{"id":180,"date":"2018-02-06T15:25:12","date_gmt":"2018-02-06T15:25:12","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/?post_type=chapter&#038;p=180"},"modified":"2018-07-04T14:49:44","modified_gmt":"2018-07-04T14:49:44","slug":"2-2-coordinate-systems-and-components-of-a-vector","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/chapter\/2-2-coordinate-systems-and-components-of-a-vector\/","title":{"raw":"2.2 Coordinate Systems and Components of a Vector","rendered":"2.2 Coordinate Systems and Components of a Vector"},"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<ul>\r\n \t<li>Describe vectors in two and three dimensions in terms of their components, using unit vectors along the axes.<\/li>\r\n \t<li>Distinguish between the vector components of a vector and the scalar components of a vector.<\/li>\r\n \t<li>Explain how the magnitude of a vector is defined in terms of the components of a vector.<\/li>\r\n \t<li>Identify the direction angle of a vector in a plane.<\/li>\r\n \t<li>Explain the connection between polar coordinates and Cartesian coordinates in a plane.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1167133446629\">Vectors are usually described in terms of their components in a <strong><span class=\"no-emphasis\">coordinate system<\/span><\/strong>. Even in everyday life we naturally invoke the concept of orthogonal projections in a rectangular coordinate system. For example, if you ask someone for directions to a particular location, you will more likely be told to go 40 km east and 30 km north than 50 km in the direction $$ 37\\text{\u00b0} $$ north of east.<\/p>\r\n<p id=\"fs-id1167133855194\">In a rectangular (Cartesian) <em>xy<\/em>-coordinate system in a plane, a point in a plane is described by a pair of coordinates (<em>x<\/em>, <em>y<\/em>). In a similar fashion, a vector $$ \\overset{\\to }{A} $$ in a plane is described by a pair of its <em>vector<\/em> coordinates. The <em>x<\/em>-coordinate of vector $$ \\overset{\\to }{A} $$ is called its <em>x<\/em>-component and the <em>y<\/em>-coordinate of vector $$ \\overset{\\to }{A} $$ is called its <em>y<\/em>-component. The vector <em>x<\/em>-component is a vector denoted by $$ {\\overset{\\to }{A}}_{x}$$. The vector <em>y<\/em>-component is a vector denoted by $$ {\\overset{\\to }{A}}_{y}$$. In the Cartesian system, the <em>x<\/em> and <em>y<\/em> <strong>vector components<\/strong> of a vector are the orthogonal projections of this vector onto the <em>x<\/em>- and <em>y<\/em>-axes, respectively. In this way, following the parallelogram rule for vector addition, each vector on a Cartesian plane can be expressed as the vector sum of its vector components:<\/p>\r\n\r\n<div id=\"fs-id1167133635848\">$$\\overset{\\to }{A}={\\overset{\\to }{A}}_{x}+{\\overset{\\to }{A}}_{y}.$$<\/div>\r\n<p id=\"fs-id1167132574978\">As illustrated in <a class=\"autogenerated-content\" href=\"#CNX_UPhysics_02_02_comp01\">(Figure)<\/a>, vector $$ \\overset{\\to }{A} $$ is the diagonal of the rectangle where the <em>x<\/em>-component $$ {\\overset{\\to }{A}}_{x} $$ is the side parallel to the <em>x<\/em>-axis and the <em>y<\/em>-component $$ {\\overset{\\to }{A}}_{y} $$ is the side parallel to the <em>y<\/em>-axis. Vector component $$ {\\overset{\\to }{A}}_{x} $$ is orthogonal to vector component $$ {\\overset{\\to }{A}}_{y}$$.<\/p>\r\n\r\n<div id=\"CNX_UPhysics_02_02_comp01\" class=\"wp-caption aligncenter\">[caption id=\"\" align=\"aligncenter\" width=\"530\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31183900\/CNX_UPhysics_02_02_comp01.jpg\" alt=\"Vector A is shown in the x y coordinate system and extends from point b at A\u2019s tail to point e and its head. Vector A points up and to the right. Unit vectors I hat and j hat are small vectors pointing in the x and y directions, respectively, and are at right angles to each other. The x component of vector A is a vector pointing horizontally from the point b to a point directly below point e at the tip of vector A. On the x axis, we see that the vector A sub x extends from x sub b to x sub e and is equal to magnitude A sub x times I hat. The magnitude A sub x equals x sub e minus x sub b. The y component of vector A is a vector pointing vertically from point b to a point directly to the left of point e at the tip of vector A. On the y axis, we see that the vector A sub y extends from y sub b to y sub e and is equal to magnitude A sub y times j hat. The magnitude A sub y equals y sub e minus y sub b.\" width=\"530\" height=\"394\" \/> <strong>Figure 2.16<\/strong> Vector $$ \\overset{\\to }{A} $$ in a plane in the Cartesian coordinate system is the vector sum of its vector x- and y-components. The x-vector component $$ {\\overset{\\to }{A}}_{x} $$ is the orthogonal projection of vector $$ \\overset{\\to }{A} $$ onto the x-axis. The y-vector component $$ {\\overset{\\to }{A}}_{y} $$ is the orthogonal projection of vector $$ \\overset{\\to }{A} $$ onto the y-axis. The numbers $$ {A}_{x} $$ and $$ {A}_{y} $$ that multiply the unit vectors are the scalar components of the vector.[\/caption]<\/div>\r\n<p id=\"fs-id1167132422244\">It is customary to denote the positive direction on the <em>x<\/em>-axis by the unit vector $$ \\hat{i} $$ and the positive direction on the <em>y<\/em>-axis by the unit vector $$ \\hat{j}$$. <strong>Unit vectors of the axes<\/strong>, $$ \\hat{i} $$ and $$ \\hat{j}$$, define two orthogonal directions in the plane. As shown in <a class=\"autogenerated-content\" href=\"#CNX_UPhysics_02_02_comp01\">(Figure)<\/a>, the <em>x<\/em>- and <em>y<\/em>- components of a vector can now be written in terms of the unit vectors of the axes:<\/p>\r\n\r\n<div id=\"fs-id1167132230873\">$$\\{\\begin{array}{c}{\\overset{\\to }{A}}_{x}={A}_{x}\\hat{i}\\\\ {\\overset{\\to }{A}}_{y}={A}_{y}\\hat{j}.\\end{array}$$<\/div>\r\n<p id=\"fs-id1167133345023\">The vectors $$ {\\overset{\\to }{A}}_{x} $$ and $$ {\\overset{\\to }{A}}_{y} $$ defined by <a class=\"autogenerated-content\" href=\"#fs-id1167132230873\">(Figure)<\/a> are the <em>vector components<\/em> of vector $$ \\overset{\\to }{A}$$. The numbers $$ {A}_{x} $$ and $$ {A}_{y} $$ that define the vector components in <a class=\"autogenerated-content\" href=\"#fs-id1167132230873\">(Figure)<\/a> are the scalar components of vector $$ \\overset{\\to }{A}$$. Combining <a class=\"autogenerated-content\" href=\"#fs-id1167133635848\">(Figure)<\/a> with <a class=\"autogenerated-content\" href=\"#fs-id1167132230873\">(Figure)<\/a>, we obtain <strong>the component form of a vector:<\/strong><\/p>\r\n\r\n<div id=\"fs-id1167132413666\" class=\"equation-callout\">\r\n<div id=\"fs-id1167132256590\">$$\\overset{\\to }{A}={A}_{x}\\hat{i}+{A}_{y}\\hat{j}.$$<\/div>\r\n<\/div>\r\n<p id=\"fs-id1167132689436\">If we know the coordinates $$ b({x}_{b},{y}_{b}) $$ of the origin point of a vector (where <em>b<\/em> stands for \u201cbeginning\u201d) and the coordinates $$ e({x}_{e},{y}_{e}) $$ of the end point of a vector (where <em>e<\/em> stands for \u201cend\u201d), we can obtain the scalar components of a vector simply by subtracting the origin point coordinates from the end point coordinates:<\/p>\r\n\r\n<div id=\"fs-id1167132581667\" class=\"equation-callout\">\r\n<div id=\"fs-id1167132197498\">$$\\{\\begin{array}{c}{A}_{x}={x}_{e}-{x}_{b}\\\\ {A}_{y}={y}_{e}-{y}_{b}.\\end{array}$$<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167132341727\" class=\"textbox examples\">\r\n<h3>Example<\/h3>\r\n<h4>Displacement of a Mouse Pointer<\/h4>\r\nA mouse pointer on the display monitor of a computer at its initial position is at point (6.0 cm, 1.6 cm) with respect to the lower left-side corner. If you move the pointer to an icon located at point (2.0 cm, 4.5 cm), what is the displacement vector of the pointer?\r\n<h4>Strategy<\/h4>\r\nThe origin of the <em>xy<\/em>-coordinate system is the lower left-side corner of the computer monitor. Therefore, the unit vector $$ \\hat{i} $$ on the <em>x<\/em>-axis points horizontally to the right and the unit vector $$ \\hat{j} $$ on the <em>y<\/em>-axis points vertically upward. The origin of the displacement vector is located at point <em>b<\/em>(6.0, 1.6) and the end of the displacement vector is located at point <em>e<\/em>(2.0, 4.5). Substitute the coordinates of these points into <a class=\"autogenerated-content\" href=\"#fs-id1167132197498\">(Figure)<\/a> to find the scalar components $$ {D}_{x} $$ and $$ {D}_{y} $$ of the displacement vector $$ \\overset{\\to }{D}$$. Finally, substitute the coordinates into <a class=\"autogenerated-content\" href=\"#fs-id1167132256590\">(Figure)<\/a> to write the displacement vector in the vector component form.\r\n<h4>Solution<\/h4>\r\n<p id=\"fs-id1167132750241\">[reveal-answer q=\"907005\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"907005\"]We identify $$ {x}_{b}=6.0$$, $$ {x}_{e}=2.0$$, $$ {y}_{b}=1.6$$, and $$ {y}_{e}=4.5$$, where the physical unit is 1 cm. The scalar x- and y-components of the displacement vector are $$ \\begin{array}{cc}\\hfill {D}_{x}&amp; ={x}_{e}-{x}_{b}=(2.0-6.0)\\text{cm}=-4.0\\,\\text{cm},\\hfill \\\\ \\hfill {D}_{y}&amp; ={y}_{e}-{y}_{b}=(4.5-1.6)\\text{cm}=+2.9\\,\\text{cm}.\\hfill \\end{array}$$<\/p>\r\nThe vector component form of the displacement vector is\r\n\r\n$$\\overset{\\to }{D}={D}_{x}\\hat{i}+{D}_{y}\\hat{j}=(-4.0\\,\\text{cm})\\hat{i}+(2.9\\,\\text{cm})\\hat{j}=(-4.0\\hat{i}+2.9\\hat{j})\\text{cm}.$$<a id=\"formula\"><\/a>This solution is shown in (Figure 2.17).\r\n<div id=\"CNX_UPhysics_02_02_comp02\" class=\"wp-caption aligncenter\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"584\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31183903\/CNX_UPhysics_02_02_comp02.jpg\" alt=\"Vector D extends from coordinates 6.0, 1.6 to coordinates 2.0, 4.5. Vector D equals vector D sub x plus vector D sub y. D sub x equals minus 4.0 I hat, and extends from x=6.0 to x =2.0. The magnitude D sub x equals 2.0-6.0 = -4.0. D sub y equals plus 2.9 j hat, and extends from y=1.6 to y=4.5. The magnitude D sub y equals 4.5 \u2212 1.6.\" width=\"584\" height=\"416\" \/> <strong>Figure 2.17<\/strong> The graph of the displacement vector. The vector points from the origin point at b to the end point at e.[\/caption]\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n<h4>Significance<\/h4>\r\nNotice that the physical unit\u2014here, 1 cm\u2014can be placed either with each component immediately before the unit vector or globally for both components, as in <a class=\"autogenerated-content\" href=\"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/chapter\/2-2-coordinate-systems-and-components-of-a-vector#formula\">(Figure)<\/a>. Often, the latter way is more convenient because it is simpler.\r\n<p id=\"fs-id1167132354421\">The vector <em>x<\/em>-component $$ {\\overset{\\to }{D}}_{x}=-4.0\\hat{i}=4.0(\\text{\u2212}\\hat{i}) $$ of the displacement vector has the magnitude $$ |{\\overset{\\to }{D}}_{x}|=|-4.0||\\hat{i}|=4.0 $$ because the magnitude of the unit vector is $$ |\\hat{i}|=1$$. Notice, too, that the direction of the <em>x<\/em>-component is $$ \\text{\u2212}\\hat{i}$$, which is antiparallel to the direction of the +<em>x<\/em>-axis; hence, the <em>x<\/em>-component vector $$ {\\overset{\\to }{D}}_{x} $$ points to the left, as shown in <a class=\"autogenerated-content\" href=\"#CNX_UPhysics_02_02_comp02\">(Figure)<\/a>. The scalar <em>x<\/em>-component of vector $$ \\overset{\\to }{D} $$ is $$ {D}_{x}=-4.0$$.<\/p>\r\n<p id=\"fs-id1167132436291\">Similarly, the vector <em>y<\/em>-component $$ {\\overset{\\to }{D}}_{y}=+2.9\\hat{j} $$ of the displacement vector has magnitude $$ |{\\overset{\\to }{D}}_{y}|=|2.9||\\hat{j}|=\\,2.9 $$ because the magnitude of the unit vector is $$ |\\hat{j}|=1$$. The direction of the <em>y<\/em>-component is $$ +\\hat{j}$$, which is parallel to the direction of the +<em>y<\/em>-axis. Therefore, the <em>y<\/em>-component vector $$ {\\overset{\\to }{D}}_{y} $$ points up, as seen in <a class=\"autogenerated-content\" href=\"#CNX_UPhysics_02_02_comp02\">(Figure)<\/a>. The scalar <em>y<\/em>-component of vector $$ \\overset{\\to }{D} $$ is $$ {D}_{y}=+2.9$$. The displacement vector $$ \\overset{\\to }{D} $$ is the resultant of its two <em>vector<\/em> components.<\/p>\r\n<p id=\"fs-id1167132579136\">The vector component form of the displacement vector <a class=\"autogenerated-content\" href=\"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/chapter\/2-2-coordinate-systems-and-components-of-a-vector#formula\">(Figure)<\/a> tells us that the mouse pointer has been moved on the monitor 4.0 cm to the left and 2.9 cm upward from its initial position.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1167132557954\" class=\"textbox exercises check-understanding\">\r\n<h3>Check Your Understanding<\/h3>\r\n<div id=\"fs-id1167132406917\" class=\"problem textbox\">\r\n<div id=\"fs-id1167133364766\">\r\n<p id=\"fs-id1167133617871\">A blue fly lands on a sheet of graph paper at a point located 10.0 cm to the right of its left edge and 8.0 cm above its bottom edge and walks slowly to a point located 5.0 cm from the left edge and 5.0 cm from the bottom edge. Choose the rectangular coordinate system with the origin at the lower left-side corner of the paper and find the displacement vector of the fly. Illustrate your solution by graphing.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1167132407980\" class=\"solution\">\r\n\r\n[reveal-answer q=\"fs-id1167132407980\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"fs-id1167132407980\"]\r\n<p id=\"fs-id1167132322353\">$$\\overset{\\to }{D}=(-5.0\\hat{i}-3.0\\hat{j})\\text{cm}$$; the fly moved 5.0 cm to the left and 3.0 cm down from its landing site.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1167132480350\">When we know the scalar components $$ {A}_{x} $$ and $$ {A}_{y} $$ of a vector $$ \\overset{\\to }{A}$$, we can find its magnitude <em>A<\/em> and its direction angle $$ {\\theta }_{A}$$. The<strong> direction angle<\/strong>\u2014or direction, for short\u2014is the angle the vector forms with the positive direction on the <em>x<\/em>-axis. The angle $$ {\\theta }_{A} $$ is measured in the <em>counterclockwise direction<\/em> from the +<em>x<\/em>-axis to the vector (<a class=\"autogenerated-content\" href=\"#CNX_UPhysics_02_02_comp03\">(Figure)<\/a>). Because the lengths <em>A<\/em>, $$ {A}_{x}$$, and $$ {A}_{y} $$ form a right triangle, they are related by the Pythagorean theorem:<\/p>\r\n\r\n<div id=\"fs-id1167132412442\" class=\"equation-callout\">\r\n<div id=\"fs-id1167132245160\">$${A}^{2}={A}_{x}^{2}+{A}_{y}^{2}\\enspace\u21d4\\enspace{A}=\\sqrt{{A}_{x}^{2}+{A}_{y}^{2}}.