lesson menu five-minute check (over lesson 8-4) then/now new vocabulary key concept:dot product and...
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Five-Minute Check (over Lesson 8-4)
Then/Now
New Vocabulary
Key Concept:Dot Product and Orthogonal Vectors in Space
Example 1:Find the Dot Product to Determine Orthogonal Vectors in Space
Example 2:Angle Between Two Vectors in Space
Key Concept:Cross Product of Vectors in Space
Example 3:Find the Cross Product of Two Vectors
Example 4:Real-World Example: Torque Using Cross Product
Example 5:Area of a Parallelogram in Space
Key Concept:Triple Scalar Product
Example 6:Volume of a Parallelepiped
Over Lesson 8-4
Find the length and the midpoint of the segmentwith endpoints at (−2, 3, 4) and (6, 1, −5).
A. 8.31;
B. 10.04;
C. 12.21;
D. 12.21;
Over Lesson 8-4
Which of the following represents 3x – 5y + zif x = 2, –7, 1, y = –5, 0, 3, andz = –1, 6, –4?
A. –23, –15, 14
B. 6, –1, –6
C. 30, −15, −16
D. 30, −15, 14
You found the dot product of two vectors in the plane. (Lesson 8-3)
• Find dot products of and angles between vectors in space.
• Find cross products of vectors in space and use cross products to find area and volume.
Find the Dot Product to Determine Orthogonal Vectors in Space
A. Find the dot product of u and v for u = –1, 6, –3 and v = 3, –1, –3. Then determine if u and v are orthogonal.
u • v = –1(3) + 6(–1) + (–3)(–3)
= –3 + (–6) + 9 or 0
Since u • v = 0, u and v are orthogonal.
Answer: 0; orthogonal
Find the Dot Product to Determine Orthogonal Vectors in Space
B. Find the dot product of u and v for u = 2, 4, –6 and v = –3, 2, 4. Then determine if u and v are orthogonal.
u • v = 2(–3) + 4(2) + (–6)(4)
= –6 + 8 + (–24) or –22
Since u • v ≠ 0, u and v are not orthogonal.
Answer: –22; not orthogonal
Find the dot product of u = –4, 5, –1 and v = 3, –3, 1. Then determine if u and v are orthogonal.
A. – 28; orthogonal
B. – 28; not orthogonal
C. – 4; orthogonal
D. – 4; not orthogonal
Angle Between Two Vectors in Space
Find the angle θ between u = –4, –1, –3 and v = 7, 3, 4 to the nearest tenth of a degree.
Angle between two vectors
u = –4, –1, –3 and v = 7, 3, 4
Evaluate the dot product and magnitudes.
Angle Between Two Vectors in Space
The measure of the angle
between u and v is about 168.6°.
Answer: 168.6°
Simplify.
Solve for θ.
Find the angle between u = –2, 3, –1 and v = –4, –3, 4 to the nearest tenth of a degree.
A. 12.0°
B. 78.0°
C. 82.8°
D. 102.0°
Find the Cross Product of Two Vectors
Find the cross product of u = 6, –1, –2 and v = –1, –4, 2. Then show that u × v is orthogonal to both u and v.
Determinant of a 3 × 3 matrix
u = 6, –1, –2 and v = –1, –4, 2
Determinants of 2 × 2 matrices
Find the Cross Product of Two Vectors
To show that u × v is orthogonal to both u and v, find the dot product of u × v with u and u × v with v.
(u × v) • u
= –10, –10, –25 • 6, –1, –2 = –10(6) + (–10)(–1) + (–25)( –2)
= –60 + 10 + 50 or 0
Component form
Simplify.
Find the Cross Product of Two Vectors
Answer: –10, –10, –25; u × v • u = –10, –10, –25 • 6, –1, –2
= – 60 + 10 + 50 = 0u × v • v = –10, –10, –25 • –1, –4, 2
= 10 + 40 – 50 = 0
(u × v) • v
= –10, –10, –25 • –1, –4, 2= –10(–1) + (–10)(–4) + (–25)( 2)
= 10 + 40 + (–50) or 0
Because both dot products are zero, the vectors are orthogonal.
Find the cross product of u = 2, 3, –1 and v = –3, 1, 4.
A. u × v = 13, 5, 11
B. u × v = 13, –5, 11
C. u × v = 12, –5, 7
D. u × v = 13, 5, 11
Torque Using Cross Product
MACHINERY A mechanic uses a 0.4-meter long wrench to tighten a nut. Find the magnitude and direction of the torque about the nut if the force is 30 newtons straight down to the end of the handle when it is 35° above the positive x-axis.
Step 1 Graph each vector in standard position.
Torque Using Cross Product
Step 2 Determine the component form of each vector.
The component form of the vector representing the directed distance from the axis of rotation to the end of the handle can be found using the triangle in the figure below and trigonometry.
Torque Using Cross Product
Vector r is therefore 0.4 cos 35°, 0, 0.4 sin 35° or about 0.33, 0, 0.23. The vector representing the force applied to the end of the handle is 30 newtons straight down, so F = 0, 0, –30. Step 3 Use the cross product of these vectors to find the vector representing the torque about the nut.
T = r × F Torque Cross Product Formula
Cross product of r and F
Torque Using Cross Product
Determinants of 2 × 2 matrices
Component form
Determinant of a 3 × 3 matrix
Answer: 9.9 N • m parallel to the positive y-axis
Torque Using Cross Product
Step 4 Find the magnitude and direction of the torque vector.
The component form of the torque vector 0, 9.9, 0 tells us that the magnitude of the vector is about 9.9 newton-meters parallel to the positive y-axis as shown below.
MACHINERY A mechanic uses a 0.3-meter-long wrench to tighten a nut. Find the magnitude and direction of the torque about the nut if the force is 35 newtons straight down to the end of the handle when it is 40° below the positive x-axis as shown below.A. 10.5 N • m parallel
to the positive y-axis
B. 8.0 N • m parallel to the positive y-axis
C. 6.7 N • m parallel to the positive y-axis
D. 4.1 N • m parallel to the positive y-axis
Area of a Parallelogram in Space
Find the area of the parallelogram with adjacent sides u = –3i – 4j +2k and v = 5i – 4j – k.
Step 1 Find u × v.
u = –3i – 4j +2k and v = 5i – 4j – k
Determinant of a 3 × 3 matrix
Area of a Parallelogram in Space
Step 2 Find the magnitude of u × v.
Determinants of 2 × 2 matrices
Magnitude of a vector in space
Simplify.
Area of a Parallelogram in Space
The area of the parallelogram shown is or about 34.9 square units.
Answer:
Find the area of a parallelogram with sides u = 4i + 5j – 2k and v = i – j + 3k.
A. about 7.3 square units
B. about 10.0 square unit
C. about 20.8 square units
D. about 21.1 square units
Find the volume of the parallelepiped with adjacent edges t = –3i + 3j + 2k, u = –3i – 4j + 2k, and v = 5i – 4j – k.
Volume of a Parallelepiped
t = –3i + 3j + 2k , u = –3i – 4j + 2k and v = 5i – 4j – k
Determinant of a 3 × 3 matrix
The volume of the parallelepiped shown below is | t ● (u × v)| or 49 cubic units.
Answer: 49 cubic units
Volume of a Parallelepiped
Determinants of 2 × 2 matricesSimplify.