angle this vector makes with the horizontal.
The vector has a length of approximately 3.9
centimeters and is at an approximate angle of 10° with the horizontal.
Drawing may not be to scale.
Practice Test - Chapter 8
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.
Find the component form and magnitude of
with the given initial and terminal points.
3. A(1, −3), B(−5, 1)
SOLUTION: First, find the component form.
1. SOLUTION: Translate q so that its tail touches the tip of p. Then
draw the resultant vector p + q as shown. Draw the
horizontal.
Next, find the magnitude. Substitute x2 − x1 = −6 and
y2 − y 1 = 4 into the formula for the magnitude of a
vector in the coordinate plane.
Measure the length of p + q. The vector has a length
of approximately 0.4 centimeter. p + q is parallel to
the horizontal. So, it is at an angle of 0° with the horizontal.
Drawing may not be to scale.
4. 2. SOLUTION: SOLUTION: Translate d so that its tail touches the tip of c. Then
draw the resultant vector c + d as shown. Draw the
horizontal.
First, find the component form.
Measure the length of c + d and then measure the
angle this vector makes with the horizontal.
The vector has a length of approximately 3.9
centimeters and is at an approximate angle of 10° with the horizontal.
Drawing may not be to scale.
Find the component form and magnitude of
with the given initial and terminal points.
3. A(1, −3), B(−5, 1)
Next, find the magnitude. Substitute x2 − x1 = −
and y 2 − y 1 =
into the formula for the magnitude of a vector in the coordinate plane.
SOLUTION: First, find the component form.
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SOLUTION: Practice Test - Chapter 8
Since Libby is moving straight forward, the
component form of her velocity v 1 is
. Use the
magnitude and direction of the ball’s velocity v 2 to
write this vector in component form.
4. SOLUTION: First, find the component form.
The component form of the vector representing
Libby’s velocity is
and the component form of
the vector representing the path of the ball is
Add the algebraic vectors representing v 1 and v 2 to
Next, find the magnitude. Substitute x2 − x1 = −
and y 2 − y 1 =
find the resultant velocity, vector r.
into the formula for the magnitude of a vector in the coordinate plane.
Find the magnitude of the resultant.
The speed of the ball is about 33.7 meters per
second.
Find the resultant direction angle θ.
5. SOFTBALL A batter on the opposing softball
team hits a ground ball that rolls out to Libby in left
field. She runs toward the ball at a velocity of 4
meters per second, scoops it, and proceeds to throw
it to the catcher at a speed of 30 meters per second
and at an angle of 25° with the horizontal in an attempt to throw out a runner. What is the resultant
speed and direction of the throw?
The speed of the ball is about 33.7 meters per
second at an angle of about 22.1° with the horizontal.
Find a unit vector in the same direction as u.
6. u =
SOLUTION: SOLUTION: Since Libby is moving straight forward, the
component form of her velocity v 1 is
. Use the
magnitude and direction of the ball’s velocity v 2 to
write this vector in component form.
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The speed
of the
ball is about833.7 meters per
Practice
Test
- Chapter
second at an angle of about 22.1° with the horizontal.
Find a unit vector in the same direction as u.
6. u =
SOLUTION: Find the dot product of u and v. Then
determine if u and v are orthogonal.
8. u =
,v=
SOLUTION: Since
9. u =
, u and v are not orthogonal.
,v=
SOLUTION: Since
, u and v are orthogonal.
10. u = 10i − 3j, v = i + 8j
SOLUTION: 7. u =
Write u and v in component form as
SOLUTION: Since
, u and v are not orthogonal.
11. MULTIPLE CHOICE Write u as the sum of
two orthogonal vectors, one of which being the
projection of u onto v if u =
and v =
.
A B C D SOLUTION: Find the projection of u onto v .
Find the dot product of u and v. Then
determine if u and v are orthogonal.
eSolutions
8. u = Manual - Powered
, v = by Cognero
SOLUTION: Page 3
Thus,
D .
Practice
Test - Chapter 8
SOLUTION: The correct answer is D.
Find the projection of u onto v .
12. MOVING Tamera is pushing a box along a level
floor with a force of 120 pounds at an angle of
depression of 20°. Determine how much work is done if the box is moved 25 feet.
SOLUTION: Diagram the situation.
To write u as the sum of two orthogonal vectors,
start by writing u as the sum of two vectors w1 and
w2, or u = w1 + w2. Since one of the vectors is the
Use the projection formula for work. The magnitude
of the projection of F onto
is The magnitude of the directed
distance
is 25.
projection of u onto v , let w1 = projvu and solve for
w2.
