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Vector Addition and Subtraction: Graphical Methods

Module by: OpenStax College. E-mail the author

Summary:

  • Understand the rules of vector addition, subtraction, and multiplication.
  • Apply graphical methods of vector addition and subtraction to determine the displacement of moving objects.

Figure 1: Displacement can be determined graphically using a scale map, such as this one of the Hawaiian Islands. A journey from Hawai’i to Moloka’i has a number of legs, or journey segments. These segments can be added graphically with a ruler to determine the total two-dimensional displacement of the journey. (credit: US Geological Survey)
Some Hawaiian Islands like Kauai Oahu, Molokai, Lanai, Maui, Kahoolawe, and Hawaii are shown. On the scale map of Hawaiian Islands the path of a journey is shown moving from Hawaii to Molokai. The path of the journey is turning at different angles and finally reaching its destination. The displacement of the journey is shown with the help of a straight line connecting its starting point and the destination.

Vectors in Two Dimensions

A vector is a quantity that has magnitude and direction. Displacement, velocity, acceleration, and force, for example, are all vectors. In one-dimensional, or straight-line, motion, the direction of a vector can be given simply by a plus or minus sign. In two dimensions (2-d), however, we specify the direction of a vector relative to some reference frame (i.e., coordinate system), using an arrow having length proportional to the vector’s magnitude and pointing in the direction of the vector.

Figure 2 shows such a graphical representation of a vector, using as an example the total displacement for the person walking in a city considered in Kinematics in Two Dimensions: An Introduction. We shall use the notation that a boldface symbol, such as DD size 12{D} {}, stands for a vector. Its magnitude is represented by the symbol in italics, DD size 12{D} {}, and its direction by θθ size 12{θ} {}.

Vectors in this Text:

In this text, we will represent a vector with a boldface variable. For example, we will represent the quantity force with the vector FF size 12{F} {}, which has both magnitude and direction. The magnitude of the vector will be represented by a variable in italics, such as FF size 12{F} {}, and the direction of the variable will be given by an angle θθ size 12{θ} {}.

Figure 2: A person walks 9 blocks east and 5 blocks north. The displacement is 10.3 blocks at an angle 29.1º29.1º size 12{"29" "." "1º"} {} north of east.
A graph is shown. On the axes the scale is set to one block is equal to one unit. A helicopter starts moving from the origin at an angle of twenty nine point one degrees above the x axis. The current position of the helicopter is ten point three blocks along its line of motion. The destination of the helicopter is the point which is nine blocks in the positive x direction and five blocks in the positive y direction. The positive direction of the x axis is east and the positive direction of the y axis is north.
Figure 3: To describe the resultant vector for the person walking in a city considered in Figure 2 graphically, draw an arrow to represent the total displacement vector DD size 12{D} {}. Using a protractor, draw a line at an angle θθ size 12{θ} {} relative to the east-west axis. The length DD size 12{D} {} of the arrow is proportional to the vector’s magnitude and is measured along the line with a ruler. In this example, the magnitude DD size 12{D} {} of the vector is 10.3 units, and the direction θθ size 12{θ} {} is 29.1º29.1º size 12{"29" "." 1 rSup { size 12{º} } } {} north of east.
On a graph a vector is shown. It is inclined at an angle theta equal to twenty nine point one degrees above the positive x axis. A protractor is shown to the right of the x axis to measure the angle. A ruler is also shown parallel to the vector to measure its length. The ruler shows that the length of the vector is ten point three units.

Vector Addition: Head-to-Tail Method

The head-to-tail method is a graphical way to add vectors, described in Figure 4 below and in the steps following. The tail of the vector is the starting point of the vector, and the head (or tip) of a vector is the final, pointed end of the arrow.

