Study Guide and Review - Chapter 8
Determine whether each statement is true or false . If false, replace the underlined term or expression to
make the statement true.
1. The terminal point of a vector is where the vector begins.
SOLUTION: False; By definition, the terminal point of a vector is where the vector ends.
2. If a =
and b =
, the dot product is calculated by –4(1) + 3(2).
SOLUTION: False; The dot product of
and is defined as . So,
.
3. The midpoint of
with A(x1, y 1, z 1) and B(x2, y 2, z 2) is given by
.
SOLUTION: True; By definition, the midpoint of
with A(x1, y 1, z 1) and B(x2, y 2, z 2) is given by
4. The magnitude of r if the initial point is A(−1, 2) and the terminal point is B(2, −4) is
.
.
SOLUTION: False; The component form of a vector with initial point A(x1, y 1) and terminal point B(x2, y 2) is given by
.
5. Two vectors are equal only if they have the same direction and magnitude.
SOLUTION: True; By definition, equivalent vectors have the same magnitude and direction.
6. When two nonzero vectors are orthogonal, the angle between them is 180°.
SOLUTION: False; When two nonzero vectors are orthogonal, they are perpendicular. Thus, the angle between them is 90°.
7. The component of u onto v is the vector with direction that is parallel to v and with length that is the component of u
along v .
SOLUTION: False; The projection of u onto v is the vector with direction that is parallel to v and with length that is the component
of u along v .
8. To find at least one vector orthogonal to any two vectors in space, calculate the cross product of the two original
vectors.
SOLUTION: True; The vector
is perpendicular to the plane containing vectors a and b.
9. When a vector is subtracted, it is equivalent to adding the opposite vector.
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True; If
and Page 1
, then
. This is equivalent to
.
8. To find at least one vector orthogonal to any two vectors in space, calculate the cross product of the two original
vectors.
SOLUTION: Study
Guide and Review - Chapter 8
True; The vector
is perpendicular to the plane containing vectors a and b.
9. When a vector is subtracted, it is equivalent to adding the opposite vector.
SOLUTION: True; If
and , then
. This is equivalent to
.
10. If v is a unit vector in the same direction as u, then v =
.
SOLUTION: False; To find a unit vector, divide the vector by its magnitude. Thus, v =
.
State whether each quantity described is a vector quantity or a scalar quantity.
11. a car driving 50 miles an hour due east
SOLUTION: This quantity has a magnitude of 50 miles an hour and a direction of due east. This is a vector quantity.
12. a gust of wind blowing 5 meters per second
SOLUTION: This quantity has a magnitude of 5 meters per second but no direction. This is a scalar quantity.
Find the resultant of each pair of vectors using either the triangle or parallelogram method. State the
magnitude of the resultant to the nearest tenth of a centimeter and its direction relative to the horizontal.
13. SOLUTION: Translate d so that its tail touches the tip of c. Then draw the resultant vector c + d as shown. Draw the horizontal.
Drawing may not be to scale.
Measure the length of c + d and then measure the angle this vector makes with the horizontal.
The vector has a length of approximately 4.0 centimeters and is at an approximate angle of 11° with the horizontal.
14. SOLUTION: Translate j so that its tail touches the tip of h. Then draw the resultant vector h + j as shown. Draw the horizontal.
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Measure the length of c + d and then measure the angle this vector makes with the horizontal.
The vector has a length of approximately 4.0 centimeters and is at an approximate angle of 11° with the horizontal.
Study
Guide and Review - Chapter 8
14. SOLUTION: Translate j so that its tail touches the tip of h. Then draw the resultant vector h + j as shown. Draw the horizontal.
Drawing may not be to scale.
Measure the length of h + j and then measure the angle this vector makes with the horizontal.
The vector has a length of approximately 0.8 centimeters and is at an approximate angle of 140° with the horizontal.
15. SOLUTION: Translate b so that its tail touches the tip of a. Then draw the resultant vector a + b as shown. Draw the horizontal.
Drawing may not be to scale.
Measure the length of a + b and then measure the angle this vector makes with the horizontal.
The vector has a length of approximately 2.8 centimeters and is at an approximate angle of 297° with the horizontal.
16. SOLUTION: Translate v so that its tail touches the tip of w. Then draw the resultant vector w + v as shown. Draw the horizontal.
