼
Copyright © 2011 Pearson, Inc. Slide 6.1 - 21
- 22. 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.
Let be the direction angle of v, then
| v | 42
12
17
4 v cos
4 17 cos
cos
4
17
360¼ cos1 4
17
180¼ 2700¼
194.04¼
Copyright © 2011 Pearson, Inc. Slide 6.1 - 22
- 23. 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.
Interpret
The direction angle for u
is 123.69¼.
The direction angle for v
is 194.04¼.
Copyright © 2011 Pearson, Inc. Slide 6.1 - 23
- 24. 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.
Copyright © 2011 Pearson, Inc. Slide 6.1 - 24
- 25. Quick Review
1. Find the values of x and y.
2. Solve for in degrees. sin-1 3
11
Copyright © 2011 Pearson, Inc. Slide 6.1 - 25
- 26. 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)
Copyright © 2011 Pearson, Inc. Slide 6.1 - 26
- 27. 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
Copyright © 2011 Pearson, Inc. Slide 6.1 - 27
- 28. 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
Copyright © 2011 Pearson, Inc. Slide 6.1 - 28