{"id":449,"date":"2014-12-11T02:30:07","date_gmt":"2014-12-11T02:30:07","guid":{"rendered":"https:\/\/courses.candelalearning.com\/colphysics\/?post_type=chapter&#038;p=449"},"modified":"2016-11-03T22:59:40","modified_gmt":"2016-11-03T22:59:40","slug":"3-2-vector-addition-and-subtraction-graphical-methods","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-physics\/chapter\/3-2-vector-addition-and-subtraction-graphical-methods\/","title":{"raw":"Vector Addition and Subtraction: Graphical Methods","rendered":"Vector Addition and Subtraction: Graphical Methods"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\nBy the end of this section, you will be able to:\r\n<div class=\"itemizedlist\">\r\n<ul class=\"itemizedlist\">\r\n \t<li class=\"listitem\">Understand the rules of vector addition, subtraction, and multiplication.<\/li>\r\n \t<li class=\"listitem\">Apply graphical methods of vector addition and subtraction to determine the displacement of moving objects.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div id=\"m42127-import-auto-id1165296227310\" class=\"figure\" title=\"Figure 3.8.\">\r\n<div class=\"body\">\r\n<div class=\"mediaobject\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"300\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101853\/Figure_03_02_00a.jpg\" alt=\"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.\" width=\"300\" height=\"403\" \/> Figure 1. Displacement can be determined graphically using a scale map, such as this one of the Hawaiian Islands. A journey from Hawai\u2019i to Moloka\u2019i 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)[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"section\" title=\"Vectors in Two Dimensions\">\r\n<div class=\"titlepage\">\r\n<div>\r\n<div>\r\n<h2 id=\"m42127-fs-id1165296240221\"><span class=\"cnx-gentext-section cnx-gentext-t\">Vectors in Two Dimensions<\/span><\/h2>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nA <em class=\"glossterm\"> vector<\/em><a id=\"id806190\" class=\"indexterm\"><\/a> 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\u2019s magnitude and pointing in the direction of the vector.\r\n\r\nFigure 2\u00a0shows such a <span class=\"emphasis\"><em>graphical representation of a vector<\/em><\/span>, using as an example the total displacement for the person walking in a city considered in <a class=\"link\" title=\"3.1. Kinematics in Two Dimensions: An Introduction\" href=\".\/chapter\/3-1-kinematics-in-two-dimensions-an-introduction\/\" target=\"_blank\">Kinematics in Two Dimensions: An Introduction<\/a>. We shall use the notation that a boldface symbol, such as<strong> D<\/strong>, stands for a vector. Its magnitude is represented by the symbol in italics, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>D<\/em><\/span><\/span>, and its direction by <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>\u03b8<\/em><\/span><\/span>.\r\n<div id=\"m42127-fs-id1165296218458\" class=\"note\">\r\n<div class=\"title textbox shaded\">\r\n<h3 class=\"title\"><strong><span class=\"cnx-gentext-tip-t\">Vectors in this Text<\/span><\/strong><\/h3>\r\n<div class=\"body\">\r\n\r\nIn this text, we will represent a vector with a boldface variable. For example, we will represent the quantity force with the vector<strong> F<\/strong>, which has both magnitude and direction. The magnitude of the vector will be represented by a variable in italics, such as <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>F<\/em><\/span><\/span>, and the direction of the variable will be given by an angle <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>\u03b8<\/em><\/span><\/span>.\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"title\"><\/div>\r\n<\/div>\r\n<div id=\"m42127-import-auto-id1165298666909\" class=\"figure\" title=\"Figure 3.9.\">\r\n<div class=\"body\">\r\n<div class=\"mediaobject\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101855\/Figure_03_02_01.jpg\" alt=\"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.\" width=\"325\" height=\"324\" \/> Figure 2. A person walks 9 blocks east and 5 blocks north. The displacement is 10.3 blocks at an angle 29.1\u00ba north of east.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"title\"><\/div>\r\n<\/div>\r\n<div id=\"m42127-import-auto-id1165298918248\" class=\"figure\" title=\"Figure 3.10.\">\r\n<div class=\"body\">\r\n<div class=\"mediaobject\">\r\n\r\n[caption id=\"attachment_12144\" align=\"aligncenter\" width=\"333\"]<img class=\"size-full wp-image-12144\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/648\/2016\/11\/03224936\/Figure_03_02_02a.jpg\" alt=\"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.\" width=\"333\" height=\"400\" \/> 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 <strong>D<\/strong>. Using a protractor, draw a line at an angle <em>\u03b8<\/em>\u00a0relative to the east-west axis. The length <em>D<\/em>\u00a0of the arrow is proportional to the vector\u2019s magnitude and is measured along the line with a ruler. In this example, the magnitude <em>D<\/em> of the vector is 10.3 units, and the direction <em>\u03b8<\/em> is 29.1\u00ba north of east.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"section\" title=\"Vector Addition: Head-to-Tail Method\">\r\n<div class=\"titlepage\">\r\n<div>\r\n<div>\r\n<h2 id=\"m42127-fs-id1165298995028\"><span class=\"cnx-gentext-section cnx-gentext-t\">Vector Addition: Head-to-Tail Method<\/span><\/h2>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nThe <em class=\"glossterm\"> head-to-tail method<\/em><a id=\"id807760\" class=\"indexterm\"><\/a> is a graphical way to add vectors, described in Figure 4\u00a0below and in the steps following. The <em class=\"glossterm\"> tail<\/em><a id=\"id807781\" class=\"indexterm\"><\/a> of the vector is the starting point of the vector, and the <em class=\"glossterm\"> head<\/em><a id=\"id807795\" class=\"indexterm\"><\/a> (or tip) of a vector is the final, pointed end of the arrow.\r\n<div id=\"m42127-import-auto-id1165298643218\" class=\"figure\" title=\"Figure 3.11.\">\r\n<div class=\"body\">\r\n<div class=\"mediaobject\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"500\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101858\/Figure_03_02_03.jpg\" alt=\"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.\" width=\"500\" height=\"407\" \/> 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 <strong>resultant vector D<\/strong>. The length of the arrow <strong>D<\/strong>\u00a0is proportional to the vector\u2019s magnitude and is measured to be 10.3 units. Its direction, described as the angle with respect to the east (or horizontal axis) \u03b8 is measured with a protractor to be 29. 1\u00ba.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<span class=\"bold\"><strong><span class=\"emphasis\"><em>Step 1.<\/em><\/span><\/strong><\/span> <span class=\"emphasis\"><em><span class=\"emphasis\"><em>Draw an arrow to represent the first vector (9 blocks to the east) using a ruler and protractor<\/em><\/span><\/em><\/span>.\r\n<div id=\"m42127-import-auto-id1165298876451\" class=\"figure\" title=\"Figure 3.12.\">\r\n<div class=\"body\">\r\n<div class=\"title\">\r\n\r\n[caption id=\"attachment_12145\" align=\"aligncenter\" width=\"344\"]<img class=\"size-full wp-image-12145\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/648\/2016\/11\/03225308\/Figure_03_02_04a.jpg\" alt=\"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.\" width=\"344\" height=\"390\" \/> Figure 5[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<span class=\"bold\"><strong><span class=\"emphasis\"><em>Step 2.<\/em><\/span><\/strong><\/span> Now draw an arrow to represent the second vector (5 blocks to the north). <span class=\"emphasis\"><em>Place the tail of the second vector at the head of the first vector<\/em><\/span>.\r\n<div id=\"m42127-import-auto-id1165298818267\" class=\"figure\" title=\"Figure 3.13.\">\r\n<div class=\"body\"><\/div>\r\n<div class=\"title\">\r\n\r\n[caption id=\"attachment_12146\" align=\"aligncenter\" width=\"343\"]<img class=\"size-full wp-image-12146\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/648\/2016\/11\/03225337\/Figure_03_02_05a.jpg\" alt=\"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.\" width=\"343\" height=\"390\" \/> Figure 6[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<span class=\"bold\"><strong><span class=\"emphasis\"><em>Step 3.<\/em><\/span><\/strong><\/span> <span class=\"emphasis\"><em>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<\/em><\/span>.\r\n\r\n<span class=\"bold\"><strong><span class=\"emphasis\"><em>Step 4.<\/em><\/span><\/strong><\/span> <span class=\"emphasis\"><em>Draw an arrow from the tail of the first vector to the head of the last vector<\/em><\/span>. This is the <em class=\"glossterm\"> resultant<\/em><a id=\"id808506\" class=\"indexterm\"><\/a>, or the sum, of the other vectors.\r\n\r\n[caption id=\"attachment_12147\" align=\"aligncenter\" width=\"323\"]<img class=\"size-full wp-image-12147\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/648\/2016\/11\/03225359\/Figure_03_02_06a.jpg\" alt=\"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.\" width=\"323\" height=\"400\" \/> Figure 7[\/caption]\r\n\r\n&nbsp;\r\n\r\n<span class=\"bold\"><strong><span class=\"emphasis\"><em>Step 5.<\/em><\/span><\/strong><\/span> To get the <em class=\"glossterm\"> magnitude<\/em><a id=\"id808567\" class=\"indexterm\"><\/a> of the resultant, <span class=\"emphasis\"><em>measure its length with a ruler. (Note that in most calculations, we will use the Pythagorean theorem to determine this length.)<\/em><\/span>\r\n\r\n<span class=\"bold\"><strong><span class=\"emphasis\"><em>Step 6. <\/em><\/span><\/strong><\/span>To get the <em class=\"glossterm\"> direction<\/em><a id=\"id808603\" class=\"indexterm\"><\/a> of the resultant, <span class=\"emphasis\"><em><span class=\"emphasis\"><em>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.)<\/em><\/span><\/em><\/span>\r\n\r\nThe 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.\r\n<div class=\"textbox examples\">\r\n<h3>Example 1.\u00a0Adding Vectors Graphically Using the Head-to-Tail Method: A Woman Takes a Walk<\/h3>\r\n<div class=\"body\">\r\n\r\nUse 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 <span class=\"token\">49.0\u00ba<\/span> north of east. Then, she walks 23.0 m heading <span class=\"token\">15.0\u00ba<\/span> north of east. Finally, she turns and walks 32.0 m in a direction 68.0\u00b0 south of east.\r\n<h4><span class=\"bold\"><strong>Strategy<\/strong><\/span><\/h4>\r\nRepresent each displacement vector graphically with an arrow, labeling the first <strong>A<\/strong>, the second <strong>B<\/strong>, and the third <strong>C<\/strong>, 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 <strong><span class=\"token\">R<\/span><\/strong>.\r\n<h4><span class=\"bold\"><strong>Solution<\/strong><\/span><\/h4>\r\n(1) Draw the three displacement vectors.\r\n<div id=\"m42127-import-auto-id1165296232338\" class=\"figure\" title=\"Figure 3.15.\">\r\n<div class=\"body\">\r\n<div class=\"mediaobject\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"458\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101905\/Figure_03_02_08.jpg\" alt=\"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.\" width=\"458\" height=\"154\" \/> Figure 8.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"title\"><\/div>\r\n<\/div>\r\n(2) Place the vectors head to tail retaining both their initial magnitude and direction.\r\n<div id=\"m42127-import-auto-id1165298788198\" class=\"figure\" title=\"Figure 3.16.\">\r\n<div class=\"body\">\r\n<div class=\"mediaobject\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"250\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101907\/Figure_03_02_09.jpg\" alt=\"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.\" width=\"250\" height=\"337\" \/> Figure 9.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"title\"><\/div>\r\n<\/div>\r\n(3) Draw the resultant vector, <strong>R<\/strong>.\r\n<div id=\"m42127-import-auto-id1165298786300\" class=\"figure\" title=\"Figure 3.17.\">\r\n<div class=\"body\">\r\n<div class=\"mediaobject\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"250\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101908\/Figure_03_02_10.jpg\" alt=\"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.\" width=\"250\" height=\"314\" \/> Figure 10.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"title\"><\/div>\r\n<\/div>\r\n(4) Use a ruler to measure the magnitude of <strong><span class=\"token\">R<\/span><\/strong>, and a protractor to measure the direction of <strong>R<\/strong>. 