Drawing may not be to scale.
Page 3
Measure the length of w + v and then measure the angle this vector makes with the horizontal.
The vector has a length of approximately 3.6 centimeters and is at an approximate angle of 196º with the horizontal.
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Drawing may not be to scale.
Measure the length of a + b and then measure the angle this vector makes with the horizontal.
The vector has a length of approximately 2.8 centimeters and is at an approximate angle of 297° with the horizontal.
Study
Guide and Review - Chapter 8
16. SOLUTION: Translate v so that its tail touches the tip of w. Then draw the resultant vector w + v as shown. Draw the horizontal.
Drawing may not be to scale.
Measure the length of w + v and then measure the angle this vector makes with the horizontal.
The vector has a length of approximately 3.6 centimeters and is at an approximate angle of 196º with the horizontal.
Determine the magnitude and direction of the resultant of each vector sum.
17. 70 meters due west and then 150 meters due east
SOLUTION: Let a = 70 meters due west and b = 150 meters due east. Draw a diagram to represent a and b using a scale of 1
cm : 20 m.
Translate b so that its tail touches the tip of a. Then draw the resultant vector a + b.
Drawings may not be to scale.
Measure the length of a + b. The length of the vector is approximately 4.0 centimeters, which is 4.0 × 20 or 80 meters. a + b is in the direction of b. Since the direction of b is due east, the resultant vector is 80 meters due east.
18. 8 newtons directly backward and then 12 newtons directly backward
SOLUTION: Let a = 8 newtons directly backward and b = 12 newtons directly backward. Draw a diagram to represent a and b
using a scale of 1 cm : 2 N. Let due west be backward.
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Measure the length of a + b. The length of the vector is approximately 4.0 centimeters, which is 4.0 × 20 or 80 meters. a + b is in the direction of b. Since the direction of b is due east, the resultant vector is 80 meters due east.
Study
Guide and Review - Chapter 8
18. 8 newtons directly backward and then 12 newtons directly backward
SOLUTION: Let a = 8 newtons directly backward and b = 12 newtons directly backward. Draw a diagram to represent a and b
using a scale of 1 cm : 2 N. Let due west be backward.
Translate b so that its tail touches the tip of a. Then draw the resultant vector a + b.
Drawings may not be to scale.
Measure the length of a + b. The length of the vector is approximately 10.0 centimeters, which is 10.0 × 2 or 20 newtons. a + b is in the direction of a and b. Since the direction of both vectors is due west, the resultant vector is 20
newtons backward.
Find the component form and magnitude of
19. A(–1, 3), B(5, 4)
with the given initial and terminal points.
SOLUTION: First, find the component form.
Next, find the magnitude. Substitute x2 − x1 = 6 and y 2 − y 1 = 1 into the formula for the magnitude of a vector in the
coordinate plane.
20. A(7, –2), B(–9, 6)
SOLUTION: First,Manual
find the
component
form.
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Study Guide and Review - Chapter 8
20. A(7, –2), B(–9, 6)
SOLUTION: First, find the component form.
Next, find the magnitude. Substitute x2 − x1 = -16 and y 2 − y 1 = 8 into the formula for the magnitude of a vector in
the coordinate plane.
21. A(–8, –4), B(6, 1)
SOLUTION: First, find the component form.
Next, find the magnitude. Substitute x2 − x1 = 14 and y 2 − y 1 = 5 into the formula for the magnitude of a vector in
the coordinate plane.
22. A(2, –10), B(3, –5)
SOLUTION: First, find the component form.
Next, find the magnitude. Substitute x2 − x1 = 1 and y 2 − y 1 = 5 into the formula for the magnitude of a vector in the
coordinate plane.
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Study Guide and Review - Chapter 8
22. A(2, –10), B(3, –5)
SOLUTION: First, find the component form.
Next, find the magnitude. Substitute x2 − x1 = 1 and y 2 − y 1 = 5 into the formula for the magnitude of a vector in the
coordinate plane.
Find each of the following for p =
23. 2q – p
,q=
, and t =
.
SOLUTION: 24. p + 2t
SOLUTION: 25. t – 3p + q
SOLUTION: 26. 2p + t – 3q
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Study Guide and Review - Chapter 8
26. 2p + t – 3q
SOLUTION: Find a unit vector u with the same direction as v.
