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 Canada2 9-2

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

Vectors Copyright © 2010 Pearson Education Canada  Applications of Vectors: Velocity & acceleration Force 9-4

Describing vectors Copyright © 2010 Pearson Education Canada  How are vectors described? A 1. A, or,  in boldface: A  9-5

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. 9-6

Vector addition Copyright © 2010 Pearson Education Canada  The tip-to-tail method: A + B = R x y O Resultant A B B 9-7

Vector addition Copyright © 2010 Pearson Education Canada  The parallelogram method: A + B = R x y O Resultant A B 9-8

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 9-9

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. 9-10

Subtracting vectors Copyright © 2010 Pearson Education Canada  An illustration: A – B = A + (- B) = R A A B -B R  9-11

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. 9-12

Vector components Copyright © 2010 Pearson Education Canada  Components of a vector: A AyAy AxAx (x, y) 9-13

Resolving vectors Copyright © 2010 Pearson Education Canada  We will use trigonometry in our calculations by means of: A x = rcos  A y = rsin  9-14

Steps used to find the x -and y - components of a vector 1. Place vector A such that  is in standard position. 2. Calculate A x, and A y from A x = A x cos  and A y = A y sin . 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. Copyright © 2010 Pearson Education Canada

Example Copyright © 2010 Pearson Education Canada  What are the rectangular components of the vector given magnitude of 75 &  = 50 °? A y = ° 75 A x = rcos  A x = 75 cos 50 ° A x = 48.2 A y = rcos  A y = 75 sin 50 ° A y = 57.4 A x = 48.2  9-16

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 -components 3. 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. Copyright © 2010 Pearson Education Canada

Example Copyright © 2010 Pearson Education Canada  Vector A : 38.6  16.6   Vector B : 28.3  58  x y O Resultant A B 9-18

Solution – finding the x-component Copyright © 2010 Pearson Education Canada  Vector A : 38.6  16.6   Vector B : 28.3  58   Ax + Bx = Rx  Ax = r cos  = 38.6 cos 16.6  =  Bx = r cos  = 28.3 cos 58  = 9-19

Solution – finding the y-component Copyright © 2010 Pearson Education Canada  Vector A : 38.6  16.6   Vector B : 28.3  58   Ay + By = Ry  Ax = r sin  = 38.6 sin 16.6  =  Bx = r sin  = 28.3 sin 58  = 9-20

Solution ( continued ) Copyright © 2010 Pearson Education Canada  Resultant Vector: (52, 35) x y O Resultant (52, 35) 9-21

Solution ( continued )  Resultant Vector: ( 52, 35 )  What is the magnitude & direction of this resultant vector? Copyright © 2010 Pearson Education Canada  = tan -1 (y/x) = tan -1 ( 35/52 ) = Answer: 62.7,  34 

Solution ( continued ) Copyright © 2010 Pearson Education Canada  Resultant Vector:  In rectangular form: ( 52, 35 )  Magnitude & direction: 62.7,  34  x y O 62.7  34   9-23