$$<\/div>\r\n<\/div>\r\n<p id=\"fs-id1167132688313\">This equation works even if the scalar components of a vector are negative. The direction angle $$ {\\theta }_{A} $$ of a vector is defined via the tangent function of angle $$ {\\theta }_{A} $$ in the triangle shown in <a class=\"autogenerated-content\" href=\"#CNX_UPhysics_02_02_comp03\">(Figure)<\/a>:<\/p>\r\n\r\n<div id=\"fs-id1167133610831\" class=\"equation-callout\">\r\n<div id=\"fs-id1167133377087\">$$\\text{tan}\\,{\\theta }_{A}=\\frac{{A}_{y}}{{A}_{x}}\\enspace\u21d2\\enspace{\\theta }_{A}={\\text{tan}}^{-1}(\\frac{{A}_{y}}{{A}_{x}}).$$<\/div>\r\n<\/div>\r\n<div id=\"CNX_UPhysics_02_02_comp03\" class=\"wp-caption aligncenter\">[caption id=\"\" align=\"aligncenter\" width=\"524\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31183906\/CNX_UPhysics_02_02_comp03.jpg\" alt=\"Vector A has horizontal x component A sub x equal to magnitude A sub x I hat and vertical y component A sub y equal to magnitude A sub y j hat. Vector A and the components form a right triangle with sides length magnitude A sub x and magnitude A sub y and hypotenuse magnitude A equal to the square root of A sub x squared plus A sub y squared. The angle between the horizontal side A sub x and the hypotenuse A is theta sub A.\" width=\"524\" height=\"378\" \/> <strong>Figure 2.18<\/strong> For vector $$ \\overset{\\to }{A}$$, its magnitude A and its direction angle $$ {\\theta }_{A} $$ are related to the magnitudes of its scalar components because A, $$ {A}_{x}$$, and $$ {A}_{y} $$ form a right triangle.[\/caption]<\/div>\r\n<p id=\"fs-id1167132436720\">When the vector lies either in the first quadrant or in the fourth quadrant, where component $$ {A}_{x} $$ is positive (<a class=\"autogenerated-content\" href=\"#CNX_UPhysics_02_02_comp04\">(Figure)<\/a>), the angle $$ \\theta  $$ in <a class=\"autogenerated-content\" href=\"#fs-id1167133377087\">(Figure)<\/a> is identical to the direction angle $$ {\\theta }_{A}$$. For vectors in the fourth quadrant, angle $$ \\theta  $$ is negative, which means that for these vectors, direction angle $$ {\\theta }_{A} $$ is measured <em>clockwise<\/em> from the positive <em>x<\/em>-axis. Similarly, for vectors in the second quadrant, angle $$ \\theta  $$ is negative. When the vector lies in either the second or third quadrant, where component $$ {A}_{x} $$ is negative, the direction angle is $$ {\\theta }_{A}=\\theta +180\\text{\u00b0} $$ (<a class=\"autogenerated-content\" href=\"#CNX_UPhysics_02_02_comp04\">(Figure)<\/a>).<\/p>\r\n\r\n<div id=\"CNX_UPhysics_02_02_comp04\" class=\"wp-caption aligncenter\">[caption id=\"\" align=\"aligncenter\" width=\"584\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31183910\/CNX_UPhysics_02_02_comp04.jpg\" alt=\"Figure I shows vector A in the first quadrant (pointing up and right.) It has positive x and y components A sub x and A sub y, and the angle theta sub A measured counterclockwise from the positive x axis is smaller than 90 degrees. Figure II shows vector A in the first second (pointing up and left.) It has negative x and positive y components A sub x and A sub y. The angle theta sub A measured counterclockwise from the positive x axis is larger than 90 degrees but less than 180 degrees. The angle theta, measured clockwise from the negative x axis, is smaller than 90 degrees. Figure III shows vector A in the third quadrant (pointing down and left.) It has negative x and y components A sub x and A sub y, and the angle theta sub A measured counterclockwise from the positive x axis is larger than 180 degrees and smaller than 270 degrees. The angle theta, measured counterclockwise from the negative x axis, is smaller than 90 degrees. Figure IV shows vector A in the fourth quadrant (pointing down and right.) It has positive x and negative y components A sub x and A sub y, and the angle theta sub A measured clockwise from the positive x axis is smaller than 90 degrees.\" width=\"584\" height=\"477\" \/> <strong>Figure 2.19<\/strong> Scalar components of a vector may be positive or negative. Vectors in the first quadrant (I) have both scalar components positive and vectors in the third quadrant have both scalar components negative. For vectors in quadrants II and III, the direction angle of a vector is $$ {\\theta }_{A}=\\theta +180\\text{\u00b0}$$.[\/caption]<\/div>\r\n<div id=\"fs-id1167133568370\" class=\"textbox examples\">\r\n<h3>Example<\/h3>\r\n<p id=\"fs-id1167132606822\">Magnitude and Direction of the Displacement VectorYou move a mouse pointer on the display monitor from its initial position at point (6.0 cm, 1.6 cm) to an icon located at point (2.0 cm, 4.5 cm). What are the magnitude and direction of the displacement vector of the pointer?<\/p>\r\n\r\n<h4>Strategy<\/h4>\r\nIn <a class=\"autogenerated-content\" href=\"#fs-id1167132341727\">(Figure)<\/a>, we found the displacement vector $$ \\overset{\\to }{D} $$ of the mouse pointer (see <a class=\"autogenerated-content\" href=\"#fs-id1167132371006\">(Figure)<\/a>). We identify its scalar components $$ {D}_{x}=-4.0\\,\\text{cm} $$ and $$ {D}_{y}=+2.9\\,\\text{cm} $$ and substitute into <a class=\"autogenerated-content\" href=\"#fs-id1167132245160\">(Figure)<\/a> and <a class=\"autogenerated-content\" href=\"#fs-id1167133377087\">(Figure)<\/a> to find the magnitude <em>D<\/em> and direction $$ {\\theta }_{D}$$, respectively.\r\n<h4>Solution<\/h4>\r\n<p id=\"fs-id1167132276264\">[reveal-answer q=\"170894\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"170894\"]The magnitude of vector $$ \\overset{\\to }{D} $$ is $$ D=\\sqrt{{D}_{x}^{2}+{D}_{y}^{2}}=\\sqrt{{(-4.0\\,\\text{cm})}^{2}+{(2.9\\,\\text{cm})}^{2}}=\\sqrt{{(4.0)}^{2}+{(2.9)}^{2}}\\,\\text{cm}=4.9\\,\\text{cm}. $$ The direction angle is $$ \\text{tan}\\,\\theta =\\frac{{D}_{y}}{{D}_{x}}=\\frac{+2.9\\,\\text{cm}}{-4.0\\,\\text{cm}}=-0.725\\enspace\u21d2\\enspace\\theta ={\\text{tan}}^{-1}(-0.725)=-35.9\\text{\u00b0}. $$ Vector $$ \\overset{\\to }{D} $$ lies in the second quadrant, so its direction angle is $$ {\\theta }_{D}=\\theta +180\\text{\u00b0}=-35.9\\text{\u00b0}+180\\text{\u00b0}=144.1\\text{\u00b0}.$$[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1167132438204\" class=\"textbox exercises check-understanding\">\r\n<h3>Check Your Understanding<\/h3>\r\n<div id=\"fs-id1167132502120\" class=\"problem textbox\">\r\n<div id=\"fs-id1167132502143\">\r\n<p id=\"fs-id1167132502219\">If the displacement vector of a blue fly walking on a sheet of graph paper is $$ \\overset{\\to }{D}=(-5.00\\hat{i}-3.00\\hat{j})\\text{cm}$$, find its magnitude and direction.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1167132332238\" class=\"solution\">\r\n\r\n[reveal-answer q=\"fs-id1167132332238\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"fs-id1167132332238\"]\r\n<p id=\"fs-id1167132335635\">5.83 cm, $$ 211\\text{\u00b0}$$<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1167132464780\">In many applications, the magnitudes and directions of vector quantities are known and we need to find the resultant of many vectors. For example, imagine 400 cars moving on the Golden Gate Bridge in San Francisco in a strong wind. Each car gives the bridge a different push in various directions and we would like to know how big the resultant push can possibly be. We have already gained some experience with the geometric construction of vector sums, so we know the task of finding the resultant by drawing the vectors and measuring their lengths and angles may become intractable pretty quickly, leading to huge errors. Worries like this do not appear when we use analytical methods. The very first step in an analytical approach is to find vector components when the direction and magnitude of a vector are known.<\/p>\r\n<p id=\"fs-id1167132464805\">Let us return to the right triangle in <a class=\"autogenerated-content\" href=\"#CNX_UPhysics_02_02_comp03\">(Figure)<\/a>. The quotient of the adjacent side $$ {A}_{x} $$ to the hypotenuse <em>A<\/em> is the cosine function of direction angle $$ {\\theta }_{A}$$, $$ {A}_{x}\\text{\/}A=\\text{cos}\\,{\\theta }_{A}$$, and the quotient of the opposite side $$ {A}_{y} $$ to the hypotenuse <em>A<\/em> is the sine function of $$ {\\theta }_{A}$$, $$ {A}_{y}\\text{\/}A=\\text{sin}\\,{\\theta }_{A}$$. When magnitude <em>A<\/em> and direction $$ {\\theta }_{A} $$ are known, we can solve these relations for the scalar components:<\/p>\r\n\r\n<div id=\"fs-id1167133740682\" class=\"equation-callout\">\r\n<div id=\"fs-id1167133740741\">$$\\{\\begin{array}{c}{A}_{x}=A\\,\\text{cos}\\,{\\theta }_{A}\\\\ {A}_{y}=A\\,\\text{sin}\\,{\\theta }_{A}\\end{array}.$$<\/div>\r\n<\/div>\r\n<p id=\"fs-id1167132266496\">When calculating vector components with <a class=\"autogenerated-content\" href=\"#fs-id1167133740741\">(Figure)<\/a>, care must be taken with the angle. The direction angle $$ {\\theta }_{A} $$ of a vector is the angle measured <em>counterclockwise<\/em> from the positive direction on the <em>x<\/em>-axis to the vector. The clockwise measurement gives a negative angle.<\/p>\r\n\r\n<div id=\"fs-id1167132576856\" class=\"textbox examples\">\r\n<h3>Example<\/h3>\r\n<h4>Components of Displacement Vectors<\/h4>\r\nA rescue party for a missing child follows a search dog named Trooper. Trooper wanders a lot and makes many trial sniffs along many different paths. Trooper eventually finds the child and the story has a happy ending, but his displacements on various legs seem to be truly convoluted. On one of the legs he walks 200.0 m southeast, then he runs north some 300.0 m. On the third leg, he examines the scents carefully for 50.0 m in the direction $$ 30\\text{\u00b0} $$ west of north. On the fourth leg, Trooper goes directly south for 80.0 m, picks up a fresh scent and turns $$ 23\\text{\u00b0} $$ west of south for 150.0 m. Find the scalar components of Trooper\u2019s displacement vectors and his displacement vectors in vector component form for each leg.\r\n<h4>Strategy<\/h4>\r\nLet\u2019s adopt a rectangular coordinate system with the positive <em>x<\/em>-axis in the direction of geographic east, with the positive <em>y<\/em>-direction pointed to geographic north. Explicitly, the unit vector $$ \\hat{i} $$ of the <em>x<\/em>-axis points east and the unit vector $$ \\hat{j} $$ of the <em>y<\/em>-axis points north. Trooper makes five legs, so there are five displacement vectors. We start by identifying their magnitudes and direction angles, then we use <a class=\"autogenerated-content\" href=\"#fs-id1167133740741\">(Figure)<\/a> to find the scalar components of the displacements and <a class=\"autogenerated-content\" href=\"#fs-id1167132256590\">(Figure)<\/a> for the displacement vectors.\r\n<h4>Solution<\/h4>\r\n<p id=\"fs-id1167132516072\">On the first leg, the displacement magnitude is $$ {L}_{1}=200.0\\,\\text{m} $$ and the direction is southeast. For direction angle $$ {\\theta }_{1} $$ we can take either $$ 45\\text{\u00b0} $$ measured clockwise from the east direction or $$ 45\\text{\u00b0}+270\\text{\u00b0} $$ measured counterclockwise from the east direction. With the first choice, $$ {\\theta }_{1}=-45\\text{\u00b0}$$. With the second choice, $$ {\\theta }_{1}=+315\\text{\u00b0}$$. We can use either one of these two angles. The components are<\/p>\r\n\r\n<div id=\"fs-id1167132433420\" class=\"unnumbered\">$$\\begin{array}{l}{L}_{1x}={L}_{1}\\,\\text{cos}\\,{\\theta }_{1}=(200.0\\,\\text{m})\\,\\text{cos}\\,315\\text{\u00b0}=141.4\\,\\text{m,}\\\\ {L}_{1y}={L}_{1}\\,\\text{sin}\\,{\\theta }_{1}=(200.0\\,\\text{m})\\,\\text{sin}\\,315\\text{\u00b0}=-141.4\\,\\text{m}.\\end{array}$$<\/div>\r\n<p id=\"fs-id1167133763578\">The displacement vector of the first leg is<\/p>\r\n\r\n<div id=\"fs-id1167133763683\" class=\"unnumbered\">$${\\overset{\\to }{L}}_{1}={L}_{1x}\\hat{i}+{L}_{1y}\\hat{j}=(141.4\\hat{i}-141.4\\hat{j})\\,\\text{m}.$$<\/div>\r\n<p id=\"fs-id1167133658240\">On the second leg of Trooper\u2019s wanderings, the magnitude of the displacement is $$ {L}_{2}=300.0\\,\\text{m} $$ and the direction is north. The direction angle is $$ {\\theta }_{2}=+90\\text{\u00b0}$$. We obtain the following results:<\/p>\r\n\r\n<div id=\"fs-id1167132541311\" class=\"unnumbered\">$$\\begin{array}{ccc}\\hfill {L}_{2x}&amp; =\\hfill &amp; {L}_{2}\\,\\text{cos}\\,{\\theta }_{2}=(300.0\\,\\text{m})\\,\\text{cos}\\,90\\text{\u00b0}=0.0\\,,\\hfill \\\\ \\hfill {L}_{2y}&amp; =\\hfill &amp; {L}_{2}\\,\\text{sin}\\,{\\theta }_{2}=(300.0\\,\\text{m})\\,\\text{sin}\\,90\\text{\u00b0}=300.0\\,\\text{m,}\\hfill \\\\ \\hfill {\\overset{\\to }{L}}_{2}&amp; =\\hfill &amp; {L}_{2x}\\hat{i}+{L}_{2y}\\hat{j}=(300.0\\,\\text{m})\\hat{j}.\\hfill \\end{array}$$<\/div>\r\n<p id=\"fs-id1167132349429\">On the third leg, the displacement magnitude is $$ {L}_{3}=50.0\\,\\text{m} $$ and the direction is $$ 30\\text{\u00b0} $$ west of north. The direction angle measured counterclockwise from the eastern direction is $$ {\\theta }_{3}=30\\text{\u00b0}+90\\text{\u00b0}=+120\\text{\u00b0}$$. This gives the following answers:<\/p>\r\n\r\n<div id=\"fs-id1167132595199\" class=\"unnumbered\">$$\\begin{array}{ccc}\\hfill {L}_{3x}&amp; =\\hfill &amp; {L}_{3}\\,\\text{cos}\\,{\\theta }_{3}=(50.0\\,\\text{m})\\,\\text{cos}\\,120\\text{\u00b0}=-25.0\\,\\text{m,}\\hfill \\\\ \\hfill {L}_{3y}&amp; =\\hfill &amp; {L}_{3}\\,\\text{sin}\\,{\\theta }_{3}=(50.0\\,\\text{m})\\,\\text{sin}\\,120\\text{\u00b0}=+43.3\\,\\text{m,}\\hfill \\\\ \\hfill {\\overset{\\to }{L}}_{3}&amp; =\\hfill &amp; {L}_{3x}\\hat{i}+{L}_{3y}\\hat{j}=(-25.0\\hat{i}+43.3\\hat{j})\\text{m}.\\hfill \\end{array}$$<\/div>\r\n<p id=\"fs-id1167132295353\">On the fourth leg of the excursion, the displacement magnitude is $$ {L}_{4}=80.0\\,\\text{m} $$ and the direction is south. The direction angle can be taken as either $$ {\\theta }_{4}=-90\\text{\u00b0} $$ or $$ {\\theta }_{4}=+270\\text{\u00b0}$$. We obtain<\/p>\r\n\r\n<div id=\"fs-id1167132408563\" class=\"unnumbered\">$$\\begin{array}{ccc}\\hfill {L}_{4x}&amp; =\\hfill &amp; {L}_{4}\\,\\text{cos}\\,{\\theta }_{4}=(80.0\\,\\text{m})\\,\\text{cos}\\,(-90\\text{\u00b0})=0\\,,\\hfill \\\\ \\hfill {L}_{4y}&amp; =\\hfill &amp; {L}_{4}\\,\\text{sin}\\,{\\theta }_{4}=(80.0\\,\\text{m})\\,\\text{sin}\\,(-90\\text{\u00b0})=-80.0\\,\\text{m,}\\hfill \\\\ \\hfill {\\overset{\\to }{L}}_{4}&amp; =\\hfill &amp; {L}_{4x}\\hat{i}+{L}_{4y}\\hat{j}=(-80.0\\,\\text{m})\\hat{j}.\\hfill \\end{array}$$<\/div>\r\n<p id=\"fs-id1167132544336\">[reveal-answer q=\"567672\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"567672\"]On the last leg, the magnitude is $$ {L}_{5}=150.0\\,\\text{m} $$ and the angle is $$ {\\theta }_{5}=-23\\text{\u00b0}+270\\text{\u00b0}=+247\\text{\u00b0} $$ $$(23\\text{\u00b0} $$ west of south), which gives $$ \\begin{array}{ccc}\\hfill {L}_{5x}&amp; =\\hfill &amp; {L}_{5}\\,\\text{cos}\\,{\\theta }_{5}=(150.0\\,\\text{m})\\,\\text{cos}\\,247\\text{\u00b0}=-58.6\\,\\text{m,}\\hfill \\\\ \\hfill {L}_{5y}&amp; =\\hfill &amp; {L}_{5}\\,\\text{sin}\\,{\\theta }_{5}=(150.0\\,\\text{m})\\,\\text{sin}\\,247\\text{\u00b0}=-138.1\\,\\text{m,}\\hfill \\\\ \\hfill {\\overset{\\to }{L}}_{5}&amp; =\\hfill &amp; {L}_{5x}\\hat{i}+{L}_{5y}\\hat{j}=(-58.6\\hat{i}-138.1\\hat{j})\\text{m}.