Five-Minute Check (over Lesson 8-2) Then/Now New Vocabulary Key Concept: Dot Product of Vectors in a Plane Key Concept: Orthogonal Vectors Example 1: Find the Dot Product to Determine Orthogonal Vectors Key Concept and Proof: Properties of the Dot Product Example 2: Use the Dot Product to Find Magnitude Key Concept and Proof: Angle Between Two Vectors Example 3: Find the Angle Between Two Vectors Key Concept and Proof: Projection of u onto v Example 4: Find the Projection of u onto v Example 5: Projection with Direction Opposite v Example 6: Real-World Example: Use a Vector Projection to Find a Force Example 7: Real-World Example: Calculate Work Lesson Menu

Find the component form and magnitude of with initial point A(−3, 7) and terminal point B(6, 2). 5–Minute Check 1

Find the component form and magnitude of with initial point A(2, 5) and terminal point B(8, –3). 5–Minute Check 2

Find 2f − 3g + 4h if f = −7, −2, g = 3, 1, and h = 9, −1. B. –11, 13 C. 41, –11 D. 41, 0 5–Minute Check 3

Find a unit vector with the same direction as v = −1, 5. B. C. 5–Minute Check 4

Which of the following represents the direction angle of the vector 2i − 8j? B. 104.04° C. 284.04° D. 345.96° 5–Minute Check 5

Find the projection of one vector onto another. You found the magnitudes of and operated with algebraic vectors. (Lesson 8-2) Find the dot product of two vectors and use the dot product to find the angle between them. Find the projection of one vector onto another. Then/Now

dot product orthogonal vector projection work Vocabulary

Key Concept 1

Key Concept 1

Since u ● v ≠ 0, u and v are not orthogonal, as illustrated. Find the Dot Product to Determine Orthogonal Vectors A. Find the dot product of u and v if u = –3, 4 and v = 3, 6. 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 Example 1

Since u ● v = 0, u and v are orthogonal, as illustrated. Find the Dot Product to Determine Orthogonal Vectors B. Find the dot product of u and v if u = 2, 7 and v = –14, 4 . 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 Example 1

Find the dot product of u = 4, –1 and v = –3, –5 Find the dot product of u = 4, –1 and v = –3, –5. Then determine if u and v are orthogonal. A. –7; yes B. –7; no C. 17; yes D. 17; no Example 1

Key Concept 2

Use the dot product to find the magnitude of a = –6, 5. Use the Dot Product to Find Magnitude Use the dot product to find the magnitude of a = –6, 5. Since |a|2 = a ● a, then |a| = . a = –6, 5 Simplify. The magnitude of a is or about 7.81. Answer: or about 7.81 Example 2

Use the dot product to find the magnitude of v = –4, –1. B. 17 C. D. 15 Example 2

Key Concept 4

Angle between two vectors Find the Angle Between Two Vectors A. Find the angle θ between u = –3, –5 and v = 2, –3 to the nearest tenth of a degree. Angle between two vectors u = –3, –5 and v = 2, –3 Evaluate. Simplify. Example 3

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

Angle between two vectors Find the Angle Between Two Vectors B. Find the angle θ between u = 1, –4 and v = 2, 6 to the nearest tenth of a degree. Angle between two vectors u = 1, –4 and v = 2, 6 Evaluate. Simplify. Example 3

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

Find the angle between vectors v = –3, –1 and w = –5, 2 to the nearest tenth of a degree. Example 3

Key Concept 3

Step 1 Find the projection of u onto v. Find the projection of u = –1, 5 onto v = 4, 6. Then write u as the sum of two orthogonal vectors, one of which is the projection of u onto v. Step 1 Find the projection of u onto v. Projection of u onto v u = –1, 5 onto v = 4, 6 Evaluate. Example 4

Scalar multiplication Find the Projection of u onto v. Scalar multiplication Step 2 Find w2. Since u = w1 + w2, w2 = u – w1. w2 = u – w1 = u – projvu = –1, 5 – 2, 3 = –3, 2 Therefore, projvu is w1 = 2, 3 as shown on the next slide, and u = 2, 3 + –3, 2 . Example 4

