lesson menu five-minute check (over lesson 8-2) then/now new vocabulary key concept: dot product of...
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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
Over Lesson 8-2
Find the component form and magnitude of with initial point A(−3, 7) and terminal point B(6, 2).
A.
B.
C.
Over Lesson 8-2
Find the component form and magnitude of with initial point A(2, 5) and terminal point B(8, –3).
A. 6, –8; 10
B. –6, 8; 10
C.
D.
Over Lesson 8-2
Find 2f − 3g + 4h if f = −7, −2, g = 3, 1, and h = 9, −1.
A. 13, –11
B. –11, 13
C. 41, –11
D. 41, 0
Over Lesson 8-2
Find a unit vector with the same direction asv = −1, 5.
A.
B.
C.
Over Lesson 8-2
Which of the following represents the directionangle of the vector 2i − 8j?
A. 75.96°
B. 104.04°
C. 284.04°
D. 345.96°
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.
• dot product
• orthogonal
• vector projection
• work
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
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
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
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| = .
Simplify.
The magnitude of a is or about 7.81.
Answer: or about 7.81
a = –6, 5
Use the dot product to find the magnitude of v = –4, –1.
A.
B. 17
C.
D. 15
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
Simplify.
u = –3, –5 and v = 2, –3
Evaluate.
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.
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.
Simplify.
Evaluate.
u = 1, –4 and v = 2, 6
Angle between two vectors
Find the Angle Between Two Vectors
Answer: 147.5°
Solve for .
The measure of the angle between u and v is about 147.5°.
Find the angle between vectors v = –3, –1 and w = –5, 2 to the nearest tenth of a degree.
A. 139.8°
B. 69.9°
C. 40.2°
D. 3.4°
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.
Evaluate.
u = –1, 5 onto v = 4, 6
Projection of u onto v
Find the Projection of u onto v.
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 .
Scalar multiplication
Answer: projvu = 2, 3; u = 2, 3 + –3, 2
Find the Projection of u onto v.
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 =
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.
Projection with Direction Opposite v
Step 1 Find the projection of u onto v.
u = 4, 2 onto v = –3, 5
Evaluate.
Scalar multiplication
Projection of u onto v
Projection with Direction Opposite v
Step 2 Find w2.
Since u = w 1 + w2, w2 = u – w 1.
w2 = u – w 1
= u – projvu
Projection with Direction Opposite v
Therefore, projvu is w1 as shown,
and u .
Answer:
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 =
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.
Use a Vector Projection to Find a Force
Step 1 Find a unit vector v in the direction of the mountain.
Use a Vector Projection to Find a Force
v = |v| (cos θ), |v| (sin θ)Component form of v in terms of |v| and θ
Use a Vector Projection to Find a Force
Step 2 Find w1, the projection of F onto unit vector v, projvF.
= 1(cos 60°), 1(sin 60°) or |v| = 1 and
θ = 60°
Since v is a unit vector, |v| = 1. Simplify.
Projection of F onto v
Answer: about 8660.3 lb
Use a Vector Projection to Find a Force
F = 0, –10,000
and v =
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.
Find the dot product.
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
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.
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
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 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, 0Substitution
= [40 cos (–45°)](12) or about 339.4Dot product
Therefore, the person does about 339.4 joules of work pushing the mower.
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
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