Section 8.2 Vector Angles

Transcription

1 Section 8.2 Vector Angles INTRODUCTION Recall that a vector has these two properties: 1. It has a certain length, called magnitude 2. It has a direction, indicated by an arrow at one end. In this section we focus on the angles related to vectors in the x-y-plane. In particular, (i) the direction angle of a single vector and (ii) the angle between two vectors. A VECTOR S DIRECTION ANGLE When placed in standard position, the angle, θ, between a vector and the positive x-axis, is called the vector s direction angle: 0 θ < 360 In the diagrams at right, we see a vector s horizontal and vertical components set up as the legs of a right triangle, with the vector acting as the hypotenuse. The magnitude of the vector is the length of the hypotenuse. This means, cosθ a which gives us a cosθ. a b V! a, b " V # a #! a, b " b Likewise, sinθ b which gives us b sinθ. This means, if we know the magnitude and direction of the vector, we can identify its components. For example, a vector, V, with magnitude 27 and a direction angle of 97 has components a cosθ and b sinθ 27 cos(97 ) 27 sin(97 ) So, we can say V -3.3, 26.8 Sec. 8.2 Vector Angles 381

2 Example 1: Given the magnitude and direction angle of a vector, find its components. Round each component to the nearest tenth. a) X 35 and θ 41.3 b) Y 14 and θ 220 c) Z 55 and θ 342 Procedure: Use the formulas for a (horizontal component) and b (vertical component): a cosθ and b sinθ Answer: a) a 35 cos(41.3 ) b) a 14 cos(220 ) c) a 55 cos(342 ) a a a b 35 sin(41.3 ) b 14 sin(220 ) b 27 sin(342 ) b b b So, X 26.3, 23.1 Y -10.7, -9.0 Z 52.3, You Try It 1 Given the magnitude and direction angle of a vector, find its components. Round each component to the nearest tenth. a) R 32 and θ 24.5 b) S 9 and θ 131 c) T 17 and θ 255 d) U 49 and θ 318 Sec. 8.2 Vector Angles 382

3 To find a vector s direction angle, θ, from its components, a, b we first identify ˆ " (in Quadrant I), by using either cos ˆ " a or sin ˆ " b In each, the numerator is the absolute value of the component and the denominator is the magnitude of the vector. To determine the actual value of θ, we find the related angle measure in the quadrant in which the vector lies, based on its components. For example, V 5, -4 lies in Quadrant IV when it is in standard position. To find θ, i) find the magnitude of the vector: (5) 2 + (-4) 2 41 ; ii) find ˆ " using either cos ˆ " a 5 41 or sin ˆ " b : Let s choose cos ˆ " a , so ˆ " cos -1 (0.7809) 38.7 iii) find the related angle in Quadrant IV by subtracting ˆ " from 360 : θ You Try It 2 Find the direction angle of the given vector. Round θ to the nearest tenth of a degree. Use the example above as a guide. a) M -9, -3 b) N -6, 8 Sec. 8.2 Vector Angles 383

4 THE DOT PRODUCT: VECTORS IN THE x-y-plane Definition: For two non-zero vectors, T x 1, y 1 and U x 2, y 2, the dot product of T and U is defined as T U x 1 x 2 + y 1 y 2. Because x 1, x 2, y 1, and y 2 are numerical components of vectors T and U, their dot product is also a number (scalar). Example 2: Find the dot product of each pair of vectors. a) R 5, 2 and S 7, 6 b) M 3, 7 and Z -8, 4 Procedure: Use the formula in the definition of the dot product. Answer: a) R S b) M Z (-32) You Try It 3 Find the dot product of each pair of vectors. a) J 1, 9 and K 4, 7 b) M -2, 4 and Z 5, -6 In and of itself, the dot product doesn t have much meaning. The main function of this scalar is to help us identify the angle measure, α, between the two vectors. We use this formula: cosα V W W The numerator is the dot product of the two vectors, and the denominator is the product of the magnitudes of the vectors. Also, 0 α 180. Note: If V is a positive scalar multiple of W, then α is 0 ; if V is a negative scalar multiple of W, then α is 180. The denominator of this formula the product of the magnitudes is always positive. The numerator the dot product can be either positive, zero, or negative. So, it is the dot product that determines whether this formula for cosα is positive, negative, or zero, and whether α is acute, obtuse, or 90 : Dot Product cosα α Positive Positive Acute Zero Zero 90 Negative Negative Obtuse Sec. 8.2 Vector Angles 384

5 Example 3: Given two vectors, find the angle, α, between them. Round α to the nearest whole number. a) V -2, 4 and W 3, 6 b) X -5, 2 and Y 3, -6 Procedure: Answer: First graph the vectors. Then calculate each of these values i) the magnitude of each vector ii) The dot product of the two vectors iii) the angle, α, between the vectors. a) -6 6! -2, 4 " -3 V ! 3, 6 " i) (-2) 2 + (4) W 3 6 W (3) 2 + (6) ii) V W (-2) iii) cosα V W W cosα 0.6 α cos -1 (0.6) α α 53 (α is acute.) b) 6 i) X (-5) 2 + (2) 2 29! -5, 2 " X Y 3 6! 3, -6 " Y (3) 2 + (-6) ii) X Y (-5) (-6) iii) cosα X Y X Y cosα α cos -1 ( ) α α 138 (α is obtuse.) Sec. 8.2 Vector Angles 385

6 Note: Finding the angle between two vectors is an algebraic process. Graphing the vectors is not required, but it is helpful to see them graphed to get an idea of the size of the angle. You Try It 4 Given two vectors, find the angle, α, between them. Round α to the nearest whole number. (Graphing the vectors is not required but might be helpful.) Use Example 3 as a guide. a) L 2, 5 and M 7, 4 b) A -4, -3 and B 6, -5 c) P 1, 3 and Q 6, -2 d) S -8, 2 and T 4, -1 Sec. 8.2 Vector Angles 386

7 You Try It Answers YTI 1: a) R 29.1, 13.3 b) S -5.9, 6.8 c) T -4.4, d) U 36.4, YTI 2: a) θ ( ˆ " 18.4 ) b) θ ( ˆ " 53.1 ) YTI 3: a) J K 67 b) M Z -34 YTI 4: a) α 38.5 b) α c) α 90 d) α 180 Section 8.2 Focus Exercises Given the magnitude and direction angle of a vector, find its components. Round each component to the nearest tenth. 1. E 22 and θ F 39 and θ G 44 and θ H 9 and θ J 10 and θ K 24 and θ L 56 and θ M 50 and θ 322 Sec. 8.2 Vector Angles 387

8 Find the direction angle of the given vector. Round θ to the nearest tenth of a degree. Use the example above as a guide. 13. A 5, B -4, C -1, D 3, W -12, X 7, Y 3.5, Z -8, -15 Find the dot product of each pair of vectors. 17. G 6, 2 and H 5, U -8, -10 and V 4, A -12, -4 and B -3, C 15, -20 and D -6, 8 Sec. 8.2 Vector Angles 388

9 Given two vectors, find the angle, α, between them. Round α to the nearest whole number. (Graphing the vectors is not required but might be helpful.) 21. M 6, 2 and N 1, P 4, 8 and Q -7, S 4, -5 and T 1, W -5, -2 and Z -4, 4 Sec. 8.2 Vector Angles 389

10 25. E -2, 8 and F -6, J -10, -2 and K 15, X 3, 6 and Y -8, Q -12, -16 and R -12, 9 Sec. 8.2 Vector Angles 390