\\hfill \\end{array}$$[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1167128918681\" class=\"textbox exercises check-understanding\">\r\n<h3>Check Your Understanding<\/h3>\r\n<div id=\"fs-id1167128918786\" class=\"problem textbox\">\r\n<div id=\"fs-id1167128918869\">\r\n<p id=\"fs-id1167128918893\">If Trooper runs 20 m west before taking a rest, what is his displacement vector?<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1167128919108\" class=\"solution\">\r\n\r\n[reveal-answer q=\"fs-id1167128919108\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"fs-id1167128919108\"]\r\n<p id=\"fs-id1167128919128\">$$\\overset{\\to }{D}=(-20\\,\\text{m})\\hat{j}$$<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167128956776\" class=\"bc-section section\">\r\n<h3>Polar Coordinates<\/h3>\r\n<p id=\"fs-id1167132336599\">To describe locations of points or vectors in a plane, we need two orthogonal directions. In the Cartesian coordinate system these directions are given by unit vectors $$ \\hat{i} $$ and $$ \\hat{j} $$ along the <em>x<\/em>-axis and the <em>y<\/em>-axis, respectively. The Cartesian coordinate system is very convenient to use in describing displacements and velocities of objects and the forces acting on them. However, it becomes cumbersome when we need to describe the rotation of objects. When describing rotation, we usually work in the<strong> polar coordinate system<\/strong>.<\/p>\r\n<p id=\"fs-id1167132747901\">In the polar coordinate system, the location of point <em>P<\/em> in a plane is given by two <strong>polar<\/strong> <strong>coordinates<\/strong> (<a class=\"autogenerated-content\" href=\"#CNX_UPhysics_02_02_polar\">(Figure)<\/a>). The first polar coordinate is the <strong>radial coordinate<\/strong> <em>r<\/em>, which is the distance of point <em>P<\/em> from the origin. The second polar coordinate is an angle $$ \\phi  $$ that the radial vector makes with some chosen direction, usually the positive <em>x<\/em>-direction. In polar coordinates, angles are measured in radians, or rads. The radial vector is attached at the origin and points away from the origin to point <em>P.<\/em> This radial direction is described by a unit radial vector $$ \\hat{r}$$. The second unit vector $$ \\hat{t} $$ is a vector orthogonal to the radial direction $$ \\hat{r}$$. The positive $$ +\\hat{t} $$ direction indicates how the angle $$ \\phi  $$ changes in the counterclockwise direction. In this way, a point <em>P<\/em> that has coordinates (<em>x<\/em>, <em>y<\/em>) in the rectangular system can be described equivalently in the polar coordinate system by the two polar coordinates $$ (r,\\phi )$$. <a class=\"autogenerated-content\" href=\"#fs-id1167133740741\">(Figure)<\/a> is valid for any vector, so we can use it to express the <em>x<\/em>- and <em>y<\/em>-coordinates of vector $$ \\overset{\\to }{r}$$. In this way, we obtain the connection between the polar coordinates and rectangular coordinates of point <em>P<\/em>:<\/p>\r\n\r\n<div id=\"fs-id1167132575879\" class=\"equation-callout\">\r\n<div id=\"fs-id1167132575978\">$$\\{\\begin{array}{c}x=r\\,\\text{cos}\\,\\phi \\\\ y=r\\,\\text{sin}\\,\\phi \\end{array}.$$<\/div>\r\n<\/div>\r\n<div id=\"CNX_UPhysics_02_02_polar\" class=\"wp-caption aligncenter\">[caption id=\"\" align=\"aligncenter\" width=\"506\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31183914\/CNX_UPhysics_02_02_polar.jpg\" alt=\"Vector r points from the origin of the x y coordinate system to point P. The angle between the vector r and the positive x direction is phi. X equals r cosine phi and y equals r sine phi. Extending a line in the direction of r vector past point P, a unit vector r hat is drawn in the same direction as r. A unit vector t hat is perpendicular to r hat, pointing 90 degrees counterclockwise to r hat.\" width=\"506\" height=\"419\" \/> <strong>Figure 2.20<\/strong> Using polar coordinates, the unit vector $$ \\hat{r} $$ defines the positive direction along the radius r (radial direction) and, orthogonal to it, the unit vector $$ \\hat{t} $$ defines the positive direction of rotation by the angle $$ \\phi $$.[\/caption]<\/div>\r\n<div id=\"fs-id1167132451446\" class=\"textbox examples\">\r\n<h3>Example<\/h3>\r\n<h4>Polar Coordinates<\/h4>\r\nA treasure hunter finds one silver coin at a location 20.0 m away from a dry well in the direction $$ 20\\text{\u00b0} $$ north of east and finds one gold coin at a location 10.0 m away from the well in the direction $$ 20\\text{\u00b0} $$ north of west. What are the polar and rectangular coordinates of these findings with respect to the well?\r\n<h4>Strategy<\/h4>\r\nThe well marks the origin of the coordinate system and east is the +<em>x<\/em>-direction. We identify radial distances from the locations to the origin, which are $$ {r}_{S}=20.0\\,\\text{m} $$ (for the silver coin) and $$ {r}_{G}=10.0\\,\\text{m} $$ (for the gold coin). To find the angular coordinates, we convert $$ 20\\text{\u00b0} $$ to radians: $$ 20\\text{\u00b0}=\\pi 20\\text{\/}180=\\pi \\text{\/}9$$. We use <a class=\"autogenerated-content\" href=\"#fs-id1167132575978\">(Figure)<\/a> to find the <em>x<\/em>- and <em>y<\/em>-coordinates of the coins.\r\n<h4>Solution<\/h4>\r\n<p id=\"fs-id1167132491818\">[reveal-answer q=\"8052\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"8052\"]The angular coordinate of the silver coin is $$ {\\phi }_{S}=\\pi \\text{\/}9$$, whereas the angular coordinate of the gold coin is $$ {\\phi }_{G}=\\pi -\\pi \\text{\/}9=8\\pi \\text{\/}9$$. Hence, the polar coordinates of the silver coin are $$ ({r}_{S},{\\phi }_{S})=(20.0\\,\\text{m},\\pi \\text{\/}9) $$ and those of the gold coin are $$ ({r}_{G},{\\phi }_{G})=(10.0\\,\\text{m},8\\pi \\text{\/}9)$$. We substitute these coordinates into (Figure) to obtain rectangular coordinates. For the gold coin, the coordinates are<\/p>\r\n$$\\{\\begin{array}{l}{x}_{G}={r}_{G}\\,\\text{cos}\\,{\\phi }_{G}=(10.0\\,\\text{m})\\,\\text{cos}\\,8\\pi \\text{\/}9=-9.4\\,\\text{m}\\\\ {y}_{G}={r}_{G}\\,\\text{sin}\\,{\\phi }_{G}=(10.0\\,\\text{m})\\,\\text{sin}\\,8\\pi \\text{\/}9=3.4\\,\\text{m}\\end{array}\\enspace\u21d2\\enspace({x}_{G},{y}_{G})=(-9.4\\,\\text{m},3.4\\,\\text{m}).$$\r\n\r\nFor the silver coin, the coordinates are\r\n\r\n$$\\{\\begin{array}{l}{x}_{S}={r}_{S}\\,\\text{cos}\\,{\\phi }_{S}=(20.0\\,\\text{m})\\,\\text{cos}\\,\\pi \\text{\/}9=18.9\\,\\text{m}\\\\ {y}_{S}={r}_{S}\\,\\text{sin}\\,{\\phi }_{S}=(20.0\\,\\text{m})\\,\\text{sin}\\,\\pi \\text{\/}9=6.8\\,\\text{m}\\end{array}\\enspace\u21d2\\enspace({x}_{S},{y}_{S})=(18.9\\,\\text{m},6.8\\,\\text{m}).$$[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167132342866\" class=\"bc-section section\">\r\n<h3>Vectors in Three Dimensions<\/h3>\r\n<p id=\"fs-id1167132741874\">To specify the location of a point in space, we need three coordinates (<em>x<\/em>, <em>y<\/em>, <em>z<\/em>), where coordinates <em>x<\/em> and <em>y<\/em> specify locations in a plane, and coordinate <em>z<\/em> gives a vertical position above or below the plane. Three-dimensional space has three orthogonal directions, so we need not two but <em>three<\/em> unit vectors to define a three-dimensional coordinate system. In the Cartesian coordinate system, the first two unit vectors are the unit vector of the <em>x<\/em>-axis $$ \\hat{i} $$ and the unit vector of the <em>y<\/em>-axis $$ \\hat{j}$$. The third unit vector $$ \\hat{k} $$ is the direction of the <em>z<\/em>-axis (<a class=\"autogenerated-content\" href=\"#CNX_UPhysics_02_02_ijknew\">(Figure)<\/a>). The order in which the axes are labeled, which is the order in which the three unit vectors appear, is important because it defines the orientation of the coordinate system. The order <em>x<\/em>-<em>y<\/em>-<em>z<\/em>, which is equivalent to the order $$ \\hat{i} $$ - $$ \\hat{j} $$ - $$ \\hat{k}$$, defines the standard right-handed coordinate system (positive orientation).<\/p>\r\n\r\n<div id=\"CNX_UPhysics_02_02_ijknew\" class=\"wp-caption aligncenter\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"484\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31183917\/CNX_UPhysics_02_02_ijknew.jpg\" alt=\"The x y z coordinate system, with unit vectors I hat, j hat and k hat respectively. I hat points out at us, j hat points to the right, and k hat points up the page. The unit vectors form the sides of a cube.\" width=\"484\" height=\"501\" \/> <strong>Figure 2.21<\/strong> Three unit vectors define a Cartesian system in three-dimensional space. The order in which these unit vectors appear defines the orientation of the coordinate system. The order shown here defines the right-handed orientation.[\/caption]\r\n\r\n<\/div>\r\n<p id=\"fs-id1167132688199\">In three-dimensional space, vector $$ \\overset{\\to }{A} $$ has three vector components: the <em>x<\/em>-component $$ {\\overset{\\to }{A}}_{x}={A}_{x}\\hat{i}$$, which is the part of vector $$ \\overset{\\to }{A} $$ along the <em>x<\/em>-axis; the <em>y<\/em>-component $$ {\\overset{\\to }{A}}_{y}={A}_{y}\\hat{j}$$, which is the part of $$ \\overset{\\to }{A} $$ along the <em>y<\/em>-axis; and the <em>z<\/em>-component $$ {\\overset{\\to }{A}}_{z}={A}_{z}\\hat{k}$$, which is the part of the vector along the <em>z<\/em>-axis. A vector in three-dimensional space is the vector sum of its three vector components (<a class=\"autogenerated-content\" href=\"#CNX_UPhysics_02_02_vector3D\">(Figure)<\/a>):<\/p>\r\n\r\n<div id=\"fs-id1167133745530\" class=\"equation-callout\">\r\n<div id=\"fs-id1167132278018\">$$\\overset{\\to }{A}={A}_{x}\\hat{i}+{A}_{y}\\hat{j}+{A}_{z}\\hat{k}.$$<\/div>\r\n<\/div>\r\n<p id=\"fs-id1167132736522\">If we know the coordinates of its origin $$ b({x}_{b},{y}_{b},{z}_{b}) $$ and of its end $$ e({x}_{e},{y}_{e},{z}_{e})$$, its scalar components are obtained by taking their differences: $$ {A}_{x} $$ and $$ {A}_{y} $$ are given by <a class=\"autogenerated-content\" href=\"#fs-id1167132197498\">(Figure)<\/a> and the <em>z<\/em>-component is given by<\/p>\r\n\r\n<div id=\"fs-id1167132432269\" class=\"equation-callout\">\r\n<div id=\"fs-id1167133636864\">$${A}_{z}={z}_{e}-{z}_{b}.$$<\/div>\r\n<\/div>\r\n<p id=\"fs-id1167132436510\">Magnitude <em>A<\/em> is obtained by generalizing <a class=\"autogenerated-content\" href=\"#fs-id1167132245160\">(Figure)<\/a> to three dimensions:<\/p>\r\n\r\n<div id=\"fs-id1167132451249\" class=\"equation-callout\">\r\n<div id=\"fs-id1167132468770\">$$A=\\sqrt{{A}_{x}^{2}+{A}_{y}^{2}+{A}_{z}^{2}}.$$<\/div>\r\n<\/div>\r\n<p id=\"fs-id1167132716914\">This expression for the vector magnitude comes from applying the Pythagorean theorem twice. As seen in <a class=\"autogenerated-content\" href=\"#CNX_UPhysics_02_02_vector3D\">(Figure)<\/a>, the diagonal in the <em>xy<\/em>-plane has length $$ \\sqrt{{A}_{x}^{2}+{A}_{y}^{2}} $$ and its square adds to the square $$ {A}_{z}^{2} $$ to give $$ {A}^{2}$$. Note that when the <em>z<\/em>-component is zero, the vector lies entirely in the <em>xy<\/em>-plane and its description is reduced to two dimensions.<\/p>\r\n\r\n<div id=\"CNX_UPhysics_02_02_vector3D\" class=\"wp-caption aligncenter\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"489\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31183920\/CNX_UPhysics_02_02_vector3D.jpg\" alt=\"Vector A in the x y z coordinate system extends from the origin. Vector A equals the sum of vectors A sub x, A sub y and A sub z. Vector A sub x is the x component along the x axis and has length A sub x I hat. Vector A sub y is the y component along the y axis and has length A sub y j hat. Vector A sub z is the z component along the z axis and has length A sub x k hat. The components form the sides of a rectangular box with sides length A sub x, A sub y, and A sub z.\" width=\"489\" height=\"450\" \/> <strong>Figure 2.22<\/strong> A vector in three-dimensional space is the vector sum of its three vector components.[\/caption]\r\n\r\n<\/div>\r\n<div id=\"fs-id1167132572875\" class=\"textbox examples\">\r\n<h3>Example<\/h3>\r\n<h4>Takeoff of a Drone<\/h4>\r\nDuring a takeoff of IAI Heron (<a class=\"autogenerated-content\" href=\"#CNX_UPhysics_02_02_heron\">(Figure)<\/a>), its position with respect to a control tower is 100 m above the ground, 300 m to the east, and 200 m to the north. One minute later, its position is 250 m above the ground, 1200 m to the east, and 2100 m to the north. What is the drone\u2019s displacement vector with respect to the control tower? What is the magnitude of its displacement vector?\r\n<div id=\"CNX_UPhysics_02_02_heron\" class=\"wp-caption aligncenter\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"465\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31183924\/CNX_UPhysics_02_02_heron.jpg\" alt=\"A photo of a drone plane.\" width=\"465\" height=\"248\" \/> <strong>Figure 2.23<\/strong> The drone IAI Heron in flight. (credit: SSgt Reynaldo Ramon, USAF)[\/caption]\r\n\r\n<\/div>\r\n<h4>Strategy<\/h4>\r\nWe take the origin of the Cartesian coordinate system as the control tower. The direction of the +<em>x<\/em>-axis is given by unit vector $$ \\hat{i} $$ to the east, the direction of the +<em>y<\/em>-axis is given by unit vector $$ \\hat{j} $$ to the north, and the direction of the +<em>z<\/em>-axis is given by unit vector $$ \\hat{k}$$, which points up from the ground. The drone\u2019s first position is the origin (or, equivalently, the beginning) of the displacement vector and its second position is the end of the displacement vector.\r\n<h4>Solution<\/h4>\r\n<p id=\"fs-id1167132742913\">[reveal-answer q=\"436377\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"436377\"]We identify b(300.0 m, 200.0 m, 100.0 m) and e(480.0 m, 370.0 m, 250.0m), and use (Figure) and (Figure) to find the scalar components of the drone\u2019s displacement vector:<\/p>\r\n$$\\{\\begin{array}{l}{D}_{x}={x}_{e}-{x}_{b}=1200.0\\,\\text{m}-300.0\\,\\text{m}=900.0\\,\\text{m},\\\\ {D}_{y}={y}_{e}-{y}_{b}=2100.0\\,\\text{m}-200.0\\,\\text{m}=1900.0\\,\\text{m,}\\\\ {D}_{z}={z}_{e}-{z}_{b}=250.0\\,\\text{m}-100.0\\,\\text{m}=150.0\\,\\text{m}.\\end{array}$$\r\n\r\nWe substitute these components into (Figure) to find the displacement vector:\r\n\r\n$$\\overset{\\to }{D}={D}_{x}\\hat{i}+{D}_{y}\\hat{j}+{D}_{z}\\hat{k}=900.0\\,\\text{m}\\hat{i}+1900.0\\,\\text{m}\\hat{j}+150.0\\,\\text{m}\\hat{k}=(0.90\\hat{i}+1.90\\hat{j}+0.15\\hat{k})\\,\\text{km}.$$\r\n\r\nWe substitute into (Figure) to find the magnitude of the displacement: $$ D=\\sqrt{{D}_{x}^{2}+{D}_{y}^{2}+{D}_{z}^{2}}=\\sqrt{{(0.90\\,\\text{km})}^{2}+{(1.90\\,\\text{km})}^{2}+{(0.15\\,\\text{km})}^{2}}=4.44\\,\\text{km}.