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.\r\n\r\n[caption id=\"attachment_12148\" align=\"aligncenter\" width=\"400\"]<img class=\"size-full wp-image-12148\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/648\/2016\/11\/03225534\/Figure_03_02_11a.jpg\" alt=\"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.\" width=\"400\" height=\"384\" \/> Figure 11[\/caption]\r\n\r\n&nbsp;\r\n\r\nIn this case, the total displacement <strong><span class=\"token\">R<\/span><\/strong> is seen to have a magnitude of 50.0 m and to lie in a direction <span class=\"token\">7.0\u00ba<\/span> south of east. By using its magnitude and direction, this vector can be expressed as <span class=\"token\"> <span class=\"emphasis mathml-mi\"><em>R<\/em><\/span> = 50.0 m <\/span> and <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>\u03b8\u00a0<\/em><\/span>= 7.0\u00ba<\/span> south of east.\r\n<h4><span class=\"bold\"><strong>Discussion<\/strong><\/span><\/h4>\r\nThe 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\u00a012\u00a0and we will still get the same solution.\r\n<div id=\"m42127-import-auto-id1165298931858\" class=\"figure\" title=\"Figure 3.19.\">\r\n<div class=\"body\">\r\n<div class=\"mediaobject\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"275\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101912\/Figure_03_02_12.jpg\" alt=\"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.\" width=\"275\" height=\"374\" \/> Figure 12.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"title\"><\/div>\r\n<\/div>\r\nHere, 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 <em class=\"glossterm\"> commutative<\/em><a id=\"id810445\" class=\"indexterm\"><\/a>. Vectors can be added in any order.\r\n<div id=\"m42127-eip-376\" class=\"equation\" style=\"text-align: center;\" title=\"Equation 3.1.\"><strong><span class=\"token\">A + B = B + A.<\/span><\/strong><\/div>\r\n<div class=\"equation\" style=\"text-align: center;\" title=\"Equation 3.1.\"><\/div>\r\n<div class=\"equation\" style=\"text-align: center;\" title=\"Equation 3.1.\"><\/div>\r\n(This is true for the addition of ordinary numbers as well\u2014you get the same result whether you add <strong><span class=\"token\">2 + 3<\/span><\/strong> or <strong><span class=\"token\">3 + 2<\/span><\/strong>, for example).\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p class=\"title\"><em>This video can be used for review.\u00a0 It includes vector basics - drawing vectors\/vector addition. You'll learn about the basic notion of a vector, how to add vectors together graphically, as well as what it means graphically to multiply a vector by a scalar.<\/em><\/p>\r\nhttps:\/\/youtu.be\/pimr9I92GZY\r\n<div class=\"section\" title=\"Vector Subtraction\">\r\n<div>\r\n<h2 id=\"m42127-fs-id1165298779158\"><strong><span class=\"cnx-gentext-section cnx-gentext-t\">Vector Subtraction<\/span><\/strong><\/h2>\r\n<\/div>\r\nVector subtraction is a straightforward extension of vector addition. To define subtraction (say we want to subtract <span class=\"token\">B<\/span> from <span class=\"token\">A<\/span> , written <span class=\"token\">A \u2013 B<\/span> , we must first define what we mean by subtraction. The <span class=\"emphasis\"><em>negative<\/em><\/span> of a vector <span class=\"token\">B<\/span> is defined to be <span class=\"token\">\u2013B<\/span>; that is, graphically <span class=\"emphasis\"><em>the negative of any vector has the same magnitude but the opposite direction<\/em><\/span>, as shown in Figure 13. In other words, <span class=\"token\">B<\/span> has the same length as <span class=\"token\">\u2013B<\/span>, but points in the opposite direction. Essentially, we just flip the vector so it points in the opposite direction.\r\n\r\n[caption id=\"attachment_12149\" align=\"aligncenter\" width=\"185\"]<img class=\"size-full wp-image-12149\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/648\/2016\/11\/03225634\/Figure_03_02_13a.jpg\" alt=\"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.\" width=\"185\" height=\"230\" \/> Figure 13. The negative of a vector is just another vector of the same magnitude but pointing in the opposite direction. So <strong>B<\/strong> is the negative of <strong>\u2013B<\/strong>; it has the same length but opposite direction.[\/caption]\r\n\r\nThe <span class=\"emphasis\"><em><span class=\"emphasis\"><em>subtraction<\/em><\/span><\/em><\/span> of vector <strong><span class=\"token\">B<\/span><\/strong> from vector <strong><span class=\"token\">A<\/span><\/strong> is then simply defined to be the addition of <strong><span class=\"token\">\u2013B<\/span><\/strong> to <strong><span class=\"token\">A<\/span><\/strong>. Note that vector subtraction is the addition of a negative vector. The order of subtraction does not affect the results.\r\n<p style=\"text-align: center;\"><strong>A - B = A + (-B)<\/strong><\/p>\r\nThis is analogous to the subtraction of scalars (where, for example, <span class=\"token\">5 \u2013 2 = 5 + (\u20132)<\/span>). 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.\r\n<div class=\"textbox examples\">\r\n<h3 id=\"ex2\">Example 2.\u00a0Subtracting Vectors Graphically: A Woman Sailing a Boat<\/h3>\r\nA woman sailing a boat at night is following directions to a dock. The instructions read to first sail 27.5 m in a direction <span class=\"token\">66.0\u00ba<\/span> north of east from her current location, and then travel 30.0 m in a direction <span class=\"token\">112\u00ba<\/span> north of east (or <span class=\"token\">22.0\u00ba<\/span> west of north). If the woman makes a mistake and travels in the <span class=\"emphasis\"><em><span class=\"emphasis\"><em>opposite<\/em><\/span><\/em><\/span> direction for the second leg of the trip, where will she end up? Compare this location with the location of the dock.\r\n<div id=\"m42127-import-auto-id1165296408744\" class=\"figure\" title=\"Figure 3.21.\">\r\n<div class=\"body\">\r\n<div class=\"mediaobject\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"450\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101914\/Figure_03_02_14.jpg\" alt=\"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.\" width=\"450\" height=\"259\" \/> Figure 14.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<h4><span class=\"bold\"><strong>Strategy<\/strong><\/span><\/h4>\r\nWe can represent the first leg of the trip with a vector <strong><span class=\"token\">A<\/span><\/strong>, and the second leg of the trip with a vector <strong><span class=\"token\">B<\/span><\/strong>. The dock is located at a location <strong><span class=\"token\">A + B<\/span><\/strong>. If the woman mistakenly travels in the <span class=\"emphasis\"><em>opposite<\/em><\/span> direction for the second leg of the journey, she will travel a distance <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>B<\/em><\/span><\/span> (30.0 m) in the direction <span class=\"token\">180\u00ba\u2013112\u00ba=68\u00ba<\/span> south of east. We represent this as <strong><span class=\"token\">\u2013B<\/span><\/strong>, as shown below. The vector <strong><span class=\"token\">\u2013B<\/span><\/strong> has the same magnitude as <span class=\"token\">B<\/span> but is in the opposite direction. Thus, she will end up at a location <strong><span class=\"token\">A + (\u2013B)<\/span><\/strong>, or <strong><span class=\"token\">A \u2013 B<\/span><\/strong>.\r\n\r\n[caption id=\"attachment_12150\" align=\"aligncenter\" width=\"330\"]<img class=\"size-full wp-image-12150\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/648\/2016\/11\/03225706\/Figure_03_02_15a.jpg\" alt=\"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.\" width=\"330\" height=\"282\" \/> Figure 15[\/caption]\r\n\r\n&nbsp;\r\n\r\nWe will perform vector addition to compare the location of the dock, <strong>A + B<\/strong>, with the location at which the woman mistakenly arrives, <strong>A + (-B)<\/strong>.\r\n<h4><span class=\"bold\"><strong>Solution<\/strong><\/span><\/h4>\r\n(1) To determine the location at which the woman arrives by accident, draw vectors <strong><span class=\"token\">A<\/span><\/strong> and <strong><span class=\"token\">\u2013B<\/span><\/strong>.\r\n\r\n(2) Place the vectors head to tail.\r\n\r\n(3) Draw the resultant vector <span class=\"token\"><span class=\"bold\"><strong>R<\/strong><\/span><\/span>.\r\n\r\n(4) Use a ruler and protractor to measure the magnitude and direction of <span class=\"token\"><span class=\"bold\"><strong>R<\/strong><\/span><\/span>.\r\n\r\n[caption id=\"attachment_12151\" align=\"aligncenter\" width=\"400\"]<img class=\"wp-image-12151 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/648\/2016\/11\/03225728\/Figure_03_02_16a.jpg\" alt=\"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.\" width=\"400\" height=\"218\" \/> Figure 16[\/caption]\r\n\r\nIn this case, <span class=\"token\"> <span class=\"emphasis mathml-mi\"><em>R<\/em><\/span> = 23 . 0 m <\/span> and <span class=\"token\"> <span class=\"emphasis mathml-mi\"><em>\u03b8<\/em><\/span> = 7 . 5\u00ba <\/span> south of east.\r\n\r\n(5) To determine the location of the dock, we repeat this method to add vectors <strong><span class=\"token\">A<\/span><\/strong> and <strong><span class=\"token\">B<\/span><\/strong>. We obtain the resultant vector <strong><span class=\"token\">R'<\/span><\/strong>:\r\n\r\n[caption id=\"attachment_12152\" align=\"aligncenter\" width=\"400\"]<img class=\"size-full wp-image-12152\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/648\/2016\/11\/03225801\/Figure_03_02_17a.jpg\" alt=\"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.\" width=\"400\" height=\"328\" \/> Figure 17[\/caption]\r\n\r\n&nbsp;\r\n\r\nIn this case <span class=\"token\"> <span class=\"emphasis mathml-mi\"><em>R<\/em><\/span> = 52.9 m <\/span> and <span class=\"token\"> <span class=\"emphasis mathml-mi\"><em>\u03b8<\/em><\/span> = 90.1\u00ba <\/span> north of east.\r\n\r\nWe 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.\r\n<h4><span class=\"bold\"><strong>Discussion<\/strong><\/span><\/h4>\r\nBecause 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.\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"section\" title=\"Multiplication of Vectors and Scalars\">\r\n<div class=\"titlepage\">\r\n<div>\r\n<div>\r\n<h2 id=\"m42127-fs-id1165298652611\"><span class=\"cnx-gentext-section cnx-gentext-t\">Multiplication of Vectors and Scalars<\/span><\/h2>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nIf we decided to walk three times as far on the first leg of the trip considered in the preceding example, then we would walk <span class=\"token\">3 \u00d7 27.5 m<\/span>, or 82.5 m, in a direction <span class=\"token\">66.0\u00ba<\/span> north of east. This is an example of multiplying a vector by a positive <em class=\"glossterm\"> scalar<\/em><a id=\"id816854\" class=\"indexterm\"><\/a>. Notice that the magnitude changes, but the direction stays the same.\r\n\r\nIf the scalar is negative, then multiplying a vector by it changes the vector\u2019s magnitude and gives the new vector the <span class=\"emphasis\"><em><span class=\"emphasis\"><em>opposite<\/em><\/span><\/em><\/span> direction. For example, if you multiply by \u20132, the magnitude doubles but the direction changes. We can summarize these rules in the following way: When vector <span class=\"token\"><span class=\"bold\"><strong>A<\/strong><\/span><\/span> is multiplied by a scalar <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>c<\/em><\/span><\/span>,\r\n<div class=\"itemizedlist\">\r\n<ul class=\"itemizedlist\">\r\n \t<li class=\"listitem\">the magnitude of the vector becomes the absolute value of <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>c<\/em><\/span><\/span><span class=\"token\"><span class=\"emphasis mathml-mi\"><em>A<\/em><\/span><\/span>,<\/li>\r\n \t<li class=\"listitem\">if <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>c<\/em><\/span><\/span> is positive, the direction of the vector does not change,<\/li>\r\n \t<li class=\"listitem\">if <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>c<\/em><\/span><\/span> is negative, the direction is reversed.<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn our case, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>c\u00a0<\/em><\/span>= 3<\/span> and <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>A\u00a0<\/em><\/span>= 27.5 m<\/span>. 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.