Module 8 Lesson 4: Applications of Vectors
So now that you have learned the basic skills necessary to understand and operate with
vectors, in this lesson, we will look at how to solve real world problems that involve
vectors. As stated earlier, vectors often involve applications of force, work, weight,
navigation, and a few other topics.
Let’s go ahead and get started. It will be best if you have your Notes from Lessons 1-3
handy in case you need to reference them.
Using Vectors to Find Speed & Direction
Ex1. An airplane is flying on a bearing of 341° at 560 mph. Find the component form of
the velocity of the plane.
Solution: It helps to sketch a picture of the situation.
If v has direction angle 𝜃, the components of v can be computed using the
formula below.
𝐯 = ⟨‖𝐯‖ cos 𝜃 , ‖𝐯‖ sin 𝜃 ⟩
Determine the magnitude of v. Here in this instance, if the plane is flying at 560
mph, then the magnitude of the plane (the length that it flies) will be 560 miles
(per hour in flight).
‖𝐯‖ = 560
To determine the direction angle, recall that this is the counterclockwise angle
between the vector v and the positive x-axis. The problem statement gives the
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bearing of v, which is the measure of the clockwise angle between v and the yaxis.
The following graph correctly demonstrates the direction of v. Notice that the
measure of the clockwise angle formed by the vector and the y-axis is the
bearing 341°.
We can determine the direction angle of v.
𝜃 = 90° + (360° − 341°) = 109°
Now, we compute the component form of the velocity of the angle.
𝐯 = ⟨‖𝐯‖ cos 𝜃 , ‖𝐯‖ sin 𝜃 ⟩ = ⟨560 cos 109° , 560 sin 109°⟩
= ⟨−182.318, 529.490⟩
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Ex2. An airplane is flying on a compass heading (bearing) of 350° at 355 mph. A wind is
blowing with the bearing 310° at 60 mph.
(a) Find the component form of the velocity of the plane.
(b) Find the actual ground speed and direction of the plane.
Solution for (a): Find vectors a and b that model the velocity of the airplane and
the velocity of the wind, then use these vectors to determine the ground speed
and bearing of the plane. Remember that the bearing is the clockwise angle of
the velocity with the positive y-axis. If a vector v has direction angle 𝜃, the
components of v can be computed using the following formula.
𝐯 = ⟨‖𝐯‖ cos 𝜃 , ‖𝐯‖ sin 𝜃 ⟩
The bearing of the airplane is 350°. Let 𝛼 be the direction angle of the airplane’s
velocity without wind. Determine the measure of 𝛼.
𝛼 = 100°
Compute the components of a, the vector of the airplane’s velocity without
wind.
𝐚 = ⟨‖𝐚‖ cos 𝛼 , ‖𝐚‖ sin 𝛼 ⟩ = ⟨335 cos 100° , 335 sin 100°⟩
= ⟨−61.645, 349.607⟩
The bearing of the wind is 310°. Let 𝛽 be the direction angle of the wind.
Determine the measure of 𝛽.
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𝛽 = 140°
Compute the component form of b, the velocity of the wind.
𝐛 = ⟨‖𝐛‖ cos 𝛽 , ‖𝐛‖ sin 𝛽 ⟩ = ⟨60 cos 140° , 60 sin 140°⟩ = ⟨−45.963, 38.567⟩
The true velocity of the plane is 𝐯 = 𝐚 + 𝐛. Add these vectors componentwise
to find v.
𝐯 = 𝐚 + 𝐛 = ⟨−61.645, 349.607⟩ + ⟨−45.963, 38.567⟩
= ⟨−107.608, 388.174⟩
Solution for (b): The ground speed of the airplane is the magnitude of the
velocity of the airplane. Recall that if, 𝐯 = ⟨a, b⟩, then ‖𝐯‖ = √𝑎2 + 𝑏 2.
Compute the ground speed of the airplane.
‖𝐯‖ ≈ √(−107.608)𝟐 + (388.174)𝟐 ≈ 402.813
The ground speed of the airplane is approximately 402.813 mph.
Now, all that is left to find is the bearing for the ground speed of the airplane.
From the formula for writing a vector in component form, we know the first
component of the velocity vector (which we found to be −107.608) can be
written as |𝐯| cos 𝜃, where 𝜃 is the directional angle. We can set up the
equation
402.813 ∙ cos 𝜃 = −107.608
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and solve for 𝜃.
cos 𝜃 =
−107.608
402.813
𝜃 = cos −1
−107.608
402.813
Compute 𝜃. Note that since the x-component of v is negative and its ycomponent is positive, this angle should occur in Quadrant II.
