{"id":5436,"date":"2022-06-02T18:14:08","date_gmt":"2022-06-02T18:14:08","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=5436"},"modified":"2022-10-20T23:55:25","modified_gmt":"2022-10-20T23:55:25","slug":"vector-components","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/vector-components\/","title":{"raw":"Vector Components","rendered":"Vector Components"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Express a vector in component form.<\/li>\r\n \t<li>Explain the formula for the magnitude of a vector.<\/li>\r\n<\/ul>\r\n<\/div>\r\nWorking with vectors in a plane is easier when we are working in a coordinate system. When the initial points and terminal points of vectors are given in Cartesian coordinates, computations become straightforward.\r\n<div id=\"fs-id1167793900968\" class=\"textbook exercises\">\r\n<h3>Example: Comparing Vectors<\/h3>\r\nAre [latex]\\bf{v}[\/latex] and [latex]\\bf{w}[\/latex] equivalent vectors?\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li style=\"list-style-type: none;\">\r\n<ol style=\"list-style-type: lower-alpha;\">a. [latex]\\bf{v}[\/latex] has initial point [latex](3,2)[\/latex] and terminal point [latex](7,2)[\/latex]<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<ol style=\"list-style-type: lower-alpha;\">\u00a0\u00a0\u00a0\u00a0[latex]\\bf{w}[\/latex] has initial point [latex](1,\u22124)[\/latex] and terminal point [latex](1,0)[\/latex]<\/ol>\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li style=\"list-style-type: none;\">\r\n<ol style=\"list-style-type: lower-alpha;\">b. [latex]\\bf{v}[\/latex] has initial point [latex](0,0)[\/latex] and terminal point [latex](1,1) [\/latex]<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<ol style=\"list-style-type: lower-alpha;\">\u00a0\u00a0\u00a0\u00a0[latex]\\bf{w}[\/latex] has initial point [latex](\u22122,2)[\/latex] and terminal point [latex](\u22121,3)[\/latex]<\/ol>\r\n<div id=\"fs-id1167793461764\" class=\"exercise\">[reveal-answer q=\"fs-id1167795055165\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167795055165\"]\r\n<ol style=\"list-style-type: lower-alpha;\">a. The vectors are each [latex]4[\/latex] units long, but they are oriented in different directions. So [latex]\\bf{v}[\/latex] and [latex]\\bf{w}[\/latex] are not equivalent (Figure 2.10).<\/ol>\r\n[caption id=\"attachment_690\" align=\"aligncenter\" width=\"342\"]<img class=\"wp-image-690 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/17190445\/2-1-10.jpeg\" alt=\"This figure is a Cartesian coordinate system with two vectors. The first vector labeled \u201cv\u201d has initial point at (3, 2) and terminal point (7, 2). It is parallel to the x-axis. The second vector is labeled \u201cw\u201d and has initial point (1, -4) and terminal point (1, 0). It is parallel to the y-axis.\" width=\"342\" height=\"384\" \/> Figure 1. These vectors are not equivalent.[\/caption]\r\n<ol style=\"list-style-type: lower-alpha;\">b. Based on Figure 2.11, and using a bit of geometry, it is clear these vectors have the same length and the same direction, so [latex]\\bf{v}[\/latex] and [latex]\\bf{w}[\/latex] are equivalent.<\/ol>\r\n[caption id=\"attachment_691\" align=\"aligncenter\" width=\"267\"]<img class=\"wp-image-691 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/17190539\/2-1-11.jpeg\" alt=\"This figure is a Cartesian coordinate system with two vectors. The first vector labeled \u201cv\u201d has initial point at (0, 0) and terminal point (1, 1). The second vector is labeled \u201cw\u201d and has initial point (-2, 2) and terminal point (-1, 3).\" width=\"267\" height=\"272\" \/> Figure 2. These vectors are equivalent.[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167793958097\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nWhich of the following vectors are equivalent?\r\n\r\n[caption id=\"attachment_692\" align=\"aligncenter\" width=\"642\"]<img class=\"wp-image-692 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/17190648\/2-1-12.jpeg\" alt=\"alt=&quot;This figure is a coordinate system with 6 vectors, each labeled a through f. Three of the vectors, \u201ca,\u201d \u201cb,\u201d and \u201ce\u201d have the same length and are pointing in the same direction.