\r\n\r\n<\/div>\r\n<div class=\"section\" title=\"Resolving a Vector into Components\">\r\n<div class=\"titlepage\">\r\n<div>\r\n<div>\r\n<h2 id=\"m42127-fs-id1165298819725\"><span class=\"cnx-gentext-section cnx-gentext-t\">Resolving a Vector into Components<\/span><\/h2>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nIn 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 <em class=\"glossterm\"> components <\/em><a id=\"id817877\" class=\"indexterm\"><\/a>of a single vector, for example the <span class=\"emphasis\"><em><span class=\"emphasis\"><em>x<\/em><\/span><\/em><\/span>-<span class=\"emphasis\"><em> and<\/em><\/span> <span class=\"emphasis\"><em><span class=\"emphasis\"><em>y<\/em><\/span><\/em><\/span>-components, or the north-south and east-west components.\r\n\r\nFor example, we may know that the total displacement of a person walking in a city is 10.3 blocks in a direction <span class=\"token\">29.0\u00ba<\/span> north of east and want to find out how many blocks east and north had to be walked. This method is called <span class=\"emphasis\"><em><span class=\"emphasis\"><em>finding the components (or parts)<\/em><\/span><\/em><\/span> 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 <a class=\"link\" title=\"3.4. Projectile Motion\" href=\".\/chapter\/3-4-projectile-motion\/\" target=\"_blank\">Projectile Motion<\/a>, and much more when we cover <span class=\"bold\"><strong>forces<\/strong><\/span> in Dynamics: Newton\u2019s 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 <a class=\"link\" title=\"3.3. Vector Addition and Subtraction: Analytical Methods\" href=\".\/chapter\/3-3-vector-addition-and-subtraction-analytical-methods\/\" target=\"_blank\">Vector Addition and Subtraction: Analytical Methods<\/a> are ideal for finding vector components.\r\n\r\n<\/div>\r\n<div id=\"m42127-eip-718\" class=\"note\">\r\n<div class=\"textbox\">\r\n<h2 class=\"title\"><span class=\"cnx-gentext-tip-t\">PhET Explorations: Maze Game<\/span><\/h2>\r\n<div class=\"body\">\r\n\r\nLearn 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.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"300\"]<a style=\"text-decoration: none;\" href=\"http:\/\/phet.colorado.edu\/sims\/maze-game\/maze-game_en.jnlp\"><img style=\"border: none;\" src=\"http:\/\/phet.colorado.edu\/sims\/maze-game\/maze-game-600.png\" alt=\"Maze Game screenshot\" width=\"300\" height=\"197\" \/><\/a> Click to download. Run using Java.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<section id=\"fs-id1165298622440\" class=\"section-summary\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Summary<\/h2>\r\n<ul id=\"fs-id1165298751188\">\r\n \t<li id=\"import-auto-id1165296253334\">The <strong>graphical method of adding vectors<\/strong>\u00a0<strong>A<\/strong> and <strong>B<\/strong> involves drawing vectors on a graph and adding them using the head-to-tail method. The resultant vector <strong>R<\/strong> is defined such that <strong>A + B = R<\/strong>. The magnitude and direction of <strong>R<\/strong> are then determined with a ruler and protractor, respectively.<\/li>\r\n \t<li id=\"import-auto-id1165298573640\">The <strong>graphical method of subtracting vector B<\/strong>\u00a0from <strong>A<\/strong> involves adding the opposite of vector <strong>B<\/strong>, which is defined as <strong>-B<\/strong>. In this case, <strong>A - B = A + (-B) = R<\/strong>. Then, the head-to-tail method of addition is followed in the usual way to obtain the resultant vector <strong>R<\/strong>.<\/li>\r\n \t<li id=\"import-auto-id1165296680072\">Addition of vectors is <span id=\"import-auto-id1165296680069\" data-type=\"term\">commutative<\/span> such that <strong>A + B = B + A<\/strong>.<\/li>\r\n \t<li id=\"import-auto-id1165296269519\">The <span id=\"import-auto-id1165298982089\" data-type=\"term\">head-to-tail method<\/span> 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.<\/li>\r\n \t<li id=\"import-auto-id1165298819524\">If a vector <strong>A<\/strong>\u00a0is multiplied by a scalar quantity <em>c<\/em>, the magnitude of the product is given by <em>cA<\/em>. If <em>c<\/em> is positive, the direction of the product points in the same direction as <strong>A<\/strong>; if <em>c<\/em> is negative, the direction of the product points in the opposite direction as <strong>A<\/strong>.<\/li>\r\n<\/ul>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Conceptual Questions<\/h3>\r\n1. Which of the following is a vector: a person\u2019s height, the altitude on Mt. Everest, the age of the Earth, the boiling point of water, the cost of this book, the Earth\u2019s population, the acceleration of gravity?\r\n\r\n2. Give a specific example of a vector, stating its magnitude, units, and direction.\r\n\r\n3. What do vectors and scalars have in common? How do they differ?\r\n\r\n4. 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?\r\n\r\n<figure id=\"import-auto-id1165298840401\"><span data-type=\"media\" data-alt=\"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.\"><span data-type=\"media\" data-alt=\"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.\">\r\n<\/span><\/span>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"275\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1322\/2015\/12\/03205655\/Figure_03_02_18a.jpg\" alt=\"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.\" width=\"275\" height=\"216\" data-media-type=\"image\/wmf\" \/> Figure 18.[\/caption]\r\n\r\n<\/figure>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\u00a019. What other information would he need to get to Sacramento?<span data-type=\"media\" data-alt=\"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.\"><span data-type=\"media\" data-alt=\"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.\">\r\n<\/span><\/span>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"303\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1322\/2015\/12\/03205657\/Figure_03_02_19a.jpg\" alt=\"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.\" width=\"303\" height=\"324\" data-media-type=\"image\/wmf\" \/> Figure 19.[\/caption]\r\n\r\n6. Suppose you take two steps <strong>A<\/strong> and <strong>B<\/strong> (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 <strong>A<\/strong> +<strong> B<\/strong> the sum of the lengths of the two steps?\r\n\r\n7. Explain why it is not possible to add a scalar to a vector.\r\n\r\n8. 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?\r\n\r\n<\/div>\r\n<\/section><section id=\"fs-id1165298586130\" class=\"problems-exercises\" data-depth=\"1\" data-element-type=\"problems-exercises\">\r\n<div class=\"textbox exercises\">\r\n<h3>Problems &amp; Exercises<\/h3>\r\n<p id=\"import-auto-id1165298672665\"><strong data-effect=\"bold\">Use graphical methods to solve these problems. You may assume data taken from graphs is accurate to three digits.<\/strong><\/p>\r\n1. Find the following for path A in Figure 20: (a) the total distance traveled, and (b) the magnitude and direction of the displacement from start to finish.\r\n<div id=\"fs-id1165298745593\" class=\"exercise\" data-type=\"exercise\" data-element-type=\"problems-exercises\">\r\n<div id=\"fs-id1165296363125\" class=\"problem\" data-type=\"problem\"><figure id=\"import-auto-id1165298872310\"><figcaption><\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"400\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1322\/2015\/12\/03205658\/Figure_03_02_20a.jpg\" alt=\"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.\" width=\"400\" height=\"201\" data-media-type=\"image\/wmf\" \/> Figure 20. The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side.[\/caption]\r\n\r\n<\/figure><\/div>\r\n<\/div>\r\n<div id=\"fs-id1165298474424\" class=\"exercise\" data-type=\"exercise\" data-element-type=\"problems-exercises\">\r\n<div id=\"fs-id1165298770529\" class=\"problem\" data-type=\"problem\">\r\n<p id=\"import-auto-id1165298723075\">2. Find the following for path B in Figure\u00a020: (a) the total distance traveled, and (b) the magnitude and direction of the displacement from start to finish.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165298867580\" class=\"exercise\" data-type=\"exercise\" data-element-type=\"problems-exercises\">\r\n<div id=\"fs-id1165296248676\" class=\"problem\" data-type=\"problem\">\r\n<p id=\"import-auto-id1165296365282\">3. Find the north and east components of the displacement for the hikers shown in\u00a0Figure\u00a020.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165296301840\" class=\"solution\" data-type=\"solution\">\r\n<p id=\"import-auto-id1165298650835\">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 <strong>A<\/strong> and <strong>B<\/strong>, as in Figure\u00a021, then this problem asks you to find their sum<strong> R = A + B<\/strong>.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165298536705\" class=\"exercise\" data-type=\"exercise\" data-element-type=\"problems-exercises\">\r\n<div id=\"fs-id1165296255759\" class=\"problem\" data-type=\"problem\"><figure id=\"import-auto-id1165296241785\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"275\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1322\/2015\/12\/03205700\/Figure_03_02_21a.jpg\" alt=\"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.\" width=\"275\" height=\"286\" data-media-type=\"image\/wmf\" \/> Figure 21. The two displacements <strong>A<\/strong> and <strong>B<\/strong> add to give a total displacement <strong>R<\/strong> having magnitude <em>R<\/em> and direction\u00a0<em>\u03b8<\/em>.[\/caption]\r\n\r\n<\/figure><\/div>\r\n<\/div>\r\n<div id=\"fs-id1165298797729\" class=\"exercise\" data-type=\"exercise\" data-element-type=\"problems-exercises\">\r\n<div id=\"fs-id1165298797730\" class=\"problem\" data-type=\"problem\">\r\n<p id=\"import-auto-id1165298942186\">5. Suppose you first walk 12.0 m in a direction 20 west of north and then 20.0 m in a direction 40.0\u00ba 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 <strong>A<\/strong> and <strong>B<\/strong>, as in Figure\u00a022, then this problem finds their sum <strong>R = A + B<\/strong>.)<\/p>\r\n\r\n<figure id=\"import-auto-id1165296430663\"><span data-type=\"media\" data-alt=\"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.\"><span data-type=\"media\" data-alt=\"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.\">\r\n<\/span><\/span>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"250\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1322\/2015\/12\/03205701\/Figure_03_02_22a.jpg\" alt=\"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.\" width=\"250\" height=\"279\" data-media-type=\"image\/wmf\" \/> Figure 22.[\/caption]\r\n\r\n<\/figure><\/div>\r\n<div id=\"fs-id1165298995751\" class=\"solution\" data-type=\"solution\">\r\n<p id=\"import-auto-id1165298542917\">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 <strong>B<\/strong>, which is 20.0 m in a direction exactly 40\u00ba south of west, and then leg <strong>A<\/strong>, which is 12.0 m in a direction exactly 12.0 west of north. (This problem shows that <strong>A + B = B + A<\/strong>.)<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165298849088\" class=\"exercise\" data-type=\"exercise\" data-element-type=\"problems-exercises\">7. (a) Repeat the problem two problems prior, but for the second leg you walk 20.0 m in a direction 40\u00ba north of east (which is equivalent to subtracting <strong>B<\/strong> from <strong>A<\/strong> \u2014that is, to finding <strong>R' = A - B<\/strong>). (b) Repeat the problem two problems prior, but now you first walk 20.0 m in a direction 40\u00ba south of west and then 12.0 m in a direction 20\u00ba east of south (which is equivalent to subtracting <strong>A<\/strong> from <strong>B<\/strong>\u2014that is, to finding\u00a0<strong>R'' = B - A = R'<\/strong>\u00a0Show that this is the case.