27. v =
SOLUTION: 28. v =
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Study Guide and Review - Chapter 8
28. v =
SOLUTION: 29. v =
SOLUTION: 30. v =
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Study Guide and Review - Chapter 8
30. v =
SOLUTION: Find the dot product of u and v. Then determine if u and v are orthogonal.
31. u =
,v=
SOLUTION: , u and v are not orthogonal.
Since
32. u =
,v =
SOLUTION: Since
33. u =
, u and v are not orthogonal.
,v =
SOLUTION: Since
34. u =
, u and v are orthogonal.
,v =
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SOLUTION: Page 10
Study Guide and Review - Chapter 8
Since
, u and v are orthogonal.
34. u =
,v =
SOLUTION: Since
, u and v are not orthogonal.
Find the angle θ between u and v to the nearest tenth of a degree.
35. u =
,v=
SOLUTION: 36. u =
,v=
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Study Guide and Review - Chapter 8
36. u =
,v=
SOLUTION: Plot each point in a three-dimensional coordinate system.
37. (1, 2, –4)
SOLUTION: Starting at the origin, move 1 unit forward along the x-axis, 2 units to the right parallel to the y-axis, and then 4 units
down parallel to the z-axis. Plot the point (1, 2, −4).
38. (3, 5, 3)
SOLUTION: Starting at the origin, move 3 units forward along the x-axis, 5 units to the right parallel to the y-axis, and then 3 units
up parallel to the z-axis. Plot the point (3, 5, 3).
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Study Guide and Review - Chapter 8
38. (3, 5, 3)
SOLUTION: Starting at the origin, move 3 units forward along the x-axis, 5 units to the right parallel to the y-axis, and then 3 units
up parallel to the z-axis. Plot the point (3, 5, 3).
39. (5, –3, –2)
SOLUTION: Starting at the origin, move 5 units forward along the x-axis, 3 units to the left parallel to the y-axis, and then 2 units
down parallel to the z-axis. Plot the point (5, −3, −2).
40. (–2, –3, –2)
SOLUTION: Starting at the origin, move 2 units backwards along the x-axis, 3 units to the left parallel to the y-axis, and then 2
units down parallel to the z-axis. Plot the point
(−2, −3, −2).
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Study Guide and Review - Chapter 8
40. (–2, –3, –2)
SOLUTION: Starting at the origin, move 2 units backwards along the x-axis, 3 units to the left parallel to the y-axis, and then 2
units down parallel to the z-axis. Plot the point
(−2, −3, −2).
Find the length and midpoint of the segment with the given endpoints.
41. (–4, 10, 4), (2, 0, 8)
SOLUTION: Use the Distance Formula for points in space to find the length of the segment.
Use the Midpoint Formula for points in space to find the midpoint of the segment.
42. (–5, 6, 4), (–9, –2, –2)
SOLUTION: Use the Distance Formula for points in space to find the length of the segment.
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Study Guide and Review - Chapter 8
42. (–5, 6, 4), (–9, –2, –2)
SOLUTION: Use the Distance Formula for points in space to find the length of the segment.
Use the Midpoint Formula for points in space to find the midpoint of the segment.
43. (3, 2, 0), (–9, –10, 4)
SOLUTION: Use the Distance Formula for points in space to find the length of the segment.
Use the Midpoint Formula for points in space to find the midpoint of the segment.
44. (8, 3, 2), (–4, –6, 6)
SOLUTION: Use the Distance Formula for points in space to find the length of the segment.
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Study Guide and Review - Chapter 8
44. (8, 3, 2), (–4, –6, 6)
SOLUTION: Use the Distance Formula for points in space to find the length of the segment.
Use the Midpoint Formula for points in space to find the midpoint of the segment.
Locate and graph each vector in space.
45. a =
SOLUTION: Start by plotting the point (0, −3, 4). From the origin, move 3 units to the left along the y-axis, and then 4 units up
parallel to the z-axis. Draw a with an initial point at the origin and terminal point (0, −3, 4).
46. b = –3i + 3j + 2k
SOLUTION: Start by writing v in component form as
Next plot the point (−3, 3, 2). From the origin, move 3 units
backwards along the x-axis, 3 units to the right parallel to the y-axis, and then 2 units up parallel to the z-axis. Draw
b with an initial point at the origin and terminal point (−3, 3, 2).
