CHAPTER 13 Vector Applications Contents: A. Problems involving vector operations B. Lines in 2-D and 3-D C. The angle between two lines D. Constant velocity problems E. The shortest distance from a line to a point F. Intersecting lines G. Relationships between lines A. PROBLEMS INVOLVING VECTOR OPERATIONS The sum of vectors is called the resultant vector. Example 1 In still water, Jacques can swim at 1.5m/s. Jacques is at point A on the edge of a canal, and considers point B directly opposite. A current is flowing from the left at a constant speed of 0.5m/s. a. If Jacques dives in straight towards B, and swims without allowing for the current, what will his actual speed and direction be? b. Jacques wants to swim directly across the canal to point B. i. At what angle should Jacques aim to swim in order that the current will correct his direction? ii What will Jacques’ actual speed be? Suppose c is the current’s velocity vector, s is the velocity vector Jacques would have if the water was still, and f = c + s is Jacques’ resultant velocity vector. a. Jacques aims directly across the river, but the current takes him downstream to the right. Applying the Pythagorean Theorem and a little trigonometry. Jacques has an actual speed of approximately 1.58m/s and his direction of motion is approximately 18.4o to the right of his intended line. b. Jacques needs to aim to the left of B so the current will correct his direction. Exercise 13A 1. An athlete can normally run with constant speed 6 m/s. Using a vector diagram to illustrate each situation, find the athlete’s speed if: a. he is assisted by a wind of 1 m/s from directly behind him b. he runs into a head wind of 1 m/s. a. 6 m/s 1 m/s 7 m/s 6 m/s b. 1 m/s 5 m/s 5 An airplane needs to fly due east from one city to another at a speed of 400 km/h. However, a 50 km/h wind blows constantly from the north-east. a. How does the wind affect the speed of the airplane? b. In what direction must the airplane head to compensate for the wind? a. x 135o 50 km/h 400 km/h Using the Law of Cosine, compute the value for x c 2 a 2 b 2 2ab cos C x 400 2 50 2 2(400)(50) cos(135) x 437 The calm airspeed of the plane 437 km/h which is slowed to 400 km/h by the northeasterly wind. b. x q 135o 50 km/h 400 km/h Determine q: use the Law of Sine. sin q sin(135) 50 437 sin(135) q sin 1 50 437 q 4.65 north of due east. LINES IN 2-D AND 3-D In both 2-D and 3-D geometry we can determine the equation of a line using its direction and any fixed point on the line. Suppose a line passes through a fixed point A with position vector a, and that the line is parallel to the vector b. Vector Equation of a line A line passes through the point A(1, 5) and has direction 3 vector . Describe the line using: 2 a. A vector equation, b. Parametric equations, c. and a Cartesian equation. Parametric equation x 1 3 t y 5 2 top : x 1 3t bottom : y 5 2t Cartesian equation: Solve both equation for t. Set them equal to one another and simplify. x 1 y 5 t 3 2 x 1 y 5 3 2 2 x 2 3 y 15 t 2 x 3 y 13 LINES IN 3-D Find a vector equation and the corresponding parametric equations of the line through (1,-2, 3) in the direction 4i + 5j – 6k. NON-UNIQUENESS OF THE VECTOR EQUATION OF A LINE Going from (5, 4) to (7, 3), go 2 to the right and 1 down. Going from (7, 3) to (5,4), go 2 to the left and 1 up. Find parametric equations of the line through A(2,-1, 4) and B(-1, 0, 2). Find the directional