Vectors

A Geometric View of Vectors

A vector is a specific quantity drawn as a line segment with an arrowhead at one end. It has an initial point, where it begins, and a terminal point, where it ends. A vector is defined by its magnitude, or the length of the line, and its direction, indicated by an arrowhead at the terminal point. Thus, a vector is a directed line segment. There are various symbols that distinguish vectors from other quantities:

Lower case, boldfaced type, with or without an arrow on top such as $v,$ $u,$ $w,$ $\vec{v},$ $\vec{u}, \vec{w} .$
Given initial point $P$ and terminal point $Q,$ a vector can be represented as $\vec{P Q} .$ The arrowhead on top is what indicates that it is not just a line, but a directed line segment.
Given an initial point of $(0, 0)$ and terminal point $(a b),$ a vector may be represented as $〈 a, b 〉 .$

This last symbol $〈 a, b 〉$ has special significance. It is called the standard position. The position vector has an initial point $(0, 0)$ and a terminal point $(a, b) .$ To change any vector into the position vector, we think about the change in the x-coordinates and the change in the y-coordinates. Thus, if the initial point of a vector $\vec{C D}$ is $C (x_{1}, y_{1})$ and the terminal point is $D (x_{2}, y_{2}),$ then the position vector is found by calculating

\begin{array}{l} \vec{A B} = 〈 x_{2} - x_{1}, y_{2} - y_{1} 〉 \\ = 〈 a, b 〉 \end{array}

In Figure 2, we see the original vector $\vec{C D}$ and the position vector $\vec{A B} .$

Plot of the original vector CD in blue and the position vector AB in orange extending from the origin. — Figure 2

Example 1

Find the Position Vector

Consider the vector whose initial point is $P (2, 3)$ and terminal point is $Q (6, 4) .$ Find the position vector.

Solution

The position vector is found by subtracting one x-coordinate from the other x-coordinate, and one y-coordinate from the other y-coordinate. Thus

\begin{array}{l} v = 〈 6 - 2, 4 - 3 〉 \\ = 〈 4, 1 〉 \end{array}

The position vector begins at $(0, 0)$ and terminates at $(4, 1) .$ The graphs of both vectors are shown in Figure 3.

Plot of the original vector in blue and the position vector in orange extending from the origin. — Figure 3

We see that the position vector is $〈 4, 1 〉 .$

Example 2

Drawing a Vector with the Given Criteria and Its Equivalent Position Vector

Find the position vector given that vector $v$ has an initial point at $(- 3, 2)$ and a terminal point at $(4, 5),$ then graph both vectors in the same plane.

Solution

The position vector is found using the following calculation:

\begin{array}{l} v = 〈 4 - (- 3), 5 - 2 〉 \\ = 〈 7, 3 〉 \end{array}

Thus, the position vector begins at $(0, 0)$ and terminates at $(7, 3) .$ See Figure 4.

Plot of the two given vectors their same position vector. — Figure 4

Finding Magnitude and Direction

To work with a vector, we need to be able to find its magnitude and its direction. We find its magnitude using the Pythagorean Theorem or the distance formula, and we find its direction using the inverse tangent function.

Example 3

Finding the Magnitude and Direction of a Vector

Find the magnitude and direction of the vector with initial point $P (- 8, 1)$ and terminal point $Q (- 2, - 5) .$ Draw the vector.

Solution

First, find the position vector.

\begin{array}{l} u = 〈 −2, - (−8), −5 −1 〉 \\ = 〈 6, - 6 〉 \end{array}

We use the Pythagorean Theorem to find the magnitude.

\begin{array}{l} | u | = \sqrt{{(6)}^{2} + {(- 6)}^{2}} \\ = \sqrt{72} \\ = 6 \sqrt{2} \end{array}

The direction is given as

\begin{array}{l} \tan θ = \frac{−6}{6} = −1 \Rightarrow θ = \tan^{−1} (−1) \\ = - 45° \end{array}

However, the angle terminates in the fourth quadrant, so we add 360° to obtain a positive angle. Thus, $- 45° + 360° = 315° .$ See Figure 6.

Plot of the position vector extending into Q4 from the origin with the magnitude 6rad2. — Figure 6

Example 4

Showing That Two Vectors Are Equal

Show that vector v with initial point at $(5, −3)$ and terminal point at $(−1, 2)$ is equal to vector u with initial point at $(−1, −3)$ and terminal point at $(−7, 2) .$ Draw the position vector on the same grid as v and u. Next, find the magnitude and direction of each vector.

Solution

As shown in Figure 7, draw the vector $v$ starting at initial $(5, −3)$ and terminal point $(−1, 2) .$ Draw the vector $u$ with initial point $(−1, −3)$ and terminal point $(−7, 2) .$ Find the standard position for each.

Next, find and sketch the position vector for v and u. We have

\begin{array}{l} v = 〈 −1 - 5, 2 - (- 3) 〉 \\ = 〈 −6, 5 〉 \\ u = 〈 −7 - (−1), 2 - (−3) 〉 \\ = 〈 −6, 5 〉 \end{array}

Since the position vectors are the same, v and u are the same.

An alternative way to check for vector equality is to show that the magnitude and direction are the same for both vectors. To show that the magnitudes are equal, use the Pythagorean Theorem.

\begin{array}{l} | v | = \sqrt{{(−1 - 5)}^{2} + {(2 - (−3))}^{2}} \\ = \sqrt{{(−6)}^{2} + {(5)}^{2}} \\ = \sqrt{36 + 25} \\ = \sqrt{61} \\ | u | = \sqrt{{(−7 - (−1))}^{2} + {(2 - (−3))}^{2}} \\ = \sqrt{{(−6)}^{2} + {(5)}^{2}} \\ = \sqrt{36 + 25} \\ = \sqrt{61} \end{array}

As the magnitudes are equal, we now need to verify the direction. Using the tangent function with the position vector gives

\begin{array}{l} \tan θ = - \frac{5}{6} \Rightarrow θ = \tan^{- 1} (- \frac{5}{6}) \\ = - 39.8° \end{array}

However, we can see that the position vector terminates in the second quadrant, so we add $180° .$ Thus, the direction is $- 39.8° + 180° = 140.2° .$

Performing Vector Addition and Scalar Multiplication

Now that we understand the properties of vectors, we can perform operations involving them. While it is convenient to think of the vector $u$ $= 〈 x, y 〉$ as an arrow or directed line segment from the origin to the point $(x, y),$ vectors can be situated anywhere in the plane. The sum of two vectors u and v, or vector addition, produces a third vector u + v, the resultant vector.

To find u + v, we first draw the vector u, and from the terminal end of u, we drawn the vector v. In other words, we have the initial point of v meet the terminal end of u. This position corresponds to the notion that we move along the first vector and then, from its terminal point, we move along the second vector. The sum u + v is the resultant vector because it results from addition or subtraction of two vectors. The resultant vector travels directly from the beginning of u to the end of v in a straight path, as shown in Figure 8.

