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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
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Find the component form and magnitude of with initial point A(−3, 7) and terminal point B(6, 2).
5–Minute Check 1
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Find the component form and magnitude of with initial point A(2, 5) and terminal point B(8, –3).
5–Minute Check 2
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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
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Find a unit vector with the same direction as v = −1, 5.
B. C. 5–Minute Check 4
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Which of the following represents the direction angle of the vector 2i − 8j?
B ° C ° D ° 5–Minute Check 5
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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
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dot product orthogonal vector projection work Vocabulary
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Key Concept 1
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Key Concept 1
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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
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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
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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
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Key Concept 2
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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
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Use the dot product to find the magnitude of v = –4, –1.
B. 17 C. D. 15 Example 2
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Key Concept 4
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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
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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
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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
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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: ° Example 3
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Find the angle between vectors v = –3, –1 and w = –5, 2 to the nearest tenth of a degree.
Example 3
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Key Concept 3
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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
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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
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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
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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
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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
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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
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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
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Therefore, projvu is w1 as shown, and u .
Projection with Direction Opposite v Therefore, projvu is w as shown, and u Answer: Example 5
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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
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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
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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
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= 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
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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 v. Since v is a unit vector, this means that this force has a magnitude of about pounds and is in the direction of the side of the mountain. Answer: about lb Example 6
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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 lb B lb C lb D lb Example 6
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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
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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
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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 Dot product Therefore, the person does about joules of work pushing the mower. Answer: about joules Example 7
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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 joules B joules C joules D joules Example 7
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