$$[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1167132218339\" class=\"textbox exercises check-understanding\">\r\n<h3>Check Your Understanding<\/h3>\r\n<div id=\"fs-id1167128919148\" class=\"problem textbox\">\r\n<div id=\"fs-id1167132230025\">\r\n<p id=\"fs-id1167132230028\">If the average velocity vector of the drone in the displacement in <a class=\"autogenerated-content\" href=\"#fs-id1167132572875\">(Figure)<\/a> is $$ \\overset{\\to }{u}=(15.0\\hat{i}+31.7\\hat{j}+2.5\\hat{k})\\text{m}\\text{\/}\\text{s}$$, what is the magnitude of the drone\u2019s velocity vector?<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1167132346740\" class=\"solution\">\r\n\r\n[reveal-answer q=\"fs-id1167132346740\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"fs-id1167132346740\"]\r\n<p id=\"fs-id1167133751212\">35.1 m\/s = 126.4 km\/h<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167132451656\" class=\"textbox key-takeaways\">\r\n<h3>Summary<\/h3>\r\n<ul id=\"fs-id1167132628006\">\r\n \t<li>Vectors are described in terms of their components in a coordinate system. In two dimensions (in a plane), vectors have two components. In three dimensions (in space), vectors have three components.<\/li>\r\n \t<li>A vector component of a vector is its part in an axis direction. The vector component is the product of the unit vector of an axis with its scalar component along this axis. A vector is the resultant of its vector components.<\/li>\r\n \t<li>Scalar components of a vector are differences of coordinates, where coordinates of the origin are subtracted from end point coordinates of a vector. In a rectangular system, the magnitude of a vector is the square root of the sum of the squares of its components.<\/li>\r\n \t<li>In a plane, the direction of a vector is given by an angle the vector has with the positive <em>x<\/em>-axis. This direction angle is measured counterclockwise. The scalar <em>x<\/em>-component of a vector can be expressed as the product of its magnitude with the cosine of its direction angle, and the scalar <em>y<\/em>-component can be expressed as the product of its magnitude with the sine of its direction angle.<\/li>\r\n \t<li>In a plane, there are two equivalent coordinate systems. The Cartesian coordinate system is defined by unit vectors $$ \\hat{i} $$ and $$ \\hat{j} $$ along the <em>x<\/em>-axis and the <em>y<\/em>-axis, respectively. The polar coordinate system is defined by the radial unit vector $$ \\hat{r}$$, which gives the direction from the origin, and a unit vector $$ \\hat{t}$$, which is perpendicular (orthogonal) to the radial direction.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-id1167132437539\" class=\"review-conceptual-questions\">\r\n<h3>Conceptual Questions<\/h3>\r\n<div id=\"fs-id1167132241622\" class=\"problem textbox\">\r\n<div id=\"fs-id1167133668968\">\r\n<p id=\"fs-id1167132345316\">Give an example of a nonzero vector that has a component of zero.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1167132706218\" class=\"solution\">\r\n\r\n[reveal-answer q=\"fs-id1167132706218\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"fs-id1167132706218\"]\r\n<p id=\"fs-id1167132328235\">a unit vector of the <em>x<\/em>-axis<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167132494452\" class=\"problem textbox\">\r\n<div id=\"fs-id1167132476171\">\r\n<p id=\"fs-id1167132476173\">Explain why a vector cannot have a component greater than its own magnitude.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167132562110\" class=\"problem textbox\">\r\n<div id=\"fs-id1167132562112\">\r\n<p id=\"fs-id1167132504879\">If two vectors are equal, what can you say about their components?<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1167133715718\" class=\"solution\">\r\n\r\n[reveal-answer q=\"fs-id1167133715718\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"fs-id1167133715718\"]\r\n<p id=\"fs-id1167133715720\">They are equal.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167132267416\" class=\"problem textbox\">\r\n<div id=\"fs-id1167132267418\">\r\n<p id=\"fs-id1167132627817\">If vectors $$ \\overset{\\to }{A} $$ and $$ \\overset{\\to }{B} $$ are orthogonal, what is the component of $$ \\overset{\\to }{B} $$ along the direction of $$ \\overset{\\to }{A}$$? What is the component of $$ \\overset{\\to }{A} $$ along the direction of $$ \\overset{\\to }{B}$$?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167132518791\" class=\"problem textbox\">\r\n<div id=\"fs-id1167132518793\">\r\n<p id=\"fs-id1167133668836\">If one of the two components of a vector is not zero, can the magnitude of the other vector component of this vector be zero?<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1167133568374\" class=\"solution\">\r\n\r\n[reveal-answer q=\"fs-id1167133568374\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"fs-id1167133568374\"]\r\n<p id=\"fs-id1167133568377\">yes<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167132535630\" class=\"problem textbox\">\r\n<div id=\"fs-id1167132535632\">\r\n<p id=\"fs-id1167132481141\">If two vectors have the same magnitude, do their components have to be the same?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167132281028\" class=\"review-problems textbox exercises\">\r\n<h3>Problems<\/h3>\r\n<div id=\"fs-id1167132612418\" class=\"problem textbox\">\r\n<div id=\"fs-id1167132612420\">\r\n<p id=\"fs-id1167132312139\">Assuming the +<em>x<\/em>-axis is horizontal and points to the right, resolve the vectors given in the following figure to their scalar components and express them in vector component form.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1167132683339\" class=\"solution\">\r\n\r\n[reveal-answer q=\"fs-id1167132683339\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"fs-id1167132683339\"]\r\n<p id=\"fs-id1167128844080\">a. $$ \\overset{\\to }{A}=+8.66\\hat{i}+5.00\\hat{j}$$, b. $$ \\overset{\\to }{B}=+30.09\\hat{i}+39.93\\hat{j}$$, c. $$ \\overset{\\to }{C}=+6.00\\hat{i}-10.39\\hat{j}$$, d. $$ \\overset{\\to }{D}=-15.97\\hat{i}+12.04\\hat{j}$$, f. $$ \\overset{\\to }{F}=-17.32\\hat{i}-10.00\\hat{j}$$<\/p>\r\n<span id=\"fs-id1167133711602\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31183852\/CNX_UPhysics_02_01_problems_img.jpg\" alt=\"The x y coordinate system has positive x to the right and positive y up. Vector A has magnitude 10.0 and points 30 degrees counterclockwise from the positive x direction. Vector B has magnitude 5.0 and points 53 degrees counterclockwise from the positive x direction. Vector C has magnitude 12.0 and points 60 degrees clockwise from the positive x direction. Vector D has magnitude 20.0 and points 37 degrees clockwise from the negative x direction. Vector F has magnitude 20.0 and points 30 degrees counterclockwise from the negative x direction.\" \/><\/span>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167132508049\" class=\"problem textbox\">\r\n<div id=\"fs-id1167132508051\">\r\n<p id=\"fs-id1167132445103\">Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point? What is your displacement vector? What is the direction of your displacement? Assume the +<em>x<\/em>-axis is horizontal to the right.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167132487189\" class=\"problem textbox\">\r\n<div id=\"fs-id1167132487191\">\r\n<p id=\"fs-id1167132366338\">You drive 7.50 km in a straight line in a direction $$ 15\\text{\u00b0} $$ east of north. (a) Find the distances you would have to drive straight east and then straight north to arrive at the same point. (b) Show that you still arrive at the same point if the east and north legs are reversed in order. Assume the +<em>x<\/em>-axis is to the east.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1167133521610\" class=\"solution\">\r\n\r\n[reveal-answer q=\"fs-id1167133521610\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"fs-id1167133521610\"]\r\n<p id=\"fs-id1167132501142\">a. 1.94 km, 7.24 km; b. proof<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167132501147\" class=\"problem textbox\">\r\n<div id=\"fs-id1167132248105\">\r\n<p id=\"fs-id1167132248107\">A sledge is being pulled by two horses on a flat terrain. The net force on the sledge can be expressed in the Cartesian coordinate system as vector $$ \\overset{\\to }{F}=(-2980.0\\hat{i}+8200.0\\hat{j})\\text{N}$$, where $$ \\hat{i} $$ and $$ \\hat{j} $$ denote directions to the east and north, respectively. Find the magnitude and direction of the pull.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167132536425\" class=\"problem textbox\">\r\n<div id=\"fs-id1167132335895\">\r\n<p id=\"fs-id1167132335897\">A trapper walks a 5.0-km straight-line distance from her cabin to the lake, as shown in the following figure. Determine the east and north components of her displacement vector. How many more kilometers would she have to walk if she walked along the component displacements? What is her displacement vector?<\/p>\r\n<span id=\"fs-id1167132294765\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31183927\/CNX_UPhysics_02_01_problem3_img.jpg\" alt=\"The vector from the cabin to the lake is vector S, magnitude 5.0 kilometers and pointing 40 degrees north of east. Two additional meandering paths are shown and labeled path 1 and path 2.\" \/><\/span>\r\n\r\n<\/div>\r\n<div id=\"fs-id1167132487930\">\r\n<p id=\"fs-id1167132487932\">3.8 km east, 3.2 km north, 2.0 km, $$ \\overset{\\to }{D}=(3.8\\hat{i}+3.2\\hat{j})\\text{km}$$<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167133544873\" class=\"problem textbox\">\r\n<div id=\"fs-id1167132628948\">\r\n<p id=\"fs-id1167132628950\">The polar coordinates of a point are $$ 4\\pi \\text{\/}3 $$ and 5.50 m. What are its Cartesian coordinates?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167132303332\" class=\"problem textbox\">\r\n<div id=\"fs-id1167132303334\">\r\n<p id=\"fs-id1167132627652\">Two points in a plane have polar coordinates $$ {P}_{1}(2.500\\,\\text{m},\\pi \\text{\/}6) $$ and $$ {P}_{2}(3.800\\,\\text{m},2\\pi \\text{\/}3)$$. Determine their Cartesian coordinates and the distance between them in the Cartesian coordinate system. Round the distance to a nearest centimeter.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1167132579170\" class=\"solution\">\r\n\r\n[reveal-answer q=\"fs-id1167132579170\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"fs-id1167132579170\"]\r\n<p id=\"fs-id1167132579172\">$${P}_{1}(2.165\\,\\text{m},1.250\\,\\text{m})$$, $$ {P}_{2}(-1.900\\,\\text{m},3.290\\,\\text{m})$$, 5.27 m<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167133676180\" class=\"problem textbox\">\r\n<div id=\"fs-id1167132458391\">\r\n<p id=\"fs-id1167132458394\">A chameleon is resting quietly on a lanai screen, waiting for an insect to come by. Assume the origin of a Cartesian coordinate system at the lower left-hand corner of the screen and the horizontal direction to the right as the +<em>x<\/em>-direction. If its coordinates are (2.000 m, 1.000 m), (a) how far is it from the corner of the screen? (b) What is its location in polar coordinates?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167132319658\" class=\"problem textbox\">\r\n<div id=\"fs-id1167132319660\">\r\n<p id=\"fs-id1167132372735\">Two points in the Cartesian plane are <em>A<\/em>(2.00 m, \u22124.00 m) and <em>B<\/em>(\u22123.00 m, 3.00 m). Find the distance between them and their polar coordinates.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1167132372708\" class=\"solution\">\r\n\r\n[reveal-answer q=\"fs-id1167132372708\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"fs-id1167132372708\"]\r\n<p id=\"fs-id1167132340314\">8.60 m, $$ A(2\\sqrt{5}\\,\\text{m},0.647\\pi )$$, $$ B(3\\sqrt{2}\\,\\text{m},0.75\\pi )$$<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167132441243\" class=\"problem textbox\">\r\n<div id=\"fs-id1167132441245\">\r\n<p id=\"fs-id1167133579494\">A fly enters through an open window and zooms around the room. In a Cartesian coordinate system with three axes along three edges of the room, the fly changes its position from point <em>b<\/em>(4.0 m, 1.5 m, 2.5 m) to point <em>e<\/em>(1.0 m, 4.5 m, 0.5 m). Find the scalar components of the fly\u2019s displacement vector and express its displacement vector in vector component form. What is its magnitude?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Glossary<\/h3>\r\n<dl id=\"fs-id1167132687581\">\r\n \t<dt><strong>component form of a vector<\/strong><\/dt>\r\n \t<dd id=\"fs-id1167132612545\">a vector written as the vector sum of its components in terms of unit vectors<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1167132612567\">\r\n \t<dt><strong>direction angle<\/strong><\/dt>\r\n \t<dd id=\"fs-id1167132612572\">in a plane, an angle between the positive direction of the <em>x<\/em>-axis and the vector, measured counterclockwise from the axis to the vector<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1167132310341\">\r\n \t<dt><strong>polar coordinate system<\/strong><\/dt>\r\n \t<dd id=\"fs-id1167132506907\">an orthogonal coordinate system where location in a plane is given by polar coordinates<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1167132506911\">\r\n \t<dt><strong>polar coordinates<\/strong><\/dt>\r\n \t<dd id=\"fs-id1167132568470\">a radial coordinate and an angle<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1167132199051\">\r\n \t<dt><strong>radial coordinate<\/strong><\/dt>\r\n \t<dd id=\"fs-id1167132541607\">distance to the origin in a polar coordinate system<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1167132541611\">\r\n \t<dt><strong>scalar component<\/strong><\/dt>\r\n \t<dd id=\"fs-id1167132584081\">a number that multiplies a unit vector in a vector component of a vector<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1167132584102\">\r\n \t<dt><strong>unit vectors of the axes<\/strong><\/dt>\r\n \t<dd id=\"fs-id1167132546147\">unit vectors that define orthogonal directions in a plane or in space<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1167132546151\">\r\n \t<dt><strong>vector components<\/strong><\/dt>\r\n \t<dd id=\"fs-id1167132266674\">orthogonal components of a vector; a vector is the vector sum of its vector components.<\/dd>\r\n<\/dl>\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<ul>\n<li>Describe vectors in two and three dimensions in terms of their components, using unit vectors along the axes.<\/li>\n<li>Distinguish between the vector components of a vector and the scalar components of a vector.<\/li>\n<li>Explain how the magnitude of a vector is defined in terms of the components of a vector.<\/li>\n<li>Identify the direction angle of a vector in a plane.<\/li>\n<li>Explain the connection between polar coordinates and Cartesian coordinates in a plane.<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1167133446629\">Vectors are usually described in terms of their components in a <strong><span class=\"no-emphasis\">coordinate system<\/span><\/strong>. Even in everyday life we naturally invoke the concept of orthogonal projections in a rectangular coordinate system. For example, if you ask someone for directions to a particular location, you will more likely be told to go 40 km east and 30 km north than 50 km in the direction $$ 37\\text{\u00b0} $$ north of east.