Figure 4: Head-to-Tail Method: The head-to-tail method of graphically adding vectors is illustrated for the two displacements of the person walking in a city considered in Figure 2. (a) Draw a vector representing the displacement to the east. (b) Draw a vector representing the displacement to the north. The tail of this vector should originate from the head of the first, east-pointing vector. (c) Draw a line from the tail of the east-pointing vector to the head of the north-pointing vector to form the sum or resultant vector DD size 12{A} {}. The length of the arrow DD size 12{A} {} is proportional to the vector’s magnitude and is measured to be 10.3 units . Its direction, described as the angle with respect to the east (or horizontal axis) θθ size 12{θ} {} is measured with a protractor to be 29 . 1º 29 . 1º size 12{"29" "." 1°} {} .
In part a, a vector of magnitude of nine units and making an angle of theta is equal to zero degrees is drawn from the origin and along the positive direction of x axis. In part b a vector of magnitude of nine units and making an angle of theta is equal to zero degree is drawn from the origin and along the positive direction of x axis. Then a vertical arrow from the head of the horizontal arrow is drawn. In part c a vector D of magnitude ten point three is drawn from the tail of the horizontal vector at an angle theta is equal to twenty nine point one degrees from the positive direction of x axis. The head of the vector D meets the head of the vertical vector. A scale is shown parallel to the vector D to measure its length. Also a protractor is shown to measure the inclination of the vectorD.

Step 1. Draw an arrow to represent the first vector (9 blocks to the east) using a ruler and protractor.

Figure 5
In part a, a vector of magnitude of nine units and making an angle theta is equal to zero degree is drawn from the origin and along the positive direction of x axis.

Step 2. Now draw an arrow to represent the second vector (5 blocks to the north). Place the tail of the second vector at the head of the first vector.

Figure 6
In part b, a vector of magnitude of nine units and making an angle theta is equal to zero degree is drawn from the origin and along the positive direction of x axis. Then a vertical vector from the head of the horizontal vector is drawn.

Step 3. If there are more than two vectors, continue this process for each vector to be added. Note that in our example, we have only two vectors, so we have finished placing arrows tip to tail.

Step 4. Draw an arrow from the tail of the first vector to the head of the last vector. This is the resultant, or the sum, of the other vectors.

Figure 7
In part c, a vector D of magnitude ten point three is drawn from the tail of the horizontal vector at an angle theta is equal to twenty nine point one degrees from the positive direction of the x axis. The head of the vector D meets the head of the vertical vector. A scale is shown parallel to the vector D to measure its length. Also a protractor is shown to measure the inclination of the vector D.

Step 5. To get the magnitude of the resultant, measure its length with a ruler. (Note that in most calculations, we will use the Pythagorean theorem to determine this length.)

Step 6. To get the direction of the resultant, measure the angle it makes with the reference frame using a protractor. (Note that in most calculations, we will use trigonometric relationships to determine this angle.)

The graphical addition of vectors is limited in accuracy only by the precision with which the drawings can be made and the precision of the measuring tools. It is valid for any number of vectors.

Example 1: Adding Vectors Graphically Using the Head-to-Tail Method: A Woman Takes a Walk

Use the graphical technique for adding vectors to find the total displacement of a person who walks the following three paths (displacements) on a flat field. First, she walks 25.0 m in a direction 49.0º49.0º size 12{"49" "." "0º"} {} north of east. Then, she walks 23.0 m heading 15.0º15.0º size 12{"15" "." "º°"} {} north of east. Finally, she turns and walks 32.0 m in a direction 68.0° south of east.

Strategy

Represent each displacement vector graphically with an arrow, labeling the first AA size 12{A} {}, the second BB size 12{B} {}, and the third CC size 12{C} {}, making the lengths proportional to the distance and the directions as specified relative to an east-west line. The head-to-tail method outlined above will give a way to determine the magnitude and direction of the resultant displacement, denoted RR size 12{R} {}.

Solution

(1) Draw the three displacement vectors.

Figure 8
On the graph a vector of magnitude twenty three meters and inclined above the x axis at an angle theta-b equal to fifteen degrees is shown. This vector is labeled as B.

(2) Place the vectors head to tail retaining both their initial magnitude and direction.

Figure 9
In this figure a vector A with a positive slope is drawn from the origin. Then from the head of the vector A another vector B with positive slope is drawn and then another vector C with negative slope from the head of the vector B is drawn which cuts the x axis.