<\/div>\r\n<div id=\"fs-id1165298560552\" class=\"exercise\" data-type=\"exercise\" data-element-type=\"problems-exercises\"><\/div>\r\n<div id=\"fs-id1165296576869\" class=\"exercise\" data-type=\"exercise\" data-element-type=\"problems-exercises\">\r\n<div id=\"fs-id1165296576870\" class=\"problem\" data-type=\"problem\">\r\n<p id=\"import-auto-id1165296261580\">8. Show that the <em data-effect=\"italics\"><em data-effect=\"italics\">order<\/em><\/em> of addition of three vectors does not affect their sum. Show this property by choosing any three vectors\u00a0<span id=\"MathJax-Element-2181-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-30267\" class=\"math\"><span id=\"MathJax-Span-30268\" class=\"mrow\"><span id=\"MathJax-Span-30269\" class=\"semantics\"><span id=\"MathJax-Span-30270\" class=\"mrow\"><span id=\"MathJax-Span-30271\" class=\"mrow\"><strong><span id=\"MathJax-Span-30272\" class=\"mrow\"><span id=\"MathJax-Span-30273\" class=\"mi\">A<\/span><\/span><\/strong><span id=\"MathJax-Span-30274\" class=\"mrow\"><\/span><\/span><\/span><\/span><\/span><\/span><\/span>, <span id=\"MathJax-Element-2182-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-30275\" class=\"math\"><span id=\"MathJax-Span-30276\" class=\"mrow\"><span id=\"MathJax-Span-30277\" class=\"semantics\"><span id=\"MathJax-Span-30278\" class=\"mrow\"><span id=\"MathJax-Span-30279\" class=\"mrow\"><strong><span id=\"MathJax-Span-30280\" class=\"mrow\"><span id=\"MathJax-Span-30281\" class=\"mi\">B<\/span><\/span><\/strong><span id=\"MathJax-Span-30282\" class=\"mrow\"><\/span><\/span><\/span><\/span><\/span><\/span><\/span>, and <strong><span id=\"MathJax-Element-2183-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-30283\" class=\"math\"><span id=\"MathJax-Span-30284\" class=\"mrow\"><span id=\"MathJax-Span-30285\" class=\"semantics\"><span id=\"MathJax-Span-30286\" class=\"mrow\"><span id=\"MathJax-Span-30287\" class=\"mrow\"><span id=\"MathJax-Span-30288\" class=\"mrow\"><span id=\"MathJax-Span-30289\" class=\"mi\">C<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/strong>, and [latex]\\mathbf{C}[\/latex] , all having different lengths and directions. Find the sum\u00a0<span id=\"MathJax-Element-2184-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-30291\" class=\"math\"><span id=\"MathJax-Span-30292\" class=\"mrow\"><span id=\"MathJax-Span-30293\" class=\"semantics\"><span id=\"MathJax-Span-30294\" class=\"mrow\"><span id=\"MathJax-Span-30295\" class=\"mrow\"><strong><span id=\"MathJax-Span-30296\" class=\"mrow\"><span id=\"MathJax-Span-30297\" class=\"mstyle\"><span id=\"MathJax-Span-30298\" class=\"mrow\"><span id=\"MathJax-Span-30299\" class=\"mrow\"><span id=\"MathJax-Span-30300\" class=\"mtext\">A\u00a0+\u00a0B\u00a0+\u00a0C<\/span><\/span><\/span><\/span><\/span><\/strong><span id=\"MathJax-Span-30301\" class=\"mrow\"><\/span><\/span><\/span><\/span><\/span><\/span><\/span> then find their sum when added in a different order and show the result is the same. (There are five other orders in which\u00a0<span id=\"MathJax-Element-2181-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-30267\" class=\"math\"><span id=\"MathJax-Span-30268\" class=\"mrow\"><span id=\"MathJax-Span-30269\" class=\"semantics\"><span id=\"MathJax-Span-30270\" class=\"mrow\"><span id=\"MathJax-Span-30271\" class=\"mrow\"><strong><span id=\"MathJax-Span-30272\" class=\"mrow\"><span id=\"MathJax-Span-30273\" class=\"mi\">A<\/span><\/span><\/strong><span id=\"MathJax-Span-30274\" class=\"mrow\"><\/span><\/span><\/span><\/span><\/span><\/span><\/span>, <span id=\"MathJax-Element-2182-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-30275\" class=\"math\"><span id=\"MathJax-Span-30276\" class=\"mrow\"><span id=\"MathJax-Span-30277\" class=\"semantics\"><span id=\"MathJax-Span-30278\" class=\"mrow\"><span id=\"MathJax-Span-30279\" class=\"mrow\"><strong><span id=\"MathJax-Span-30280\" class=\"mrow\"><span id=\"MathJax-Span-30281\" class=\"mi\">B<\/span><\/span><\/strong><span id=\"MathJax-Span-30282\" class=\"mrow\"><\/span><\/span><\/span><\/span><\/span><\/span><\/span>, and <strong><span id=\"MathJax-Element-2183-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-30283\" class=\"math\"><span id=\"MathJax-Span-30284\" class=\"mrow\"><span id=\"MathJax-Span-30285\" class=\"semantics\"><span id=\"MathJax-Span-30286\" class=\"mrow\"><span id=\"MathJax-Span-30287\" class=\"mrow\"><span id=\"MathJax-Span-30288\" class=\"mrow\"><span id=\"MathJax-Span-30289\" class=\"mi\">C<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/strong>,\u00a0can be added; choose only one.)<\/p>\r\n9. Show that the sum of the vectors discussed in <a title=\"jump to learn more\" href=\"#ex2\">Example 2<\/a>\u00a0gives the result shown in Figure 17.\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165298704732\" class=\"exercise\" data-type=\"exercise\" data-element-type=\"problems-exercises\">\r\n<div id=\"fs-id1165298994968\" class=\"problem\" data-type=\"problem\">\r\n<p id=\"import-auto-id1165296335201\">10. Find the magnitudes of velocities <em>V<\/em><sub>A<\/sub>\u00a0and <em>V<\/em><sub>B<\/sub>\u00a0in Figure\u00a023.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"250\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1322\/2015\/12\/03205702\/Figure_03_02_23a.jpg\" alt=\"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.\" width=\"250\" height=\"318\" data-media-type=\"image\/wmf\" \/> Figure 23. The two velocities <em>V<sub>A<\/sub><\/em>\u00a0and V<sub><em>B<\/em><\/sub>\u00a0add to give a total V<em><sub>tot<\/sub>.<\/em>[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165298735190\" class=\"exercise\" data-type=\"exercise\" data-element-type=\"problems-exercises\">\r\n<div id=\"fs-id1165296330739\" class=\"problem\" data-type=\"problem\">\r\n<p id=\"import-auto-id1165296227138\">11. Find the components of <em>v<\/em><sub>tot<\/sub>\u00a0along the <em data-effect=\"italics\"><em data-effect=\"italics\">x<\/em><\/em>- and <em data-effect=\"italics\"><em data-effect=\"italics\">y<\/em><\/em>-axes in Figure 23.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165296227129\" class=\"exercise\" data-type=\"exercise\" data-element-type=\"problems-exercises\">\r\n<div id=\"fs-id1165296227130\" class=\"problem\" data-type=\"problem\">\r\n<p id=\"import-auto-id1165298978795\">12. Find the components of <em>v<\/em><sub>tot<\/sub>\u00a0along a set of perpendicular axes rotated 30\u00ba\u00a0counterclockwise relative to those in Figure 23.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section>\r\n<div data-type=\"glossary\">\r\n<h2 data-type=\"glossary-title\">Glossary<\/h2>\r\n<dl id=\"import-auto-id1165296580417\" class=\"definition\">\r\n \t<dt>component (of a 2-d vector):<\/dt>\r\n \t<dd id=\"fs-id1165296249638\">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<\/dd>\r\n<\/dl>\r\n<dl id=\"import-auto-id1165298785711\" class=\"definition\">\r\n \t<dt>commutative:<\/dt>\r\n \t<dd id=\"fs-id1165296218975\">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<\/dd>\r\n<\/dl>\r\n<dl id=\"import-auto-id1165298785715\" class=\"definition\">\r\n \t<dt>direction (of a vector):<\/dt>\r\n \t<dd id=\"fs-id1165296376119\">the orientation of a vector in space<\/dd>\r\n<\/dl>\r\n<dl id=\"import-auto-id1165298785717\" class=\"definition\">\r\n \t<dt>head (of a vector):<\/dt>\r\n \t<dd id=\"fs-id1165298657414\">the end point of a vector; the location of the tip of the vector\u2019s arrowhead; also referred to as the \"tip\"<\/dd>\r\n<\/dl>\r\n<dl id=\"import-auto-id1165298837859\" class=\"definition\">\r\n \t<dt>head-to-tail method:<\/dt>\r\n \t<dd id=\"fs-id1165298948455\">a method of adding vectors in which the tail of each vector is placed at the head of the previous vector<\/dd>\r\n<\/dl>\r\n<dl id=\"import-auto-id1165298837863\" class=\"definition\">\r\n \t<dt>magnitude (of a vector):<\/dt>\r\n \t<dd id=\"fs-id1165298543790\">the length or size of a vector; magnitude is a scalar quantity<\/dd>\r\n<\/dl>\r\n<dl id=\"import-auto-id1165298837865\" class=\"definition\">\r\n \t<dt>resultant:<\/dt>\r\n \t<dd id=\"fs-id1165298458853\">the sum of two or more vectors<\/dd>\r\n<\/dl>\r\n<dl id=\"import-auto-id1165298863819\" class=\"definition\">\r\n \t<dt>resultant vector:<\/dt>\r\n \t<dd id=\"fs-id1165298621908\">the vector sum of two or more vectors<\/dd>\r\n<\/dl>\r\n<dl id=\"import-auto-id1165298863821\" class=\"definition\">\r\n \t<dt>scalar:<\/dt>\r\n \t<dd id=\"fs-id1165298797862\">a quantity with magnitude but no direction<\/dd>\r\n<\/dl>\r\n<dl id=\"import-auto-id1165298863823\" class=\"definition\">\r\n \t<dt>tail:<\/dt>\r\n \t<dd id=\"fs-id1165298761576\">the start point of a vector; opposite to the head or tip of the arrow<\/dd>\r\n<\/dl>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Selected Solutions to Problems &amp; Exercises<\/h3>\r\n1. (a) <span id=\"MathJax-Element-1825-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-26895\" class=\"math\"><span id=\"MathJax-Span-26896\" class=\"mrow\"><span id=\"MathJax-Span-26897\" class=\"semantics\"><span id=\"MathJax-Span-26898\" class=\"mrow\"><span id=\"MathJax-Span-26899\" class=\"mrow\"><span id=\"MathJax-Span-26900\" class=\"mrow\"><span id=\"MathJax-Span-26901\" class=\"mtext\">480 m\u00a0<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>(b) <span id=\"MathJax-Element-1826-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-26903\" class=\"math\"><span id=\"MathJax-Span-26904\" class=\"mrow\"><span id=\"MathJax-Span-26905\" class=\"semantics\"><span id=\"MathJax-Span-26906\" class=\"mrow\"><span id=\"MathJax-Span-26907\" class=\"mrow\"><span id=\"MathJax-Span-26908\" class=\"mrow\"><span id=\"MathJax-Span-26909\" class=\"mtext\">379 m<\/span><\/span><span id=\"MathJax-Span-26910\" class=\"mrow\"><\/span><\/span><\/span><\/span><\/span><\/span><\/span>, <span id=\"MathJax-Element-1827-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-26911\" class=\"math\"><span id=\"MathJax-Span-26912\" class=\"mrow\"><span id=\"MathJax-Span-26913\" class=\"semantics\"><span id=\"MathJax-Span-26914\" class=\"mrow\"><span id=\"MathJax-Span-26915\" class=\"mrow\"><span id=\"MathJax-Span-26916\" class=\"mrow\"><span id=\"MathJax-Span-26917\" class=\"mrow\"><span id=\"MathJax-Span-26918\" class=\"mtext\">18.4\u00ba<\/span><\/span><\/span><span id=\"MathJax-Span-26919\" class=\"mrow\"><\/span><\/span><\/span><\/span><\/span><\/span><\/span> east of north\r\n\r\n3.\u00a0north component 3.21 km, east component 3.83 km\r\n\r\n5. 19.5 m, 4.65\u00ba south of west\r\n\r\n7. (a) 26.6 m, 65.1\u00ba north of east (b) 26.6 m, 65.1\u00ba south of west\r\n\r\n9. 52.9 m,\u00a0<span id=\"MathJax-Element-2189-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-30337\" class=\"math\"><span id=\"MathJax-Span-30338\" class=\"mrow\"><span id=\"MathJax-Span-30339\" class=\"semantics\"><span id=\"MathJax-Span-30340\" class=\"mrow\"><span id=\"MathJax-Span-30341\" class=\"mrow\"><span id=\"MathJax-Span-30342\" class=\"mrow\"><span id=\"MathJax-Span-30343\" class=\"mrow\"><span id=\"MathJax-Span-30344\" class=\"mtext\">90<\/span><span id=\"MathJax-Span-30345\" class=\"mtext\">.<\/span><span id=\"MathJax-Span-30346\" class=\"mtext\">1\u00ba<\/span><\/span><\/span><span id=\"MathJax-Span-30347\" class=\"mrow\"><\/span><\/span><\/span><\/span><\/span><\/span><\/span> with respect to the <em data-effect=\"italics\">x<\/em>-axis.\r\n\r\n11.\u00a0<em data-effect=\"italics\">x<\/em>-component 4.41 m\/s,\u00a0<em data-effect=\"italics\">y<\/em>-component 5.07 m\/s\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>By the end of this section, you will be able to:<\/p>\n<div class=\"itemizedlist\">\n<ul class=\"itemizedlist\">\n<li class=\"listitem\">Understand the rules of vector addition, subtraction, and multiplication.<\/li>\n<li class=\"listitem\">Apply graphical methods of vector addition and subtraction to determine the displacement of moving objects.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div id=\"m42127-import-auto-id1165296227310\" class=\"figure\" title=\"Figure 3.8.\">\n<div class=\"body\">\n<div class=\"mediaobject\">\n<div style=\"width: 310px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101853\/Figure_03_02_00a.jpg\" alt=\"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.\" width=\"300\" height=\"403\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. Displacement can be determined graphically using a scale map, such as this one of the Hawaiian Islands. A journey from Hawai\u2019i to Moloka\u2019i 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)<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"section\" title=\"Vectors in Two Dimensions\">\n<div class=\"titlepage\">\n<div>\n<div>\n<h2 id=\"m42127-fs-id1165296240221\"><span class=\"cnx-gentext-section cnx-gentext-t\">Vectors in Two Dimensions<\/span><\/h2>\n<\/div>\n<\/div>\n<\/div>\n<p>A <em class=\"glossterm\"> vector<\/em><a id=\"id806190\" class=\"indexterm\"><\/a> 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\u2019s magnitude and pointing in the direction of the vector.<\/p>\n<p>Figure 2\u00a0shows such a <span class=\"emphasis\"><em>graphical representation of a vector<\/em><\/span>, using as an example the total displacement for the person walking in a city considered in <a class=\"link\" title=\"3.1. Kinematics in Two Dimensions: An Introduction\" href=\".