&quot;\" width=\"642\" height=\"384\" \/> Figure 3. Determine which vectors are equivalent.[\/caption]\r\n\r\n<div id=\"fs-id1167793940339\" class=\"exercise\">[reveal-answer q=\"fs-id1167794933124\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794933124\"]\r\nVectors [latex]\\bf{a}[\/latex], [latex]\\bf{b}[\/latex], and [latex]\\bf{e}[\/latex] are equivalent.\r\n[\/hidden-answer]<\/div>\r\n<\/div>\r\nWe have seen how to plot a vector when we are given an initial point and a terminal point. However, because a vector can be placed anywhere in a plane, it may be easier to perform calculations with a vector when its initial point coincides with the origin. We call a vector with its initial point at the origin a <strong>standard-position vector<\/strong>. Because the initial point of any vector in standard position is known to be [latex](0,0)[\/latex], we can describe the vector by looking at the coordinates of its terminal point. Thus, if vector [latex]\\bf{v}[\/latex] has its initial point at the origin and its terminal point at [latex](x,y)[\/latex], we write the vector in component form as\r\n\r\n<center>[latex]\\bf{v}[\/latex] [latex]= \\langle x,y \\rangle[\/latex].<\/center>When a vector is written in component form like this, the scalars [latex]x[\/latex] and [latex]y[\/latex] are called the <strong>components <\/strong>of [latex]\\bf{v}[\/latex].\r\n<div id=\"fs-id1167793372221\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\r\n\r\n<hr \/>\r\n\r\nThe vector with initial point [latex](0,0)[\/latex] and terminal point [latex](x,y)[\/latex] can be written in component form as\r\n\r\n<center>[latex]\\bf{v}[\/latex] [latex]= \\langle x,y \\rangle[\/latex].<\/center>\r\nThe scalars [latex]x[\/latex] and [latex]y[\/latex] are called the components of [latex]\\bf{v}[\/latex].\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nRecall that vectors are named with lowercase letters in bold type or by drawing an arrow over their name. We have also learned that we can name a vector by its component form, with the coordinates of its terminal point in angle brackets. However, when writing the component form of a vector, it is important to distinguish between [latex]\\langle x,y \\rangle[\/latex] and [latex](x,y)[\/latex]. The first ordered pair uses angle brackets to describe a vector, whereas the second uses parentheses to describe a point in a plane. The initial point of [latex]\\langle x,y \\rangle[\/latex] is [latex](0,0)[\/latex]; the terminal point of [latex]\\langle x,y \\rangle[\/latex] is [latex](x,y)[\/latex].\r\n\r\nWhen we have a vector not already in standard position, we can determine its component form in one of two ways. We can use a geometric approach, in which we sketch the vector in the coordinate plane, and then sketch an equivalent standard-position vector. Alternatively, we can find it algebraically, using the coordinates of the initial point and the terminal point. To find it algebraically, we subtract the [latex]x[\/latex]-coordinate of the initial point from the [latex]x[\/latex]-coordinate of the terminal point to get the [latex]x[\/latex] component, and we subtract the [latex]y[\/latex]-coordinate of the initial point from the [latex]y[\/latex]-coordinate of the terminal point to get the [latex]y[\/latex] component.\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Component Form of a Vector<\/h3>\r\n\r\n<hr \/>\r\n\r\nLet [latex]\\bf{v}[\/latex] be a vector with initial point [latex](x_i,y_i)[\/latex] and terminal point [latex](x_t,y_t)[\/latex]. Then we can express [latex]\\bf{v}[\/latex] in component form as\r\n[latex]\\bf{v}[\/latex] [latex]=\\langle x_t \u2212 x_i,y_t \u2212 y_i\\rangle[\/latex].\r\n\r\n<\/div>\r\n<div id=\"fs-id1167793900968\" class=\"textbook exercises\">\r\n<h3>Example: Expressing Vectors in Component Form<\/h3>\r\nExpress vector [latex]\\bf{v}[\/latex] with initial point [latex](\u22123,4)[\/latex] and terminal point [latex](1,2)[\/latex] in component form.\r\n<div id=\"fs-id1167893461764\" class=\"exercise\">[reveal-answer q=\"fs-id1167895055165\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167895055165\"]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li style=\"list-style-type: none;\">\r\n<ol style=\"list-style-type: lower-alpha;\">a. Geometric\r\n \t<li style=\"list-style-type: none;\">\r\n<ol style=\"list-style-type: decimal;\">1. Sketch the vector in the coordinate plane (Figure 13).