(3) Draw the resultant vector, RR size 12{R} {}.

Figure 10
In this figure a vector A with a positive slope is drawn from the origin. Then from the head of the vector A another vector B with positive slope is drawn and then another vector C with negative slope from the head of the vector B is drawn which cuts the x axis. From the tail of the vector A a vector R of magnitude of fifty point zero meters and with negative slope of seven degrees is drawn. The head of this vector R meets the head of the vector C. The vector R is known as the resultant vector.

(4) Use a ruler to measure the magnitude of RR size 12{R} {}, and a protractor to measure the direction of RR size 12{R} {}. While the direction of the vector can be specified in many ways, the easiest way is to measure the angle between the vector and the nearest horizontal or vertical axis. Since the resultant vector is south of the eastward pointing axis, we flip the protractor upside down and measure the angle between the eastward axis and the vector.

Figure 11
In this figure a vector A with a positive slope is drawn from the origin. Then from the head of the vector A another vector B with positive slope is drawn and then another vector C with negative slope from the head of the vector B is drawn which cuts the x axis. From the tail of the vector A a vector R of magnitude of fifty meter and with negative slope of seven degrees is drawn. The head of this vector R meets the head of the vector C. The vector R is known as the resultant vector. A ruler is placed along the vector R to measure it. Also there is a protractor to measure the angle.

In this case, the total displacement RR size 12{R} {} is seen to have a magnitude of 50.0 m and to lie in a direction 7.0º7.0º size 12{7 "." 0°} {} south of east. By using its magnitude and direction, this vector can be expressed as R = 50.0 m R = 50.0 m size 12{R" = 50" "." "0 m"} {} and θ=7.0ºθ=7.0º size 12{θ=7 "." "0°"} {} south of east.

Discussion

The head-to-tail graphical method of vector addition works for any number of vectors. It is also important to note that the resultant is independent of the order in which the vectors are added. Therefore, we could add the vectors in any order as illustrated in Figure 12 and we will still get the same solution.

Figure 12
In this figure a vector C with a negative slope is drawn from the origin. Then from the head of the vector C another vector A with positive slope is drawn and then another vector B with negative slope from the head of the vector A is drawn. From the tail of the vector C a vector R of magnitude of fifty point zero meters and with negative slope of seven degrees is drawn. The head of this vector R meets the head of the vector B. The vector R is known as the resultant vector.

Here, we see that when the same vectors are added in a different order, the result is the same. This characteristic is true in every case and is an important characteristic of vectors. Vector addition is commutative. Vectors can be added in any order.

A+B=B+A.A+B=B+A. size 12{"A+B=B+A"} {}
(1)

(This is true for the addition of ordinary numbers as well—you get the same result whether you add 2+32+3 size 12{"2+3"} {} or 3+23+2 size 12{"3+2"} {}, for example).

Vector Subtraction

Vector subtraction is a straightforward extension of vector addition. To define subtraction (say we want to subtract BB size 12{B} {} from AA size 12{A} {} , written A – BA – B size 12{ "A" "-B"} {} , we must first define what we mean by subtraction. The negative of a vector BB is defined to be –B–B; that is, graphically the negative of any vector has the same magnitude but the opposite direction, as shown in Figure 13. In other words, BB size 12{B} {} has the same length as –B–B size 12{"-" "B"} {}, but points in the opposite direction. Essentially, we just flip the vector so it points in the opposite direction.

Figure 13: The negative of a vector is just another vector of the same magnitude but pointing in the opposite direction. So BB size 12{B} {} is the negative of –B–B size 12{ ital "-B"} {}; it has the same length but opposite direction.
Two vectors are shown. One of the vectors is labeled as vector in north east direction. The other vector is of the same magnitude and is in the opposite direction to that of vector B. This vector is denoted as negative B.

The subtraction of vector BB from vector AA is then simply defined to be the addition of –B–B to AA. Note that vector subtraction is the addition of a negative vector. The order of subtraction does not affect the results.