<\/p>\n<p id=\"fs-id1167133855194\">In a rectangular (Cartesian) <em>xy<\/em>-coordinate system in a plane, a point in a plane is described by a pair of coordinates (<em>x<\/em>, <em>y<\/em>). In a similar fashion, a vector $$ \\overset{\\to }{A} $$ in a plane is described by a pair of its <em>vector<\/em> coordinates. The <em>x<\/em>-coordinate of vector $$ \\overset{\\to }{A} $$ is called its <em>x<\/em>-component and the <em>y<\/em>-coordinate of vector $$ \\overset{\\to }{A} $$ is called its <em>y<\/em>-component. The vector <em>x<\/em>-component is a vector denoted by $$ {\\overset{\\to }{A}}_{x}$$. The vector <em>y<\/em>-component is a vector denoted by $$ {\\overset{\\to }{A}}_{y}$$. In the Cartesian system, the <em>x<\/em> and <em>y<\/em> <strong>vector components<\/strong> of a vector are the orthogonal projections of this vector onto the <em>x<\/em>&#8211; and <em>y<\/em>-axes, respectively. In this way, following the parallelogram rule for vector addition, each vector on a Cartesian plane can be expressed as the vector sum of its vector components:<\/p>\n<div id=\"fs-id1167133635848\">$$\\overset{\\to }{A}={\\overset{\\to }{A}}_{x}+{\\overset{\\to }{A}}_{y}.$$<\/div>\n<p id=\"fs-id1167132574978\">As illustrated in <a class=\"autogenerated-content\" href=\"#CNX_UPhysics_02_02_comp01\">(Figure)<\/a>, vector $$ \\overset{\\to }{A} $$ is the diagonal of the rectangle where the <em>x<\/em>-component $$ {\\overset{\\to }{A}}_{x} $$ is the side parallel to the <em>x<\/em>-axis and the <em>y<\/em>-component $$ {\\overset{\\to }{A}}_{y} $$ is the side parallel to the <em>y<\/em>-axis. Vector component $$ {\\overset{\\to }{A}}_{x} $$ is orthogonal to vector component $$ {\\overset{\\to }{A}}_{y}$$.<\/p>\n<div id=\"CNX_UPhysics_02_02_comp01\" class=\"wp-caption aligncenter\">\n<div style=\"width: 540px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31183900\/CNX_UPhysics_02_02_comp01.jpg\" alt=\"Vector A is shown in the x y coordinate system and extends from point b at A\u2019s tail to point e and its head. Vector A points up and to the right. Unit vectors I hat and j hat are small vectors pointing in the x and y directions, respectively, and are at right angles to each other. The x component of vector A is a vector pointing horizontally from the point b to a point directly below point e at the tip of vector A. On the x axis, we see that the vector A sub x extends from x sub b to x sub e and is equal to magnitude A sub x times I hat. The magnitude A sub x equals x sub e minus x sub b. The y component of vector A is a vector pointing vertically from point b to a point directly to the left of point e at the tip of vector A. On the y axis, we see that the vector A sub y extends from y sub b to y sub e and is equal to magnitude A sub y times j hat. The magnitude A sub y equals y sub e minus y sub b.\" width=\"530\" height=\"394\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 2.16<\/strong> Vector $$ \\overset{\\to }{A} $$ in a plane in the Cartesian coordinate system is the vector sum of its vector x- and y-components. The x-vector component $$ {\\overset{\\to }{A}}_{x} $$ is the orthogonal projection of vector $$ \\overset{\\to }{A} $$ onto the x-axis. The y-vector component $$ {\\overset{\\to }{A}}_{y} $$ is the orthogonal projection of vector $$ \\overset{\\to }{A} $$ onto the y-axis. The numbers $$ {A}_{x} $$ and $$ {A}_{y} $$ that multiply the unit vectors are the scalar components of the vector.<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1167132422244\">It is customary to denote the positive direction on the <em>x<\/em>-axis by the unit vector $$ \\hat{i} $$ and the positive direction on the <em>y<\/em>-axis by the unit vector $$ \\hat{j}$$. <strong>Unit vectors of the axes<\/strong>, $$ \\hat{i} $$ and $$ \\hat{j}$$, define two orthogonal directions in the plane. As shown in <a class=\"autogenerated-content\" href=\"#CNX_UPhysics_02_02_comp01\">(Figure)<\/a>, the <em>x<\/em>&#8211; and <em>y<\/em>&#8211; components of a vector can now be written in terms of the unit vectors of the axes:<\/p>\n<div id=\"fs-id1167132230873\">$$\\{\\begin{array}{c}{\\overset{\\to }{A}}_{x}={A}_{x}\\hat{i}\\\\ {\\overset{\\to }{A}}_{y}={A}_{y}\\hat{j}.\\end{array}$$<\/div>\n<p id=\"fs-id1167133345023\">The vectors $$ {\\overset{\\to }{A}}_{x} $$ and $$ {\\overset{\\to }{A}}_{y} $$ defined by <a class=\"autogenerated-content\" href=\"#fs-id1167132230873\">(Figure)<\/a> are the <em>vector components<\/em> of vector $$ \\overset{\\to }{A}$$. The numbers $$ {A}_{x} $$ and $$ {A}_{y} $$ that define the vector components in <a class=\"autogenerated-content\" href=\"#fs-id1167132230873\">(Figure)<\/a> are the scalar components of vector $$ \\overset{\\to }{A}$$. Combining <a class=\"autogenerated-content\" href=\"#fs-id1167133635848\">(Figure)<\/a> with <a class=\"autogenerated-content\" href=\"#fs-id1167132230873\">(Figure)<\/a>, we obtain <strong>the component form of a vector:<\/strong><\/p>\n<div id=\"fs-id1167132413666\" class=\"equation-callout\">\n<div id=\"fs-id1167132256590\">$$\\overset{\\to }{A}={A}_{x}\\hat{i}+{A}_{y}\\hat{j}.$$<\/div>\n<\/div>\n<p id=\"fs-id1167132689436\">If we know the coordinates $$ b({x}_{b},{y}_{b}) $$ of the origin point of a vector (where <em>b<\/em> stands for \u201cbeginning\u201d) and the coordinates $$ e({x}_{e},{y}_{e}) $$ of the end point of a vector (where <em>e<\/em> stands for \u201cend\u201d), we can obtain the scalar components of a vector simply by subtracting the origin point coordinates from the end point coordinates:<\/p>\n<div id=\"fs-id1167132581667\" class=\"equation-callout\">\n<div id=\"fs-id1167132197498\">$$\\{\\begin{array}{c}{A}_{x}={x}_{e}-{x}_{b}\\\\ {A}_{y}={y}_{e}-{y}_{b}.\\end{array}$$<\/div>\n<\/div>\n<div id=\"fs-id1167132341727\" class=\"textbox examples\">\n<h3>Example<\/h3>\n<h4>Displacement of a Mouse Pointer<\/h4>\n<p>A mouse pointer on the display monitor of a computer at its initial position is at point (6.0 cm, 1.6 cm) with respect to the lower left-side corner. If you move the pointer to an icon located at point (2.0 cm, 4.5 cm), what is the displacement vector of the pointer?<\/p>\n<h4>Strategy<\/h4>\n<p>The origin of the <em>xy<\/em>-coordinate system is the lower left-side corner of the computer monitor. Therefore, the unit vector $$ \\hat{i} $$ on the <em>x<\/em>-axis points horizontally to the right and the unit vector $$ \\hat{j} $$ on the <em>y<\/em>-axis points vertically upward. The origin of the displacement vector is located at point <em>b<\/em>(6.0, 1.6) and the end of the displacement vector is located at point <em>e<\/em>(2.0, 4.5). Substitute the coordinates of these points into <a class=\"autogenerated-content\" href=\"#fs-id1167132197498\">(Figure)<\/a> to find the scalar components $$ {D}_{x} $$ and $$ {D}_{y} $$ of the displacement vector $$ \\overset{\\to }{D}$$. Finally, substitute the coordinates into <a class=\"autogenerated-content\" href=\"#fs-id1167132256590\">(Figure)<\/a> to write the displacement vector in the vector component form.<\/p>\n<h4>Solution<\/h4>\n<p id=\"fs-id1167132750241\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q907005\">Show Answer<\/span><\/p>\n<div id=\"q907005\" class=\"hidden-answer\" style=\"display: none\">We identify $$ {x}_{b}=6.0$$, $$ {x}_{e}=2.0$$, $$ {y}_{b}=1.6$$, and $$ {y}_{e}=4.5$$, where the physical unit is 1 cm. The scalar x- and y-components of the displacement vector are $$ \\begin{array}{cc}\\hfill {D}_{x}&amp; ={x}_{e}-{x}_{b}=(2.0-6.0)\\text{cm}=-4.0\\,\\text{cm},\\hfill \\\\ \\hfill {D}_{y}&amp; ={y}_{e}-{y}_{b}=(4.5-1.6)\\text{cm}=+2.9\\,\\text{cm}.\\hfill \\end{array}$$<\/p>\n<p>The vector component form of the displacement vector is<\/p>\n<p>$$\\overset{\\to }{D}={D}_{x}\\hat{i}+{D}_{y}\\hat{j}=(-4.0\\,\\text{cm})\\hat{i}+(2.9\\,\\text{cm})\\hat{j}=(-4.0\\hat{i}+2.9\\hat{j})\\text{cm}.$$<a id=\"formula\"><\/a>This solution is shown in (Figure 2.17).<\/p>\n<div id=\"CNX_UPhysics_02_02_comp02\" class=\"wp-caption aligncenter\">\n<div style=\"width: 594px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31183903\/CNX_UPhysics_02_02_comp02.jpg\" alt=\"Vector D extends from coordinates 6.0, 1.6 to coordinates 2.0, 4.5. Vector D equals vector D sub x plus vector D sub y. D sub x equals minus 4.0 I hat, and extends from x=6.0 to x =2.0. The magnitude D sub x equals 2.0-6.0 = -4.0. D sub y equals plus 2.9 j hat, and extends from y=1.6 to y=4.5. The magnitude D sub y equals 4.5 \u2212 1.6.\" width=\"584\" height=\"416\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 2.17<\/strong> The graph of the displacement vector. The vector points from the origin point at b to the end point at e.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h4>Significance<\/h4>\n<p>Notice that the physical unit\u2014here, 1 cm\u2014can be placed either with each component immediately before the unit vector or globally for both components, as in <a class=\"autogenerated-content\" href=\"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/chapter\/2-2-coordinate-systems-and-components-of-a-vector#formula\">(Figure)<\/a>. Often, the latter way is more convenient because it is simpler.<\/p>\n<p id=\"fs-id1167132354421\">The vector <em>x<\/em>-component $$ {\\overset{\\to }{D}}_{x}=-4.0\\hat{i}=4.0(\\text{\u2212}\\hat{i}) $$ of the displacement vector has the magnitude $$ |{\\overset{\\to }{D}}_{x}|=|-4.0||\\hat{i}|=4.0 $$ because the magnitude of the unit vector is $$ |\\hat{i}|=1$$. Notice, too, that the direction of the <em>x<\/em>-component is $$ \\text{\u2212}\\hat{i}$$, which is antiparallel to the direction of the +<em>x<\/em>-axis; hence, the <em>x<\/em>-component vector $$ {\\overset{\\to }{D}}_{x} $$ points to the left, as shown in <a class=\"autogenerated-content\" href=\"#CNX_UPhysics_02_02_comp02\">(Figure)<\/a>. The scalar <em>x<\/em>-component of vector $$ \\overset{\\to }{D} $$ is $$ {D}_{x}=-4.0$$.<\/p>\n<p id=\"fs-id1167132436291\">Similarly, the vector <em>y<\/em>-component $$ {\\overset{\\to }{D}}_{y}=+2.9\\hat{j} $$ of the displacement vector has magnitude $$ |{\\overset{\\to }{D}}_{y}|=|2.9||\\hat{j}|=\\,2.9 $$ because the magnitude of the unit vector is $$ |\\hat{j}|=1$$. The direction of the <em>y<\/em>-component is $$ +\\hat{j}$$, which is parallel to the direction of the +<em>y<\/em>-axis. Therefore, the <em>y<\/em>-component vector $$ {\\overset{\\to }{D}}_{y} $$ points up, as seen in <a class=\"autogenerated-content\" href=\"#CNX_UPhysics_02_02_comp02\">(Figure)<\/a>. The scalar <em>y<\/em>-component of vector $$ \\overset{\\to }{D} $$ is $$ {D}_{y}=+2.9$$. The displacement vector $$ \\overset{\\to }{D} $$ is the resultant of its two <em>vector<\/em> components.<\/p>\n<p id=\"fs-id1167132579136\">The vector component form of the displacement vector <a class=\"autogenerated-content\" href=\"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/chapter\/2-2-coordinate-systems-and-components-of-a-vector#formula\">(Figure)<\/a> tells us that the mouse pointer has been moved on the monitor 4.0 cm to the left and 2.9 cm upward from its initial position.<\/p>\n<\/div>\n<div id=\"fs-id1167132557954\" class=\"textbox exercises check-understanding\">\n<h3>Check Your Understanding<\/h3>\n<div id=\"fs-id1167132406917\" class=\"problem textbox\">\n<div id=\"fs-id1167133364766\">\n<p id=\"fs-id1167133617871\">A blue fly lands on a sheet of graph paper at a point located 10.0 cm to the right of its left edge and 8.0 cm above its bottom edge and walks slowly to a point located 5.0 cm from the left edge and 5.0 cm from the bottom edge. Choose the rectangular coordinate system with the origin at the lower left-side corner of the paper and find the displacement vector of the fly. Illustrate your solution by graphing.<\/p>\n<\/div>\n<div id=\"fs-id1167132407980\" class=\"solution\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167132407980\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167132407980\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167132322353\">$$\\overset{\\to }{D}=(-5.0\\hat{i}-3.0\\hat{j})\\text{cm}$$; the fly moved 5.0 cm to the left and 3.0 cm down from its landing site.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1167132480350\">When we know the scalar components $$ {A}_{x} $$ and $$ {A}_{y} $$ of a vector $$ \\overset{\\to }{A}$$, we can find its magnitude <em>A<\/em> and its direction angle $$ {\\theta }_{A}$$. The<strong> direction angle<\/strong>\u2014or direction, for short\u2014is the angle the vector forms with the positive direction on the <em>x<\/em>-axis. The angle $$ {\\theta }_{A} $$ is measured in the <em>counterclockwise direction<\/em> from the +<em>x<\/em>-axis to the vector (<a class=\"autogenerated-content\" href=\"#CNX_UPhysics_02_02_comp03\">(Figure)<\/a>). Because the lengths <em>A<\/em>, $$ {A}_{x}$$, and $$ {A}_{y} $$ form a right triangle, they are related by the Pythagorean theorem:<\/p>\n<div id=\"fs-id1167132412442\" class=\"equation-callout\">\n<div id=\"fs-id1167132245160\">$${A}^{2}={A}_{x}^{2}+{A}_{y}^{2}\\enspace\u21d4\\enspace{A}=\\sqrt{{A}_{x}^{2}+{A}_{y}^{2}}.$$<\/div>\n<\/div>\n<p id=\"fs-id1167132688313\">This equation works even if the scalar components of a vector are negative. The direction angle $$ {\\theta }_{A} $$ of a vector is defined via the tangent function of angle $$ {\\theta }_{A} $$ in the triangle shown in <a class=\"autogenerated-content\" href=\"#CNX_UPhysics_02_02_comp03\">(Figure)<\/a>:<\/p>\n<div id=\"fs-id1167133610831\" class=\"equation-callout\">\n<div id=\"fs-id1167133377087\">$$\\text{tan}\\,{\\theta }_{A}=\\frac{{A}_{y}}{{A}_{x}}\\enspace\u21d2\\enspace{\\theta }_{A}={\\text{tan}}^{-1}(\\frac{{A}_{y}}{{A}_{x}}).$$<\/div>\n<\/div>\n<div id=\"CNX_UPhysics_02_02_comp03\" class=\"wp-caption aligncenter\">\n<div style=\"width: 534px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31183906\/CNX_UPhysics_02_02_comp03.jpg\" alt=\"Vector A has horizontal x component A sub x equal to magnitude A sub x I hat and vertical y component A sub y equal to magnitude A sub y j hat. Vector A and the components form a right triangle with sides length magnitude A sub x and magnitude A sub y and hypotenuse magnitude A equal to the square root of A sub x squared plus A sub y squared. The angle between the horizontal side A sub x and the hypotenuse A is theta sub A.\" width=\"524\" height=\"378\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 2.18<\/strong> For vector $$ \\overset{\\to }{A}$$, its magnitude A and its direction angle $$ {\\theta }_{A} $$ are related to the magnitudes of its scalar components because A, $$ {A}_{x}$$, and $$ {A}_{y} $$ form a right triangle.