Diagrams of vector addition and subtraction. — Figure 8

Vector subtraction is similar to vector addition. To find u − v, view it as u + (−v). Adding −v is reversing direction of v and adding it to the end of u. The new vector begins at the start of u and stops at the end point of −v. See Figure 9 for a visual that compares vector addition and vector subtraction using parallelograms.

Showing vector addition and subtraction with parallelograms. For addition, the base is u, the side is v, the diagonal connecting the start of the base to the end of the side is u+v. For subtraction, thetop is u, the side is -v, and the diagonal connecting the start of the top to the end of the side is u-v. — Figure 9

Example 5

Adding and Subtracting Vectors

Given $u$ $= 〈 3, - 2 〉$ and $v$ $= 〈 −1, 4 〉,$ find two new vectors u + v, and u − v.

Solution

To find the sum of two vectors, we add the components. Thus,

\begin{array}{l} u + v = 〈 3, - 2 〉 + 〈 - 1, 4 〉 \\ = 〈 3 + (- 1), - 2 + 4 〉 \\ = 〈 2, 2 〉 \end{array}

See Figure 10(a).

To find the difference of two vectors, add the negative components of $v$ to $u .$ Thus,

\begin{array}{l} u + (- v) = 〈 3, - 2 〉 + 〈 1, - 4 〉 \\ = 〈 3 + 1, - 2 + (- 4) 〉 \\ = 〈 4, - 6 〉 \end{array}

See Figure 10(b).

Further diagrams of vector addition and subtraction. — Figure 10 (a) Sum of two vectors (b) Difference of two vectors

Multiplying By a Scalar

While adding and subtracting vectors gives us a new vector with a different magnitude and direction, the process of multiplying a vector by a scalar, a constant, changes only the magnitude of the vector or the length of the line. Scalar multiplication has no effect on the direction unless the scalar is negative, in which case the direction of the resulting vector is opposite the direction of the original vector.

Example 6

Performing Scalar Multiplication

Given vector $v$ $= 〈 3, 1 〉,$ find 3v, $\frac{1}{2}$ $v,$ and −v.

Solution

See Figure 11 for a geometric interpretation. If $v$ $= 〈 3, 1 〉,$ then

\begin{array}{l} 3 v = 〈 3 \cdot 3, 3 \cdot 1 〉 \\ = 〈 9, 3 〉 \\ \frac{1}{2} v = 〈 \frac{1}{2} \cdot 3, \frac{1}{2} \cdot 1 〉 \\ = 〈 \frac{3}{2}, \frac{1}{2} 〉 \\ - v = 〈 −3, −1 〉 \end{array}

Showing the effect of scaling a vector: 3x, 1x, .5x, and -1x. The 3x is three times as long, the 1x stays the same, the .5x halves the length, and the -1x reverses the direction of the vector but keeps the length the same. The rest keep the same direction; only the magnitude changes. — Figure 11

Analysis

Notice that the vector 3v is three times the length of v, $\frac{1}{2}$ $v$ is half the length of v, and –v is the same length of v, but in the opposite direction.

Example 7

Using Vector Addition and Scalar Multiplication to Find a New Vector

Given $u$ $= 〈 3, - 2 〉$ and $v$ $= 〈 - 1, 4 〉,$ find a new vector w = 3u + 2v.

Solution

First, we must multiply each vector by the scalar.

\begin{array}{l} 3 u = 3 〈 3, - 2 〉 \\ = 〈 9, - 6 〉 \\ 2 v = 2 〈 - 1, 4 〉 \\ = 〈 - 2, 8 〉 \end{array}

Then, add the two together.

\begin{array}{l} w = 3 u + 2 v \\ = 〈 9, - 6 〉 + 〈 - 2, 8 〉 \\ = 〈 9 - 2, - 6 + 8 〉 \\ = 〈 7, 2 〉 \end{array}

So, $w$ $= 〈 7, 2 〉 .$

Finding Component Form

In some applications involving vectors, it is helpful for us to be able to break a vector down into its components. Vectors are comprised of two components: the horizontal component is the $x$ direction, and the vertical component is the $y$ direction. For example, we can see in the graph in Figure 12 that the position vector $〈 2, 3 〉$ comes from adding the vectors v₁ and v₂. We have v₁ with initial point $(0, 0)$ and terminal point $(2, 0) .$

\begin{array}{l} v_{1} = 〈 2 - 0, 0 - 0 〉 \\ = 〈 2, 0 〉 \end{array}

We also have v₂ with initial point $(0, 0)$ and terminal point $(0, 3) .$

\begin{array}{l} v_{2} = 〈 0 - 0, 3 - 0 〉 \\ = 〈 0, 3 〉 \end{array}

Therefore, the position vector is

\begin{array}{l} v = 〈 2 + 0, 3 + 0 〉 \\ = 〈 2, 3 〉 \end{array}

Using the Pythagorean Theorem, the magnitude of v₁ is 2, and the magnitude of v₂ is 3. To find the magnitude of v, use the formula with the position vector.

\begin{array}{l} | v | = \sqrt{| v_{1} |^{2} + | v_{2} |^{2}} \\ \begin{array}{l} = \sqrt{2^{2} + 3^{2}} \\ = \sqrt{13} \end{array} \end{array}

The magnitude of v is $\sqrt{13} .$ To find the direction, we use the tangent function $\tan θ = \frac{y}{x} .$

\begin{array}{l} \tan θ = \frac{|v_{2}|}{|v_{1}|} \\ \tan θ = \frac{3}{2} \\ θ = \tan^{- 1} (\frac{3}{2}) = 56.3° \end{array}

Diagram of a vector in root position with its horizontal and vertical components. — Figure 12

Thus, the magnitude of $v$ is $\sqrt{13}$ and the direction is ${56.3}^{\circ}$ off the horizontal.

Example 8

Finding the Components of the Vector

Find the components of the vector $v$ with initial point $(3, 2)$ and terminal point $(7, 4) .$

Solution

First find the standard position.

\begin{array}{l} v = 〈 7 - 3, 4 - 2 〉 \\ = 〈 4, 2 〉 \end{array}

See the illustration in Figure 13.

Diagram of a vector in root position with its horizontal (4,0) and vertical (0,2) components. — Figure 13

The horizontal component is $v_{1}$ $= 〈 4, 0 〉$ and the vertical component is $v_{2}$ $= 〈 0, 2 〉.$

Finding the Unit Vector in the Direction of v

In addition to finding a vector’s components, it is also useful in solving problems to find a vector in the same direction as the given vector, but of magnitude 1. We call a vector with a magnitude of 1 a unit vector. We can then preserve the direction of the original vector while simplifying calculations.