A – B = A + (–B).A – B = A + (–B). size 12{ bold "A – B = A + " \( bold "–B" \) } {}
(2)

This is analogous to the subtraction of scalars (where, for example, 5 – 2 = 5 + (–2)5 – 2 = 5 + (–2) size 12{"5 – 2 = 5 + " \( "–2" \) } {}). Again, the result is independent of the order in which the subtraction is made. When vectors are subtracted graphically, the techniques outlined above are used, as the following example illustrates.

Example 2: Subtracting Vectors Graphically: A Woman Sailing a Boat

A woman sailing a boat at night is following directions to a dock. The instructions read to first sail 27.5 m in a direction 66.0º66.0º size 12{"66" "." 0º} {} north of east from her current location, and then travel 30.0 m in a direction 112º112º size 12{"112"º} {} north of east (or 22.0º22.0º size 12{"22" "." 0º} {} west of north). If the woman makes a mistake and travels in the opposite direction for the second leg of the trip, where will she end up? Compare this location with the location of the dock.

Figure 14
A vector of magnitude twenty seven point five meters is shown. It is inclined to the horizontal at an angle of sixty six degrees. Another vector of magnitude thirty point zero meters is shown. It is inclined to the horizontal at an angle of one hundred and twelve degrees.

Strategy

We can represent the first leg of the trip with a vector AA, and the second leg of the trip with a vector BB size 12{B} {}. The dock is located at a location A + BA + B. If the woman mistakenly travels in the opposite direction for the second leg of the journey, she will travel a distance BB (30.0 m) in the direction 180º–112º=68º180º–112º=68º south of east. We represent this as –B–B, as shown below. The vector –B–B has the same magnitude as BB but is in the opposite direction. Thus, she will end up at a location A+(–B)A+(–B), or A–BA–B.

Figure 15
A vector labeled negative B is inclined at an angle of sixty-eight degrees below a horizontal line. A dotted line in the reverse direction inclined at one hundred and twelve degrees above the horizontal line is also shown.

We will perform vector addition to compare the location of the dock, A + BA + B size 12{ ital "A ""+ "B} {}, with the location at which the woman mistakenly arrives, A + (–B)A + (–B) size 12{ bold "A + " \( bold "–B" \) } {}.

Solution

(1) To determine the location at which the woman arrives by accident, draw vectors AA size 12{A} {} and –B–B.

(2) Place the vectors head to tail.

(3) Draw the resultant vector RR size 12{R} {}.

(4) Use a ruler and protractor to measure the magnitude and direction of RR size 12{R} {}.

Figure 16
Vectors A and negative B are connected in head to tail method. Vector A is inclined with horizontal with positive slope and vector negative B with a negative slope. The resultant of these two vectors is shown as a vector R from tail of A to the head of negative B. The length of the resultant is twenty three point zero meters and has a negative slope of seven point five degrees.

In this case, R = 23 . 0 m R = 23 . 0 m size 12{R"=23" "." "0 m"} {} and θ = 7 . 5º θ = 7 . 5º size 12{θ=7 "." "5° south of east"} {} south of east.

(5) To determine the location of the dock, we repeat this method to add vectors AA size 12{A} {} and BB size 12{B} {}. We obtain the resultant vector R'R' size 12{R'} {}:

Figure 17
A vector A inclined at sixty six degrees with horizontal is shown. From the head of this vector another vector B is started. Vector B is inclined at one hundred and twelve degrees with the horizontal. Another vector labeled as R prime from the tail of vector A to the head of vector B is drawn. The length of this vector is fifty two point nine meters and its inclination with the horizontal is shown as ninety point one degrees. Vector R prime is equal to the sum of vectors A and B.

In this case R  = 52.9 m R  = 52.9 m size 12{R" = 52" "." "9 m"} {} and θ = 90.1º θ = 90.1º size 12{θ="90" "." "1° north of east "} {}  north of east.

We can see that the woman will end up a significant distance from the dock if she travels in the opposite direction for the second leg of the trip.

Discussion

Because subtraction of a vector is the same as addition of a vector with the opposite direction, the graphical method of subtracting vectors works the same as for addition.