<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1167132436720\">When the vector lies either in the first quadrant or in the fourth quadrant, where component $$ {A}_{x} $$ is positive (<a class=\"autogenerated-content\" href=\"#CNX_UPhysics_02_02_comp04\">(Figure)<\/a>), the angle $$ \\theta  $$ in <a class=\"autogenerated-content\" href=\"#fs-id1167133377087\">(Figure)<\/a> is identical to the direction angle $$ {\\theta }_{A}$$. For vectors in the fourth quadrant, angle $$ \\theta  $$ is negative, which means that for these vectors, direction angle $$ {\\theta }_{A} $$ is measured <em>clockwise<\/em> from the positive <em>x<\/em>-axis. Similarly, for vectors in the second quadrant, angle $$ \\theta  $$ is negative. When the vector lies in either the second or third quadrant, where component $$ {A}_{x} $$ is negative, the direction angle is $$ {\\theta }_{A}=\\theta +180\\text{\u00b0} $$ (<a class=\"autogenerated-content\" href=\"#CNX_UPhysics_02_02_comp04\">(Figure)<\/a>).<\/p>\n<div id=\"CNX_UPhysics_02_02_comp04\" class=\"wp-caption aligncenter\">\n<div style=\"width: 594px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31183910\/CNX_UPhysics_02_02_comp04.jpg\" alt=\"Figure I shows vector A in the first quadrant (pointing up and right.) It has positive x and y components A sub x and A sub y, and the angle theta sub A measured counterclockwise from the positive x axis is smaller than 90 degrees. Figure II shows vector A in the first second (pointing up and left.) It has negative x and positive y components A sub x and A sub y. The angle theta sub A measured counterclockwise from the positive x axis is larger than 90 degrees but less than 180 degrees. The angle theta, measured clockwise from the negative x axis, is smaller than 90 degrees. Figure III shows vector A in the third quadrant (pointing down and left.) It has negative x and y components A sub x and A sub y, and the angle theta sub A measured counterclockwise from the positive x axis is larger than 180 degrees and smaller than 270 degrees. The angle theta, measured counterclockwise from the negative x axis, is smaller than 90 degrees. Figure IV shows vector A in the fourth quadrant (pointing down and right.) It has positive x and negative y components A sub x and A sub y, and the angle theta sub A measured clockwise from the positive x axis is smaller than 90 degrees.\" width=\"584\" height=\"477\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 2.19<\/strong> Scalar components of a vector may be positive or negative. Vectors in the first quadrant (I) have both scalar components positive and vectors in the third quadrant have both scalar components negative. For vectors in quadrants II and III, the direction angle of a vector is $$ {\\theta }_{A}=\\theta +180\\text{\u00b0}$$.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1167133568370\" class=\"textbox examples\">\n<h3>Example<\/h3>\n<p id=\"fs-id1167132606822\">Magnitude and Direction of the Displacement VectorYou move a mouse pointer on the display monitor from its initial position at point (6.0 cm, 1.6 cm) to an icon located at point (2.0 cm, 4.5 cm). What are the magnitude and direction of the displacement vector of the pointer?<\/p>\n<h4>Strategy<\/h4>\n<p>In <a class=\"autogenerated-content\" href=\"#fs-id1167132341727\">(Figure)<\/a>, we found the displacement vector $$ \\overset{\\to }{D} $$ of the mouse pointer (see <a class=\"autogenerated-content\" href=\"#fs-id1167132371006\">(Figure)<\/a>). We identify its scalar components $$ {D}_{x}=-4.0\\,\\text{cm} $$ and $$ {D}_{y}=+2.9\\,\\text{cm} $$ and substitute into <a class=\"autogenerated-content\" href=\"#fs-id1167132245160\">(Figure)<\/a> and <a class=\"autogenerated-content\" href=\"#fs-id1167133377087\">(Figure)<\/a> to find the magnitude <em>D<\/em> and direction $$ {\\theta }_{D}$$, respectively.<\/p>\n<h4>Solution<\/h4>\n<p id=\"fs-id1167132276264\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q170894\">Show Answer<\/span><\/p>\n<div id=\"q170894\" class=\"hidden-answer\" style=\"display: none\">The magnitude of vector $$ \\overset{\\to }{D} $$ is $$ D=\\sqrt{{D}_{x}^{2}+{D}_{y}^{2}}=\\sqrt{{(-4.0\\,\\text{cm})}^{2}+{(2.9\\,\\text{cm})}^{2}}=\\sqrt{{(4.0)}^{2}+{(2.9)}^{2}}\\,\\text{cm}=4.9\\,\\text{cm}. $$ The direction angle is $$ \\text{tan}\\,\\theta =\\frac{{D}_{y}}{{D}_{x}}=\\frac{+2.9\\,\\text{cm}}{-4.0\\,\\text{cm}}=-0.725\\enspace\u21d2\\enspace\\theta ={\\text{tan}}^{-1}(-0.725)=-35.9\\text{\u00b0}. $$ Vector $$ \\overset{\\to }{D} $$ lies in the second quadrant, so its direction angle is $$ {\\theta }_{D}=\\theta +180\\text{\u00b0}=-35.9\\text{\u00b0}+180\\text{\u00b0}=144.1\\text{\u00b0}.$$<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167132438204\" class=\"textbox exercises check-understanding\">\n<h3>Check Your Understanding<\/h3>\n<div id=\"fs-id1167132502120\" class=\"problem textbox\">\n<div id=\"fs-id1167132502143\">\n<p id=\"fs-id1167132502219\">If the displacement vector of a blue fly walking on a sheet of graph paper is $$ \\overset{\\to }{D}=(-5.00\\hat{i}-3.00\\hat{j})\\text{cm}$$, find its magnitude and direction.<\/p>\n<\/div>\n<div id=\"fs-id1167132332238\" class=\"solution\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167132332238\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167132332238\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167132335635\">5.83 cm, $$ 211\\text{\u00b0}$$<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1167132464780\">In many applications, the magnitudes and directions of vector quantities are known and we need to find the resultant of many vectors. For example, imagine 400 cars moving on the Golden Gate Bridge in San Francisco in a strong wind. Each car gives the bridge a different push in various directions and we would like to know how big the resultant push can possibly be. We have already gained some experience with the geometric construction of vector sums, so we know the task of finding the resultant by drawing the vectors and measuring their lengths and angles may become intractable pretty quickly, leading to huge errors. Worries like this do not appear when we use analytical methods. The very first step in an analytical approach is to find vector components when the direction and magnitude of a vector are known.<\/p>\n<p id=\"fs-id1167132464805\">Let us return to the right triangle in <a class=\"autogenerated-content\" href=\"#CNX_UPhysics_02_02_comp03\">(Figure)<\/a>. The quotient of the adjacent side $$ {A}_{x} $$ to the hypotenuse <em>A<\/em> is the cosine function of direction angle $$ {\\theta }_{A}$$, $$ {A}_{x}\\text{\/}A=\\text{cos}\\,{\\theta }_{A}$$, and the quotient of the opposite side $$ {A}_{y} $$ to the hypotenuse <em>A<\/em> is the sine function of $$ {\\theta }_{A}$$, $$ {A}_{y}\\text{\/}A=\\text{sin}\\,{\\theta }_{A}$$. When magnitude <em>A<\/em> and direction $$ {\\theta }_{A} $$ are known, we can solve these relations for the scalar components:<\/p>\n<div id=\"fs-id1167133740682\" class=\"equation-callout\">\n<div id=\"fs-id1167133740741\">$$\\{\\begin{array}{c}{A}_{x}=A\\,\\text{cos}\\,{\\theta }_{A}\\\\ {A}_{y}=A\\,\\text{sin}\\,{\\theta }_{A}\\end{array}.$$<\/div>\n<\/div>\n<p id=\"fs-id1167132266496\">When calculating vector components with <a class=\"autogenerated-content\" href=\"#fs-id1167133740741\">(Figure)<\/a>, care must be taken with the angle. The direction angle $$ {\\theta }_{A} $$ of a vector is the angle measured <em>counterclockwise<\/em> from the positive direction on the <em>x<\/em>-axis to the vector. The clockwise measurement gives a negative angle.<\/p>\n<div id=\"fs-id1167132576856\" class=\"textbox examples\">\n<h3>Example<\/h3>\n<h4>Components of Displacement Vectors<\/h4>\n<p>A rescue party for a missing child follows a search dog named Trooper. Trooper wanders a lot and makes many trial sniffs along many different paths. Trooper eventually finds the child and the story has a happy ending, but his displacements on various legs seem to be truly convoluted. On one of the legs he walks 200.0 m southeast, then he runs north some 300.0 m. On the third leg, he examines the scents carefully for 50.0 m in the direction $$ 30\\text{\u00b0} $$ west of north. On the fourth leg, Trooper goes directly south for 80.0 m, picks up a fresh scent and turns $$ 23\\text{\u00b0} $$ west of south for 150.0 m. Find the scalar components of Trooper\u2019s displacement vectors and his displacement vectors in vector component form for each leg.<\/p>\n<h4>Strategy<\/h4>\n<p>Let\u2019s adopt a rectangular coordinate system with the positive <em>x<\/em>-axis in the direction of geographic east, with the positive <em>y<\/em>-direction pointed to geographic north. Explicitly, the unit vector $$ \\hat{i} $$ of the <em>x<\/em>-axis points east and the unit vector $$ \\hat{j} $$ of the <em>y<\/em>-axis points north. Trooper makes five legs, so there are five displacement vectors. We start by identifying their magnitudes and direction angles, then we use <a class=\"autogenerated-content\" href=\"#fs-id1167133740741\">(Figure)<\/a> to find the scalar components of the displacements and <a class=\"autogenerated-content\" href=\"#fs-id1167132256590\">(Figure)<\/a> for the displacement vectors.<\/p>\n<h4>Solution<\/h4>\n<p id=\"fs-id1167132516072\">On the first leg, the displacement magnitude is $$ {L}_{1}=200.0\\,\\text{m} $$ and the direction is southeast. For direction angle $$ {\\theta }_{1} $$ we can take either $$ 45\\text{\u00b0} $$ measured clockwise from the east direction or $$ 45\\text{\u00b0}+270\\text{\u00b0} $$ measured counterclockwise from the east direction. With the first choice, $$ {\\theta }_{1}=-45\\text{\u00b0}$$. With the second choice, $$ {\\theta }_{1}=+315\\text{\u00b0}$$. We can use either one of these two angles. The components are<\/p>\n<div id=\"fs-id1167132433420\" class=\"unnumbered\">$$\\begin{array}{l}{L}_{1x}={L}_{1}\\,\\text{cos}\\,{\\theta }_{1}=(200.0\\,\\text{m})\\,\\text{cos}\\,315\\text{\u00b0}=141.4\\,\\text{m,}\\\\ {L}_{1y}={L}_{1}\\,\\text{sin}\\,{\\theta }_{1}=(200.0\\,\\text{m})\\,\\text{sin}\\,315\\text{\u00b0}=-141.4\\,\\text{m}.\\end{array}$$<\/div>\n<p id=\"fs-id1167133763578\">The displacement vector of the first leg is<\/p>\n<div id=\"fs-id1167133763683\" class=\"unnumbered\">$${\\overset{\\to }{L}}_{1}={L}_{1x}\\hat{i}+{L}_{1y}\\hat{j}=(141.4\\hat{i}-141.4\\hat{j})\\,\\text{m}.$$<\/div>\n<p id=\"fs-id1167133658240\">On the second leg of Trooper\u2019s wanderings, the magnitude of the displacement is $$ {L}_{2}=300.0\\,\\text{m} $$ and the direction is north. The direction angle is $$ {\\theta }_{2}=+90\\text{\u00b0}$$. We obtain the following results:<\/p>\n<div id=\"fs-id1167132541311\" class=\"unnumbered\">$$\\begin{array}{ccc}\\hfill {L}_{2x}&amp; =\\hfill &amp; {L}_{2}\\,\\text{cos}\\,{\\theta }_{2}=(300.0\\,\\text{m})\\,\\text{cos}\\,90\\text{\u00b0}=0.0\\,,\\hfill \\\\ \\hfill {L}_{2y}&amp; =\\hfill &amp; {L}_{2}\\,\\text{sin}\\,{\\theta }_{2}=(300.0\\,\\text{m})\\,\\text{sin}\\,90\\text{\u00b0}=300.0\\,\\text{m,}\\hfill \\\\ \\hfill {\\overset{\\to }{L}}_{2}&amp; =\\hfill &amp; {L}_{2x}\\hat{i}+{L}_{2y}\\hat{j}=(300.0\\,\\text{m})\\hat{j}.\\hfill \\end{array}$$<\/div>\n<p id=\"fs-id1167132349429\">On the third leg, the displacement magnitude is $$ {L}_{3}=50.0\\,\\text{m} $$ and the direction is $$ 30\\text{\u00b0} $$ west of north. The direction angle measured counterclockwise from the eastern direction is $$ {\\theta }_{3}=30\\text{\u00b0}+90\\text{\u00b0}=+120\\text{\u00b0}$$. This gives the following answers:<\/p>\n<div id=\"fs-id1167132595199\" class=\"unnumbered\">$$\\begin{array}{ccc}\\hfill {L}_{3x}&amp; =\\hfill &amp; {L}_{3}\\,\\text{cos}\\,{\\theta }_{3}=(50.0\\,\\text{m})\\,\\text{cos}\\,120\\text{\u00b0}=-25.0\\,\\text{m,}\\hfill \\\\ \\hfill {L}_{3y}&amp; =\\hfill &amp; {L}_{3}\\,\\text{sin}\\,{\\theta }_{3}=(50.0\\,\\text{m})\\,\\text{sin}\\,120\\text{\u00b0}=+43.3\\,\\text{m,}\\hfill \\\\ \\hfill {\\overset{\\to }{L}}_{3}&amp; =\\hfill &amp; {L}_{3x}\\hat{i}+{L}_{3y}\\hat{j}=(-25.0\\hat{i}+43.3\\hat{j})\\text{m}.\\hfill \\end{array}$$<\/div>\n<p id=\"fs-id1167132295353\">On the fourth leg of the excursion, the displacement magnitude is $$ {L}_{4}=80.0\\,\\text{m} $$ and the direction is south. The direction angle can be taken as either $$ {\\theta }_{4}=-90\\text{\u00b0} $$ or $$ {\\theta }_{4}=+270\\text{\u00b0}$$. We obtain<\/p>\n<div id=\"fs-id1167132408563\" class=\"unnumbered\">$$\\begin{array}{ccc}\\hfill {L}_{4x}&amp; =\\hfill &amp; {L}_{4}\\,\\text{cos}\\,{\\theta }_{4}=(80.0\\,\\text{m})\\,\\text{cos}\\,(-90\\text{\u00b0})=0\\,,\\hfill \\\\ \\hfill {L}_{4y}&amp; =\\hfill &amp; {L}_{4}\\,\\text{sin}\\,{\\theta }_{4}=(80.0\\,\\text{m})\\,\\text{sin}\\,(-90\\text{\u00b0})=-80.0\\,\\text{m,}\\hfill \\\\ \\hfill {\\overset{\\to }{L}}_{4}&amp; =\\hfill &amp; {L}_{4x}\\hat{i}+{L}_{4y}\\hat{j}=(-80.0\\,\\text{m})\\hat{j}.\\hfill \\end{array}$$<\/div>\n<p id=\"fs-id1167132544336\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q567672\">Show Answer<\/span><\/p>\n<div id=\"q567672\" class=\"hidden-answer\" style=\"display: none\">On the last leg, the magnitude is $$ {L}_{5}=150.0\\,\\text{m} $$ and the angle is $$ {\\theta }_{5}=-23\\text{\u00b0}+270\\text{\u00b0}=+247\\text{\u00b0} $$ $$(23\\text{\u00b0} $$ west of south), which gives $$ \\begin{array}{ccc}\\hfill {L}_{5x}&amp; =\\hfill &amp; {L}_{5}\\,\\text{cos}\\,{\\theta }_{5}=(150.0\\,\\text{m})\\,\\text{cos}\\,247\\text{\u00b0}=-58.6\\,\\text{m,}\\hfill \\\\ \\hfill {L}_{5y}&amp; =\\hfill &amp; {L}_{5}\\,\\text{sin}\\,{\\theta }_{5}=(150.0\\,\\text{m})\\,\\text{sin}\\,247\\text{\u00b0}=-138.1\\,\\text{m,}\\hfill \\\\ \\hfill {\\overset{\\to }{L}}_{5}&amp; =\\hfill &amp; {L}_{5x}\\hat{i}+{L}_{5y}\\hat{j}=(-58.6\\hat{i}-138.1\\hat{j})\\text{m}.\\hfill \\end{array}$$<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167128918681\" class=\"textbox exercises check-understanding\">\n<h3>Check Your Understanding<\/h3>\n<div id=\"fs-id1167128918786\" class=\"problem textbox\">\n<div id=\"fs-id1167128918869\">\n<p id=\"fs-id1167128918893\">If Trooper runs 20 m west before taking a rest, what is his displacement vector?<\/p>\n<\/div>\n<div id=\"fs-id1167128919108\" class=\"solution\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167128919108\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167128919108\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167128919128\">$$\\overset{\\to }{D}=(-20\\,\\text{m})\\hat{j}$$<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167128956776\" class=\"bc-section section\">\n<h3>Polar Coordinates<\/h3>\n<p id=\"fs-id1167132336599\">To describe locations of points or vectors in a plane, we need two orthogonal directions. In the Cartesian coordinate system these directions are given by unit vectors $$ \\hat{i} $$ and $$ \\hat{j} $$ along the <em>x<\/em>-axis and the <em>y<\/em>-axis, respectively. The Cartesian coordinate system is very convenient to use in describing displacements and velocities of objects and the forces acting on them. However, it becomes cumbersome when we need to describe the rotation of objects. When describing rotation, we usually work in the<strong> polar coordinate system<\/strong>.