Unit vectors are defined in terms of components. The horizontal unit vector is written as $i$ $= 〈 1, 0 〉$ and is directed along the positive horizontal axis. The vertical unit vector is written as $j$ $= 〈 0, 1 〉$ and is directed along the positive vertical axis. See Figure 14.

Plot showing the unit vectors i=91,0) and j=(0,1) — Figure 14

Example 9

Finding the Unit Vector in the Direction of v

Find a unit vector in the same direction as $v$ $= 〈 −5, 12 〉.$

Solution

First, we will find the magnitude.

\begin{array}{l} | v | = \sqrt{{(- 5)}^{2} + {(12)}^{2}} \\ = \sqrt{25 + 144} \\ = \sqrt{169} \\ = 13 \end{array}

Then we divide each component by $| v |,$ which gives a unit vector in the same direction as v:

\frac{v}{| v |} = - \frac{5}{13} i + \frac{12}{13} j

or, in component form

\frac{v}{| v |} = 〈 - \frac{5}{13}, \frac{12}{13} 〉

See Figure 15.

Plot showing the unit vector (-5/13, 12/13) in the direction of (-5, 12) — Figure 15

Verify that the magnitude of the unit vector equals 1. The magnitude of $- \frac{5}{13} i + \frac{12}{13} j$ is given as

\begin{array}{l} \sqrt{{(- \frac{5}{13})}^{2} + {(\frac{12}{13})}^{2}} = \sqrt{\frac{25}{169} + \frac{144}{169}} \\ = \sqrt{\frac{169}{169}} = 1 \end{array}

The vector u $= \frac{5}{13}$ i $+ \frac{12}{13}$ j is the unit vector in the same direction as v $= 〈 - 5, 12 〉 .$

Performing Operations with Vectors in Terms of i and j

So far, we have investigated the basics of vectors: magnitude and direction, vector addition and subtraction, scalar multiplication, the components of vectors, and the representation of vectors geometrically. Now that we are familiar with the general strategies used in working with vectors, we will represent vectors in rectangular coordinates in terms of i and j.

a general note label

Vectors in the Rectangular Plane

Given a vector $v$ with initial point $P = (x_{1}, y_{1})$ and terminal point $Q = (x_{2}, y_{2}),$ v is written as

v = (x_{2} - x_{1}) i + (y_{2} - y_{1}) j

The position vector from $(0, 0)$ to $(a, b),$ where $(x_{2} - x_{1}) = a$ and $(y_{2} - y_{1}) = b,$ is written as v = ai + bj. This vector sum is called a linear combination of the vectors i and j.

The magnitude of v = ai + bj is given as $| v | = \sqrt{a^{2} + b^{2}} .$ See Figure 16.

Figure 16

Example 10

Writing a Vector in Terms of i and j

Given a vector $v$ with initial point $P = (2, −6)$ and terminal point $Q = (−6, 6),$ write the vector in terms of $i$ and $j .$

Solution

Begin by writing the general form of the vector. Then replace the coordinates with the given values.

\begin{array}{l} v = (x_{2} - x_{1}) i + (y_{2} - y_{1}) j \\ = (- 6 - 2) i + (6 - (- 6)) j \\ = - 8 i + 12 j \end{array}

Example 11

Writing a Vector in Terms of i and j Using Initial and Terminal Points

Given initial point $P_{1} = (- 1, 3)$ and terminal point $P_{2} = (2, 7),$ write the vector $v$ in terms of $i$ and $j .$

Solution

Begin by writing the general form of the vector. Then replace the coordinates with the given values.

\begin{array}{l} v = (x_{2} - x_{1}) i + (y_{2} - y_{1}) j \\ v = (2 - (- 1)) i + (7 - 3) j \\ = 3 i + 4 j \end{array}

Performing Operations on Vectors in Terms of i and j

When vectors are written in terms of $i$ and $j,$ we can carry out addition, subtraction, and scalar multiplication by performing operations on corresponding components.

Example 12

Finding the Sum of the Vectors

Find the sum of $v_{1} = 2 i - 3 j$ and $v_{2} = 4 i + 5 j .$

Solution

According to the formula, we have

\begin{array}{l} v_{1} + v_{2} = (2 + 4) i + (- 3 + 5) j \\ = 6 i + 2 j \end{array}

Calculating the Component Form of a Vector: Direction

We have seen how to draw vectors according to their initial and terminal points and how to find the position vector. We have also examined notation for vectors drawn specifically in the Cartesian coordinate plane using $i and j .$ For any of these vectors, we can calculate the magnitude. Now, we want to combine the key points, and look further at the ideas of magnitude and direction.

Calculating direction follows the same straightforward process we used for polar coordinates. We find the direction of the vector by finding the angle to the horizontal. We do this by using the basic trigonometric identities, but with $| v |$ replacing $r .$

Example 13

Writing a Vector in Component Form When It Is Given in Magnitude and Direction Form

Given a vector with length 7 and an angle of 135°, write it in component form.

Solution

Using the conversion formulas $x = | v | \cos θ$ and $y = | v | \sin θ,$ we find that

\begin{array}{l} x = 7 \cos (135°) \\ = - \frac{7 \sqrt{2}}{2} \\ y = 7 \sin (135°) \\ = \frac{7 \sqrt{2}}{2} \end{array}

This vector can be written as $v = 7 \cos (135°) + 7 \sin (135°)$ or simplified as

v = - \frac{7 \sqrt{2}}{2} i + \frac{7 \sqrt{2}}{2} j

Finding the Dot Product of Two Vectors

As we discussed earlier in the section, scalar multiplication involves multiplying a vector by a scalar, and the result is a vector. As we have seen, multiplying a vector by a number is called scalar multiplication. If we multiply a vector by a vector, there are two possibilities: the dot product and the cross product. We will only examine the dot product here; you may encounter the cross product in more advanced mathematics courses.

The dot product of two vectors involves multiplying two vectors together, and the result is a scalar.

Example 14

Finding the Dot Product of Two Vectors

Find the dot product of $v = 〈 5, 12 〉$ and $u = 〈 −3, 4 〉 .$

Solution

Using the formula, we have

\begin{array}{l} v \cdot u = 〈 5, 12 〉 \cdot 〈 - 3, 4 〉 \\ = 5 \cdot (- 3) + 12 \cdot 4 \\ = - 15 + 48 \\ = 33 \end{array}

Example 15

Finding the Dot Product of Two Vectors and the Angle between Them

Find the dot product of v₁ = 5i + 2j and v₂ = 3i + 7j. Then, find the angle between the two vectors.