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Study Guide and Review - Chapter 8
46. b = –3i + 3j + 2k
SOLUTION: Start by writing v in component form as
Next plot the point (−3, 3, 2). From the origin, move 3 units
backwards along the x-axis, 3 units to the right parallel to the y-axis, and then 2 units up parallel to the z-axis. Draw
b with an initial point at the origin and terminal point (−3, 3, 2).
47. c = –2i – 3j + 5k
SOLUTION: Start by writing v in component form as
Next plot the point (−2, −3, 5). From the origin, move 2 units
backwards along the x-axis, 3 units to the left parallel to the y-axis, and then 5 units up parallel to the z-axis. Draw c
with an initial point at the origin and terminal point (−2, −3, 5).
48. d =
SOLUTION: Start by plotting the point (−4, −5, −3). From the origin, move 4 units backwards along the x-axis, 5 units to the left
along the y-axis, and then 3 units down parallel to the z-axis. Draw d with an initial point at the origin and terminal
point (−4, −5, −3).
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Study Guide and Review - Chapter 8
48. d =
SOLUTION: Start by plotting the point (−4, −5, −3). From the origin, move 4 units backwards along the x-axis, 5 units to the left
along the y-axis, and then 3 units down parallel to the z-axis. Draw d with an initial point at the origin and terminal
point (−4, −5, −3).
Find the dot product of u and v. Then determine if u and v are orthogonal.
,v=
49. u =
SOLUTION: Since
, u and v are orthogonal.
50. u =
,v=
SOLUTION: Since
, u and v are not orthogonal.
Find the cross product of u and v. Then show that u × v is orthogonal to both u and v.
,v=
51. u =
SOLUTION: Find the cross product of u and v .
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Study Guide and Review - Chapter 8
Since
, u and v are not orthogonal.
Find the cross product of u and v. Then show that u × v is orthogonal to both u and v.
,v=
51. u =
SOLUTION: Find the cross product of u and v .
To show that
is orthogonal to both u and v , find the dot product of
with u and
with v .
with u and
with v .
Because both dot products are zero, the vectors are orthogonal.
52. u =
,v=
SOLUTION: Find the cross product of u and v .
To show that
is orthogonal to both u and v , find the dot product of
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Study
Guideboth
and dot
Review
- Chapter
8 the vectors are orthogonal.
Because
products
are zero,
52. u =
,v=
SOLUTION: Find the cross product of u and v .
To show that
is orthogonal to both u and v , find the dot product of
with u and
with v .
Because both dot products are zero, the vectors are orthogonal.
53. BASEBALL A player throws a baseball with an initial velocity of 55 feet per second at an angle of 25° above the
horizontal, as shown below. Find the magnitude of the horizontal and vertical components.
SOLUTION: The horizontal and vertical components of the vector form a right triangle. Use the sine or cosine ratios to find the
magnitude of each component.
The Manual
magnitude
of the
horizontal
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component is about 49.8 feet per second and the magnitude of the vertical
component is about 23.2 feet per second.
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54. STROLLER Miriam is pushing a stroller with a force of 200 newtons at an angle of 20° below the horizontal. Find
Study
Guide and Review - Chapter 8
Because both dot products are zero, the vectors are orthogonal.
53. BASEBALL A player throws a baseball with an initial velocity of 55 feet per second at an angle of 25° above the
horizontal, as shown below. Find the magnitude of the horizontal and vertical components.
SOLUTION: The horizontal and vertical components of the vector form a right triangle. Use the sine or cosine ratios to find the
magnitude of each component.
The magnitude of the horizontal component is about 49.8 feet per second and the magnitude of the vertical
component is about 23.2 feet per second.
54. STROLLER Miriam is pushing a stroller with a force of 200 newtons at an angle of 20° below the horizontal. Find
the magnitude of the horizontal and vertical components of the force.
SOLUTION: Diagram the situation.
The horizontal and vertical components of the vector form a right triangle. Use the sine or cosine ratios to find the
magnitude of each component.
The magnitude of the horizontal component is about 187.9 newtons and the magnitude of the vertical component is
about 68.4 newtons.