<\/ol>\r\n<ol style=\"list-style-type: decimal;\">2. The terminal point is [latex]4[\/latex] units to the right and [latex]2[\/latex] units down from the initial point.<\/ol>\r\n<ol style=\"list-style-type: decimal;\">3. Find the point that is [latex]4[\/latex] units to the right and [latex]2[\/latex] units down from the origin.<\/ol>\r\n<ol style=\"list-style-type: decimal;\">4. In standard position, this vector has initial point [latex](0,0)[\/latex] and terminal point [latex](4,\u22122)[\/latex]:<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<center>[latex]\\bf{v}[\/latex] [latex] = \\langle4,\u22122\\rangle.[\/latex]<\/center>\r\n\r\n[caption id=\"attachment_693\" align=\"aligncenter\" width=\"417\"]<img class=\"size-full wp-image-693\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/17190801\/2-1-13.jpeg\" alt=\"This figure is a coordinate system. There are two vectors on the graph. The first vector has initial point at the origin and terminal point at (4, -2). The horizontal distance from the initial to the terminal point for the vector is labeled as \u201c4 units.\u201d The vertical distance from the initial to the terminal point is labeled as \u201c2 units.\u201d The second vector has initial point at (-3, 4) and terminal point at (1, 2). The horizontal distance from the initial to the terminal point for the vector is labeled as \u201c4 units.\u201d The vertical distance from the initial to the terminal point is labeled as \u201c2 units.\u201d\" width=\"417\" height=\"384\" \/> Figure 4.\u00a0These vectors are equivalent.[\/caption]\r\n\r\n&nbsp;\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li style=\"list-style-type: none;\">\r\n<ol style=\"list-style-type: lower-alpha;\">b. Algebraic<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li style=\"list-style-type: none;\">\r\n<ol style=\"list-style-type: lower-alpha;\">In the first solution, we used a sketch of the vector to see that the terminal point lies [latex]4[\/latex] units to the right. We can accomplish this algebraically by finding the difference of the [latex]x[\/latex]-coordinates:<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<center>[latex]x_t - x_i = 1 - (-3) = 4[\/latex]<\/center>\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li style=\"list-style-type: none;\">\r\n<ol style=\"list-style-type: lower-alpha;\">Similarly, the difference of the y-coordinates shows the vertical length of the vector.<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<center>[latex]y_t - y_i = 2 - 4 = -2[\/latex]<\/center>\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li style=\"list-style-type: none;\">\r\n<ol style=\"list-style-type: lower-alpha;\">So, in component form,<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<center>[latex]\\bf{v}[\/latex] [latex]=\\langle x_t \u2212 x_i,y_t \u2212 y_i\\rangle[\/latex]<\/center>&nbsp;\r\n\r\n<center>[latex]=\\langle 1 - (-3), 2 - 4\\rangle[\/latex]<\/center>&nbsp;\r\n\r\n<center>[latex]=\\langle 4, -2\\rangle[\/latex]<\/center>[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167793958097\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nVector [latex]\\bf{w}[\/latex] has initial point [latex](\u22124,\u22125)[\/latex] and terminal point [latex](\u22121,2)[\/latex]. Express [latex]\\bf{w}[\/latex] in component form.\r\n<div id=\"fs-id1167793940339\" class=\"exercise\">[reveal-answer q=\"fs-id1168794933124\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1168794933124\"]\r\n[latex]\\langle 3,7 \\rangle[\/latex]\r\n[\/hidden-answer]<\/div>\r\n<\/div>\r\nTo find the magnitude of a vector, we calculate the distance between its initial point and its terminal point. The magnitude of vector\r\n[latex]\\bf{v}[\/latex][latex]=\\langle x,y\\rangle[\/latex] is denoted [latex]\\bf{||v||}[\/latex], or [latex]\\bf{|v|}[\/latex], and can be computed using the formula\r\n\r\n<center>[latex]\\bf{||v||}[\/latex] [latex]=\\sqrt{x^2 + y^2}[\/latex]<\/center>.