Solution

Finding the dot product, we multiply corresponding components.

\begin{array}{l} v_{1} \cdot v_{2} = 〈 5, 2 〉 \cdot 〈 3, 7 〉 \\ = 5 \cdot 3 + 2 \cdot 7 \\ = 15 + 14 \\ = 29 \end{array}

To find the angle between them, we use the formula $\cos θ = \frac{v}{| v |} \cdot \frac{u}{| u |} .$

\begin{array}{l} \frac{v}{| v |} \cdot \frac{u}{| u |} = 〈 \frac{5}{\sqrt{29}}, \frac{2}{\sqrt{29}} 〉 \cdot 〈 \frac{3}{\sqrt{58}}, \frac{7}{\sqrt{58}} 〉 \\ = \frac{5}{\sqrt{29}} \cdot \frac{3}{\sqrt{58}} + \frac{2}{\sqrt{29}} \cdot \frac{7}{\sqrt{58}} \\ = \frac{15}{\sqrt{1682}} + \frac{14}{\sqrt{1682}} = \frac{29}{\sqrt{1682}} \\ \begin{array}{l} = 0.707107 \\ \cos^{- 1} (0.707107) = 45° \end{array} \end{array}

See Figure 17.

Plot showing the two position vectors (3,7) and (5,2) and the 45 degree angle between them. — Figure 17

Example 16

Finding the Angle between Two Vectors

Find the angle between $u = 〈 - 3, 4 〉$ and $v = 〈 5, 12 〉 .$

Solution

Using the formula, we have

\begin{array}{l} θ = \cos^{- 1} (\frac{u}{| u |} \cdot \frac{v}{| v |}) \\ (\frac{u}{| u |} \cdot \frac{v}{| v |}) = \frac{- 3 i + 4 j}{5} \cdot \frac{5 i + 12 j}{13} \\ = (- \frac{3}{5} \cdot \frac{5}{13}) + (\frac{4}{5} \cdot \frac{12}{13}) \\ = - \frac{15}{65} + \frac{48}{65} \\ = \frac{33}{65} \\ θ = \cos^{- 1} (\frac{33}{65}) \\ = {59.5}^{\circ} \end{array}

See Figure 18.

Plot showing the two position vectors (-3,4) and (5,12) and the 59.5 degree angle between them. — Figure 18

Example 17

Finding Ground Speed and Bearing Using Vectors

We now have the tools to solve the problem we introduced in the opening of the section.

An airplane is flying at an airspeed of 200 miles per hour headed on a SE bearing of 140°. A north wind (from north to south) is blowing at 16.2 miles per hour. What are the ground speed and actual bearing of the plane? See Figure 19.

Image of a plan flying SE at 140 degrees and the north wind blowing. — Figure 19

Solution

The ground speed is represented by $x$ in the diagram, and we need to find the angle $α$ in order to calculate the adjusted bearing, which will be $140° + α .$

Notice in Figure 19, that angle $B C O$ must be equal to angle $A O C$ by the rule of alternating interior angles, so angle $B C O$ is 140°. We can find $x$ by the Law of Cosines:

\begin{array}{l} x^{2} = {(16.2)}^{2} + {(200)}^{2} - 2 (16.2) (200) \cos (140°) \\ x^{2} = 45, 226.41 \\ x = \sqrt{45, 226.41} \\ x = 212.7 \end{array}

The ground speed is approximately 213 miles per hour. Now we can calculate the bearing using the Law of Sines.

\begin{array}{l} \frac{\sin α}{16.2} = \frac{\sin (140°)}{212.7} \\ \sin α = \frac{16.2 \sin (140°)}{212.7} \\ = 0.04896 \\ \sin^{- 1} (0.04896) = 2.8° \end{array}

Therefore, the plane has a SE bearing of 140°+2.8°=142.8°. The ground speed is 212.7 miles per hour.

Key Concepts

The position vector has its initial point at the origin. See Example 1.
If the position vector is the same for two vectors, they are equal. See Example 2.
Vectors are defined by their magnitude and direction. See Example 3.
If two vectors have the same magnitude and direction, they are equal. See Example 4.
Vector addition and subtraction result in a new vector found by adding or subtracting corresponding elements. See Example 5.
Scalar multiplication is multiplying a vector by a constant. Only the magnitude changes; the direction stays the same. See Example 6 and Example 7.
Vectors are comprised of two components: the horizontal component along the positive x-axis, and the vertical component along the positive y-axis. See Example 8.
The unit vector in the same direction of any nonzero vector is found by dividing the vector by its magnitude.
The magnitude of a vector in the rectangular coordinate system is $| v | = \sqrt{a^{2} + b^{2}} .$ See Example 9.
In the rectangular coordinate system, unit vectors may be represented in terms of $i$ and $j$ where $i$ represents the horizontal component and $j$ represents the vertical component. Then, v = ai + bj is a scalar multiple of $v$ by real numbers $a and b .$ See Example 10 and Example 11.
Adding and subtracting vectors in terms of i and j consists of adding or subtracting corresponding coefficients of i and corresponding coefficients of j. See Example 12.
A vector v = ai + bj is written in terms of magnitude and direction as $v = | v | \cos θ i + | v | \sin θ j .$ See Example 13.
The dot product of two vectors is the product of the $i$ terms plus the product of the $j$ terms. See Example 14.
We can use the dot product to find the angle between two vectors. Example 15 and Example 16.
Dot products are useful for many types of physics applications. See Example 17.

Section Exercises

Verbal

Exercise 1

What are the characteristics of the letters that are commonly used to represent vectors?

Solution

lowercase, bold letter, usually $u, v, w$

Exercise 2

How is a vector more specific than a line segment?

Exercise 3

What are $i$ and $j,$ and what do they represent?

Solution

They are unit vectors. They are used to represent the horizontal and vertical components of a vector. They each have a magnitude of 1.

Exercise 4

What is component form?

Exercise 5

When a unit vector is expressed as $〈 a, b 〉,$ which letter is the coefficient of the $i$ and which the $j ?$

Solution

The first number always represents the coefficient of the $i,$ and the second represents the $j .$

Algebraic

Exercise 6

Given a vector with initial point $(5, 2)$ and terminal point $(- 1, - 3),$ find an equivalent vector whose initial point is $(0, 0) .$ Write the vector in component form $〈 a, b 〉 .$

Exercise 7

Given a vector with initial point $(- 4, 2)$ and terminal point $(3, - 3),$ find an equivalent vector whose initial point is $(0, 0) .$ Write the vector in component form $〈 a, b 〉 .$

Solution

$〈 7, - 5 〉$

Exercise 8

Given a vector with initial point $(7, - 1)$ and terminal point $(- 1, - 7),$ find an equivalent vector whose initial point is $(0, 0) .$ Write the vector in component form $〈 a, b 〉 .$

For the following exercises, determine whether the two vectors $u$ and $v$ are equal, where $u$ has an initial point $P_{1}$ and a terminal point $P_{2}$ and $v$ has an initial point $P_{3}$ and a terminal point $P_{4}$ .