Multiplication of Vectors and Scalars

If we decided to walk three times as far on the first leg of the trip considered in the preceding example, then we would walk 3 × 27.5 m3 × 27.5 m size 12{"3 " times " 27" "." "5 m"} {}, or 82.5 m, in a direction 66.0º66.0º size 12{"66" "." 0 { size 12{º} } } {} north of east. This is an example of multiplying a vector by a positive scalar. Notice that the magnitude changes, but the direction stays the same.

If the scalar is negative, then multiplying a vector by it changes the vector’s magnitude and gives the new vector the opposite direction. For example, if you multiply by –2, the magnitude doubles but the direction changes. We can summarize these rules in the following way: When vector AA size 12{A} {} is multiplied by a scalar cc size 12{c} {},

  • the magnitude of the vector becomes the absolute value of cc size 12{c} {}AA size 12{A} {},
  • if cc size 12{A} {} is positive, the direction of the vector does not change,
  • if cc size 12{A} {} is negative, the direction is reversed.

In our case, c=3c=3size 12{c=3} and A=27.5 mA=27.5 msize 12{"A= 27.5 m"}. Vectors are multiplied by scalars in many situations. Note that division is the inverse of multiplication. For example, dividing by 2 is the same as multiplying by the value (1/2). The rules for multiplication of vectors by scalars are the same for division; simply treat the divisor as a scalar between 0 and 1.

Resolving a Vector into Components

In the examples above, we have been adding vectors to determine the resultant vector. In many cases, however, we will need to do the opposite. We will need to take a single vector and find what other vectors added together produce it. In most cases, this involves determining the perpendicular components of a single vector, for example the x- and y-components, or the north-south and east-west components.

For example, we may know that the total displacement of a person walking in a city is 10.3 blocks in a direction 29.0º29.0º size 12{"29" "." 0º} } {} north of east and want to find out how many blocks east and north had to be walked. This method is called finding the components (or parts) of the displacement in the east and north directions, and it is the inverse of the process followed to find the total displacement. It is one example of finding the components of a vector. There are many applications in physics where this is a useful thing to do. We will see this soon in Projectile Motion, and much more when we cover forces in Dynamics: Newton’s Laws of Motion. Most of these involve finding components along perpendicular axes (such as north and east), so that right triangles are involved. The analytical techniques presented in Vector Addition and Subtraction: Analytical Methods are ideal for finding vector components.

PhET Explorations: Maze Game:

Learn about position, velocity, and acceleration in the "Arena of Pain". Use the green arrow to move the ball. Add more walls to the arena to make the game more difficult. Try to make a goal as fast as you can.

Figure 18: Maze Game
Figure 18 (maze-game_en.jar)

Summary

  • The graphical method of adding vectors AA size 12{A} {} and BB size 12{B} {} involves drawing vectors on a graph and adding them using the head-to-tail method. The resultant vector RR size 12{A} {} is defined such that A+B=RA+B=R. The magnitude and direction of RR size 12{A} {} are then determined with a ruler and protractor, respectively.
  • The graphical method of subtracting vector B B from AA involves adding the opposite of vector BB, which is defined as −B−B size 12{ - B} {}. In this case, A–B=A+(–B)=RA–B=A+(–B)=R. Then, the head-to-tail method of addition is followed in the usual way to obtain the resultant vector RR.
  • Addition of vectors is commutative such that A+B=B+AA+B=B+A size 12{"A + B = B + A"} {} .
  • The head-to-tail method of adding vectors involves drawing the first vector on a graph and then placing the tail of each subsequent vector at the head of the previous vector. The resultant vector is then drawn from the tail of the first vector to the head of the final vector.
  • If a vector AA size 12{A} {} is multiplied by a scalar quantity cc size 12{A} {}, the magnitude of the product is given by cAcA size 12{ ital "cA"} {}. If cc size 12{c} {} is positive, the direction of the product points in the same direction as AA size 12{A} {}; if cc size 12{c} {} is negative, the direction of the product points in the opposite direction as AA size 12{A} {}.