\/chapter\/3-1-kinematics-in-two-dimensions-an-introduction\/\" target=\"_blank\">Kinematics in Two Dimensions: An Introduction<\/a>. We shall use the notation that a boldface symbol, such as<strong> D<\/strong>, stands for a vector. Its magnitude is represented by the symbol in italics, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>D<\/em><\/span><\/span>, and its direction by <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>\u03b8<\/em><\/span><\/span>.<\/p>\n<div id=\"m42127-fs-id1165296218458\" class=\"note\">\n<div class=\"title textbox shaded\">\n<h3 class=\"title\"><strong><span class=\"cnx-gentext-tip-t\">Vectors in this Text<\/span><\/strong><\/h3>\n<div class=\"body\">\n<p>In this text, we will represent a vector with a boldface variable. For example, we will represent the quantity force with the vector<strong> F<\/strong>, which has both magnitude and direction. The magnitude of the vector will be represented by a variable in italics, such as <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>F<\/em><\/span><\/span>, and the direction of the variable will be given by an angle <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>\u03b8<\/em><\/span><\/span>.<\/p>\n<\/div>\n<\/div>\n<div class=\"title\"><\/div>\n<\/div>\n<div id=\"m42127-import-auto-id1165298666909\" class=\"figure\" title=\"Figure 3.9.\">\n<div class=\"body\">\n<div class=\"mediaobject\">\n<div style=\"width: 335px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101855\/Figure_03_02_01.jpg\" alt=\"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.\" width=\"325\" height=\"324\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. A person walks 9 blocks east and 5 blocks north. The displacement is 10.3 blocks at an angle 29.1\u00ba north of east.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"title\"><\/div>\n<\/div>\n<div id=\"m42127-import-auto-id1165298918248\" class=\"figure\" title=\"Figure 3.10.\">\n<div class=\"body\">\n<div class=\"mediaobject\">\n<div id=\"attachment_12144\" style=\"width: 343px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-12144\" class=\"size-full wp-image-12144\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/648\/2016\/11\/03224936\/Figure_03_02_02a.jpg\" alt=\"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.\" width=\"333\" height=\"400\" \/><\/p>\n<p id=\"caption-attachment-12144\" class=\"wp-caption-text\">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 <strong>D<\/strong>. Using a protractor, draw a line at an angle <em>\u03b8<\/em>\u00a0relative to the east-west axis. The length <em>D<\/em>\u00a0of the arrow is proportional to the vector\u2019s magnitude and is measured along the line with a ruler. In this example, the magnitude <em>D<\/em> of the vector is 10.3 units, and the direction <em>\u03b8<\/em> is 29.1\u00ba north of east.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"section\" title=\"Vector Addition: Head-to-Tail Method\">\n<div class=\"titlepage\">\n<div>\n<div>\n<h2 id=\"m42127-fs-id1165298995028\"><span class=\"cnx-gentext-section cnx-gentext-t\">Vector Addition: Head-to-Tail Method<\/span><\/h2>\n<\/div>\n<\/div>\n<\/div>\n<p>The <em class=\"glossterm\"> head-to-tail method<\/em><a id=\"id807760\" class=\"indexterm\"><\/a> is a graphical way to add vectors, described in Figure 4\u00a0below and in the steps following. The <em class=\"glossterm\"> tail<\/em><a id=\"id807781\" class=\"indexterm\"><\/a> of the vector is the starting point of the vector, and the <em class=\"glossterm\"> head<\/em><a id=\"id807795\" class=\"indexterm\"><\/a> (or tip) of a vector is the final, pointed end of the arrow.<\/p>\n<div id=\"m42127-import-auto-id1165298643218\" class=\"figure\" title=\"Figure 3.11.\">\n<div class=\"body\">\n<div class=\"mediaobject\">\n<div style=\"width: 510px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101858\/Figure_03_02_03.jpg\" alt=\"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.\" width=\"500\" height=\"407\" \/><\/p>\n<p class=\"wp-caption-text\">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 <strong>resultant vector D<\/strong>. The length of the arrow <strong>D<\/strong>\u00a0is proportional to the vector\u2019s magnitude and is measured to be 10.3 units. Its direction, described as the angle with respect to the east (or horizontal axis) \u03b8 is measured with a protractor to be 29. 1\u00ba.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p><span class=\"bold\"><strong><span class=\"emphasis\"><em>Step 1.<\/em><\/span><\/strong><\/span> <span class=\"emphasis\"><em><span class=\"emphasis\"><em>Draw an arrow to represent the first vector (9 blocks to the east) using a ruler and protractor<\/em><\/span><\/em><\/span>.<\/p>\n<div id=\"m42127-import-auto-id1165298876451\" class=\"figure\" title=\"Figure 3.12.\">\n<div class=\"body\">\n<div class=\"title\">\n<div id=\"attachment_12145\" style=\"width: 354px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-12145\" class=\"size-full wp-image-12145\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/648\/2016\/11\/03225308\/Figure_03_02_04a.jpg\" alt=\"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.\" width=\"344\" height=\"390\" \/><\/p>\n<p id=\"caption-attachment-12145\" class=\"wp-caption-text\">Figure 5<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><span class=\"bold\"><strong><span class=\"emphasis\"><em>Step 2.<\/em><\/span><\/strong><\/span> Now draw an arrow to represent the second vector (5 blocks to the north). <span class=\"emphasis\"><em>Place the tail of the second vector at the head of the first vector<\/em><\/span>.<\/p>\n<div id=\"m42127-import-auto-id1165298818267\" class=\"figure\" title=\"Figure 3.13.\">\n<div class=\"body\"><\/div>\n<div class=\"title\">\n<div id=\"attachment_12146\" style=\"width: 353px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-12146\" class=\"size-full wp-image-12146\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/648\/2016\/11\/03225337\/Figure_03_02_05a.jpg\" alt=\"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.\" width=\"343\" height=\"390\" \/><\/p>\n<p id=\"caption-attachment-12146\" class=\"wp-caption-text\">Figure 6<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><span class=\"bold\"><strong><span class=\"emphasis\"><em>Step 3.<\/em><\/span><\/strong><\/span> <span class=\"emphasis\"><em>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<\/em><\/span>.<\/p>\n<p><span class=\"bold\"><strong><span class=\"emphasis\"><em>Step 4.<\/em><\/span><\/strong><\/span> <span class=\"emphasis\"><em>Draw an arrow from the tail of the first vector to the head of the last vector<\/em><\/span>. This is the <em class=\"glossterm\"> resultant<\/em><a id=\"id808506\" class=\"indexterm\"><\/a>, or the sum, of the other vectors.<\/p>\n<div id=\"attachment_12147\" style=\"width: 333px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-12147\" class=\"size-full wp-image-12147\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/648\/2016\/11\/03225359\/Figure_03_02_06a.jpg\" alt=\"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.\" width=\"323\" height=\"400\" \/><\/p>\n<p id=\"caption-attachment-12147\" class=\"wp-caption-text\">Figure 7<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p><span class=\"bold\"><strong><span class=\"emphasis\"><em>Step 5.<\/em><\/span><\/strong><\/span> To get the <em class=\"glossterm\"> magnitude<\/em><a id=\"id808567\" class=\"indexterm\"><\/a> of the resultant, <span class=\"emphasis\"><em>measure its length with a ruler. (Note that in most calculations, we will use the Pythagorean theorem to determine this length.)<\/em><\/span><\/p>\n<p><span class=\"bold\"><strong><span class=\"emphasis\"><em>Step 6. <\/em><\/span><\/strong><\/span>To get the <em class=\"glossterm\"> direction<\/em><a id=\"id808603\" class=\"indexterm\"><\/a> of the resultant, <span class=\"emphasis\"><em><span class=\"emphasis\"><em>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.)<\/em><\/span><\/em><\/span><\/p>\n<p>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.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 1.\u00a0Adding Vectors Graphically Using the Head-to-Tail Method: A Woman Takes a Walk<\/h3>\n<div class=\"body\">\n<p>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 <span class=\"token\">49.0\u00ba<\/span> north of east. Then, she walks 23.0 m heading <span class=\"token\">15.0\u00ba<\/span> north of east. Finally, she turns and walks 32.0 m in a direction 68.0\u00b0 south of east.<\/p>\n<h4><span class=\"bold\"><strong>Strategy<\/strong><\/span><\/h4>\n<p>Represent each displacement vector graphically with an arrow, labeling the first <strong>A<\/strong>, the second <strong>B<\/strong>, and the third <strong>C<\/strong>, 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 <strong><span class=\"token\">R<\/span><\/strong>.<\/p>\n<h4><span class=\"bold\"><strong>Solution<\/strong><\/span><\/h4>\n<p>(1) Draw the three displacement vectors.<\/p>\n<div id=\"m42127-import-auto-id1165296232338\" class=\"figure\" title=\"Figure 3.15.\">\n<div class=\"body\">\n<div class=\"mediaobject\">\n<div style=\"width: 468px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101905\/Figure_03_02_08.jpg\" alt=\"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.\" width=\"458\" height=\"154\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 8.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"title\"><\/div>\n<\/div>\n<p>(2) Place the vectors head to tail retaining both their initial magnitude and direction.<\/p>\n<div id=\"m42127-import-auto-id1165298788198\" class=\"figure\" title=\"Figure 3.16.\">\n<div class=\"body\">\n<div class=\"mediaobject\">\n<div style=\"width: 260px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101907\/Figure_03_02_09.jpg\" alt=\"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.\" width=\"250\" height=\"337\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 9.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"title\"><\/div>\n<\/div>\n<p>(3) Draw the resultant vector, <strong>R<\/strong>.<\/p>\n<div id=\"m42127-import-auto-id1165298786300\" class=\"figure\" title=\"Figure 3.17.\">\n<div class=\"body\">\n<div class=\"mediaobject\">\n<div style=\"width: 260px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101908\/Figure_03_02_10.jpg\" alt=\"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.\" width=\"250\" height=\"314\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 10.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"title\"><\/div>\n<\/div>\n<p>(4) Use a ruler to measure the magnitude of <strong><span class=\"token\">R<\/span><\/strong>, and a protractor to measure the direction of <strong>R<\/strong>. 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.<\/p>\n<div id=\"attachment_12148\" style=\"width: 410px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-12148\" class=\"size-full wp-image-12148\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/648\/2016\/11\/03225534\/Figure_03_02_11a.jpg\" alt=\"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.\" width=\"400\" height=\"384\" \/><\/p>\n<p id=\"caption-attachment-12148\" class=\"wp-caption-text\">Figure 11<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>In this case, the total displacement <strong><span class=\"token\">R<\/span><\/strong> is seen to have a magnitude of 50.0 m and to lie in a direction <span class=\"token\">7.0\u00ba<\/span> south of east. By using its magnitude and direction, this vector can be expressed as <span class=\"token\"> <span class=\"emphasis mathml-mi\"><em>R<\/em><\/span> = 50.0 m <\/span> and <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>\u03b8\u00a0<\/em><\/span>= 7.0\u00ba<\/span> south of east.<\/p>\n<h4><span class=\"bold\"><strong>Discussion<\/strong><\/span><\/h4>\n<p>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\u00a012\u00a0and we will still get the same solution.<\/p>\n<div id=\"m42127-import-auto-id1165298931858\" class=\"figure\" title=\"Figure 3.19.\">\n<div class=\"body\">\n<div class=\"mediaobject\">\n<div style=\"width: 285px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101912\/Figure_03_02_12.jpg\" alt=\"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.\" width=\"275\" height=\"374\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 12.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"title\"><\/div>\n<\/div>\n<p>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 <em class=\"glossterm\"> commutative<\/em><a id=\"id810445\" class=\"indexterm\"><\/a>. Vectors can be added in any order.<\/p>\n<div id=\"m42127-eip-376\" class=\"equation\" style=\"text-align: center;\" title=\"Equation 3.1.\"><strong><span class=\"token\">A + B = B + A.