<\/p>\n<p id=\"fs-id1167132747901\">In the polar coordinate system, the location of point <em>P<\/em> in a plane is given by two <strong>polar<\/strong> <strong>coordinates<\/strong> (<a class=\"autogenerated-content\" href=\"#CNX_UPhysics_02_02_polar\">(Figure)<\/a>). The first polar coordinate is the <strong>radial coordinate<\/strong> <em>r<\/em>, which is the distance of point <em>P<\/em> from the origin. The second polar coordinate is an angle $$ \\phi  $$ that the radial vector makes with some chosen direction, usually the positive <em>x<\/em>-direction. In polar coordinates, angles are measured in radians, or rads. The radial vector is attached at the origin and points away from the origin to point <em>P.<\/em> This radial direction is described by a unit radial vector $$ \\hat{r}$$. The second unit vector $$ \\hat{t} $$ is a vector orthogonal to the radial direction $$ \\hat{r}$$. The positive $$ +\\hat{t} $$ direction indicates how the angle $$ \\phi  $$ changes in the counterclockwise direction. In this way, a point <em>P<\/em> that has coordinates (<em>x<\/em>, <em>y<\/em>) in the rectangular system can be described equivalently in the polar coordinate system by the two polar coordinates $$ (r,\\phi )$$. <a class=\"autogenerated-content\" href=\"#fs-id1167133740741\">(Figure)<\/a> is valid for any vector, so we can use it to express the <em>x<\/em>&#8211; and <em>y<\/em>-coordinates of vector $$ \\overset{\\to }{r}$$. In this way, we obtain the connection between the polar coordinates and rectangular coordinates of point <em>P<\/em>:<\/p>\n<div id=\"fs-id1167132575879\" class=\"equation-callout\">\n<div id=\"fs-id1167132575978\">$$\\{\\begin{array}{c}x=r\\,\\text{cos}\\,\\phi \\\\ y=r\\,\\text{sin}\\,\\phi \\end{array}.$$<\/div>\n<\/div>\n<div id=\"CNX_UPhysics_02_02_polar\" class=\"wp-caption aligncenter\">\n<div style=\"width: 516px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31183914\/CNX_UPhysics_02_02_polar.jpg\" alt=\"Vector r points from the origin of the x y coordinate system to point P. The angle between the vector r and the positive x direction is phi. X equals r cosine phi and y equals r sine phi. Extending a line in the direction of r vector past point P, a unit vector r hat is drawn in the same direction as r. A unit vector t hat is perpendicular to r hat, pointing 90 degrees counterclockwise to r hat.\" width=\"506\" height=\"419\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 2.20<\/strong> Using polar coordinates, the unit vector $$ \\hat{r} $$ defines the positive direction along the radius r (radial direction) and, orthogonal to it, the unit vector $$ \\hat{t} $$ defines the positive direction of rotation by the angle $$ \\phi $$.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1167132451446\" class=\"textbox examples\">\n<h3>Example<\/h3>\n<h4>Polar Coordinates<\/h4>\n<p>A treasure hunter finds one silver coin at a location 20.0 m away from a dry well in the direction $$ 20\\text{\u00b0} $$ north of east and finds one gold coin at a location 10.0 m away from the well in the direction $$ 20\\text{\u00b0} $$ north of west. What are the polar and rectangular coordinates of these findings with respect to the well?<\/p>\n<h4>Strategy<\/h4>\n<p>The well marks the origin of the coordinate system and east is the +<em>x<\/em>-direction. We identify radial distances from the locations to the origin, which are $$ {r}_{S}=20.0\\,\\text{m} $$ (for the silver coin) and $$ {r}_{G}=10.0\\,\\text{m} $$ (for the gold coin). To find the angular coordinates, we convert $$ 20\\text{\u00b0} $$ to radians: $$ 20\\text{\u00b0}=\\pi 20\\text{\/}180=\\pi \\text{\/}9$$. We use <a class=\"autogenerated-content\" href=\"#fs-id1167132575978\">(Figure)<\/a> to find the <em>x<\/em>&#8211; and <em>y<\/em>-coordinates of the coins.<\/p>\n<h4>Solution<\/h4>\n<p id=\"fs-id1167132491818\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q8052\">Show Answer<\/span><\/p>\n<div id=\"q8052\" class=\"hidden-answer\" style=\"display: none\">The angular coordinate of the silver coin is $$ {\\phi }_{S}=\\pi \\text{\/}9$$, whereas the angular coordinate of the gold coin is $$ {\\phi }_{G}=\\pi -\\pi \\text{\/}9=8\\pi \\text{\/}9$$. Hence, the polar coordinates of the silver coin are $$ ({r}_{S},{\\phi }_{S})=(20.0\\,\\text{m},\\pi \\text{\/}9) $$ and those of the gold coin are $$ ({r}_{G},{\\phi }_{G})=(10.0\\,\\text{m},8\\pi \\text{\/}9)$$. We substitute these coordinates into (Figure) to obtain rectangular coordinates. For the gold coin, the coordinates are<\/p>\n<p>$$\\{\\begin{array}{l}{x}_{G}={r}_{G}\\,\\text{cos}\\,{\\phi }_{G}=(10.0\\,\\text{m})\\,\\text{cos}\\,8\\pi \\text{\/}9=-9.4\\,\\text{m}\\\\ {y}_{G}={r}_{G}\\,\\text{sin}\\,{\\phi }_{G}=(10.0\\,\\text{m})\\,\\text{sin}\\,8\\pi \\text{\/}9=3.4\\,\\text{m}\\end{array}\\enspace\u21d2\\enspace({x}_{G},{y}_{G})=(-9.4\\,\\text{m},3.4\\,\\text{m}).$$<\/p>\n<p>For the silver coin, the coordinates are<\/p>\n<p>$$\\{\\begin{array}{l}{x}_{S}={r}_{S}\\,\\text{cos}\\,{\\phi }_{S}=(20.0\\,\\text{m})\\,\\text{cos}\\,\\pi \\text{\/}9=18.9\\,\\text{m}\\\\ {y}_{S}={r}_{S}\\,\\text{sin}\\,{\\phi }_{S}=(20.0\\,\\text{m})\\,\\text{sin}\\,\\pi \\text{\/}9=6.8\\,\\text{m}\\end{array}\\enspace\u21d2\\enspace({x}_{S},{y}_{S})=(18.9\\,\\text{m},6.8\\,\\text{m}).$$<\/p><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167132342866\" class=\"bc-section section\">\n<h3>Vectors in Three Dimensions<\/h3>\n<p id=\"fs-id1167132741874\">To specify the location of a point in space, we need three coordinates (<em>x<\/em>, <em>y<\/em>, <em>z<\/em>), where coordinates <em>x<\/em> and <em>y<\/em> specify locations in a plane, and coordinate <em>z<\/em> gives a vertical position above or below the plane. Three-dimensional space has three orthogonal directions, so we need not two but <em>three<\/em> unit vectors to define a three-dimensional coordinate system. In the Cartesian coordinate system, the first two unit vectors are the unit vector of the <em>x<\/em>-axis $$ \\hat{i} $$ and the unit vector of the <em>y<\/em>-axis $$ \\hat{j}$$. The third unit vector $$ \\hat{k} $$ is the direction of the <em>z<\/em>-axis (<a class=\"autogenerated-content\" href=\"#CNX_UPhysics_02_02_ijknew\">(Figure)<\/a>). The order in which the axes are labeled, which is the order in which the three unit vectors appear, is important because it defines the orientation of the coordinate system. The order <em>x<\/em>&#8211;<em>y<\/em>&#8211;<em>z<\/em>, which is equivalent to the order $$ \\hat{i} $$ &#8211; $$ \\hat{j} $$ &#8211; $$ \\hat{k}$$, defines the standard right-handed coordinate system (positive orientation).<\/p>\n<div id=\"CNX_UPhysics_02_02_ijknew\" class=\"wp-caption aligncenter\">\n<div style=\"width: 494px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31183917\/CNX_UPhysics_02_02_ijknew.jpg\" alt=\"The x y z coordinate system, with unit vectors I hat, j hat and k hat respectively. I hat points out at us, j hat points to the right, and k hat points up the page. The unit vectors form the sides of a cube.\" width=\"484\" height=\"501\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 2.21<\/strong> Three unit vectors define a Cartesian system in three-dimensional space. The order in which these unit vectors appear defines the orientation of the coordinate system. The order shown here defines the right-handed orientation.<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1167132688199\">In three-dimensional space, vector $$ \\overset{\\to }{A} $$ has three vector components: the <em>x<\/em>-component $$ {\\overset{\\to }{A}}_{x}={A}_{x}\\hat{i}$$, which is the part of vector $$ \\overset{\\to }{A} $$ along the <em>x<\/em>-axis; the <em>y<\/em>-component $$ {\\overset{\\to }{A}}_{y}={A}_{y}\\hat{j}$$, which is the part of $$ \\overset{\\to }{A} $$ along the <em>y<\/em>-axis; and the <em>z<\/em>-component $$ {\\overset{\\to }{A}}_{z}={A}_{z}\\hat{k}$$, which is the part of the vector along the <em>z<\/em>-axis. A vector in three-dimensional space is the vector sum of its three vector components (<a class=\"autogenerated-content\" href=\"#CNX_UPhysics_02_02_vector3D\">(Figure)<\/a>):<\/p>\n<div id=\"fs-id1167133745530\" class=\"equation-callout\">\n<div id=\"fs-id1167132278018\">$$\\overset{\\to }{A}={A}_{x}\\hat{i}+{A}_{y}\\hat{j}+{A}_{z}\\hat{k}.$$<\/div>\n<\/div>\n<p id=\"fs-id1167132736522\">If we know the coordinates of its origin $$ b({x}_{b},{y}_{b},{z}_{b}) $$ and of its end $$ e({x}_{e},{y}_{e},{z}_{e})$$, its scalar components are obtained by taking their differences: $$ {A}_{x} $$ and $$ {A}_{y} $$ are given by <a class=\"autogenerated-content\" href=\"#fs-id1167132197498\">(Figure)<\/a> and the <em>z<\/em>-component is given by<\/p>\n<div id=\"fs-id1167132432269\" class=\"equation-callout\">\n<div id=\"fs-id1167133636864\">$${A}_{z}={z}_{e}-{z}_{b}.$$<\/div>\n<\/div>\n<p id=\"fs-id1167132436510\">Magnitude <em>A<\/em> is obtained by generalizing <a class=\"autogenerated-content\" href=\"#fs-id1167132245160\">(Figure)<\/a> to three dimensions:<\/p>\n<div id=\"fs-id1167132451249\" class=\"equation-callout\">\n<div id=\"fs-id1167132468770\">$$A=\\sqrt{{A}_{x}^{2}+{A}_{y}^{2}+{A}_{z}^{2}}.$$<\/div>\n<\/div>\n<p id=\"fs-id1167132716914\">This expression for the vector magnitude comes from applying the Pythagorean theorem twice. As seen in <a class=\"autogenerated-content\" href=\"#CNX_UPhysics_02_02_vector3D\">(Figure)<\/a>, the diagonal in the <em>xy<\/em>-plane has length $$ \\sqrt{{A}_{x}^{2}+{A}_{y}^{2}} $$ and its square adds to the square $$ {A}_{z}^{2} $$ to give $$ {A}^{2}$$. Note that when the <em>z<\/em>-component is zero, the vector lies entirely in the <em>xy<\/em>-plane and its description is reduced to two dimensions.<\/p>\n<div id=\"CNX_UPhysics_02_02_vector3D\" class=\"wp-caption aligncenter\">\n<div style=\"width: 499px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31183920\/CNX_UPhysics_02_02_vector3D.jpg\" alt=\"Vector A in the x y z coordinate system extends from the origin. Vector A equals the sum of vectors A sub x, A sub y and A sub z. Vector A sub x is the x component along the x axis and has length A sub x I hat. Vector A sub y is the y component along the y axis and has length A sub y j hat. Vector A sub z is the z component along the z axis and has length A sub x k hat. The components form the sides of a rectangular box with sides length A sub x, A sub y, and A sub z.\" width=\"489\" height=\"450\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 2.22<\/strong> A vector in three-dimensional space is the vector sum of its three vector components.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1167132572875\" class=\"textbox examples\">\n<h3>Example<\/h3>\n<h4>Takeoff of a Drone<\/h4>\n<p>During a takeoff of IAI Heron (<a class=\"autogenerated-content\" href=\"#CNX_UPhysics_02_02_heron\">(Figure)<\/a>), its position with respect to a control tower is 100 m above the ground, 300 m to the east, and 200 m to the north. One minute later, its position is 250 m above the ground, 1200 m to the east, and 2100 m to the north. What is the drone\u2019s displacement vector with respect to the control tower? What is the magnitude of its displacement vector?<\/p>\n<div id=\"CNX_UPhysics_02_02_heron\" class=\"wp-caption aligncenter\">\n<div style=\"width: 475px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31183924\/CNX_UPhysics_02_02_heron.jpg\" alt=\"A photo of a drone plane.\" width=\"465\" height=\"248\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 2.23<\/strong> The drone IAI Heron in flight. (credit: SSgt Reynaldo Ramon, USAF)<\/p>\n<\/div>\n<\/div>\n<h4>Strategy<\/h4>\n<p>We take the origin of the Cartesian coordinate system as the control tower. The direction of the +<em>x<\/em>-axis is given by unit vector $$ \\hat{i} $$ to the east, the direction of the +<em>y<\/em>-axis is given by unit vector $$ \\hat{j} $$ to the north, and the direction of the +<em>z<\/em>-axis is given by unit vector $$ \\hat{k}$$, which points up from the ground. The drone\u2019s first position is the origin (or, equivalently, the beginning) of the displacement vector and its second position is the end of the displacement vector.<\/p>\n<h4>Solution<\/h4>\n<p id=\"fs-id1167132742913\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q436377\">Show Answer<\/span><\/p>\n<div id=\"q436377\" class=\"hidden-answer\" style=\"display: none\">We identify b(300.0 m, 200.0 m, 100.0 m) and e(480.0 m, 370.0 m, 250.0m), and use (Figure) and (Figure) to find the scalar components of the drone\u2019s displacement vector:<\/p>\n<p>$$\\{\\begin{array}{l}{D}_{x}={x}_{e}-{x}_{b}=1200.0\\,\\text{m}-300.0\\,\\text{m}=900.0\\,\\text{m},\\\\ {D}_{y}={y}_{e}-{y}_{b}=2100.0\\,\\text{m}-200.0\\,\\text{m}=1900.0\\,\\text{m,}\\\\ {D}_{z}={z}_{e}-{z}_{b}=250.0\\,\\text{m}-100.0\\,\\text{m}=150.0\\,\\text{m}.\\end{array}$$<\/p>\n<p>We substitute these components into (Figure) to find the displacement vector:<\/p>\n<p>$$\\overset{\\to }{D}={D}_{x}\\hat{i}+{D}_{y}\\hat{j}+{D}_{z}\\hat{k}=900.0\\,\\text{m}\\hat{i}+1900.0\\,\\text{m}\\hat{j}+150.0\\,\\text{m}\\hat{k}=(0.90\\hat{i}+1.90\\hat{j}+0.15\\hat{k})\\,\\text{km}.$$<\/p>\n<p>We substitute into (Figure) to find the magnitude of the displacement: $$ D=\\sqrt{{D}_{x}^{2}+{D}_{y}^{2}+{D}_{z}^{2}}=\\sqrt{{(0.90\\,\\text{km})}^{2}+{(1.90\\,\\text{km})}^{2}+{(0.15\\,\\text{km})}^{2}}=4.44\\,\\text{km}.$$<\/p><\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167132218339\" class=\"textbox exercises check-understanding\">\n<h3>Check Your Understanding<\/h3>\n<div id=\"fs-id1167128919148\" class=\"problem textbox\">\n<div id=\"fs-id1167132230025\">\n<p id=\"fs-id1167132230028\">If the average velocity vector of the drone in the displacement in <a class=\"autogenerated-content\" href=\"#fs-id1167132572875\">(Figure)<\/a> is $$ \\overset{\\to }{u}=(15.0\\hat{i}+31.7\\hat{j}+2.5\\hat{k})\\text{m}\\text{\/}\\text{s}$$, what is the magnitude of the drone\u2019s velocity vector?<\/p>\n<\/div>\n<div id=\"fs-id1167132346740\" class=\"solution\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167132346740\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167132346740\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167133751212\">35.1 m\/s = 126.4 km\/h<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167132451656\" class=\"textbox key-takeaways\">\n<h3>Summary<\/h3>\n<ul id=\"fs-id1167132628006\">\n<li>Vectors are described in terms of their components in a coordinate system. In two dimensions (in a plane), vectors have two components. In three dimensions (in space), vectors have three components.