Conceptual Questions

Exercise 1

Which of the following is a vector: a person’s height, the altitude on Mt. Everest, the age of the Earth, the boiling point of water, the cost of this book, the Earth’s population, the acceleration of gravity?

Exercise 2

Give a specific example of a vector, stating its magnitude, units, and direction.

Exercise 3

What do vectors and scalars have in common? How do they differ?

Exercise 4

Two campers in a national park hike from their cabin to the same spot on a lake, each taking a different path, as illustrated below. The total distance traveled along Path 1 is 7.5 km, and that along Path 2 is 8.2 km. What is the final displacement of each camper?

Figure 19
At the southwest corner of the figure is a cabin and in the northeast corner is a lake. A vector S with a length five point zero kilometers connects the cabin to the lake at an angle of 40 degrees north of east. Two winding paths labeled Path 1 and Path 2 represent the routes travelled from the cabin to the lake.

Exercise 5

If an airplane pilot is told to fly 123 km in a straight line to get from San Francisco to Sacramento, explain why he could end up anywhere on the circle shown in Figure 20. What other information would he need to get to Sacramento?

Figure 20
A map of northern California with a circle with a radius of one hundred twenty three kilometers centered on San Francisco. Sacramento lies on the circumference of this circle in a direction forty-five degrees north of east from San Francisco.

Exercise 6

Suppose you take two steps AA and BB (that is, two nonzero displacements). Under what circumstances can you end up at your starting point? More generally, under what circumstances can two nonzero vectors add to give zero? Is the maximum distance you can end up from the starting point A+BA+B the sum of the lengths of the two steps?

Exercise 7

Explain why it is not possible to add a scalar to a vector.

Exercise 8

If you take two steps of different sizes, can you end up at your starting point? More generally, can two vectors with different magnitudes ever add to zero? Can three or more?

Problems & Exercises

Use graphical methods to solve these problems. You may assume data taken from graphs is accurate to three digits.

Exercise 1

Find the following for path A in Figure 21: (a) the total distance traveled, and (b) the magnitude and direction of the displacement from start to finish.

Figure 21: The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side.
A map of city is shown. The houses are in form of square blocks of side one hundred and twenty meters each. The path of A extends to three blocks towards north and then one block towards east. It is asked to find out the total distance traveled the magnitude and the direction of the displacement from start to finish.
[ Show Solution ]

Exercise 2

Find the following for path B in Figure 21: (a) the total distance traveled, and (b) the magnitude and direction of the displacement from start to finish.

Exercise 3

Find the north and east components of the displacement for the hikers shown in Figure 19.

[ Show Solution ]

Exercise 4

Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements AA size 12{A} {} and BB size 12{B} {}, as in Figure 22, then this problem asks you to find their sum R=A+BR=A+B size 12{"R = A + B"} {}.)

Figure 22: The two displacements AA size 12{A} {} and BB size 12{B} {} add to give a total displacement RR size 12{R} {} having magnitude RR size 12{R} {} and direction θθ size 12{θ} {}.
In this figure coordinate axes are shown. Vector A from the origin towards the negative of x axis is shown. From the head of the vector A another vector B is drawn towards the positive direction of y axis. The resultant R of these two vectors is shown as a vector from the tail of vector A to the head of vector B. This vector R is inclined at an angle theta with the negative x axis.

Exercise 5

Suppose you first walk 12.0 m in a direction 20º20º size 12{"20" { size 12{°} } } {} west of north and then 20.0 m in a direction 40.0º40.0º size 12{"40" { size 12{°} } } {} south of west. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements AA size 12{A} {} and BB size 12{B} {}, as in Figure 23, then this problem finds their sum R = A + BR = A + B size 12{ bold "R = A + B"} {}.)