𝜃 = 105.494°
Determine the bearing to which 105° corresponds.
The bearing = 360° − (105.494° − 90°) = 345.506°
Therefore, the ground speed is approximately 403 mph with bearing 346°.
Ex3. A basketball is shot at a 60° angle with the horizontal direction with an initial
speed of 42 feet per second. Find the component form of the initial velocity.
Solution: Let v be a nonzero vector. If 𝜃 is the direction angle measured from
the positive x-axis to v, then the vector can be expressed in terms of its
magnitude and direction angle by the formula below.
𝐯 = ⟨‖𝐯‖ cos 𝜃 , ‖𝐯‖ sin 𝜃 ⟩
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Let’s find the value of the direction angle measured from the positive x-axis to
the vector v.
𝜃 = 60°
Since the basketball is released with a speed of 42 feet per second, the
magnitude of v is 42 feet per second.
‖𝐯‖ = 42
Now, substitute the values for ‖𝐯‖ = 42 and 𝜃 = 60° into our formula to
express v in terms of its magnitude and directional angle.
𝐯 = ⟨‖𝐯‖ cos 𝜃 , ‖𝐯‖ sin 𝜃 ⟩ = ⟨42 cos 60° , 42 sin 60° ⟩ = ⟨21, 36.373⟩
Thus, the vector v is expressed in component form ⟨21, 36.373⟩.
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Finding a Force
The previous examples are applications in which two vectors are added to produce a
resultant vector. Many applications in physics and engineering pose the reverse
problem – decomposing a given vector into the sum of two vector components.
Consider a boat being pulled on an inclined ramp (shown above). The force F due to
gravity pulls the boat down and against the ramp. These two orthogonal
(perpendicular) vectors, 𝐰1 + 𝐰2, are vector components of F.
𝐅 = 𝐰1 + 𝐰2
The negative of component 𝐰1represents the force needed to keep the boat from
rolling down the ramp, and 𝐰2 represents the force that the tires must withstand
against the ramp.
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To find the force required in this problem, we will use the formula
𝐅∙𝐯
proj𝐯 𝐅 = (
)𝐯
‖𝐯‖2
Ex4. A 600 pound boat sits on a ramp inclined at 30°. What force is required to keep
the boat from rolling down the hill? (In other words, we are only looking for 𝐰1.)
Solution: Because the force due to gravity is vertical and downward and has a
magnitude equal to the combined weight, you can represent the gravitational
force by the vector
𝐅 = −600𝐣
The ramp is inclined at 30°. To find the force required to keep the boat from
rolling down the ramp, we can project F onto a unit vector v in the direction of
the ramp, as follows.
𝐯 = (cos 30°)𝐢 + (sin 30°)𝐣 =
1
√3
𝐢+ 𝐣
2
2
Therefore the projection of F onto v is as follows. Remember
that ‖𝐯‖ = 1 if it v is a unit vector.
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Note: In this example,
we will refrain from
using our calculators
and therefore practice
our “unit circle”
knowledge to evaluate
our trigonometric
functions.
𝐰1 = proj𝐯 𝐅 = (
𝐅∙𝐯
) 𝐯 = (𝐅 ∙ 𝐯)𝐯
‖𝐯‖2
It’s easiest to find (𝐅 ∙ 𝐯) first using our Dot Product knowledge.
1
√3
(𝐅 ∙ 𝐯) = (−600𝐣) ∙ ( 𝐢 + 𝐣)
2
2
= (0 ∙
1
√3
) + (−600 ∙ )
2
2
= 0 − 300
= −300
So now we can substitute this value
(𝐅 ∙ 𝐯)𝐯 = −300 (
1
√3
𝐢 + 𝐣)
2
2
The magnitude of this force is 300, and therefore a force of 300 pounds is
required to keep the boat from rolling down the hill.
Finding Work
⃑⃑⃑⃑⃑ is
The work W done by a force F as its point of application moves along the vector 𝑃𝑄
given by
𝑾 = 𝐅 ∙ ⃑⃑⃑⃑⃑⃑
𝑷𝑸
Ex5. Find the work done by a force F of 12 pounds acting in the direction ⟨2,2⟩ in
moving an object 7 feet from (0, 0) to (7, 0).