Exercise 9

$P_{1} = (5, 1), P_{2} = (3, - 2), P_{3} = (- 1, 3),$ and $P_{4} = (9, - 4)$

Solution

not equal

Exercise 10

$P_{1} = (2, - 3), P_{2} = (5, 1), P_{3} = (6, - 1),$ and $P_{4} = (9, 3)$

Exercise 11

$P_{1} = (- 1, - 1), P_{2} = (- 4, 5), P_{3} = (- 10, 6),$ and $P_{4} = (- 13, 12)$

Solution

equal

Exercise 12

$P_{1} = (3, 7), P_{2} = (2, 1), P_{3} = (1, 2),$ and $P_{4} = (- 1, - 4)$

Exercise 13

$P_{1} = (8, 3), P_{2} = (6, 5), P_{3} = (11, 8),$ and $P_{4} = (9, 10)$

Solution

equal

Exercise 14

Given initial point $P_{1} = (- 3, 1)$ and terminal point $P_{2} = (5, 2),$ write the vector $v$ in terms of $i$ and $j .$

Exercise 15

Given initial point $P_{1} = (6, 0)$ and terminal point $P_{2} = (- 1, - 3),$ write the vector $v$ in terms of $i$ and $j .$

Solution

$- 7 i - 3 j$

For the following exercises, use the vectors u = i + 5j, v = −2i− 3j, and w = 4i − j.

Exercise 16

Find u + (v − w)

Exercise 17

Find 4v + 2u

Solution

$- 6 i - 2 j$

For the following exercises, use the given vectors to compute u + v, u − v, and 2u − 3v.

Exercise 18

$u = 〈 2, - 3 〉, v = 〈 1, 5 〉$

Exercise 19

$u = 〈 - 3, 4 〉, v = 〈 - 2, 1 〉$

Solution

$u + v = 〈 - 5, 5 〉, u - v = 〈 - 1, 3 〉, 2 u - 3 v = 〈 0, 5 〉$

Exercise 20

Let v = −4i + 3j. Find a vector that is half the length and points in the same direction as $v .$

Exercise 21

Let v = 5i + 2j. Find a vector that is twice the length and points in the opposite direction as $v .$

Solution

$- 10 i - 4 j$

For the following exercises, find a unit vector in the same direction as the given vector.

Exercise 22

a = 3i + 4j

Exercise 23

b = −2i + 5j

Solution

$- \frac{2 \sqrt{29}}{29} i + \frac{5 \sqrt{29}}{29} j$

Exercise 24

c = 10i – j

Exercise 25

$d = - \frac{1}{3} i + \frac{5}{2} j$

Solution

$- \frac{2 \sqrt{229}}{229} i + \frac{15 \sqrt{229}}{229} j$

Exercise 26

u = 100i + 200j

Exercise 27

u = −14i + 2j

Solution

$- \frac{7 \sqrt{2}}{10} i + \frac{\sqrt{2}}{10} j$

For the following exercises, find the magnitude and direction of the vector, $0 \leq θ < 2 π .$

Exercise 28

$〈 0, 4 〉$

Exercise 29

$〈 6, 5 〉$

Solution

$| v | = 7.810, θ = 39.806 °$

Exercise 30

$〈 2, −5 〉$

Exercise 31

$〈 −4, −6 〉$

Solution

$| v | = 7.211, θ = 236.310°$

Exercise 32

Given u = 3i − 4j and v = −2i + 3j, calculate $u \cdot v .$

Exercise 33

Given u = −i − j and v = i + 5j, calculate $u \cdot v .$

Solution

$- 6$

Exercise 34

Given $u = 〈 - 2, 4 〉$ and $v = 〈 - 3, 1 〉,$ calculate $u \cdot v .$

Exercise 35

Given u $= 〈 - 1, 6 〉$ and v $= 〈 6, - 1 〉,$ calculate $u \cdot v .$

Solution

$- 12$

Graphical

For the following exercises, given $v,$ draw $v,$ 3v and $\frac{1}{2} v .$

Exercise 36

$〈 2, −1 〉$

Exercise 37

$〈 −1, 4 〉$

Solution

Three graphs illustrate scalar multiplication of a vector. The first graph shows vector v. The second shows vector 3v, which is three times longer than v. The third shows vector 1/2v, half the length of v.

Exercise 38

$〈 −3, −2 〉$

For the following exercises, use the vectors shown to sketch u + v, u − v, and 2u.

Exercise 39

Plot of vectors u and v extending from the same origin point. In terms of that point, u goes to (1,1) and v goes to (-1,2).

Solution

Plot of u+v, u-v, and 2u based on the above vectors. In relation to the same origin point, u+v goes to (0,3), u-v goes to (2,-1), and 2u goes to (2,2).

Exercise 40

Plot of vectors u and v extending from the same origin point. In terms of that point, u goes to (1,2) and v goes to (1,-1).

Exercise 41

Plot of vectors u and v located head to tail. Take u's start point as the origin. In terms of that, u goes from the origin to (3,-2), and v goes from (3,-2) to (2,-3)

Solution

Plot of vectors u+v, u-v, and 2u based on the above vectors.Given that u's start point was the origin, u+v starts at the origin and goes to (2,-3); u-v starts at the origin and goes to (4,-1); 2u goes from the origin to (6,-4).

For the following exercises, use the vectors shown to sketch 2u + v.

Exercise 42

Plot of the vectors u and v extending from the same point. Taking that base point as the origin, u goes from the origin to (3,1) and v goes from the origin to (2,-2).

Exercise 43

Plot of the vectors u and v extending from the same point. Taking that base point as the origin, u goes from the origin to (1,-2) and v goes from the origin to (-3,-2).

Solution

Plot of a single vector. Taking the start point of the vector as (0,0) from the above set up, the vector goes from the origin to (-1,-6).

For the following exercises, use the vectors shown to sketch u − 3v.

Exercise 44

Plot of the vectors u and v extending from the same point. Taking that base point as the origin, u goes from the origin to (-4,0) and v goes from the origin to (1,-1).

Exercise 45

Solution

Vector extending from the origin to (7,5), taking the base as the origin.

For the following exercises, write the vector shown in component form.

Exercise 46

Exercise 47

Insert figure(table) alt text: Vector going from the origin to (4,1).

Solution

$〈 4, 1 〉$

Exercise 48

Given initial point $P_{1} = (2, 1)$ and terminal point $P_{2} = (- 1, 2),$ write the vector $v$ in terms of $i$ and $j,$ then draw the vector on the graph.

Exercise 49

Given initial point $P_{1} = (4, - 1)$ and terminal point $P_{2} = (- 3, 2),$ write the vector $v$ in terms of $i$ and $j .$ Draw the points and the vector on the graph.

Solution

$v = - 7 i + 3 j$

Exercise 50

Given initial point $P_{1} = (3, 3)$ and terminal point $P_{2} = (- 3, 3),$ write the vector $v$ in terms of $i$ and $j .$ Draw the points and the vector on the graph.