Figure 23
In the given figure coordinates axes are shown. Vector A with tail at origin is inclined at an angle of twenty degrees with the positive direction of x axis. The magnitude of vector A is twelve meters. Another vector B is starts from the head of vector A and inclined at an angle of forty degrees with the horizontal. The resultant R of the vectors A and B is also drawn from the tail of vector A to the head of vector B. The inclination of vector R is theta with the horizontal.
[ Show Solution ]

Exercise 6

Repeat the problem above, but reverse the order of the two legs of the walk; show that you get the same final result. That is, you first walk leg BB size 12{B} {}, which is 20.0 m in a direction exactly 40º40º size 12{"20" { size 12{°} } } {} south of west, and then leg AA size 12{A} {}, which is 12.0 m in a direction exactly 20º20º size 12{"20" { size 12{°} } } {} west of north. (This problem shows that A+B=B+AA+B=B+A size 12{A+B=B+A} {}.)

Exercise 7

(a) Repeat the problem two problems prior, but for the second leg you walk 20.0 m in a direction 40.0º40.0º size 12{"40.0" { size 12{°} } } north of east (which is equivalent to subtracting BB size 12{B} {} from AA size 12{A} {} —that is, to finding R′=A−BR′=A−B size 12{ bold "R'"=A - B} {}). (b) Repeat the problem two problems prior, but now you first walk 20.0 m in a direction 40.0º40.0º size 12{"40.0" { size 12{°} } } south of west and then 12.0 m in a direction 20.0º20.0º size 12{"20.0" { size 12{ ° } } } {} east of south (which is equivalent to subtracting AA size 12{A} {} from BB size 12{B} {} —that is, to finding R′′= B-A=-R′R′′= B-A=-R′ size 12{R''= B – A = -R' } {}). Show that this is the case.

[ Show Solution ]

Exercise 8

Show that the order of addition of three vectors does not affect their sum. Show this property by choosing any three vectors AA size 12{A} {}, BB size 12{B} {}, and CC size 12{C} {}, all having different lengths and directions. Find the sum A + B + CA + B + C size 12{ bold "A + B + C"} {} then find their sum when added in a different order and show the result is the same. (There are five other orders in which AA size 12{A} {}, BB size 12{B} {}, and CC size 12{C} {} can be added; choose only one.)

Exercise 9

Show that the sum of the vectors discussed in Example 2 gives the result shown in Figure 17.

[ Show Solution ]

Exercise 10

Find the magnitudes of velocities vAvA size 12{v rSub { size 8{A} } } {} and vBvB size 12{v rSub { size 8{B} } } {} in Figure 24

Figure 24: The two velocities vAvA size 12{v rSub { size 8{A} } } {} and vBvB size 12{v rSub { size 8{B} } } {} add to give a total vtotvtot size 12{v rSub { size 8{"tot"} } } {}.
On the graph velocity vector V sub A begins at the origin and is inclined to x axis at an angle of twenty two point five degrees. From the head of vector V sub A another vector V sub B begins. The resultant of the two vectors, labeled V sub tot, is inclined to vector V sub A at twenty six point five degrees and to the vector V sub B at twenty three point zero degrees. V sub tot has a magnitude of 6.72 meters per second.

Exercise 11

Find the components of vtotvtot size 12{v rSub { size 8{"tot"} } } {} along the x- and y-axes in Figure 24.

[ Show Solution ]

Exercise 12

Find the components of vtotvtot size 12{v rSub { size 8{"tot"} } } {} along a set of perpendicular axes rotated 30º30º size 12{"30º"} {} counterclockwise relative to those in Figure 24.

Glossary

component (of a 2-d vector):
a piece of a vector that points in either the vertical or the horizontal direction; every 2-d vector can be expressed as a sum of two vertical and horizontal vector components
commutative:
refers to the interchangeability of order in a function; vector addition is commutative because the order in which vectors are added together does not affect the final sum
direction (of a vector):
the orientation of a vector in space
head (of a vector):
the end point of a vector; the location of the tip of the vector’s arrowhead; also referred to as the “tip”
head-to-tail method:
a method of adding vectors in which the tail of each vector is placed at the head of the previous vector
magnitude (of a vector):
the length or size of a vector; magnitude is a scalar quantity
resultant:
the sum of two or more vectors
resultant vector:
the vector sum of two or more vectors
scalar:
a quantity with magnitude but no direction
tail:
the start point of a vector; opposite to the head or tip of the arrow

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