Vectors

Learning Objectives

In this section you will:

• View vectors geometrically.
• Find magnitude and direction.
• Perform vector addition and scalar multiplication.
• Find the component form of a vector.
• Find the unit vector in the direction of  v𝑣.
• Perform operations with vectors in terms of  i𝑖  and  j𝑗.
• Find the dot product of two vectors.

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, as shown in Figure 1. What are the ground speed and actual bearing of the plane?

Figure 1

Ground speed refers to the speed of a plane relative to the ground. Airspeed refers to the speed a plane can travel relative to its surrounding air mass. These two quantities are not the same because of the effect of wind. In an earlier section, we used triangles to solve a similar problem involving the movement of boats. Later in this section, we will find the airplane’s groundspeed and bearing, while investigating another approach to problems of this type. First, however, let’s examine the basics of 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,𝑣, u,𝑢, w,𝑤, v⃗ ,𝑣→, u⃗ ,w⃗ .𝑢→,𝑤→.
• Given initial point P𝑃 and terminal point Q,𝑄, a vector can be represented as PQ−→−.𝑃𝑄→. 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)(0,0) and terminal point (a,b),(𝑎,𝑏), a vector may be represented as ⟨a,b⟩.〈𝑎,𝑏〉.

This last symbol ⟨a,b⟩〈𝑎,𝑏〉 has special significance. It is called the standard position. The position vector has an initial point (0,0)(0,0) and a terminal point (a,b).(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 CD−→−𝐶𝐷→ is C(x1,y1)𝐶(𝑥1,𝑦1) and the terminal point is D(x2,y2),𝐷(𝑥2,𝑦2), then the position vector is found by calculating

Ab−→=⟨x2−x1,y2−y1⟩=⟨a,b⟩𝐴𝑏→=〈𝑥2−𝑥1,𝑦2−𝑦1〉=〈𝑎,𝑏〉

In Figure 2, we see the original vector CD−→−𝐶𝐷→ and the position vector Ab−→.𝐴𝑏→.

Figure 2

PROPERTIES OF VECTORS

A vector is a directed line segment with an initial point and a terminal point. Vectors are identified by magnitude, or the length of the line, and direction, represented by the arrowhead pointing toward the terminal point. The position vector has an initial point at (0,0)(0,0) and is identified by its terminal point (a,b).(𝑎,𝑏).

EXAMPLE 1

Find the Position Vector

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

v=⟨6−2,4−3⟩=⟨4,1⟩𝑣=〈6−2,4−3〉=〈4,1〉

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

Figure 3

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

EXAMPLE 2

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

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

Solution

The position vector is found using the following calculation:

v=⟨4−(−3),5−2⟩  =⟨7,3⟩𝑣=〈4−(−3),5−2〉  =〈7,3〉

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

Figure 4

TRY IT #1

Draw a vector v𝑣 that connects from the origin to the point (3,5).(3,5).

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.

MAGNITUDE AND DIRECTION OF A VECTOR

Given a position vector v𝑣 =⟨a,b⟩,=〈𝑎,𝑏〉, the magnitude is found by |v|=a2+b2−−−−−−√.|𝑣|=𝑎2+𝑏2. The direction is equal to the angle formed with the x-axis, or with the y-axis, depending on the application. For a position vector, the direction is found by tanθ=(ba)⇒θ=tan−1(ba),tan𝜃=(𝑏𝑎)⇒𝜃=tan−1(𝑏𝑎), as illustrated in Figure 5.

Figure 5

Two vectors v and u are considered equal if they have the same magnitude and the same direction. Additionally, if both vectors have the same position vector, they are equal.

EXAMPLE 3

Finding the Magnitude and Direction of a Vector

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

Solution

First, find the position vector.

u=⟨−2,−(−8),−5−1⟩  =⟨6,−6⟩𝑢=〈−2,−(−8),−5−1〉  =〈6,−6〉

We use the Pythagorean Theorem to find the magnitude.