<\/span><\/strong><\/div>\n<div class=\"equation\" style=\"text-align: center;\" title=\"Equation 3.1.\"><\/div>\n<div class=\"equation\" style=\"text-align: center;\" title=\"Equation 3.1.\"><\/div>\n<p>(This is true for the addition of ordinary numbers as well\u2014you get the same result whether you add <strong><span class=\"token\">2 + 3<\/span><\/strong> or <strong><span class=\"token\">3 + 2<\/span><\/strong>, for example).<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p class=\"title\"><em>This video can be used for review.\u00a0 It includes vector basics &#8211; drawing vectors\/vector addition. You&#8217;ll learn about the basic notion of a vector, how to add vectors together graphically, as well as what it means graphically to multiply a vector by a scalar.<\/em><\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Vector Basics - Drawing Vectors\/ Vector Addition\" width=\"500\" height=\"375\" src=\"https:\/\/www.youtube.com\/embed\/pimr9I92GZY?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"section\" title=\"Vector Subtraction\">\n<div>\n<h2 id=\"m42127-fs-id1165298779158\"><strong><span class=\"cnx-gentext-section cnx-gentext-t\">Vector Subtraction<\/span><\/strong><\/h2>\n<\/div>\n<p>Vector subtraction is a straightforward extension of vector addition. To define subtraction (say we want to subtract <span class=\"token\">B<\/span> from <span class=\"token\">A<\/span> , written <span class=\"token\">A \u2013 B<\/span> , we must first define what we mean by subtraction. The <span class=\"emphasis\"><em>negative<\/em><\/span> of a vector <span class=\"token\">B<\/span> is defined to be <span class=\"token\">\u2013B<\/span>; that is, graphically <span class=\"emphasis\"><em>the negative of any vector has the same magnitude but the opposite direction<\/em><\/span>, as shown in Figure 13. In other words, <span class=\"token\">B<\/span> has the same length as <span class=\"token\">\u2013B<\/span>, but points in the opposite direction. Essentially, we just flip the vector so it points in the opposite direction.<\/p>\n<div id=\"attachment_12149\" style=\"width: 195px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-12149\" class=\"size-full wp-image-12149\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/648\/2016\/11\/03225634\/Figure_03_02_13a.jpg\" alt=\"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.\" width=\"185\" height=\"230\" \/><\/p>\n<p id=\"caption-attachment-12149\" class=\"wp-caption-text\">Figure 13. The negative of a vector is just another vector of the same magnitude but pointing in the opposite direction. So <strong>B<\/strong> is the negative of <strong>\u2013B<\/strong>; it has the same length but opposite direction.<\/p>\n<\/div>\n<p>The <span class=\"emphasis\"><em><span class=\"emphasis\"><em>subtraction<\/em><\/span><\/em><\/span> of vector <strong><span class=\"token\">B<\/span><\/strong> from vector <strong><span class=\"token\">A<\/span><\/strong> is then simply defined to be the addition of <strong><span class=\"token\">\u2013B<\/span><\/strong> to <strong><span class=\"token\">A<\/span><\/strong>. Note that vector subtraction is the addition of a negative vector. The order of subtraction does not affect the results.<\/p>\n<p style=\"text-align: center;\"><strong>A &#8211; B = A + (-B)<\/strong><\/p>\n<p>This is analogous to the subtraction of scalars (where, for example, <span class=\"token\">5 \u2013 2 = 5 + (\u20132)<\/span>). 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.<\/p>\n<div class=\"textbox examples\">\n<h3 id=\"ex2\">Example 2.\u00a0Subtracting Vectors Graphically: A Woman Sailing a Boat<\/h3>\n<p>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 <span class=\"token\">66.0\u00ba<\/span> north of east from her current location, and then travel 30.0 m in a direction <span class=\"token\">112\u00ba<\/span> north of east (or <span class=\"token\">22.0\u00ba<\/span> west of north). If the woman makes a mistake and travels in the <span class=\"emphasis\"><em><span class=\"emphasis\"><em>opposite<\/em><\/span><\/em><\/span> direction for the second leg of the trip, where will she end up? Compare this location with the location of the dock.<\/p>\n<div id=\"m42127-import-auto-id1165296408744\" class=\"figure\" title=\"Figure 3.21.\">\n<div class=\"body\">\n<div class=\"mediaobject\">\n<div style=\"width: 460px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101914\/Figure_03_02_14.jpg\" alt=\"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.\" width=\"450\" height=\"259\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 14.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h4><span class=\"bold\"><strong>Strategy<\/strong><\/span><\/h4>\n<p>We can represent the first leg of the trip with a vector <strong><span class=\"token\">A<\/span><\/strong>, and the second leg of the trip with a vector <strong><span class=\"token\">B<\/span><\/strong>. The dock is located at a location <strong><span class=\"token\">A + B<\/span><\/strong>. If the woman mistakenly travels in the <span class=\"emphasis\"><em>opposite<\/em><\/span> direction for the second leg of the journey, she will travel a distance <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>B<\/em><\/span><\/span> (30.0 m) in the direction <span class=\"token\">180\u00ba\u2013112\u00ba=68\u00ba<\/span> south of east. We represent this as <strong><span class=\"token\">\u2013B<\/span><\/strong>, as shown below. The vector <strong><span class=\"token\">\u2013B<\/span><\/strong> has the same magnitude as <span class=\"token\">B<\/span> but is in the opposite direction. Thus, she will end up at a location <strong><span class=\"token\">A + (\u2013B)<\/span><\/strong>, or <strong><span class=\"token\">A \u2013 B<\/span><\/strong>.<\/p>\n<div id=\"attachment_12150\" style=\"width: 340px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-12150\" class=\"size-full wp-image-12150\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/648\/2016\/11\/03225706\/Figure_03_02_15a.jpg\" alt=\"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.\" width=\"330\" height=\"282\" \/><\/p>\n<p id=\"caption-attachment-12150\" class=\"wp-caption-text\">Figure 15<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>We will perform vector addition to compare the location of the dock, <strong>A + B<\/strong>, with the location at which the woman mistakenly arrives, <strong>A + (-B)<\/strong>.<\/p>\n<h4><span class=\"bold\"><strong>Solution<\/strong><\/span><\/h4>\n<p>(1) To determine the location at which the woman arrives by accident, draw vectors <strong><span class=\"token\">A<\/span><\/strong> and <strong><span class=\"token\">\u2013B<\/span><\/strong>.<\/p>\n<p>(2) Place the vectors head to tail.<\/p>\n<p>(3) Draw the resultant vector <span class=\"token\"><span class=\"bold\"><strong>R<\/strong><\/span><\/span>.<\/p>\n<p>(4) Use a ruler and protractor to measure the magnitude and direction of <span class=\"token\"><span class=\"bold\"><strong>R<\/strong><\/span><\/span>.<\/p>\n<div id=\"attachment_12151\" style=\"width: 410px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-12151\" class=\"wp-image-12151 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/648\/2016\/11\/03225728\/Figure_03_02_16a.jpg\" alt=\"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.\" width=\"400\" height=\"218\" \/><\/p>\n<p id=\"caption-attachment-12151\" class=\"wp-caption-text\">Figure 16<\/p>\n<\/div>\n<p>In this case, <span class=\"token\"> <span class=\"emphasis mathml-mi\"><em>R<\/em><\/span> = 23 . 0 m <\/span> and <span class=\"token\"> <span class=\"emphasis mathml-mi\"><em>\u03b8<\/em><\/span> = 7 . 5\u00ba <\/span> south of east.<\/p>\n<p>(5) To determine the location of the dock, we repeat this method to add vectors <strong><span class=\"token\">A<\/span><\/strong> and <strong><span class=\"token\">B<\/span><\/strong>. We obtain the resultant vector <strong><span class=\"token\">R&#8217;<\/span><\/strong>:<\/p>\n<div id=\"attachment_12152\" style=\"width: 410px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-12152\" class=\"size-full wp-image-12152\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/648\/2016\/11\/03225801\/Figure_03_02_17a.jpg\" alt=\"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.\" width=\"400\" height=\"328\" \/><\/p>\n<p id=\"caption-attachment-12152\" class=\"wp-caption-text\">Figure 17<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>In this case <span class=\"token\"> <span class=\"emphasis mathml-mi\"><em>R<\/em><\/span> = 52.9 m <\/span> and <span class=\"token\"> <span class=\"emphasis mathml-mi\"><em>\u03b8<\/em><\/span> = 90.1\u00ba <\/span> north of east.<\/p>\n<p>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.<\/p>\n<h4><span class=\"bold\"><strong>Discussion<\/strong><\/span><\/h4>\n<p>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.<\/p>\n<\/div>\n<\/div>\n<div class=\"section\" title=\"Multiplication of Vectors and Scalars\">\n<div class=\"titlepage\">\n<div>\n<div>\n<h2 id=\"m42127-fs-id1165298652611\"><span class=\"cnx-gentext-section cnx-gentext-t\">Multiplication of Vectors and Scalars<\/span><\/h2>\n<\/div>\n<\/div>\n<\/div>\n<p>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 <span class=\"token\">3 \u00d7 27.5 m<\/span>, or 82.5 m, in a direction <span class=\"token\">66.0\u00ba<\/span> north of east. This is an example of multiplying a vector by a positive <em class=\"glossterm\"> scalar<\/em><a id=\"id816854\" class=\"indexterm\"><\/a>. Notice that the magnitude changes, but the direction stays the same.<\/p>\n<p>If the scalar is negative, then multiplying a vector by it changes the vector\u2019s magnitude and gives the new vector the <span class=\"emphasis\"><em><span class=\"emphasis\"><em>opposite<\/em><\/span><\/em><\/span> direction. For example, if you multiply by \u20132, the magnitude doubles but the direction changes. We can summarize these rules in the following way: When vector <span class=\"token\"><span class=\"bold\"><strong>A<\/strong><\/span><\/span> is multiplied by a scalar <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>c<\/em><\/span><\/span>,<\/p>\n<div class=\"itemizedlist\">\n<ul class=\"itemizedlist\">\n<li class=\"listitem\">the magnitude of the vector becomes the absolute value of <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>c<\/em><\/span><\/span><span class=\"token\"><span class=\"emphasis mathml-mi\"><em>A<\/em><\/span><\/span>,<\/li>\n<li class=\"listitem\">if <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>c<\/em><\/span><\/span> is positive, the direction of the vector does not change,<\/li>\n<li class=\"listitem\">if <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>c<\/em><\/span><\/span> is negative, the direction is reversed.<\/li>\n<\/ul>\n<\/div>\n<p>In our case, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>c\u00a0<\/em><\/span>= 3<\/span> and <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>A\u00a0<\/em><\/span>= 27.5 m<\/span>. 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.<\/p>\n<\/div>\n<div class=\"section\" title=\"Resolving a Vector into Components\">\n<div class=\"titlepage\">\n<div>\n<div>\n<h2 id=\"m42127-fs-id1165298819725\"><span class=\"cnx-gentext-section cnx-gentext-t\">Resolving a Vector into Components<\/span><\/h2>\n<\/div>\n<\/div>\n<\/div>\n<p>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 <em class=\"glossterm\"> components <\/em><a id=\"id817877\" class=\"indexterm\"><\/a>of a single vector, for example the <span class=\"emphasis\"><em><span class=\"emphasis\"><em>x<\/em><\/span><\/em><\/span>&#8211;<span class=\"emphasis\"><em> and<\/em><\/span> <span class=\"emphasis\"><em><span class=\"emphasis\"><em>y<\/em><\/span><\/em><\/span>-components, or the north-south and east-west components.<\/p>\n<p>For example, we may know that the total displacement of a person walking in a city is 10.3 blocks in a direction <span class=\"token\">29.0\u00ba<\/span> north of east and want to find out how many blocks east and north had to be walked. This method is called <span class=\"emphasis\"><em><span class=\"emphasis\"><em>finding the components (or parts)<\/em><\/span><\/em><\/span> 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 <a class=\"link\" title=\"3.4. Projectile Motion\" href=\".\/chapter\/3-4-projectile-motion\/\" target=\"_blank\">Projectile Motion<\/a>, and much more when we cover <span class=\"bold\"><strong>forces<\/strong><\/span> in Dynamics: Newton\u2019s 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 <a class=\"link\" title=\"3.3. Vector Addition and Subtraction: Analytical Methods\" href=\".