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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

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

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 ° C ° D ° 4 of 21

Key Concept 1 5 of 21

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

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

Key Concept 4 8 of 21

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

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

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

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

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 lb 13 of 21

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 lb B lb C lb D.5000 lb 14 of 21

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

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

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

Example 7 Calculate Work Answer: about 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 Therefore, the person does about joules of work pushing the mower. 18 of 21

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 joules C joules D joules 19 of 21

End of the Lesson 20 of 21 HW: PG odds, odds, 33-36