∣∣∣u∣∣∣=(6)2+(−6)2−−−−−−−−−−√=72−−√=62–√|𝑢|=(6)2+(−6)2=72=62

The direction is given as

tanθ=−66=−1⇒θ=tan−1(−1)=−45°tan𝜃=−66=−1⇒𝜃=tan−1(−1)=−45°

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

Figure 6

EXAMPLE 4

Showing That Two Vectors Are Equal

Show that vector v with initial point at (5,−3)(5,−3) and terminal point at (−1,2)(−1,2) is equal to vector u with initial point at (−1,−3)(−1,−3) and terminal point at (−7,2).(−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 v𝑣 starting at initial (5,−3)(5,−3) and terminal point (−1,2).(−1,2). Draw the vector u𝑢 with initial point (−1,−3)(−1,−3) and terminal point (−7,2).(−7,2). Find the standard position for each.

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

v=⟨−1−5,2−(−3)⟩  =⟨−6,5⟩u=⟨−7−(−1),2−(−3)⟩  =⟨−6,5⟩𝑣=〈−1−5,2−(−3)〉  =〈−6,5〉𝑢=〈−7−(−1),2−(−3)〉  =〈−6,5〉

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.

∣∣∣v∣∣∣=(−1−5)2+(2−(−3))2−−−−−−−−−−−−−−−−−−−√=(−6)2+(5)2−−−−−−−−−−√=36+25−−−−−−√=61−−√∣∣∣u∣∣∣=(−7−(−1))2+(2−(−3))2−−−−−−−−−−−−−−−−−−−−−−√=(−6)2+(5)2−−−−−−−−−−√=36+25−−−−−−√=61−−√|𝑣|=(−1−5)2+(2−(−3))2=(−6)2+(5)2=36+25=61|𝑢|=(−7−(−1))2+(2−(−3))2=(−6)2+(5)2=36+25=61

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

tanθ=−56⇒θ=tan−1(−56)=−39.8°tan𝜃=−56⇒𝜃=tan−1(−56)=−39.8°

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

Figure 7

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 u𝑢 =⟨x,y⟩=〈𝑥,𝑦〉 as an arrow or directed line segment from the origin to the point (x,y),(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.

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.

Figure 9

EXAMPLE 5

Adding and Subtracting Vectors

Given u𝑢 =⟨3,−2⟩=〈3,−2〉 and v𝑣 =⟨−1,4⟩,=〈−1,4〉, find two new vectors u + v, and u − v.

Solution

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

u+v=⟨3,−2⟩+⟨−1,4⟩=⟨3+(−1),−2+4⟩=⟨2,2⟩𝑢+𝑣=〈3,−2〉+〈−1,4〉=〈3+(−1),−2+4〉=〈2,2〉

See Figure 10(a).

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

u+(−v)=⟨3,−2⟩+⟨1,−4⟩=⟨3+1,−2+(−4)⟩=⟨4,−6⟩𝑢+(−𝑣)=〈3,−2〉+〈1,−4〉=〈3+1,−2+(−4)〉=〈4,−6〉

See Figure 10(b).

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.

SCALAR MULTIPLICATION

Scalar multiplication involves the product of a vector and a scalar. Each component of the vector is multiplied by the scalar. Thus, to multiply v𝑣 =⟨a,b⟩=〈𝑎,𝑏〉 by k𝑘, we have

kv=⟨ka,kb⟩𝑘𝑣=〈𝑘𝑎,𝑘𝑏〉

Only the magnitude changes, unless k𝑘 is negative, and then the vector reverses direction.

EXAMPLE 6

Performing Scalar Multiplication

Given vector v𝑣 =⟨3,1⟩,=〈3,1〉, find 3v, 1212 v,𝑣, and −v.

Solution

See Figure 11 for a geometric interpretation. If v𝑣 =⟨3,1⟩,=〈3,1〉, then

3v=⟨3⋅3,3⋅1⟩=⟨9,3⟩12v=⟨12⋅3,12⋅1⟩=⟨32,12⟩−v=⟨−3,−1⟩3𝑣=〈3⋅3,3⋅1〉=〈9,3〉12𝑣=〈12⋅3,12⋅1〉=〈32,12〉−𝑣=〈−3,−1〉

Figure 11

Analysis

Notice that the vector 3v is three times the length of v, 1212 v𝑣 is half the length of v, and –v is the same length of v, but in the opposite direction.

