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Section 6.3. A ball flies through the air at a certain speed and in a particular direction. The speed and direction are the velocity of the ball. The.

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Presentation on theme: "Section 6.3. A ball flies through the air at a certain speed and in a particular direction. The speed and direction are the velocity of the ball. The."— Presentation transcript:

1 Section 6.3

2 A ball flies through the air at a certain speed and in a particular direction. The speed and direction are the velocity of the ball. The velocity is a vector quantity since it has both a magnitude and a direction. Vectors are used to represent velocity, force, tension, and many other quantities. A vector is a quantity with both a magnitude and a direction.

3 A quantity with magnitude and direction is represented by a directed line segment PQ with initial point P and terminal point Q. The vector v = PQ is the set of all directed line segments of length ||PQ|| which are parallel to PQ. P Q

4 Two vectors, u and v, are equal if the line segments representing them go in the same direction and have the same length or magnitude. u v

5 Scalar multiplication is the product of a scalar, or real number, times a vector. For example, the scalar 3 times v results in the vector 3v, three times as long and in the same direction as v. v 3v3v

6 v The product of -½ and v gives a vector half as long as and in the opposite direction to v. - v

7 Vector Addition To add vectors v + u: 1. Place the initial point of u at the terminal point of v. 2. Draw the vector with the same initial point as v and the same terminal point as u. u v u + v v u v u

8 Vector Subtraction To subtract vectors u − v: 1. Place the initial point of −v at the terminal point of u. 2. Draw the vector u  v from the initial point of u to the terminal point of −v. v u −v u u  v −v u

9 A vector with initial point (0, 0) is in standard position and is represented uniquely by its terminal point (u 1, u 2 ). x y (u 1, u 2 )

10 If v is a vector with initial point P = (p 1, p 2 ) and terminal point Q = (q 1, q 2 ), then 1. The component form of v is v = 2. The magnitude (or length) of v is ||v|| = x y P (p 1, p 2 ) Q (q 1, q 2 )

11 Example 1 Find the component form and magnitude of the vector v with initial point P = (3,  2) and terminal point Q = (  1, 1).

12 The magnitude of v is (p 1, p 2 ) = (3,  2) (q 1, q 2 ) = (  1, 1) So, v 1 =  1  3 =  4 and v 2 = 1  (  2) = 3. Therefore, the component form of v is

13 Two vectors u = and v = are equal if and only if u 1 = v 1 and u 2 = v 2.

14 Example 2

15 Calculate the component form for each vector Therefore v = w but v ≠ u and w ≠ u.

16 Operations on Vectors in the Coordinate Plane Let u =, v =, and let c be a scalar. 1. Scalar multiplication cu = 2. Addition u + v = 3. Subtraction u  v =

17 Example 3 Given vectors u = and v = Find -2u, u + v, and u − v. Draw each on a coordinate plane.

18 x y u -2u -2u = 4, 2 -8, -4

19 u + v x y 6, 7 2, 5 4, 2 v u u + v =

20 u  v 2, -3 x y 2, 5 4, 2 v u u  v =

21 Unit Vector A unit vector is a vector whose magnitude is 1. In many applications of vectors, it is useful to find a unit vector that has the same direction as a given nonzero vector v.

22 Example 4 Find a unit vector u in the direction of v = and verify that the result has magnitude 1.

23

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25 This sum is called a linear combination of vectors i and j.

26 Example 5 Let u be the vector with initial point (-2, 6) and terminal point (-8, 3). Write u as a linear combination of the standard unit vectors i and j.

27 Example 6 Let u = i + j and v = 5i – 3j. Find 2u – 3v. 2(i + j) – 3(5i – 3j) = 2i + 2j – 15i + 9j = -13i + 11j THE ANSWER TO ALL VECTOR PROBLEMS WILL BE IN THE FORM GIVEN IN THE PROBLEM.

28 x y The direction angle  of a vector v is the angle formed by the positive half of the x-axis and the ray along which v lies. x y v θ v θ x y v x y (x, y) If v =, then tan  =.

29 Example 7 Find exact value of the direction angle in degrees.

30 a.v = -6i + 6j Tangent is negative in the 2 nd and 4 th quadrant. This vector’s terminal point is in the 2 nd quadrant. So θ = 135°

31 b.v = -7i − 4j Round to the nearest hundredth. Tangent is positive in the 1 st and 3 rd quadrant. This vector’s terminal point is in the 3 rd quadrant. So θ ≈ 29.74 + 180 ≈ 209.74°

32 Using Direction Angles in Application Problems

33 If u is a unit vector such that θ is the angle (measured counterclockwise) from the positive x- axis to u, the terminal point of u lies on the unit circle. Write the component form of u in terms of θ. θ is the direction angle of the vector u. θ u (x, y)

34 Any vector v that goes in the same direction of a unit vector u has the same direction angle θ. This means that v = can be written in terms of cos θ and sin θ in the following way.

35 Example 8 Find the exact value of the component form of the vector that represents the velocity of an airplane descending at a speed of 100 miles per hour at a bearing of S 45° W. 45° θ = 45° + 180° = 225°

36 Example 9 An airplane is traveling at a speed of 724 kilometers per hour at a bearing of 100°. If the wind velocity is 32 kilometers per hour from the west, find the ground speed to the nearest tenth and bearing of the plane to the nearest hundredth.

37 100° = 90° + 10° The direction angle of the airplane vector is either -10° or 350°. The direction angle of the wind vector makes is 180°.

38

39 θ, the direction angle of a + w is in the 4 th quadrant.


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