No expression.
55. LIGHTS A traffic light at an intersection is hanging from two wires of equal length at 15° below the horizontal as
shown. If the traffic light weighs 560 pounds, what is the tension in each wire keeping the light at equilibrium?
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SOLUTION: Page 21
The
magnitude
of the horizontal
is about 187.9 newtons and the magnitude of the vertical component is
Study
Guide
and Review
- Chaptercomponent
8
about 68.4 newtons.
No expression.
55. LIGHTS A traffic light at an intersection is hanging from two wires of equal length at 15° below the horizontal as
shown. If the traffic light weighs 560 pounds, what is the tension in each wire keeping the light at equilibrium?
SOLUTION: Diagram the situation using vectors to represent the tension in each wire.
Translate T2 so that its tail touches the tip of T1 and draw T1 + T2.
Since the equilibrant of T1 + T2 and the vector representing the weight of the light are equal, T1 + T2 is exerting a
force of 560 lb upward. The angle created by T1 and the horizontal is 15°. Since the direction of T1 + T2 is upward,
T1 + T2 creates a right angle with the horizontal. Thus, the angle created by T1 and T1 + T2 is 90° − 15° or 75°. Since T1 = T2, the triangle is isosceles and the angle created by T2 and T1 + T2 is also 75°. The remaining angle of the triangle is 180° − 75° − 75° or 30°.
Use the Law of Sines to find the magnitude of T1.
Since T1 and T2 are equal in length, they both have the same magnitude. Thus, the tension in each wire is about
1081.8 pounds.
56. AIRPLANE An airplane is descending at a speed of 110 miles per hour at an angle of 10° below the horizontal.
Find the component form of the vector that represents the velocity of the airplane.
SOLUTION: Use the magnitude and direction of the airplane’s velocity v to write this vector in component form. Since the
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airplane is descending, the vertical component is negative.
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Since T1 and T2 are equal in length, they both have the same magnitude. Thus, the tension in each wire is about
Study
Guide
and Review - Chapter 8
1081.8
pounds.
56. AIRPLANE An airplane is descending at a speed of 110 miles per hour at an angle of 10° below the horizontal.
Find the component form of the vector that represents the velocity of the airplane.
SOLUTION: Use the magnitude and direction of the airplane’s velocity v to write this vector in component form. Since the
airplane is descending, the vertical component is negative.
57. LIFEGUARD A lifeguard at a wave pool swims at a speed of 4 miles per hour at a 60° angle to the side of the
pool as shown.
a. At what speed is the lifeguard traveling if the current in the pool is 2 miles per hour parallel to the side of the pool
as shown?
b. At what angle is the lifeguard traveling with respect to the starting side of the pool?
SOLUTION: a. Find the component form of each vector. Since the current of the pool is 2 miles, let
.
Use the magnitude and direction of the swimmer’s velocity v to write this vector in component form.
Find the resultant vector r by calculating c + v .
Find the magnitude of the resultant vector r.
The resultant speed of the lifeguard is about 5.3 miles per hour.
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b. The
vector
representing
the
resultant velocity of the lifeguard is
Find the resultant direction angle θ.
.
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Study Guide and Review - Chapter 8
57. LIFEGUARD A lifeguard at a wave pool swims at a speed of 4 miles per hour at a 60° angle to the side of the
pool as shown.
a. At what speed is the lifeguard traveling if the current in the pool is 2 miles per hour parallel to the side of the pool
as shown?
b. At what angle is the lifeguard traveling with respect to the starting side of the pool?
SOLUTION: a. Find the component form of each vector. Since the current of the pool is 2 miles, let
.
Use the magnitude and direction of the swimmer’s velocity v to write this vector in component form.
Find the resultant vector r by calculating c + v .
Find the magnitude of the resultant vector r.
The resultant speed of the lifeguard is about 5.3 miles per hour.
b. The vector representing the resultant velocity of the lifeguard is
.
Find the resultant direction angle θ.
The lifeguard is traveling at an angle of about 40.9° with respect to the starting side of the pool.
58. TRAFFIC A 1500-pound car is stopped in traffic on a hill that is at an incline of 10°. Determine the force that is required to keep the car from rolling down the hill.
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The lifeguard is traveling at an angle of about 40.9° with respect to the starting side of the pool.