TRY IT #2

Find the scalar multiple 3 u𝑢 given u𝑢 =⟨5,4⟩.=〈5,4〉.

EXAMPLE 7

Using Vector Addition and Scalar Multiplication to Find a New Vector

Given u𝑢 =⟨3,−2⟩=〈3,−2〉 and v𝑣 =⟨−1,4⟩,=〈−1,4〉, find a new vector w = 3u + 2v.

Solution

First, we must multiply each vector by the scalar.

3u=3⟨3,−2⟩=⟨9,−6⟩2v=2⟨−1,4⟩=⟨−2,8⟩3𝑢=3〈3,−2〉=〈9,−6〉2𝑣=2〈−1,4〉=〈−2,8〉

Then, add the two together.

w=3u+2v=⟨9,−6⟩+⟨−2,8⟩=⟨9−2,−6+8⟩=⟨7,2⟩𝑤=3𝑢+2𝑣=〈9,−6〉+〈−2,8〉=〈9−2,−6+8〉=〈7,2〉

So, w𝑤 =⟨7,2⟩.=〈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 x𝑥 direction, and the vertical component is the y𝑦 direction. For example, we can see in the graph in Figure 12 that the position vector ⟨2,3⟩〈2,3〉 comes from adding the vectors v₁ and v₂. We have v₁ with initial point (0,0)(0,0) and terminal point (2,0).(2,0).

v1=⟨2−0,0−0⟩=⟨2,0⟩𝑣1=〈2−0,0−0〉=〈2,0〉

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

v2=⟨0−0,3−0⟩=⟨0,3⟩𝑣2=〈0−0,3−0〉=〈0,3〉

Therefore, the position vector is

v=⟨2+0,3+0⟩=⟨2,3⟩𝑣=〈2+0,3+0〉=〈2,3〉

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.

∣∣∣v∣∣∣=|v1|2+|v2|2−−−−−−−−−−√=22+32−−−−−−√=13−−√|𝑣|=|𝑣1|2+|𝑣2|2=22+32=13

The magnitude of v is 13−−√.13. To find the direction, we use the tangent function tanθ=yx.tan𝜃=𝑦𝑥.

tanθ=v2v1tanθ=32θ=tan−1(32)=56.3°tan𝜃=𝑣2𝑣1tan𝜃=32𝜃=tan−1(32)=56.3°

Figure 12

Thus, the magnitude of v𝑣 is 13−−√13 and the direction is 56.3∘56.3∘ off the horizontal.

EXAMPLE 8

Finding the Components of the Vector

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

Solution

First find the standard position.

v=⟨7−3,4−2⟩=⟨4,2⟩𝑣=〈7−3,4−2〉=〈4,2〉

See the illustration in Figure 13.

Figure 13

The horizontal component is v1𝑣1 =⟨4,0⟩=〈4,0〉 and the vertical component is v2𝑣2 =⟨0,2⟩.=〈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 i𝑖 =⟨1,0⟩=〈1,0〉 and is directed along the positive horizontal axis. The vertical unit vector is written as j𝑗 =⟨0,1⟩=〈0,1〉 and is directed along the positive vertical axis. See Figure 14.

Figure 14

THE UNIT VECTORS

If v𝑣 is a nonzero vector, then v|v|𝑣|𝑣| is a unit vector in the direction of v.𝑣. Any vector divided by its magnitude is a unit vector. Notice that magnitude is always a scalar, and dividing by a scalar is the same as multiplying by the reciprocal of the scalar.

EXAMPLE 9

Finding the Unit Vector in the Direction of v

Find a unit vector in the same direction as v𝑣 =⟨−5,12⟩.=〈−5,12〉.

Solution

First, we will find the magnitude.

∣∣∣v∣∣∣=(−5)2+(12)2−−−−−−−−−−−√=25+144−−−−−−−√=169−−−√=13|𝑣|=(−5)2+(12)2=25+144=169=13

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

v|v|=−513i+1213j𝑣|𝑣|=−513𝑖+1213𝑗

or, in component form

v|v|=⟨−513,1213⟩𝑣|𝑣|=〈−513,1213〉

See Figure 15.