<\/li>\n<li>A vector component of a vector is its part in an axis direction. The vector component is the product of the unit vector of an axis with its scalar component along this axis. A vector is the resultant of its vector components.<\/li>\n<li>Scalar components of a vector are differences of coordinates, where coordinates of the origin are subtracted from end point coordinates of a vector. In a rectangular system, the magnitude of a vector is the square root of the sum of the squares of its components.<\/li>\n<li>In a plane, the direction of a vector is given by an angle the vector has with the positive <em>x<\/em>-axis. This direction angle is measured counterclockwise. The scalar <em>x<\/em>-component of a vector can be expressed as the product of its magnitude with the cosine of its direction angle, and the scalar <em>y<\/em>-component can be expressed as the product of its magnitude with the sine of its direction angle.<\/li>\n<li>In a plane, there are two equivalent coordinate systems. The Cartesian coordinate system is defined by unit vectors $$ \\hat{i} $$ and $$ \\hat{j} $$ along the <em>x<\/em>-axis and the <em>y<\/em>-axis, respectively. The polar coordinate system is defined by the radial unit vector $$ \\hat{r}$$, which gives the direction from the origin, and a unit vector $$ \\hat{t}$$, which is perpendicular (orthogonal) to the radial direction.<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1167132437539\" class=\"review-conceptual-questions\">\n<h3>Conceptual Questions<\/h3>\n<div id=\"fs-id1167132241622\" class=\"problem textbox\">\n<div id=\"fs-id1167133668968\">\n<p id=\"fs-id1167132345316\">Give an example of a nonzero vector that has a component of zero.<\/p>\n<\/div>\n<div id=\"fs-id1167132706218\" class=\"solution\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167132706218\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167132706218\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167132328235\">a unit vector of the <em>x<\/em>-axis<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167132494452\" class=\"problem textbox\">\n<div id=\"fs-id1167132476171\">\n<p id=\"fs-id1167132476173\">Explain why a vector cannot have a component greater than its own magnitude.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1167132562110\" class=\"problem textbox\">\n<div id=\"fs-id1167132562112\">\n<p id=\"fs-id1167132504879\">If two vectors are equal, what can you say about their components?<\/p>\n<\/div>\n<div id=\"fs-id1167133715718\" class=\"solution\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167133715718\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167133715718\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167133715720\">They are equal.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167132267416\" class=\"problem textbox\">\n<div id=\"fs-id1167132267418\">\n<p id=\"fs-id1167132627817\">If vectors $$ \\overset{\\to }{A} $$ and $$ \\overset{\\to }{B} $$ are orthogonal, what is the component of $$ \\overset{\\to }{B} $$ along the direction of $$ \\overset{\\to }{A}$$? What is the component of $$ \\overset{\\to }{A} $$ along the direction of $$ \\overset{\\to }{B}$$?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1167132518791\" class=\"problem textbox\">\n<div id=\"fs-id1167132518793\">\n<p id=\"fs-id1167133668836\">If one of the two components of a vector is not zero, can the magnitude of the other vector component of this vector be zero?<\/p>\n<\/div>\n<div id=\"fs-id1167133568374\" class=\"solution\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167133568374\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167133568374\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167133568377\">yes<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167132535630\" class=\"problem textbox\">\n<div id=\"fs-id1167132535632\">\n<p id=\"fs-id1167132481141\">If two vectors have the same magnitude, do their components have to be the same?<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167132281028\" class=\"review-problems textbox exercises\">\n<h3>Problems<\/h3>\n<div id=\"fs-id1167132612418\" class=\"problem textbox\">\n<div id=\"fs-id1167132612420\">\n<p id=\"fs-id1167132312139\">Assuming the +<em>x<\/em>-axis is horizontal and points to the right, resolve the vectors given in the following figure to their scalar components and express them in vector component form.<\/p>\n<\/div>\n<div id=\"fs-id1167132683339\" class=\"solution\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167132683339\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167132683339\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167128844080\">a. $$ \\overset{\\to }{A}=+8.66\\hat{i}+5.00\\hat{j}$$, b. $$ \\overset{\\to }{B}=+30.09\\hat{i}+39.93\\hat{j}$$, c. $$ \\overset{\\to }{C}=+6.00\\hat{i}-10.39\\hat{j}$$, d. $$ \\overset{\\to }{D}=-15.97\\hat{i}+12.04\\hat{j}$$, f. $$ \\overset{\\to }{F}=-17.32\\hat{i}-10.00\\hat{j}$$<\/p>\n<p><span id=\"fs-id1167133711602\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31183852\/CNX_UPhysics_02_01_problems_img.jpg\" alt=\"The x y coordinate system has positive x to the right and positive y up. Vector A has magnitude 10.0 and points 30 degrees counterclockwise from the positive x direction. Vector B has magnitude 5.0 and points 53 degrees counterclockwise from the positive x direction. Vector C has magnitude 12.0 and points 60 degrees clockwise from the positive x direction. Vector D has magnitude 20.0 and points 37 degrees clockwise from the negative x direction. Vector F has magnitude 20.0 and points 30 degrees counterclockwise from the negative x direction.\" \/><\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167132508049\" class=\"problem textbox\">\n<div id=\"fs-id1167132508051\">\n<p id=\"fs-id1167132445103\">Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point? What is your displacement vector? What is the direction of your displacement? Assume the +<em>x<\/em>-axis is horizontal to the right.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1167132487189\" class=\"problem textbox\">\n<div id=\"fs-id1167132487191\">\n<p id=\"fs-id1167132366338\">You drive 7.50 km in a straight line in a direction $$ 15\\text{\u00b0} $$ east of north. (a) Find the distances you would have to drive straight east and then straight north to arrive at the same point. (b) Show that you still arrive at the same point if the east and north legs are reversed in order. Assume the +<em>x<\/em>-axis is to the east.<\/p>\n<\/div>\n<div id=\"fs-id1167133521610\" class=\"solution\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167133521610\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167133521610\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167132501142\">a. 1.94 km, 7.24 km; b. proof<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167132501147\" class=\"problem textbox\">\n<div id=\"fs-id1167132248105\">\n<p id=\"fs-id1167132248107\">A sledge is being pulled by two horses on a flat terrain. The net force on the sledge can be expressed in the Cartesian coordinate system as vector $$ \\overset{\\to }{F}=(-2980.0\\hat{i}+8200.0\\hat{j})\\text{N}$$, where $$ \\hat{i} $$ and $$ \\hat{j} $$ denote directions to the east and north, respectively. Find the magnitude and direction of the pull.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1167132536425\" class=\"problem textbox\">\n<div id=\"fs-id1167132335895\">\n<p id=\"fs-id1167132335897\">A trapper walks a 5.0-km straight-line distance from her cabin to the lake, as shown in the following figure. Determine the east and north components of her displacement vector. How many more kilometers would she have to walk if she walked along the component displacements? What is her displacement vector?<\/p>\n<p><span id=\"fs-id1167132294765\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31183927\/CNX_UPhysics_02_01_problem3_img.jpg\" alt=\"The vector from the cabin to the lake is vector S, magnitude 5.0 kilometers and pointing 40 degrees north of east. Two additional meandering paths are shown and labeled path 1 and path 2.\" \/><\/span><\/p>\n<\/div>\n<div id=\"fs-id1167132487930\">\n<p id=\"fs-id1167132487932\">3.8 km east, 3.2 km north, 2.0 km, $$ \\overset{\\to }{D}=(3.8\\hat{i}+3.2\\hat{j})\\text{km}$$<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1167133544873\" class=\"problem textbox\">\n<div id=\"fs-id1167132628948\">\n<p id=\"fs-id1167132628950\">The polar coordinates of a point are $$ 4\\pi \\text{\/}3 $$ and 5.50 m. What are its Cartesian coordinates?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1167132303332\" class=\"problem textbox\">\n<div id=\"fs-id1167132303334\">\n<p id=\"fs-id1167132627652\">Two points in a plane have polar coordinates $$ {P}_{1}(2.500\\,\\text{m},\\pi \\text{\/}6) $$ and $$ {P}_{2}(3.800\\,\\text{m},2\\pi \\text{\/}3)$$. Determine their Cartesian coordinates and the distance between them in the Cartesian coordinate system. Round the distance to a nearest centimeter.<\/p>\n<\/div>\n<div id=\"fs-id1167132579170\" class=\"solution\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167132579170\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167132579170\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167132579172\">$${P}_{1}(2.165\\,\\text{m},1.250\\,\\text{m})$$, $$ {P}_{2}(-1.900\\,\\text{m},3.290\\,\\text{m})$$, 5.27 m<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167133676180\" class=\"problem textbox\">\n<div id=\"fs-id1167132458391\">\n<p id=\"fs-id1167132458394\">A chameleon is resting quietly on a lanai screen, waiting for an insect to come by. Assume the origin of a Cartesian coordinate system at the lower left-hand corner of the screen and the horizontal direction to the right as the +<em>x<\/em>-direction. If its coordinates are (2.000 m, 1.000 m), (a) how far is it from the corner of the screen? (b) What is its location in polar coordinates?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1167132319658\" class=\"problem textbox\">\n<div id=\"fs-id1167132319660\">\n<p id=\"fs-id1167132372735\">Two points in the Cartesian plane are <em>A<\/em>(2.00 m, \u22124.00 m) and <em>B<\/em>(\u22123.00 m, 3.00 m). Find the distance between them and their polar coordinates.<\/p>\n<\/div>\n<div id=\"fs-id1167132372708\" class=\"solution\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167132372708\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167132372708\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167132340314\">8.60 m, $$ A(2\\sqrt{5}\\,\\text{m},0.647\\pi )$$, $$ B(3\\sqrt{2}\\,\\text{m},0.75\\pi )$$<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167132441243\" class=\"problem textbox\">\n<div id=\"fs-id1167132441245\">\n<p id=\"fs-id1167133579494\">A fly enters through an open window and zooms around the room. In a Cartesian coordinate system with three axes along three edges of the room, the fly changes its position from point <em>b<\/em>(4.0 m, 1.5 m, 2.5 m) to point <em>e<\/em>(1.0 m, 4.5 m, 0.5 m). Find the scalar components of the fly\u2019s displacement vector and express its displacement vector in vector component form. What is its magnitude?<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Glossary<\/h3>\n<dl id=\"fs-id1167132687581\">\n<dt><strong>component form of a vector<\/strong><\/dt>\n<dd id=\"fs-id1167132612545\">a vector written as the vector sum of its components in terms of unit vectors<\/dd>\n<\/dl>\n<dl id=\"fs-id1167132612567\">\n<dt><strong>direction angle<\/strong><\/dt>\n<dd id=\"fs-id1167132612572\">in a plane, an angle between the positive direction of the <em>x<\/em>-axis and the vector, measured counterclockwise from the axis to the vector<\/dd>\n<\/dl>\n<dl id=\"fs-id1167132310341\">\n<dt><strong>polar coordinate system<\/strong><\/dt>\n<dd id=\"fs-id1167132506907\">an orthogonal coordinate system where location in a plane is given by polar coordinates<\/dd>\n<\/dl>\n<dl id=\"fs-id1167132506911\">\n<dt><strong>polar coordinates<\/strong><\/dt>\n<dd id=\"fs-id1167132568470\">a radial coordinate and an angle<\/dd>\n<\/dl>\n<dl id=\"fs-id1167132199051\">\n<dt><strong>radial coordinate<\/strong><\/dt>\n<dd id=\"fs-id1167132541607\">distance to the origin in a polar coordinate system<\/dd>\n<\/dl>\n<dl id=\"fs-id1167132541611\">\n<dt><strong>scalar component<\/strong><\/dt>\n<dd id=\"fs-id1167132584081\">a number that multiplies a unit vector in a vector component of a vector<\/dd>\n<\/dl>\n<dl id=\"fs-id1167132584102\">\n<dt><strong>unit vectors of the axes<\/strong><\/dt>\n<dd id=\"fs-id1167132546147\">unit vectors that define orthogonal directions in a plane or in space<\/dd>\n<\/dl>\n<dl id=\"fs-id1167132546151\">\n<dt><strong>vector components<\/strong><\/dt>\n<dd id=\"fs-id1167132266674\">orthogonal components of a vector; a vector is the vector sum of its vector components.<\/dd>\n<\/dl>\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-180\">\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>OpenStax University Physics. <strong>Authored by<\/strong>: OpenStax CNX. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/cnx.org\/contents\/1Q9uMg_a@10.16:Gofkr9Oy@15\">https:\/\/cnx.org\/contents\/1Q9uMg_a@10.16:Gofkr9Oy@15<\/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>: Download for free at http:\/\/cnx.org\/contents\/d50f6e32-0fda-46ef-a362-9bd36ca7c97d@10.16<\/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":311,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"OpenStax University Physics\",\"author\":\"OpenStax CNX\",\"organization\":\"\",\"url\":\"https:\/\/cnx.org\/contents\/1Q9uMg_a@10.16:Gofkr9Oy@15\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/d50f6e32-0fda-46ef-a362-9bd36ca7c97d@10.16\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":"all-rights-reserved"},"chapter-type":[],"contributor":[],"license":[56],"class_list":["post-180","chapter","type-chapter","status-publish","hentry","license-all-rights-reserved"],"part":164,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/pressbooks\/v2\/chapters\/180","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/wp\/v2\/users\/311"}],"version-history":[{"count":9,"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/pressbooks\/v2\/chapters\/180\/revisions"}],"predecessor-version":[{"id":2211,"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/pressbooks\/v2\/chapters\/180\/revisions\/2211"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/pressbooks\/v2\/parts\/164"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/pressbooks\/v2\/chapters\/180\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/wp\/v2\/media?parent=180"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/pressbooks\/v2\/chapter-type?post=180"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/wp\/v2\/contributor?post=180"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/wp\/v2\/license?post=180"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}