Solution: The work done by a force F moving an object from P to Q is
⃑⃑⃑⃑⃑ .
𝑾 = 𝐅 ∙ 𝑃𝑄
If F has a magnitude 12 and acts in the direction ⟨2,2⟩, then the following is true
⟨2,2⟩
12
12
12
⟨2,2⟩ =
⟨2,2⟩ =
⟨2,2⟩
𝐅 = 12
=
‖⟨2,2⟩‖ ‖⟨2,2⟩‖
2√2
√22 + 22
⃑⃑⃑⃑⃑ = ⟨7,0⟩.
Since the object is being moved from point (0, 0) to (7, 0), then PQ
𝐖 = 𝐅 ∙ ⃑⃑⃑⃑⃑
𝐏𝐐
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=(
12
2√2
⟨2,2⟩) (⟨7,0⟩)
Use the Dot Product to find ⟨2,2⟩ ∙ ⟨7,0⟩.
⟨2,2⟩ ∙ ⟨7,0⟩ = 2 ∙ 7 + 2 ∙ 0 = 14 + 0 = 14
Therefore, we have
12
2√2
∙ 14 ≈ 59.397 foot-pounds.
Geometric Applications: Area
For applications of 3D vectors, we will be calculating the area of figures in the 3D plane.
Here in this example, we will find the area of a parallelogram.
Ex6. Find the area of a parallelogram with vertices 𝐴(5, 2, 0), 𝐵(2, 6, 1), 𝐶(2, 4, 7), and
𝐷(5, 0, 6).
Solution: Let’s first sketch these points in the xyz plane.
The area formula for a parallelogram is 𝐴 = 𝑙 ∙ ℎ. In this parallelogram, the
⃑⃑⃑⃑⃑ (or you could choose 𝐵𝐶
⃑⃑⃑⃑⃑ ) and the height is 𝐴𝐵
⃑⃑⃑⃑⃑ (or you
length is the vector 𝐴𝐷
⃑⃑⃑⃑⃑ ).
could choose 𝐶𝐷
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⃑⃑⃑⃑⃑ × ⃑⃑⃑⃑⃑
Therefore the area of the parallelogram will be 𝐴 = ‖𝐴𝐷
𝐴𝐵 ‖.
Step 1: Find ⃑⃑⃑⃑⃑
𝐴𝐷 and ⃑⃑⃑⃑⃑
𝐴𝐵 .
⃑⃑⃑⃑⃑ = (5 − 5)𝐢 + (0 − 2)𝐣 + (6 − 0)𝐤 = 0𝐢 − 2𝐣 + 6𝐤
𝐴𝐷
⃑⃑⃑⃑⃑ = (2 − 5)𝐢 + (6 − 2)𝐣 + (1 − 0)𝐤 = −3𝐢 + 4𝐣 + 1𝐤
𝐴𝐵
Step 2: Find ⃑⃑⃑⃑⃑
𝐴𝐷 × ⃑⃑⃑⃑⃑
𝐴𝐵 .
𝐢
𝐣 𝐤
⃑⃑⃑⃑⃑ × 𝐴𝐵
⃑⃑⃑⃑⃑ = | 0 −2 6|
𝐴𝐷
−3 4 1
0 6
0 −2
−2 6
=|
|𝐢 − |
|𝐣 + |
|𝐤
−3 1
−3 4
4 1
Using our TI Graphing Calculator (see Lesson 3) we have…
⃑⃑⃑⃑⃑ × 𝐴𝐵
⃑⃑⃑⃑⃑ = −26𝐢 − 18𝐣 − 6𝐤
Therefore, 𝐴𝐷
Don’t forget
the minus
sign for the j
component!
⃑⃑⃑⃑⃑ × ⃑⃑⃑⃑⃑
Step 3: Find 𝐴 = ‖𝐴𝐷
𝐴𝐵 ‖.
⃑⃑⃑⃑⃑ × ⃑⃑⃑⃑⃑
‖𝐴𝐷
𝐴𝐵 ‖ = √(−26)2 + (−18)2 + (−6)2 = √1036 ≈ 32.187
So the area of this parallelogram is approximately 32.187 square units.
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Finding the Area of Any Figure
We can find the area of any shape in the 3D plane. First, simply state your area formula
to determine which sides of the figure you will need to find.
Second, find the vectors of the corresponding vertices necessary for your sides.
Third, find the length (or magnitude) of each of your vectors (sides). Use these values in
your area formula.
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