copyright 2010 pearson education canada 9-1 chapter 9: vectors and oblique triangles

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Copyright © 2010 Pearson Education Canada 9-1 CHAPTER 9: VECTORS AND OBLIQUE TRIANGLES

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Ch. 9.1: Introduction to Vectors Copyright © 2010 Pearson Education Canada  Definition of a vector: - a directed line segment  Components of a vector:  Direction  Magnitude 9-3

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Page 1: Copyright  2010 Pearson Education Canada 9-1 CHAPTER 9: VECTORS AND OBLIQUE TRIANGLES

Copyright © 2010 Pearson Education Canada 9-1

CHAPTER 9:VECTORS AND OBLIQUE TRIANGLES

Page 2: Copyright  2010 Pearson Education Canada 9-1 CHAPTER 9: VECTORS AND OBLIQUE TRIANGLES

Learning Outcomes

At the end of this chapter the student will: describe the oblique triangle, recognize when to use the Law of Sines

and the Law of Cosines to solve the oblique triangle,

apply the Law of Sines and the Law of Cosines.

Copyright © 2010 Pearson Education Canada 2

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Page 3: Copyright  2010 Pearson Education Canada 9-1 CHAPTER 9: VECTORS AND OBLIQUE TRIANGLES

Ch. 9.1: Introduction to Vectors

Copyright © 2010 Pearson Education Canada

Definition of a vector: - a directed line segment

  Components of a vector:

Direction Magnitude

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Vectors

Copyright © 2010 Pearson Education Canada

Applications of Vectors:

Velocity & acceleration

Force

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Describing vectors

Copyright © 2010 Pearson Education Canada

How are vectors described?

A

1. A, or,

2. in boldface: A

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Describing vectors

Copyright © 2010 Pearson Education Canada

How is the magnitude or the size of a vector described?

|AB| or AB

Note the boldface lettering within the

absolute value bars.

Note the italicized lettering.

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Vector addition

Copyright © 2010 Pearson Education Canada

The tip-to-tail method: A + B = R

x

y

O

Resultant

A

B B

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Vector addition

Copyright © 2010 Pearson Education Canada

The parallelogram method: A + B = R

x

y

O

Resultant

A

B

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Scalar multiple of vector

Copyright © 2010 Pearson Education Canada

If vector C is in the same direction as vector A and C has a magnitude n times that of A, then C = nA.

x

y

O

A

C

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Subtracting vectors

Copyright © 2010 Pearson Education Canada

To subtract vectors, the direction of the vector is reversed.

Therefore, A – B = A + (-B), where the minus sign indicates that vector -B has the opposite direction of vector B.

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Page 11: Copyright  2010 Pearson Education Canada 9-1 CHAPTER 9: VECTORS AND OBLIQUE TRIANGLES

Subtracting vectors

Copyright © 2010 Pearson Education Canada

An illustration: A – B = A + (- B) = R

AA

B-B

R

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Ch. 9.2: Components of Vectors

Copyright © 2010 Pearson Education Canada

Two vectors that, when added together, have a resultant equal to the original vector, are called components.

Any vector can be replaced by its x- & y- components.

Finding these component vectors is called resolving the vector.

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Vector components

Copyright © 2010 Pearson Education Canada

Components of a vector:

AAAy

Ax

(x, y)

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Page 14: Copyright  2010 Pearson Education Canada 9-1 CHAPTER 9: VECTORS AND OBLIQUE TRIANGLES

Resolving vectors

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We will use trigonometry in our calculations by means of:

Ax = rcos Ay = rsin

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Steps used to find the x-and y-components of a vector

1. Place vector A such that is in standard position.

2. Calculate Ax, and Ay from Ax = Axcos and Ay = Aysin . We may use the reference angle if we note the direction of the component.

3. Check the components to see if each is in the correct direction and has a magnitude that is proper for the reference angle.

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Example

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What are the rectangular components of the vector given magnitude of 75 & = 50°?

Ay =

57.

4

50°75

Ax = rcos Ax = 75 cos 50°Ax = 48.2

Ay = rcos Ay = 75 sin 50° Ay = 57.4

Ax = 48.2

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Ch. 9.3: Vector Addition by Components To add two vectors:

1. Place each vector with its tail at the origin.2. Resolve each vector into its x- and y-

components3. Add the x-components of each vector together. 4. Add the y-components of each vector together.5. Using the Pythagorean Theorem, find the

magnitude of the resultant vector.6. Using the tangent ratio, find the direction of

the resultant vector.

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Example

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Vector A: 38.616.6 Vector B: 28.358

x

y

O

Resultant

A

B

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Solution – finding the x-component

Copyright © 2010 Pearson Education Canada

Vector A: 38.616.6 Vector B: 28.358 Ax + Bx = Rx Ax = rcos = 38.6cos16.6 =

Bx = rcos = 28.3cos58=

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Solution – finding the y-component

Copyright © 2010 Pearson Education Canada

Vector A: 38.616.6 Vector B: 28.358 Ay + By = Ry Ax = rsin = 38.6sin16.6 =

Bx = rsin = 28.3sin58=

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Solution (continued)

Copyright © 2010 Pearson Education Canada

Resultant Vector: (52, 35)

x

y

O

Resultant

(52, 35)

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Solution (continued)

Resultant Vector: (52, 35) What is the magnitude & direction of

this resultant vector?

2222 3552yxr

Copyright © 2010 Pearson Education Canada

= tan-1 (y/x) = tan-1 (35/52) = Answer: 62.7, 34

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Solution (continued)

Copyright © 2010 Pearson Education Canada

Resultant Vector: In rectangular form: (52, 35) Magnitude & direction: 62.7, 34

x

y

O

62.734

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