Figure 15

Verify that the magnitude of the unit vector equals 1. The magnitude of −513i+1213j−513𝑖+1213𝑗 is given as

(−513)2+(1213)2−−−−−−−−−−−−−√=25169+144169−−−−−−−−√                            =169169−−−√=1(−513)2+(1213)2=25169+144169                            =169169=1

The vector u =513=513 i +1213+1213 j is the unit vector in the same direction as v =⟨−5,12⟩.=〈−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.

VECTORS IN THE RECTANGULAR PLANE

Given a vector v𝑣 with initial point P=(x1,y1)𝑃=(𝑥1,𝑦1) and terminal point Q=(x2,y2),𝑄=(𝑥2,𝑦2), v is written as

v=(x2−x1)i+(y2−y1)j𝑣=(𝑥2−𝑥1)𝑖+(𝑦2−𝑦1)𝑗

The position vector from (0,0)(0,0) to (a,b),(𝑎,𝑏), where (x2−x1)=a(𝑥2−𝑥1)=𝑎 and (y2−y1)=b,(𝑦2−𝑦1)=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|=a2+b2−−−−−−√.|𝑣|=𝑎2+𝑏2. See Figure 16.

Figure 16

EXAMPLE 10

Writing a Vector in Terms of i and j

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

Solution

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

v=(x2−x1)i+(y2−y1)j=(−6−2)i+(6−(−6))j=−8i+12j𝑣=(𝑥2−𝑥1)𝑖+(𝑦2−𝑦1)𝑗=(−6−2)𝑖+(6−(−6))𝑗=−8𝑖+12𝑗

EXAMPLE 11

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

Given initial point P1=(−1,3)𝑃1=(−1,3) and terminal point P2=(2,7),𝑃2=(2,7), write the vector v𝑣 in terms of i𝑖 and j.𝑗.

Solution

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

v=(x2−x1)i+(y2−y1)jv=(2−(−1))i+(7−3)j=3i+4j𝑣=(𝑥2−𝑥1)𝑖+(𝑦2−𝑦1)𝑗𝑣=(2−(−1))𝑖+(7−3)𝑗=3𝑖+4𝑗

TRY IT #3

Write the vector u𝑢 with initial point P=(−1,6)𝑃=(−1,6) and terminal point Q=(7,−5)𝑄=(7,−5) in terms of i𝑖 and j.𝑗.

Performing Operations on Vectors in Terms of i and j

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

ADDING AND SUBTRACTING VECTORS IN RECTANGULAR COORDINATES

Given v = ai + bj and u = ci + dj, then

v+u=(a+c)i+(b+d)jv−u=(a−c)i+(b−d)j𝑣+𝑢=(𝑎+𝑐)𝑖+(𝑏+𝑑)𝑗𝑣−𝑢=(𝑎−𝑐)𝑖+(𝑏−𝑑)𝑗

EXAMPLE 12

Finding the Sum of the Vectors

Find the sum of v1=2i−3j𝑣1=2𝑖−3𝑗 and v2=4i+5j.𝑣2=4𝑖+5𝑗.

Solution

According to the formula, we have

v1+v2=(2+4)i+(−3+5)j=6i+2j𝑣1+𝑣2=(2+4)𝑖+(−3+5)𝑗=6𝑖+2𝑗

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 iandj.𝑖and𝑗. 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||𝑣| replacing r.𝑟.

VECTOR COMPONENTS IN TERMS OF MAGNITUDE AND DIRECTION

Given a position vector v=⟨x,y⟩𝑣=〈𝑥,𝑦〉 and a direction angle θ,𝜃,

cosθ=x|v|x=|v|cosθandsinθ=y|v|y=|v|sinθcos𝜃=𝑥|𝑣|andsin𝜃=𝑦|𝑣|𝑥=|𝑣|cos𝜃𝑦=|𝑣|sin𝜃

Thus, v=xi+yj=|v|cosθi+|v|sinθj,𝑣=𝑥𝑖+𝑦𝑗=|𝑣|cos𝜃𝑖+|𝑣|sin𝜃𝑗, and magnitude is expressed as |v|=x2+y2−−−−−−√.|𝑣|=𝑥2+𝑦2.