58. TRAFFIC A 1500-pound car is stopped in traffic on a hill that is at an incline of 10°. Determine the force that is required to keep the car from rolling down the hill.
SOLUTION: The weight of the car is the force exerted due to gravity,
. To find the force −w1 required to keep the
car from rolling down the hill, project F onto a unit vector v in the direction of the side of the hill.
Find a unit vector v in the direction of the side of the hill.
Find w1, the projection of F onto the unit vector v , projvF.
Since w1 points down the hill, the force required is −w1 = −(−260.5v ) or 260.5v . Since v is a unit vector, about 260.5
pounds represents the magnitude of the force required to keep the car from rolling down the hill.
59. WORK At a warehouse, Phil pushes a box on sliders with a constant force of 80 newtons up a ramp that has an incline of 15° with the horizontal. Determine the amount of work in joules that Phil does if he pushes the dolly 10
meters.
SOLUTION: In this case, Phil is actually pushing 80 N in the exact direction that the box is moving. Therefore, there is no
projection of the force and he is not losing any of the force that he is exerting on the box. The work is simply his force multiplied by the distance. 80 N × 10 m = 800 joules
60. SATELLITES
The positions of two satellites that are in orbit can be represented by the coordinates (28,625, eSolutions
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32,461, −38,426) and (−31,613, −29,218, 43,015), where (0, 0, 0) represents the center of Earth and the coordinates
are given in miles. The radius of Earth is about 3963 miles.
The work is simply his force multiplied by the distance. 80 N × 10 m = 800 joules
Guide and Review - Chapter 8
Study
60. SATELLITES The positions of two satellites that are in orbit can be represented by the coordinates (28,625, 32,461, −38,426) and (−31,613, −29,218, 43,015), where (0, 0, 0) represents the center of Earth and the coordinates
are given in miles. The radius of Earth is about 3963 miles.
a. Determine the distance between the two satellites.
b. If a third satellite were to be placed directly between the two satellites, what would the coordinates be?
c. Can a third satellite be placed at the coordinates found in part b? Explain your reasoning.
SOLUTION: a. Use the Distance Formula for points in space to find the distance between the two satellites.
The distance between the two satellites is about 118,598 miles.
b. Use the Midpoint Formula for points in space to find the halfway point between the two satellites.
So, the third satellite would have the coordinates (−1494, 1621.5, 2294.5).
c. No; sample answer: The third satellite cannot exist, because its coordinates are inside Earth. Use the Distance
Formula for points in space to find the distance between the third satellite and the center of Earth.
The distance from the center of Earth to the third satellite is about 3182 miles. Since the radius of Earth is about
3962 miles, the coordinates of the third satellite would have it placed inside Earth.
61. BICYCLES A bicyclist applies 18 newtons of force down on the pedal to put the bicycle in motion. The pedal has an initial position of 47º above the y-axis, and a length of 0.19 meters to the pedal’s axle, as shown.
a. Find the vector representing the torque about the axle of the bicycle pedal in component form.
b. Find the magnitude and direction of the torque.
SOLUTION: a. The component form of the vector representing the directed distance from the axle to the pedal can be found
using the triangle shown and trigonometric ratios.
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Guide
andfrom
Review
- Chapter
8 to the third satellite is about 3182 miles. Since the radius of Earth is about
The
distance
the center
of Earth
3962 miles, the coordinates of the third satellite would have it placed inside Earth.
61. BICYCLES A bicyclist applies 18 newtons of force down on the pedal to put the bicycle in motion. The pedal has an initial position of 47º above the y-axis, and a length of 0.19 meters to the pedal’s axle, as shown.
a. Find the vector representing the torque about the axle of the bicycle pedal in component form.
b. Find the magnitude and direction of the torque.
SOLUTION: a. The component form of the vector representing the directed distance from the axle to the pedal can be found
using the triangle shown and trigonometric ratios.
Vector r is therefore
or about to the pedal is 18 newtons down, so
.
. The vector representing the force applied
Use the cross product of these vectors to find the vector representing the torque about the hinge.
The component form of the torque vector is
.
b. The component form of the torque vector
newton-meters parallel to the negative x-axis.
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tells us that the magnitude of the vector is about 2.3 Page 27