Extensions

For the following exercises, use the given magnitude and direction in standard position, write the vector in component form.

Exercise 51

$| v | = 6, θ = 45 °$

Solution

$3 \sqrt{2} i + 3 \sqrt{2} j$

Exercise 52

$| v | = 8, θ = 220°$

Exercise 53

$| v | = 2, θ = 300°$

Solution

$i - \sqrt{3} j$

Exercise 54

$| v | = 5, θ = 135°$

Exercise 55

A 60-pound box is resting on a ramp that is inclined 12°. Rounding to the nearest tenth,

ⓐ Find the magnitude of the normal (perpendicular) component of the force.
ⓑ Find the magnitude of the component of the force that is parallel to the ramp.

Solution

ⓐ 58.7
ⓑ 12.5

Exercise 56

A 25-pound box is resting on a ramp that is inclined 8°. Rounding to the nearest tenth,

ⓐ Find the magnitude of the normal (perpendicular) component of the force.
ⓑ Find the magnitude of the component of the force that is parallel to the ramp.

Exercise 57

Find the magnitude of the horizontal and vertical components of a vector with magnitude 8 pounds pointed in a direction of 27° above the horizontal. Round to the nearest hundredth.

Solution

$x = 7.13$ pounds, $y = 3.63$ pounds

Exercise 58

Find the magnitude of the horizontal and vertical components of the vector with magnitude 4 pounds pointed in a direction of 127° above the horizontal. Round to the nearest hundredth.

Exercise 59

Find the magnitude of the horizontal and vertical components of a vector with magnitude 5 pounds pointed in a direction of 55° above the horizontal. Round to the nearest hundredth.

Solution

$x = 2.87$ pounds, $y = 4.10$ pounds

Exercise 60

Find the magnitude of the horizontal and vertical components of the vector with magnitude 1 pound pointed in a direction of 8° above the horizontal. Round to the nearest hundredth.

Real-World Applications

Exercise 61

A woman leaves home and walks 3 miles west, then 2 miles southwest. How far from home is she, and in what direction must she walk to head directly home?

Solution

4.635 miles, 17.764° N of E

Exercise 62

A boat leaves the marina and sails 6 miles north, then 2 miles northeast. How far from the marina is the boat, and in what direction must it sail to head directly back to the marina?

Exercise 63

A man starts walking from home and walks 4 miles east, 2 miles southeast, 5 miles south, 4 miles southwest, and 2 miles east. How far has he walked? If he walked straight home, how far would he have to walk?

Solution

17 miles. 10.318 miles

Exercise 64

A woman starts walking from home and walks 4 miles east, 7 miles southeast, 6 miles south, 5 miles southwest, and 3 miles east. How far has she walked? If she walked straight home, how far would she have to walk?

Exercise 65

A man starts walking from home and walks 3 miles at 20° north of west, then 5 miles at 10° west of south, then 4 miles at 15° north of east. If he walked straight home, how far would he have to the walk, and in what direction?

Solution

Distance: 2.868. Direction: 86.474° North of West, or 3.526° West of North

Exercise 66

A woman starts walking from home and walks 6 miles at 40° north of east, then 2 miles at 15° east of south, then 5 miles at 30° south of west. If she walked straight home, how far would she have to walk, and in what direction?

Exercise 67

An airplane is heading north at an airspeed of 600 km/hr, but there is a wind blowing from the southwest at 80 km/hr. How many degrees off course will the plane end up flying, and what is the plane’s speed relative to the ground?

Solution

4.924°. 659 km/hr

Exercise 68

An airplane is heading north at an airspeed of 500 km/hr, but there is a wind blowing from the northwest at 50 km/hr. How many degrees off course will the plane end up flying, and what is the plane’s speed relative to the ground?

Exercise 69

An airplane needs to head due north, but there is a wind blowing from the southwest at 60 km/hr. The plane flies with an airspeed of 550 km/hr. To end up flying due north, how many degrees west of north will the pilot need to fly the plane?

Solution

4.424°

Exercise 70

An airplane needs to head due north, but there is a wind blowing from the northwest at 80 km/hr. The plane flies with an airspeed of 500 km/hr. To end up flying due north, how many degrees west of north will the pilot need to fly the plane?

Exercise 71

As part of a video game, the point $(5, 7)$ is rotated counterclockwise about the origin through an angle of 35°. Find the new coordinates of this point.

Solution

$(0.081, 8.602)$

Exercise 72

As part of a video game, the point $(7, 3)$ is rotated counterclockwise about the origin through an angle of 40°. Find the new coordinates of this point.

Exercise 73

Two children are throwing a ball back and forth straight across the back seat of a car. The ball is being thrown 10 mph relative to the car, and the car is traveling 25 mph down the road. If one child doesn't catch the ball, and it flies out the window, in what direction does the ball fly (ignoring wind resistance)?

Solution

21.801°, relative to the car’s forward direction

Exercise 74

Two children are throwing a ball back and forth straight across the back seat of a car. The ball is being thrown 8 mph relative to the car, and the car is traveling 45 mph down the road. If one child doesn't catch the ball, and it flies out the window, in what direction does the ball fly (ignoring wind resistance)?

Exercise 75

A 50-pound object rests on a ramp that is inclined 19°. Find the magnitude of the components of the force parallel to and perpendicular to (normal) the ramp to the nearest tenth of a pound.

Solution

parallel: 16.28, perpendicular: 47.28 pounds

Exercise 76

Suppose a body has a force of 10 pounds acting on it to the right, 25 pounds acting on it upward, and 5 pounds acting on it 45° from the horizontal. What single force is the resultant force acting on the body?

Exercise 77

Suppose a body has a force of 10 pounds acting on it to the right, 25 pounds acting on it ─135° from the horizontal, and 5 pounds acting on it directed 150° from the horizontal. What single force is the resultant force acting on the body?

Solution

19.35 pounds, 231.54° from the horizontal

Exercise 78

The condition of equilibrium is when the sum of the forces acting on a body is the zero vector. Suppose a body has a force of 2 pounds acting on it to the right, 5 pounds acting on it upward, and 3 pounds acting on it 45° from the horizontal. What single force is needed to produce a state of equilibrium on the body?

Exercise 79

Suppose a body has a force of 3 pounds acting on it to the left, 4 pounds acting on it upward, and 2 pounds acting on it 30° from the horizontal. What single force is needed to produce a state of equilibrium on the body? Draw the vector.

Solution

5.1583 pounds, 75.8° from the horizontal

Chapter Review Exercises

Non-right Triangles: Law of Sines

For the following exercises, assume $α$ is opposite side $a, β$ is opposite side $b,$ and $γ$ is opposite side $c .$ Solve each triangle, if possible. Round each answer to the nearest tenth.

$β = 50°, a = 105, b = 45$

Solution

Not possible

$α = 43.1°, a = 184.2, b = 242.8$

Solve the triangle.