vector AB or BA 1 2 3 AB 0 1 1 2 4 2 Using point A 2 1 3 BA 1 0 1 42 2 Using point B x 2 3 y 1 t 1 z 4 2 x 1 3 y 0 s 1 z 2 2 x 2 3t , y 1 t , z 4 2t , t R x 1 3s, y s, z 2 2 s, s R Exercise 13B x 3 1 1ai. t , t R y 4 4 1aii. x 3 t , y 4 4t y4 1aiii. x 3 4 4 x 12 y 4 4 x y 16 x 5 2 1bi. t , t R y 2 5 1bii. x 5 2t , y 2 5t x 5 y 2 1biii. 2 5 5 x 25 2 y 4 5 x 2 y 29 x 6 3 1ci. t , t R y 0 7 1cii. x 6 3t , y 7t x6 y 1ciii. 3 7 7 x 42 3 y 7 x 3 y 42 x 1 2 1di. t , t R y 11 1 1dii. x 1 2t , y 11 t x 1 1diii. y 11 2 x 1 2 y 22 x 2 y 21 3 2 1 3a t , t R 2 1 3 3 2t 2 1 3t t 1 t 1 same t therefore yes Solve for t, set equal and simplify x 1 3b. t 2 x 1 y 1 2 x 1 2 y 2 k 1 2(4) 2 k 8 2 1 5 y 1 t Sub in (k, 4) THE ANGLE BETWEEN TWO LINES Exercise 13C Find the angle between the lines L1: x = -4 + 12t, y = 3 + 5t L2: x = 3s, y = -6 – 4s 12 3 b1 , b2 5 4 | 36 (20) | 16 cos q 65 169 25 1 16 q cos 75.7 65 3 b1 16 , 7 3 b2 8 5 | 9 (128) (35) | 154 cos q 314 98 30772 154 q cos 28.6 30772 1 b. 3 0 16 3 0 7 x 3 0 16 3 7 x 0 48 7 x 0 48 x 7 Find the measure of the acute angle between the lines 2x + y = 5 and 3x – 2y = 8. 2x y 5 2 slope 1 3x 2 y 8 3 slope 2 1 2 b1 , b2 2 3 | 2 (6) | 4 cos q 5 13 65 4 q cos 1 60.3 65 1 direction vector 2 2 direction vector 3 Find the measure of the angle between the lines: y = 2 – x and x – 2y = 7 Answer: 71.6o CONSTANT VELOCITY PROBLEMS a a. The initial position of the object occurs at t = 0 x 1 3 t y 9 4 x 1 3 0 at t 0, y 9 4 therefore x 1, y 9 or (1, 9) x 1 y 9 x 1 3 t y 9 4 x 1 3 0 at t 0, y 9 4 at t 1, at t 2, at t 3, x 1 3 1 y 9 4 x 1 3 2 y 9 4 x 1 3 3 y 9 4 x 1 y 9 x 4 y 5 x 7 y 1 x 10 y 3 x 1 3 t y 9 4 velocity vector 3 4 Is the velocity vector speed | b | 3 4 32 (4) 2 5 m/s To find the position of the object at time t, we need to find the parametric equation for x and y. The coordinates of (x, y) will give us our position. Step 1: set up the vector equation of the object. Step 2: find the parametric equation from the vector equation Step 3: write the coordinate as an ordered pair of parametric equations. b. To determine the speed of the object, we need to find the magnitude or length of the vector. c. Plug in the value for t into our ordered pair. d. Due east of (0, 0) means when y = 0 (aka x-intercept) 22. EXERCISE 13d 1a. The initial position occurs at t = 0. x(0) = 1 + 2(0) = 1 y(0) = 2 – 5(0) = 2 the initial position (1, 2) 1b. x = 1 + 2t y = 2 - 5t (x, y) 0 1 2 (1, 2) 1 3 -3 (3, -3) 2 5 -8 (5, -8) 3 7 -13 (7, -13) 4 2 0 0 2 4 6 8 -2 -4 Series1 -6 -8 -10 -12 -14 x 1 2 1c . t y 2 5 2 velocity 5 1d . 2 2 29 2 5 THE SHORTEST DISTANCE FROM A LINE TO A POINT Perpendicular/orthogonal EXERCISE 13E a. 2 x 3 y 36 y int : 3 y 36 y 12 A (0, 12) x int : 2 x 36 B (18, 0) x 18 b. 2 x 3 y 36 2 y x 12 3 2 R ( x, x 12) 3 0 18 18 c. AB 12 0 12 x4 x4 2 PR 2 x 12 0 x 12 3 3 d .PR AB 0 x 4 18 36 2 x 0 12 3 36 2 x 18( x 4) 12 0 3 18 x 72 144 8 x 0 26 x 216 108 x 13 108 36 2 36 2 x 13 84 y 3 3 13 1 2t a. R 1 3t 2t 1 2t 1 2t PR 1 3t 1 3t 2 2 t 3 t 1 2 2t 2 PR 3 3t 2 3 4t 9t 6 t 1 0 1 t 1 1 1 2 1 1 2 2 2 1 R 1 3 1 2 2 5 2 21 2 t 1 2 2 1 2 t 2 1 1 PR 3t 2 3 2 1 2 1 t 1 2 1 2 2 1 1 1 2 6 2 INTERSECTING LINES EXERCISE 13F 0 2 0 1 a. r t 2 1 5 1 2r t 2 r 5t sub 2r for t , 2 r 5 2r r 1 t 2r 2(1) 2 x 0 2 A 1 (2, 3) y 2 1 B (8, 6) C (5, 0) AB (2 8) (3 6) 3 5 2 2 AC (2 5) 2 (3 0) 2 3 BC (8 5) 2 (6 0) 2 3 5 AB BC RELATIONSHIPS BETWEEN LINES LINE CLASSIFICATION IN 3 DIMENSION