\/chapter\/3-3-vector-addition-and-subtraction-analytical-methods\/\" target=\"_blank\">Vector Addition and Subtraction: Analytical Methods<\/a> are ideal for finding vector components.<\/p>\n<\/div>\n<div id=\"m42127-eip-718\" class=\"note\">\n<div class=\"textbox\">\n<h2 class=\"title\"><span class=\"cnx-gentext-tip-t\">PhET Explorations: Maze Game<\/span><\/h2>\n<div class=\"body\">\n<p>Learn about position, velocity, and acceleration in the &#8220;Arena of Pain&#8221;. 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.<\/p>\n<div style=\"width: 310px\" class=\"wp-caption aligncenter\"><a style=\"text-decoration: none;\" href=\"http:\/\/phet.colorado.edu\/sims\/maze-game\/maze-game_en.jnlp\"><img loading=\"lazy\" decoding=\"async\" style=\"border: none;\" src=\"http:\/\/phet.colorado.edu\/sims\/maze-game\/maze-game-600.png\" alt=\"Maze Game screenshot\" width=\"300\" height=\"197\" \/><\/a><\/p>\n<p class=\"wp-caption-text\">Click to download. Run using Java.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<section id=\"fs-id1165298622440\" class=\"section-summary\" data-depth=\"1\">\n<h2 data-type=\"title\">Summary<\/h2>\n<ul id=\"fs-id1165298751188\">\n<li id=\"import-auto-id1165296253334\">The <strong>graphical method of adding vectors<\/strong>\u00a0<strong>A<\/strong> and <strong>B<\/strong> involves drawing vectors on a graph and adding them using the head-to-tail method. The resultant vector <strong>R<\/strong> is defined such that <strong>A + B = R<\/strong>. The magnitude and direction of <strong>R<\/strong> are then determined with a ruler and protractor, respectively.<\/li>\n<li id=\"import-auto-id1165298573640\">The <strong>graphical method of subtracting vector B<\/strong>\u00a0from <strong>A<\/strong> involves adding the opposite of vector <strong>B<\/strong>, which is defined as <strong>-B<\/strong>. In this case, <strong>A &#8211; B = A + (-B) = R<\/strong>. Then, the head-to-tail method of addition is followed in the usual way to obtain the resultant vector <strong>R<\/strong>.<\/li>\n<li id=\"import-auto-id1165296680072\">Addition of vectors is <span id=\"import-auto-id1165296680069\" data-type=\"term\">commutative<\/span> such that <strong>A + B = B + A<\/strong>.<\/li>\n<li id=\"import-auto-id1165296269519\">The <span id=\"import-auto-id1165298982089\" data-type=\"term\">head-to-tail method<\/span> 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.<\/li>\n<li id=\"import-auto-id1165298819524\">If a vector <strong>A<\/strong>\u00a0is multiplied by a scalar quantity <em>c<\/em>, the magnitude of the product is given by <em>cA<\/em>. If <em>c<\/em> is positive, the direction of the product points in the same direction as <strong>A<\/strong>; if <em>c<\/em> is negative, the direction of the product points in the opposite direction as <strong>A<\/strong>.<\/li>\n<\/ul>\n<div class=\"textbox key-takeaways\">\n<h3>Conceptual Questions<\/h3>\n<p>1. Which of the following is a vector: a person\u2019s height, the altitude on Mt. Everest, the age of the Earth, the boiling point of water, the cost of this book, the Earth\u2019s population, the acceleration of gravity?<\/p>\n<p>2. Give a specific example of a vector, stating its magnitude, units, and direction.<\/p>\n<p>3. What do vectors and scalars have in common? How do they differ?<\/p>\n<p>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?<\/p>\n<figure id=\"import-auto-id1165298840401\"><span data-type=\"media\" data-alt=\"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.\"><span data-type=\"media\" data-alt=\"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.\"><br \/>\n<\/span><\/span><\/p>\n<div style=\"width: 285px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1322\/2015\/12\/03205655\/Figure_03_02_18a.jpg\" alt=\"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.\" width=\"275\" height=\"216\" data-media-type=\"image\/wmf\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 18.<\/p>\n<\/div>\n<\/figure>\n<p>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\u00a019. What other information would he need to get to Sacramento?<span data-type=\"media\" data-alt=\"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.\"><span data-type=\"media\" data-alt=\"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.\"><br \/>\n<\/span><\/span><\/p>\n<div style=\"width: 313px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1322\/2015\/12\/03205657\/Figure_03_02_19a.jpg\" alt=\"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.\" width=\"303\" height=\"324\" data-media-type=\"image\/wmf\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 19.<\/p>\n<\/div>\n<p>6. Suppose you take two steps <strong>A<\/strong> and <strong>B<\/strong> (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 <strong>A<\/strong> +<strong> B<\/strong> the sum of the lengths of the two steps?<\/p>\n<p>7. Explain why it is not possible to add a scalar to a vector.<\/p>\n<p>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?<\/p>\n<\/div>\n<\/section>\n<section id=\"fs-id1165298586130\" class=\"problems-exercises\" data-depth=\"1\" data-element-type=\"problems-exercises\">\n<div class=\"textbox exercises\">\n<h3>Problems &amp; Exercises<\/h3>\n<p id=\"import-auto-id1165298672665\"><strong data-effect=\"bold\">Use graphical methods to solve these problems. You may assume data taken from graphs is accurate to three digits.<\/strong><\/p>\n<p>1. Find the following for path A in Figure 20: (a) the total distance traveled, and (b) the magnitude and direction of the displacement from start to finish.<\/p>\n<div id=\"fs-id1165298745593\" class=\"exercise\" data-type=\"exercise\" data-element-type=\"problems-exercises\">\n<div id=\"fs-id1165296363125\" class=\"problem\" data-type=\"problem\">\n<figure id=\"import-auto-id1165298872310\"><figcaption><\/figcaption><div style=\"width: 410px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1322\/2015\/12\/03205658\/Figure_03_02_20a.jpg\" alt=\"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.\" width=\"400\" height=\"201\" data-media-type=\"image\/wmf\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 20. The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side.<\/p>\n<\/div>\n<\/figure>\n<\/div>\n<\/div>\n<div id=\"fs-id1165298474424\" class=\"exercise\" data-type=\"exercise\" data-element-type=\"problems-exercises\">\n<div id=\"fs-id1165298770529\" class=\"problem\" data-type=\"problem\">\n<p id=\"import-auto-id1165298723075\">2. Find the following for path B in Figure\u00a020: (a) the total distance traveled, and (b) the magnitude and direction of the displacement from start to finish.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165298867580\" class=\"exercise\" data-type=\"exercise\" data-element-type=\"problems-exercises\">\n<div id=\"fs-id1165296248676\" class=\"problem\" data-type=\"problem\">\n<p id=\"import-auto-id1165296365282\">3. Find the north and east components of the displacement for the hikers shown in\u00a0Figure\u00a020.<\/p>\n<\/div>\n<div id=\"fs-id1165296301840\" class=\"solution\" data-type=\"solution\">\n<p id=\"import-auto-id1165298650835\">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 <strong>A<\/strong> and <strong>B<\/strong>, as in Figure\u00a021, then this problem asks you to find their sum<strong> R = A + B<\/strong>.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165298536705\" class=\"exercise\" data-type=\"exercise\" data-element-type=\"problems-exercises\">\n<div id=\"fs-id1165296255759\" class=\"problem\" data-type=\"problem\">\n<figure id=\"import-auto-id1165296241785\">\n<div style=\"width: 285px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1322\/2015\/12\/03205700\/Figure_03_02_21a.jpg\" alt=\"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.\" width=\"275\" height=\"286\" data-media-type=\"image\/wmf\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 21. The two displacements <strong>A<\/strong> and <strong>B<\/strong> add to give a total displacement <strong>R<\/strong> having magnitude <em>R<\/em> and direction\u00a0<em>\u03b8<\/em>.<\/p>\n<\/div>\n<\/figure>\n<\/div>\n<\/div>\n<div id=\"fs-id1165298797729\" class=\"exercise\" data-type=\"exercise\" data-element-type=\"problems-exercises\">\n<div id=\"fs-id1165298797730\" class=\"problem\" data-type=\"problem\">\n<p id=\"import-auto-id1165298942186\">5. Suppose you first walk 12.0 m in a direction 20 west of north and then 20.0 m in a direction 40.0\u00ba 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 <strong>A<\/strong> and <strong>B<\/strong>, as in Figure\u00a022, then this problem finds their sum <strong>R = A + B<\/strong>.)<\/p>\n<figure id=\"import-auto-id1165296430663\"><span data-type=\"media\" data-alt=\"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.\"><span data-type=\"media\" data-alt=\"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.\"><br \/>\n<\/span><\/span><\/p>\n<div style=\"width: 260px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1322\/2015\/12\/03205701\/Figure_03_02_22a.jpg\" alt=\"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.\" width=\"250\" height=\"279\" data-media-type=\"image\/wmf\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 22.<\/p>\n<\/div>\n<\/figure>\n<\/div>\n<div id=\"fs-id1165298995751\" class=\"solution\" data-type=\"solution\">\n<p id=\"import-auto-id1165298542917\">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 <strong>B<\/strong>, which is 20.0 m in a direction exactly 40\u00ba south of west, and then leg <strong>A<\/strong>, which is 12.0 m in a direction exactly 12.0 west of north. (This problem shows that <strong>A + B = B + A<\/strong>.)<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165298849088\" class=\"exercise\" data-type=\"exercise\" data-element-type=\"problems-exercises\">7. (a) Repeat the problem two problems prior, but for the second leg you walk 20.0 m in a direction 40\u00ba north of east (which is equivalent to subtracting <strong>B<\/strong> from <strong>A<\/strong> \u2014that is, to finding <strong>R&#8217; = A &#8211; B<\/strong>). (b) Repeat the problem two problems prior, but now you first walk 20.0 m in a direction 40\u00ba south of west and then 12.0 m in a direction 20\u00ba east of south (which is equivalent to subtracting <strong>A<\/strong> from <strong>B<\/strong>\u2014that is, to finding\u00a0<strong>R&#8221; = B &#8211; A = R&#8217;<\/strong>\u00a0Show that this is the case.<\/div>\n<div id=\"fs-id1165298560552\" class=\"exercise\" data-type=\"exercise\" data-element-type=\"problems-exercises\"><\/div>\n<div id=\"fs-id1165296576869\" class=\"exercise\" data-type=\"exercise\" data-element-type=\"problems-exercises\">\n<div id=\"fs-id1165296576870\" class=\"problem\" data-type=\"problem\">\n<p id=\"import-auto-id1165296261580\">8. Show that the <em data-effect=\"italics\"><em data-effect=\"italics\">order<\/em><\/em> of addition of three vectors does not affect their sum. Show this property by choosing any three vectors\u00a0<span id=\"MathJax-Element-2181-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-30267\" class=\"math\"><span id=\"MathJax-Span-30268\" class=\"mrow\"><span id=\"MathJax-Span-30269\" class=\"semantics\"><span id=\"MathJax-Span-30270\" class=\"mrow\"><span id=\"MathJax-Span-30271\" class=\"mrow\"><strong><span id=\"MathJax-Span-30272\" class=\"mrow\"><span id=\"MathJax-Span-30273\" class=\"mi\">A<\/span><\/span><\/strong><span id=\"MathJax-Span-30274\" class=\"mrow\"><\/span><\/span><\/span><\/span><\/span><\/span><\/span>, <span id=\"MathJax-Element-2182-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-30275\" class=\"math\"><span id=\"MathJax-Span-30276\" class=\"mrow\"><span id=\"MathJax-Span-30277\" class=\"semantics\"><span id=\"MathJax-Span-30278\" class=\"mrow\"><span id=\"MathJax-Span-30279\" class=\"mrow\"><strong><span id=\"MathJax-Span-30280\" class=\"mrow\"><span id=\"MathJax-Span-30281\" class=\"mi\">B<\/span><\/span><\/strong><span id=\"MathJax-Span-30282\" class=\"mrow\"><\/span><\/span><\/span><\/span><\/span><\/span><\/span>, and <strong><span id=\"MathJax-Element-2183-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-30283\" class=\"math\"><span id=\"MathJax-Span-30284\" class=\"mrow\"><span id=\"MathJax-Span-30285\" class=\"semantics\"><span id=\"MathJax-Span-30286\" class=\"mrow\"><span id=\"MathJax-Span-30287\" class=\"mrow\"><span id=\"MathJax-Span-30288\" class=\"mrow\"><span id=\"MathJax-Span-30289\" class=\"mi\">C<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/strong>, and [latex]\\mathbf{C}[\/latex] , all having different lengths and directions. Find the sum\u00a0<span id=\"MathJax-Element-2184-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-30291\" class=\"math\"><span