Therefore, Tamera is doing about 2819.1 footpounds of work.
Find each of the following for a =
, and c =
.
13. 2a + 5b − 3c
,b=
SOLUTION: Thus,
.
The correct answer is D.
12. MOVING Tamera is pushing a box along a level
floor with a force of 120 pounds at an angle of
depression of 20°. Determine how much work is done if the box is moved 25 feet.
14. b − 6a + 2c
SOLUTION: SOLUTION: Diagram the situation.
15. HOT AIR BALLOONS During a festival, twelve
hot air balloons take off. A few minutes later, the
coordinates of the first two balloons are (20, 25, 30)
and (−29, 15, 10) as shown, where the coordinates
are given in feet.
Use the projection formula for work. The magnitude
F Cognero
of the
projection
of by
onto
is eSolutions
Manual
- Powered
The magnitude of the directed
distance
is 25.
Page 4
15. HOT AIR BALLOONS During a festival, twelve
Use the component form to find the magnitude of v .
hot air balloons take off. A few minutes later, the
coordinates
first two balloons
are (20, 25, 30)
Practice
Testof -the
Chapter
8
and (−29, 15, 10) as shown, where the coordinates
are given in feet.
Using this magnitude and component form, find a
unit vector u in the direction of v .
a. Determine the distance between the first two
balloons that took off.
b. A third balloon is halfway between the first two
balloons. Determine the coordinates of the third
balloon.
c. Find a unit vector in the direction of the first balloon if it took off at (0, 0, 0).
SOLUTION: a. . Use the Distance Formula for points in space to
find the distance between the two balloons.
Find the angle θ between vectors u and v to
the nearest tenth of a degree.
16. u =
,v=
SOLUTION: The distance between the two balloons is about 53.9
feet.
b. Use the Midpoint Formula for points in space to
find the halfway point between the first two balloons.
So, the coordinates of the third balloon are (−4.5, 20,
20).
c. Find the component form of the vector v
representing the path of the first balloon.
17. u = −9i + 5j + 11k , v = −5i − 7j − 6k
SOLUTION: Use the component form to find the magnitude of v .
Using this magnitude and component form, find a
unit vector u in the direction of v .
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Write u and v in component form as
Page 5
Practice Test - Chapter 8
17. u = −9i + 5j + 11k , v = −5i − 7j − 6k
SOLUTION: Write u and v in component form as
Find the cross product of u and v. Then show
that u × v is orthogonal to both u and v.
18. u =
,v=
SOLUTION: Find the cross product of u and v .
Find the cross product of u and v. Then show
that u × v is orthogonal to both u and v.
18. u =
,v=
To show that
is orthogonal to both u and v ,
find the dot product of
with u and
with v.
SOLUTION: Find the cross product of u and v .
Because both dot products are zero, the vectors are
orthogonal.
To show that
is orthogonal to both u and v ,
find the dot product of
with u and
with v.
19. u = −6i + 2j − k , v = 5i − 3j − 2k
SOLUTION: Write u and v in component form as
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Because both dot products are zero, the vectors are
orthogonal.
To show that
, 6
is orthogonal to both u and vPage
u
find the dot product of
with and
with v.
Because both dot products are zero, the vectors are
Practice
Test - Chapter 8
orthogonal.
that is developed about the axle of the tiller if 50
newtons of force is applied in a direction parallel to
the positive y-axis.
SOLUTION: Diagram the situation.
19. u = −6i + 2j − k , v = 5i − 3j − 2k
SOLUTION: Write u and v in component form as
To show that
is orthogonal to both u and v ,
find the dot product of
with u and
with v.
The component form of the vector representing the
directed distance from the axle to the end of the tiller
can be found using the triangle shown and
trigonometric ratios.
Vector r is therefore
or about
. The vector representing the
force applied to the tiller is 50 newtons parallel to the
positive y-axis, so
.
Because both dot products are zero, the vectors are
orthogonal.
Use the cross product of these vectors to find the
vector representing the torque about the hinge.
20. BOATING The tiller is a lever that controls the
position of the rudder on a boat. When force is
applied to the tiller, the boat will turn. Suppose the
tiller on a certain boat is 0.75 meter in length and is
currently resting in the xy-plane at a 15° angle from the positive x-axis. Find the magnitude of the torque
that is developed about the axle of the tiller if 50
newtons of force is applied in a direction parallel to
the positive y-axis.
SOLUTION: Diagram the situation.
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The component form of the torque vector is
. This tells us that the magnitude of the
vector is about 36.2 newton-meters.
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