Answer: projvu = 2, 3; u = 2, 3 + –3, 2 Find the Projection of u onto v. Answer: projvu = 2, 3; u = 2, 3 + –3, 2 Example 4

Find the projection of u = 4, –3 onto v = 1, 1 Find the projection of u = 4, –3 onto v = 1, 1. Then write u as a sum of two orthogonal vectors, one of which is the projection of u onto v. A. projvu = B. projvu = C. projvu = D. projvu = Example 4

Projection with Direction Opposite v Find the projection of u = 4, 2 onto v = –3, 5. Then write u as the sum of two orthogonal vectors, one of which is the projection of u onto v. Notice that the angle between u and v is obtuse, so the projection of u onto v lies on the vector opposite v or –v, as shown above. Example 5

Step 1 Find the projection of u onto v. Projection with Direction Opposite v Step 1 Find the projection of u onto v. Projection of u onto v u = 4, 2 onto v = –3, 5 Evaluate. Scalar multiplication Example 5

Step 2 Find w2. Since u = w 1 + w2, w2 = u – w 1. w2 = u – w 1 Projection with Direction Opposite v Step 2 Find w2. Since u = w 1 + w2, w2 = u – w 1. w2 = u – w 1 = u – projvu Example 5

Therefore, projvu is w1 as shown, and u . Projection with Direction Opposite v Therefore, projvu is w1 as shown, and u . Answer: Example 5

Find the projection of u = 3, –2 onto v = 4, 3 Find the projection of u = 3, –2 onto v = 4, 3. Then write u as the sum of two orthogonal vectors, one of which is the projection of u onto v. A. projvu = B. projvu = C. projvu = D. projvu = Example 5

Use a Vector Projection to Find a Force 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? The weight of the boulder is the force exerted due to gravity, F = 0, –10,000. To find the force –w1 required to keep the boulder from rolling down the mountain, project F onto a unit vector v in the direction of the side of the mountain. Example 6

Step 1 Find a unit vector v in the direction of the mountain. Use a Vector Projection to Find a Force Step 1 Find a unit vector v in the direction of the mountain. v = |v| (cos θ), |v| (sin θ) Component form of v in terms of |v| and θ Example 6

= 1(cos 60°), 1(sin 60°) or |v| = 1 and θ = 60° Use a Vector Projection to Find a Force = 1(cos 60°), 1(sin 60°) or |v| = 1 and θ = 60° Step 2 Find w1, the projection of F onto unit vector v, projvF. Projection of F onto v Since v is a unit vector, |v| = 1. Simplify. Example 6

F = 0, –10,000 and v = Find the dot product. Use a Vector Projection to Find a Force F = 0, –10,000 and v = Find the dot product. The force required is –w1 = –(8660.3v) or 8660.3v. Since v is a unit vector, this means that this force has a magnitude of about 8660.3 pounds and is in the direction of the side of the mountain. Answer: about 8660.3 lb Example 6

TRUCKS A 5000-pound truck sits on a hill inclined at a 15° angle 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 Example 6

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. Example 7

Method 1 Use the projection formula for work. Calculate Work Method 1 Use the projection formula for work. The magnitude of the projection of F onto is |F| cos θ = 40 cos 45°. The magnitude of , the directed distance, is 12. W = Projection formula for work = (40 cos 45°)(12) Substitution or about 339.4 Example 7

Method 2 Use the dot product formula for work. Calculate Work 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 40 cos (–45°), 40 sin (–45°). The component form of the directed distance the mower is moved is 12, 0. W = F ● Dot product formula for work = 40 cos (–45°), 40 sin (–45°) ● 12, 0 Substitution = [40 cos (–45°)](12) or about 339.4 Dot product Therefore, the person does about 339.4 joules of work pushing the mower. Answer: about 339.4 joules 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 Example 7