Triangle with standard labels. Angle A is 36 degrees with opposite side a unknown. Angle B is 24 degrees with opposite side b = 16. Angle C and side c are unknown.

Solution

$C = 120°, a = 23.1, c = 34.1$

Find the area of the triangle.

A triangle. One angle is 75 degrees with opposite side unknown. The adjacent sides to the 75 degree angle are 8 and 11.

A pilot is flying over a straight highway. He determines the angles of depression to two mileposts, 2.1 km apart, to be 25° and 49°, as shown in Figure 20. Find the distance of the plane from point $A$ and the elevation of the plane.

Diagram of a plane flying over a highway. It is to the left and above points A and B on the ground in that order. There is a horizontal line going through the plan parallel to the ground. The angle formed by the horizontal line, the plane, and the line from the plane to point B is 25 degrees. The angle formed by the horizontal line, the plane, and point A is 49 degrees. — Figure 20

Solution

distance of the plane from point $A :$ 2.2 km, elevation of the plane: 1.6 km

Non-right Triangles: Law of Cosines

Solve the triangle, rounding to the nearest tenth, assuming $α$ is opposite side $a, β$ is opposite side $b,$ and $γ$ is opposite side $c : a = 4, b = 6, c = 8.$

Solve the triangle in Figure 21, rounding to the nearest tenth.

A standardly labeled triangle. Angle A is 54 degrees with opposite side a unknown. Angle B is unknown with opposite side b=15. Angle C is unknown with opposite side C=13. — Figure 21

Solution

$b = 71.0°, C = 55.0°, a = 12.8$

Find the area of a triangle with sides of length 8.3, 6.6, and 9.1.

To find the distance between two cities, a satellite calculates the distances and angle shown in Figure 22 (not to scale). Find the distance between the cities. Round answers to the nearest tenth.

Diagram of a satellite above and to the right of two cities. The distance from the satellite to the closer city is 210 km. The distance from the satellite to the further city is 250 km. The angle formed by the closer city, the satellite, and the other city is 1.8 degrees. — Figure 22

Solution

40.6 km

Polar Coordinates

Plot the point with polar coordinates $(3, \frac{π}{6}) .$

Plot the point with polar coordinates $(5, - \frac{2 π}{3})$

Solution

Polar coordinate grid with a point plotted on the fifth concentric circle 2/3 the way between pi and 3pi/2 (closer to 3pi/2).

Convert $(6, - \frac{3 π}{4})$ to rectangular coordinates.

Convert $(- 2, \frac{3 π}{2})$ to rectangular coordinates.

Solution

$(0, 2)$

Convert $(7, - 2)$ to polar coordinates.

Convert $(- 9, - 4)$ to polar coordinates.

Solution

$(9.8489, 203.96°)$

For the following exercises, convert the given Cartesian equation to a polar equation.

$x = - 2$

$x^{2} + y^{2} = 64$

Solution

$r = 8$

$x^{2} + y^{2} = - 2 y$

For the following exercises, convert the given polar equation to a Cartesian equation.

$r = 7 cos θ$

Solution

$x^{2} + y^{2} = 7 x$

$r = \frac{- 2}{4 \cos θ + \sin θ}$

For the following exercises, convert to rectangular form and graph.

$θ = \frac{3 π}{4}$

Solution

$y = - x$

Plot of the function y=-x in rectangular coordinates.

$r = 5 \sec θ$

Polar Coordinates: Graphs

For the following exercises, test each equation for symmetry.

$r = 4 + 4 \sin θ$

Solution

symmetric with respect to the line $θ = \frac{π}{2}$

$r = 7$

Sketch a graph of the polar equation $r = 1 - 5 \sin θ .$ Label the axis intercepts.

Solution

Graph of the given polar equation - an inner loop limaçon.

Sketch a graph of the polar equation $r = 5 \sin (7 θ) .$

Sketch a graph of the polar equation $r = 3 - 3 \cos θ$

Solution

Graph of the given polar equation - a cardioid.

Polar Form of Complex Numbers

For the following exercises, find the absolute value of each complex number.

$- 2 + 6 i$

$4 - 3 i$

Solution

5

Write the complex number in polar form.

$5 + 9 i$

$\frac{1}{2} - \frac{\sqrt{3}}{2} i$

Solution

$cis (- \frac{π}{3})$

For the following exercises, convert the complex number from polar to rectangular form.

$z = 5 cis (\frac{5 π}{6})$

$z = 3 cis (40°)$

Solution

$2.3 + 1.9 i$

For the following exercises, find the product $z_{1} z_{2}$ in polar form.

$z_{1} = 2 cis (89°)$

$z_{2} = 5 cis (23°)$

$z_{1} = 10 cis (\frac{π}{6})$

$z_{2} = 6 cis (\frac{π}{3})$

Solution

$60 cis (\frac{π}{2})$

For the following exercises, find the quotient $\frac{z_{1}}{z_{2}}$ in polar form.

$z_{1} = 12 cis (55°)$

$z_{2} = 3 cis (18°)$

$z_{1} = 27 cis (\frac{5 π}{3})$

$z_{2} = 9 cis (\frac{π}{3})$

Solution

$3 cis (\frac{4 π}{3})$

For the following exercises, find the powers of each complex number in polar form.

Find $z^{4}$ when $z = 2 cis (70°)$

Find $z^{2}$ when $z = 5 cis (\frac{3 π}{4})$

Solution

$25 cis (\frac{3 π}{2})$

For the following exercises, evaluate each root.

Evaluate the cube root of $z$ when $z = 64 cis (210°) .$

Evaluate the square root of $z$ when $z = 25 cis (\frac{3 π}{2}) .$

Solution

$5 cis (\frac{3 π}{4}), 5 cis (\frac{7 π}{4})$

For the following exercises, plot the complex number in the complex plane.

$6 - 2 i$

$- 1 + 3 i$

Solution

Plot of -1 + 3i in the complex plane (-1 along the real axis, 3 along the imaginary).

Parametric Equations

For the following exercises, eliminate the parameter $t$ to rewrite the parametric equation as a Cartesian equation.

${\begin{array}{l} x (t) = 3 t - 1 \\ y (t) = \sqrt{t} \end{array}$

${\begin{cases} x (t) = - \cos t \\ y (t) = 2 \sin^{2} t \end{cases}$

Solution

$x^{2} + \frac{1}{2} y = 1$

Parameterize (write a parametric equation for) each Cartesian equation by using $x (t) = a \cos t$ and $y (t) = b \sin t$ for $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1.$

Parameterize the line from $(- 2, 3)$ to $(4, 7)$ so that the line is at $(- 2, 3)$ at $t = 0$ and $(4, 7)$ at $t = 1.$

Solution

${\begin{array}{l} x (t) = - 2 + 6 t \\ y (t) = 3 + 4 t \end{array}$

Parametric Equations: Graphs

For the following exercises, make a table of values for each set of parametric equations, graph the equations, and include an orientation; then write the Cartesian equation.

