1. Splash Screen

2. Over Lesson 8-2 5–Minute Check 1 Find the component form and magnitude of with initial point A (−3, 7) and terminal point B (6, 2). A. B. C. D. 2 of 21

3. Over Lesson 8-2 5–Minute Check 4 Find a unit vector with the same direction as v =. 3 of 21

4. Over Lesson 8-2 5–Minute Check 5 Which of the following represents the direction angle of the vector 2i − 8j? A.75.96° B.104.04° C.284.04° D.345.96° 4 of 21

5. Key Concept 1 5 of 21

6. Example 1 Find the Dot Product to Determine Orthogonal Vectors A. Find the dot product of u and v if u = and v =. Then determine if u and v are orthogonal. u ● v= –3(3) + 4(6) = 15 Since u ● v ≠ 0, u and v are not orthogonal, as illustrated. Answer: 15; not orthogonal 6 of 21

7. Example 1 Find the Dot Product to Determine Orthogonal Vectors B. Find the dot product of u and v if u = and v =. Then determine if u and v are orthogonal. u ● v= 2(–14) + 7(4) = 0 Since u ● v = 0, u and v are orthogonal, as illustrated. Answer: 0; orthogonal 7 of 21

8. Key Concept 4 8 of 21

9. Example 3 Find the Angle Between Two Vectors A. Find the angle θ between u = and v = to the nearest tenth of a degree. 9 of 21

10. Example 3 Find the Angle Between Two Vectors Answer: 64.7° The measure of the angle between u and v is about 64.7°. Solve for θ. 10 of 21

11. Example 3 Find the Angle Between Two Vectors B. Find the angle θ between u = and v = to the nearest tenth of a degree. 11 of 21

12. Example 3 Find the Angle Between Two Vectors Answer: 147.5° Solve for θ. The measure of the angle between u and v is about 147.5°. 12 of 21

13. Example 6 BOULDERS A 10,000-pound boulder sits on a mountain at an incline of 60°. Ignoring the force of friction, what force is required to keep the boulder from rolling down the mountain? Use a Vector Projection to Find a Force Answer: about 8660.3 lb 13 of 21

14. Example 6 TRUCKS A 5000-pound truck sits on a hill inclined at a 15° angle. Ignoring the force of friction, what force is required to keep the truck from rolling down the hill? A.1294.1 lb B.2588.2 lb C.4829.6 lb D.5000 lb 14 of 21

15. Work The work W done by a constant force F acting along the line of motion of an object is given by W = (magnitude of force)(distance) = ||F|| Force acts along the line of motion. 15 of 21

16. If the constant force F is not directed along the line of motion (see Figure 6.45), then the work W done by the force is given by Figure 6.45 Force acts at angle  with the line of motion. Dot product form for work is force dot with distance 16 of 21

17. Example 7 Calculate Work MOWING A person pushes a reel mower with a constant force of 40 newtons at a constant angle of 45°. Find the work done in joules moving the mower 12 meters. 17 of 21

18. Example 7 Calculate Work Answer: about 339.4 joules Method 2 Use the dot product formula for work. The component form of the force vector F in terms of magnitude and direction angle given is. The component form of the directed distance the mower is moved is. W= F ● = ● = [40 cos (–45°)](12) or about 339.4 Therefore, the person does about 339.4 joules of work pushing the mower. 18 of 21

19. Example 7 CRATE A person pushes a crate along the floor with a constant force of 30 newtons at a constant angle of 30°. Find the work done in joules moving the crate 8 meters. A.26.0 joules B.120.0 joules C.207.8 joules D.678.8 joules 19 of 21

20. End of the Lesson 20 of 21 HW: PG. 506 1-5 odds, 17-21 odds, 33-36