id=\"MathJax-Span-30292\" class=\"mrow\"><span id=\"MathJax-Span-30293\" class=\"semantics\"><span id=\"MathJax-Span-30294\" class=\"mrow\"><span id=\"MathJax-Span-30295\" class=\"mrow\"><strong><span id=\"MathJax-Span-30296\" class=\"mrow\"><span id=\"MathJax-Span-30297\" class=\"mstyle\"><span id=\"MathJax-Span-30298\" class=\"mrow\"><span id=\"MathJax-Span-30299\" class=\"mrow\"><span id=\"MathJax-Span-30300\" class=\"mtext\">A\u00a0+\u00a0B\u00a0+\u00a0C<\/span><\/span><\/span><\/span><\/span><\/strong><span id=\"MathJax-Span-30301\" class=\"mrow\"><\/span><\/span><\/span><\/span><\/span><\/span><\/span> then find their sum when added in a different order and show the result is the same. (There are five other orders in which\u00a0<span id=\"MathJax-Element-2181-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-30267\" class=\"math\"><span id=\"MathJax-Span-30268\" class=\"mrow\"><span id=\"MathJax-Span-30269\" class=\"semantics\"><span id=\"MathJax-Span-30270\" class=\"mrow\"><span id=\"MathJax-Span-30271\" class=\"mrow\"><strong><span id=\"MathJax-Span-30272\" class=\"mrow\"><span id=\"MathJax-Span-30273\" class=\"mi\">A<\/span><\/span><\/strong><span id=\"MathJax-Span-30274\" class=\"mrow\"><\/span><\/span><\/span><\/span><\/span><\/span><\/span>, <span id=\"MathJax-Element-2182-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-30275\" class=\"math\"><span id=\"MathJax-Span-30276\" class=\"mrow\"><span id=\"MathJax-Span-30277\" class=\"semantics\"><span id=\"MathJax-Span-30278\" class=\"mrow\"><span id=\"MathJax-Span-30279\" class=\"mrow\"><strong><span id=\"MathJax-Span-30280\" class=\"mrow\"><span id=\"MathJax-Span-30281\" class=\"mi\">B<\/span><\/span><\/strong><span id=\"MathJax-Span-30282\" class=\"mrow\"><\/span><\/span><\/span><\/span><\/span><\/span><\/span>, and <strong><span id=\"MathJax-Element-2183-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-30283\" class=\"math\"><span id=\"MathJax-Span-30284\" class=\"mrow\"><span id=\"MathJax-Span-30285\" class=\"semantics\"><span id=\"MathJax-Span-30286\" class=\"mrow\"><span id=\"MathJax-Span-30287\" class=\"mrow\"><span id=\"MathJax-Span-30288\" class=\"mrow\"><span id=\"MathJax-Span-30289\" class=\"mi\">C<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/strong>,\u00a0can be added; choose only one.)<\/p>\n<p>9. Show that the sum of the vectors discussed in <a title=\"jump to learn more\" href=\"#ex2\">Example 2<\/a>\u00a0gives the result shown in Figure 17.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165298704732\" class=\"exercise\" data-type=\"exercise\" data-element-type=\"problems-exercises\">\n<div id=\"fs-id1165298994968\" class=\"problem\" data-type=\"problem\">\n<p id=\"import-auto-id1165296335201\">10. Find the magnitudes of velocities <em>V<\/em><sub>A<\/sub>\u00a0and <em>V<\/em><sub>B<\/sub>\u00a0in Figure\u00a023.<\/p>\n<div style=\"width: 260px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1322\/2015\/12\/03205702\/Figure_03_02_23a.jpg\" alt=\"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.\" width=\"250\" height=\"318\" data-media-type=\"image\/wmf\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 23. The two velocities <em>V<sub>A<\/sub><\/em>\u00a0and V<sub><em>B<\/em><\/sub>\u00a0add to give a total V<em><sub>tot<\/sub>.<\/em><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165298735190\" class=\"exercise\" data-type=\"exercise\" data-element-type=\"problems-exercises\">\n<div id=\"fs-id1165296330739\" class=\"problem\" data-type=\"problem\">\n<p id=\"import-auto-id1165296227138\">11. Find the components of <em>v<\/em><sub>tot<\/sub>\u00a0along the <em data-effect=\"italics\"><em data-effect=\"italics\">x<\/em><\/em>&#8211; and <em data-effect=\"italics\"><em data-effect=\"italics\">y<\/em><\/em>-axes in Figure 23.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165296227129\" class=\"exercise\" data-type=\"exercise\" data-element-type=\"problems-exercises\">\n<div id=\"fs-id1165296227130\" class=\"problem\" data-type=\"problem\">\n<p id=\"import-auto-id1165298978795\">12. Find the components of <em>v<\/em><sub>tot<\/sub>\u00a0along a set of perpendicular axes rotated 30\u00ba\u00a0counterclockwise relative to those in Figure 23.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<div data-type=\"glossary\">\n<h2 data-type=\"glossary-title\">Glossary<\/h2>\n<dl id=\"import-auto-id1165296580417\" class=\"definition\">\n<dt>component (of a 2-d vector):<\/dt>\n<dd id=\"fs-id1165296249638\">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<\/dd>\n<\/dl>\n<dl id=\"import-auto-id1165298785711\" class=\"definition\">\n<dt>commutative:<\/dt>\n<dd id=\"fs-id1165296218975\">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<\/dd>\n<\/dl>\n<dl id=\"import-auto-id1165298785715\" class=\"definition\">\n<dt>direction (of a vector):<\/dt>\n<dd id=\"fs-id1165296376119\">the orientation of a vector in space<\/dd>\n<\/dl>\n<dl id=\"import-auto-id1165298785717\" class=\"definition\">\n<dt>head (of a vector):<\/dt>\n<dd id=\"fs-id1165298657414\">the end point of a vector; the location of the tip of the vector\u2019s arrowhead; also referred to as the &#8220;tip&#8221;<\/dd>\n<\/dl>\n<dl id=\"import-auto-id1165298837859\" class=\"definition\">\n<dt>head-to-tail method:<\/dt>\n<dd id=\"fs-id1165298948455\">a method of adding vectors in which the tail of each vector is placed at the head of the previous vector<\/dd>\n<\/dl>\n<dl id=\"import-auto-id1165298837863\" class=\"definition\">\n<dt>magnitude (of a vector):<\/dt>\n<dd id=\"fs-id1165298543790\">the length or size of a vector; magnitude is a scalar quantity<\/dd>\n<\/dl>\n<dl id=\"import-auto-id1165298837865\" class=\"definition\">\n<dt>resultant:<\/dt>\n<dd id=\"fs-id1165298458853\">the sum of two or more vectors<\/dd>\n<\/dl>\n<dl id=\"import-auto-id1165298863819\" class=\"definition\">\n<dt>resultant vector:<\/dt>\n<dd id=\"fs-id1165298621908\">the vector sum of two or more vectors<\/dd>\n<\/dl>\n<dl id=\"import-auto-id1165298863821\" class=\"definition\">\n<dt>scalar:<\/dt>\n<dd id=\"fs-id1165298797862\">a quantity with magnitude but no direction<\/dd>\n<\/dl>\n<dl id=\"import-auto-id1165298863823\" class=\"definition\">\n<dt>tail:<\/dt>\n<dd id=\"fs-id1165298761576\">the start point of a vector; opposite to the head or tip of the arrow<\/dd>\n<\/dl>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Selected Solutions to Problems &amp; Exercises<\/h3>\n<p>1. (a) <span id=\"MathJax-Element-1825-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-26895\" class=\"math\"><span id=\"MathJax-Span-26896\" class=\"mrow\"><span id=\"MathJax-Span-26897\" class=\"semantics\"><span id=\"MathJax-Span-26898\" class=\"mrow\"><span id=\"MathJax-Span-26899\" class=\"mrow\"><span id=\"MathJax-Span-26900\" class=\"mrow\"><span id=\"MathJax-Span-26901\" class=\"mtext\">480 m\u00a0<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>(b) <span id=\"MathJax-Element-1826-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-26903\" class=\"math\"><span id=\"MathJax-Span-26904\" class=\"mrow\"><span id=\"MathJax-Span-26905\" class=\"semantics\"><span id=\"MathJax-Span-26906\" class=\"mrow\"><span id=\"MathJax-Span-26907\" class=\"mrow\"><span id=\"MathJax-Span-26908\" class=\"mrow\"><span id=\"MathJax-Span-26909\" class=\"mtext\">379 m<\/span><\/span><span id=\"MathJax-Span-26910\" class=\"mrow\"><\/span><\/span><\/span><\/span><\/span><\/span><\/span>, <span id=\"MathJax-Element-1827-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-26911\" class=\"math\"><span id=\"MathJax-Span-26912\" class=\"mrow\"><span id=\"MathJax-Span-26913\" class=\"semantics\"><span id=\"MathJax-Span-26914\" class=\"mrow\"><span id=\"MathJax-Span-26915\" class=\"mrow\"><span id=\"MathJax-Span-26916\" class=\"mrow\"><span id=\"MathJax-Span-26917\" class=\"mrow\"><span id=\"MathJax-Span-26918\" class=\"mtext\">18.4\u00ba<\/span><\/span><\/span><span id=\"MathJax-Span-26919\" class=\"mrow\"><\/span><\/span><\/span><\/span><\/span><\/span><\/span> east of north<\/p>\n<p>3.\u00a0north component 3.21 km, east component 3.83 km<\/p>\n<p>5. 19.5 m, 4.65\u00ba south of west<\/p>\n<p>7. (a) 26.6 m, 65.1\u00ba north of east (b) 26.6 m, 65.1\u00ba south of west<\/p>\n<p>9. 52.9 m,\u00a0<span id=\"MathJax-Element-2189-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-30337\" class=\"math\"><span id=\"MathJax-Span-30338\" class=\"mrow\"><span id=\"MathJax-Span-30339\" class=\"semantics\"><span id=\"MathJax-Span-30340\" class=\"mrow\"><span id=\"MathJax-Span-30341\" class=\"mrow\"><span id=\"MathJax-Span-30342\" class=\"mrow\"><span id=\"MathJax-Span-30343\" class=\"mrow\"><span id=\"MathJax-Span-30344\" class=\"mtext\">90<\/span><span id=\"MathJax-Span-30345\" class=\"mtext\">.<\/span><span id=\"MathJax-Span-30346\" class=\"mtext\">1\u00ba<\/span><\/span><\/span><span id=\"MathJax-Span-30347\" class=\"mrow\"><\/span><\/span><\/span><\/span><\/span><\/span><\/span> with respect to the <em data-effect=\"italics\">x<\/em>-axis.<\/p>\n<p>11.\u00a0<em data-effect=\"italics\">x<\/em>-component 4.41 m\/s,\u00a0<em data-effect=\"italics\">y<\/em>-component 5.07 m\/s<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-449\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Physics. <strong>Authored by<\/strong>: OpenStax College. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/031da8d3-b525-429c-80cf-6c8ed997733a\/College_Physics\">http:\/\/cnx.org\/contents\/031da8d3-b525-429c-80cf-6c8ed997733a\/College_Physics<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Located at License<\/li><li>PhET Interactive Simulations . <strong>Provided by<\/strong>: University of Colorado Boulder. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/phet.colorado.edu\">http:\/\/phet.colorado.edu<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">All rights reserved content<\/div><ul class=\"citation-list\"><li>Vector Basics - Drawing Vectors\/ Vector Addition . <strong>Authored by<\/strong>: patrickJMT. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/www.youtube.com\/watch?v=XoTx7Rt4dig&#038;index=40&#038;list=PL8dPuuaLjXtOPRKzVLY0jJY-uHOH9KVU6\">https:\/\/www.youtube.com\/watch?v=XoTx7Rt4dig&#038;index=40&#038;list=PL8dPuuaLjXtOPRKzVLY0jJY-uHOH9KVU6<\/a>. <strong>License<\/strong>: <em>All Rights Reserved<\/em>. <strong>License Terms<\/strong>: Standard YouTube license<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":1,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"College Physics\",\"author\":\"OpenStax College\",\"organization\":\"\",\"url\":\"http:\/\/cnx.org\/contents\/031da8d3-b525-429c-80cf-6c8ed997733a\/College_Physics\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Located at License\"},{\"type\":\"copyrighted_video\",\"description\":\"Vector Basics - Drawing Vectors\/ Vector Addition \",\"author\":\"patrickJMT\",\"organization\":\"\",\"url\":\"https:\/\/www.youtube.com\/watch?v=XoTx7Rt4dig&index=40&list=PL8dPuuaLjXtOPRKzVLY0jJY-uHOH9KVU6\",\"project\":\"\",\"license\":\"arr\",\"license_terms\":\"Standard YouTube license\"},{\"type\":\"cc\",\"description\":\"PhET Interactive Simulations \",\"author\":\"\",\"organization\":\"University of Colorado Boulder\",\"url\":\" http:\/\/phet.colorado.edu\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-449","chapter","type-chapter","status-publish","hentry"],"part":7457,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-physics\/wp-json\/pressbooks\/v2\/chapters\/449","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-physics\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-physics\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-physics\/wp-json\/wp\/v2\/users\/1"}],"version-history":[{"count":31,"href":"https:\/\/courses.lumenlearning.com\/suny-physics\/wp-json\/pressbooks\/v2\/chapters\/449\/revisions"}],"predecessor-version":[{"id":12153,"href":"https:\/\/courses.lumenlearning.com\/suny-physics\/wp-json\/pressbooks\/v2\/chapters\/449\/revisions\/12153"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-physics\/wp-json\/pressbooks\/v2\/parts\/7457"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-physics\/wp-json\/pressbooks\/v2\/chapters\/449\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-physics\/wp-json\/wp\/v2\/media?parent=449"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-physics\/wp-json\/pressbooks\/v2\/chapter-type?post=449"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-physics\/wp-json\/wp\/v2\/contributor?post=449"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-physics\/wp-json\/wp\/v2\/license?post=449"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}