${\begin{array}{l} x (t) = 3 t^{2} \\ y (t) = 2 t - 1 \end{array}$

${\begin{array}{l} x (t) = e^{t} \\ y (t) = - 2 e^{5 t} \end{array}$

Solution

$y = - 2 x^{5}$

${\begin{array}{l} x (t) = 3 \cos t \\ y (t) = 2 \sin t \end{array}$

A ball is launched with an initial velocity of 80 feet per second at an angle of 40° to the horizontal. The ball is released at a height of 4 feet above the ground.

ⓐ Find the parametric equations to model the path of the ball.
ⓑ Where is the ball after 3 seconds?

Solution

${\begin{cases} x (t) = (80 \cos (40°)) t \\ y (t) = - 16 t^{2} + (80 \sin (40°)) t + 4 \end{cases}$
The ball is 14 feet high and 184 feet from where it was launched.
3.3 seconds

For the following exercises, determine whether the two vectors, $u$ and $v,$ are equal, where $u$ has an initial point $P_{1}$ and a terminal point $P_{2},$ and $v$ has an initial point $P_{3}$ and a terminal point $P_{4} .$

$P_{1} = (- 1, 4), P_{2} = (3, 1), P_{3} = (5, 5)$ and $P_{4} = (9, 2)$

$P_{1} = (6, 11), P_{2} = (- 2, 8), P_{3} = (0, - 1)$ and $P_{4} = (- 8, 2)$

Solution

not equal

For the following exercises, use the vectors $u = 2 i - j, v = 4 i - 3 j,$ and $w = - 2 i + 5 j$ to evaluate the expression.

u − v

2v − u + w

Solution

4i

For the following exercises, find a unit vector in the same direction as the given vector.

a = 8i − 6j

b = −3i − j

Solution

$- \frac{3 \sqrt{10}}{10}$ i $- \frac{\sqrt{10}}{10}$ j

For the following exercises, find the magnitude and direction of the vector.

$〈 6, −2 〉$

$〈 −3, −3 〉$

Solution

Magnitude: $3 \sqrt{2},$ Direction: $225°$

For the following exercises, calculate $u \cdot v .$

u = −2i + j and v = 3i + 7j

u = i + 4j and v = 4i + 3j

Solution

$16$

Given v $= 〈 −3, 4 〉$ draw v, 2v, and $\frac{1}{2}$ v.

Given the vectors shown in Figure 23, sketch u + v, u − v and 3v.

Diagram of vectors v, 2v, and 1/2 v. The 2v vector is in the same direction as v but has twice the magnitude. The 1/2 v vector is in the same direction as v but has half the magnitude. — Figure 23

Solution

Diagram of vectors u and v. Taking u's starting point as the origin, u goes from the origin to (4,1), and v goes from (4,1) to (6,0).

Given initial point $P_{1} = (3, 2)$ and terminal point $P_{2} = (- 5, - 1),$ write the vector $v$ in terms of $i$ and $j .$ Draw the points and the vector on the graph.

Practice Test

Assume $α$ is opposite side $a, β$ is opposite side $b,$ and $γ$ is opposite side $c .$ Solve the triangle, if possible, and round each answer to the nearest tenth, given $β = 68°, b = 21, c = 16.$

Solution

$α = 67.1°, γ = 44.9°, a = 20.9$

Find the area of the triangle in Figure 24. Round each answer to the nearest tenth.

A triangle. One angle is 60 degrees with opposite side 6.25. The other two sides are 5 and 7. — Figure 24

A pilot flies in a straight path for 2 hours. He then makes a course correction, heading 15° to the right of his original course, and flies 1 hour in the new direction. If he maintains a constant speed of 575 miles per hour, how far is he from his starting position?

Solution

$1712 miles$

Convert $(2, 2)$ to polar coordinates, and then plot the point.

Convert $(2, \frac{π}{3})$ to rectangular coordinates.

Solution

$(1, \sqrt{3})$

Convert the polar equation to a Cartesian equation: $x^{2} + y^{2} = 5 y.$

Convert to rectangular form and graph: $r = - 3 \csc θ .$

Solution

$y = - 3$

Plot of the given equation in rectangular form - line y=-3.

Test the equation for symmetry: $r = - 4 \sin (2 θ).$

Graph $r = 3 + 3 \cos θ .$

Solution

Graph of the given equations - a cardioid.

Graph $r = 3 - 5 sin θ .$

Find the absolute value of the complex number $5 - 9 i .$

Solution

$\sqrt{106}$

Write the complex number in polar form: $4 + i .$

Convert the complex number from polar to rectangular form: $z = 5 cis (\frac{2 π}{3}) .$

Solution

$\frac{- 5}{2} + i \frac{5 \sqrt{3}}{2}$

Given $z_{1} = 8 cis (36°)$ and $z_{2} = 2 cis (15°),$ evaluate each expression.

$z_{1} z_{2}$

$\frac{z_{1}}{z_{2}}$

Solution

$4 cis (21°)$

${(z_{2})}^{3}$

$\sqrt{z_{1}}$

Solution

$2 \sqrt{2} cis (18°), 2 \sqrt{2} cis (198°)$

Plot the complex number $−5 - i$ in the complex plane.

Eliminate the parameter $t$ to rewrite the following parametric equations as a Cartesian equation: ${\begin{array}{l} x (t) = t + 1 \\ y (t) = 2 t^{2} \end{array} .$

Solution

$y = 2 {(x - 1)}^{2}$

Parameterize (write a parametric equation for) the following Cartesian equation by using $x (t) = a \cos t$ and $y (t) = b \sin t :$ $\frac{x^{2}}{36} + \frac{y^{2}}{100} = 1.$

Graph the set of parametric equations and find the Cartesian equation: ${\begin{array}{l} x (t) = - 2 \sin t \\ y (t) = 5 \cos t \end{array} .$

Solution

Graph of the given equations - a vertical ellipse.

A ball is launched with an initial velocity of 95 feet per second at an angle of 52° to the horizontal. The ball is released at a height of 3.5 feet above the ground.

ⓐFind the parametric equations to model the path of the ball.
ⓑWhere is the ball after 2 seconds?

For the following exercises, use the vectors u = i − 3j and v = 2i + 3j.

Find 2u − 3v.

Solution

−4i − 15j

Calculate $u \cdot v .$

Find a unit vector in the same direction as $v .$

Solution

$\frac{2 \sqrt{13}}{13} i + \frac{3 \sqrt{13}}{13} j$

Given vector $v$ has an initial point $P_{1} = (2, 2)$ and terminal point $P_{2} = (- 1, 0),$ write the vector $v$ in terms of $i$ and $j .$ On the graph, draw $v,$ and $- v .$