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|cosθ𝑥=|𝑣|cos𝜃 and y=|v|sinθ,𝑦=|𝑣|sin𝜃, we find that

x=7cos(135°)=−72√2y=7sin(135°)=72√2𝑥=7cos(135°)=−722𝑦=7sin(135°)=722

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

v=−72–√2i+72–√2j𝑣=−722𝑖+722𝑗

TRY IT #4

A vector travels from the origin to the point (3,5).(3,5). Write the vector in terms of magnitude and direction.

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.

DOT PRODUCT

The dot product of two vectors v=⟨a,b⟩𝑣=〈𝑎,𝑏〉 and u=⟨c,d⟩𝑢=〈𝑐,𝑑〉 is the sum of the product of the horizontal components and the product of the vertical components.

v⋅u=ac+bd𝑣⋅𝑢=𝑎𝑐+𝑏𝑑

To find the angle between the two vectors, use the formula below.

cosθ=v|v|⋅u|u|cos𝜃=𝑣|𝑣|⋅𝑢|𝑢|

EXAMPLE 14

Finding the Dot Product of Two Vectors

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

Solution

Using the formula, we have

v⋅u=⟨5,12⟩⋅⟨−3,4⟩=5⋅(−3)+12⋅4=−15+48=33𝑣⋅𝑢=〈5,12〉⋅〈−3,4〉=5⋅(−3)+12⋅4=−15+48=33

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.

v1⋅v2=⟨5,2⟩⋅⟨3,7⟩=5⋅3+2⋅7=15+14=29𝑣1⋅𝑣2=〈5,2〉⋅〈3,7〉=5⋅3+2⋅7=15+14=29

To find the angle between them, we use the formula cosθ=v|v|⋅u|u|.cos𝜃=𝑣|𝑣|⋅𝑢|𝑢|.

v|v|⋅u|u|=⟨529√,229√⟩⋅⟨358√,758√⟩=529√⋅358√+229√⋅758√=151682√+141682√=291682√=0.707107cos−1(0.707107)=45°𝑣|𝑣|⋅𝑢|𝑢|=〈529,229〉⋅〈358,758〉=529⋅358+229⋅758=151682+141682=291682=0.707107cos−1(0.707107)=45°

See Figure 17.

Figure 17

EXAMPLE 16

Finding the Angle between Two Vectors

Find the angle between u=⟨−3,4⟩𝑢=〈−3,4〉 and v=⟨5,12⟩.𝑣=〈5,12〉.

Solution

Using the formula, we have

θ=cos−1(u|u|⋅v|v|)(u|u|⋅v|v|)=−3i+4j5⋅5i+12j13=(−35⋅513)+(45⋅1213)=−1565+4865=3365θ=cos−1(3365)=59.5∘𝜃=cos−1(𝑢|𝑢|⋅𝑣|𝑣|)(𝑢|𝑢|⋅𝑣|𝑣|)=−3𝑖+4𝑗5⋅5𝑖+12𝑗13=(−35⋅513)+(45⋅1213)=−1565+4865=3365𝜃=cos−1(3365)=59.5∘

See Figure 18.

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.

Figure 19

Solution

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

Notice in Figure 19, that angle bCO𝑏𝐶𝑂 must be equal to angle AOC𝐴𝑂𝐶 by the rule of alternating interior angles, so angle bCO𝑏𝐶𝑂 is 140°. We can find x𝑥 by the Law of Cosines:

x2=(16.2)2+(200)2−2(16.2)(200)cos(140°)x2=45,226.41x=45,226.41−−−−−−−−√x=212.7𝑥2=(16.2)2+(200)2−2(16.2)(200)cos(140°)𝑥2=45,226.41𝑥=45,226.41𝑥=212.7

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

sinα16.2=sin(140°)212.7sinα=16.2sin(140°)212.7=0.04896sin−1(0.04896)=2.8°sin𝛼16.2=sin(140°)212.7sin𝛼=16.2sin(140°)212.7=0.04896sin−1(0.04896)=2.8°

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