\r\n\r\nNote that because this vector is written in component form, it is equivalent to a vector in standard position, with its initial point at the origin and terminal point [latex](x,y)[\/latex]. Thus, it suffices to calculate the magnitude of the vector in standard position. Using the distance formula to calculate the distance between initial point [latex](0,0)[\/latex] and terminal point [latex](x,y)[\/latex], we have\r\n\r\n<center>[latex]\\bf{||v||}[\/latex][latex]=\\sqrt{(x - 0)^2 + (y - 0)^2}[\/latex]<\/center>&nbsp;\r\n\r\n<center>[latex]=\\sqrt{x^2 + y^2}[\/latex]<\/center>&nbsp;\r\n\r\nBased on this formula, it is clear that for any vector [latex]\\bf{v}[\/latex], [latex]\\bf{||v||}[\/latex][latex] \\geq 0[\/latex], and [latex]\\bf{||v||}[\/latex][latex] = 0[\/latex] if and only if [latex]\\bf{v}[\/latex][latex] = 0[\/latex].\r\n\r\nThe magnitude of a vector can also be derived using the Pythagorean theorem, as in the following figure.\r\n\r\n[caption id=\"attachment_694\" align=\"aligncenter\" width=\"282\"]<img class=\"size-full wp-image-694\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/17190913\/2-1-14.jpeg\" alt=\"This figure is a right triangle. The two sides are labeled \u201cx\u201d and \u201cy.\u201d The hypotenuse is represented as a vector and is labeled \u201csquare root (x^2 + y^2).\u201d\" width=\"282\" height=\"119\" \/> Figure 5. If you use the components of a vector to define a right triangle, the magnitude of the vector is the length of the triangle\u2019s hypotenuse.[\/caption]\r\n\r\nWe have defined scalar multiplication and vector addition geometrically. Expressing vectors in component form allows us to perform these same operations algebraically.\r\n<div id=\"fs-id1167793372221\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\r\n\r\n<hr \/>\r\n\r\nLet [latex]\\bf{v}[\/latex][latex] = \\langle x_1, y_1\\rangle.[\/latex] and [latex]\\bf{w}[\/latex][latex] = \\langle x_2, y_2\\rangle[\/latex] be vectors, and let [latex]k[\/latex] be a scalar.\r\n\r\n<strong>Scalar multiplication<\/strong>: [latex]k\\bf{v}[\/latex][latex]=\\langle kx_1, ky_1\\rangle[\/latex]\r\n\r\n<strong>Vector addition<\/strong>: [latex]\\bf{v + w}[\/latex][latex]=\\langle x_1, y_1\\rangle + \\langle x_2, y_2\\rangle = \\langle x_1 + x_2, y_1 + y_2\\rangle[\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167793905960\" class=\"textbook exercises\">\r\n<h3>Example: Performing Operations in Component Form<\/h3>\r\nLet [latex]\\bf{v}[\/latex] be the vector with initial point [latex](2,5)[\/latex] and terminal point [latex](8,13)[\/latex], and let [latex]\\bf{w}[\/latex][latex] = \\langle\u22122,4\\rangle[\/latex].\r\n<ol style=\"list-style-type: lower-alpha;\">a. Express [latex]\\bf{v}[\/latex] in component form and find [latex]\\bf{||v||}[\/latex]. Then, using algebra, find<\/ol>\r\n<ol style=\"list-style-type: lower-alpha;\">b. [latex]\\bf{v + w}[\/latex],<\/ol>\r\n<ol style=\"list-style-type: lower-alpha;\">c. [latex]3[\/latex][latex]\\bf{v}[\/latex], and<\/ol>\r\n<ol style=\"list-style-type: lower-alpha;\">d. [latex]\\bf{v}[\/latex][latex]\u22122[\/latex][latex]\\bf{w}[\/latex].<\/ol>\r\n<div id=\"fs-id1167793461764\" class=\"exercise\">[reveal-answer q=\"fs-id1167794055754\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794055754\"]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li style=\"list-style-type: none;\">\r\n<ol style=\"list-style-type: lower-alpha;\">a. To place the initial point of [latex]\\bf{v}[\/latex] at the origin, we must translate the vector [latex]2[\/latex] units to the left and [latex]5[\/latex] units down (Figure 2.15). Using the algebraic method, we can express [latex]\\bf{v}[\/latex] as [latex]\\bf{v}[\/latex] [latex] = \\langle 8 \u2212 2,13 \u2212 5\\rangle = \\langle6,8\\rangle[\/latex]:<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<center>[latex]\\bf{||v||}[\/latex] [latex]=\\sqrt{6^2 + 8^2} = \\sqrt{36+64} = \\sqrt{100} = 10[\/latex].<\/center><center><\/center><center>[caption id=\"attachment_696\" align=\"alignnone\" width=\"267\"]<img class=\"wp-image-696 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/17191050\/2-1-15.jpeg\" alt=\"This figure is the first quadrant of a coordinate system. It has two vectors. The first vector has initial point at (2, 5) and terminal point (8, 13). The second vector has initial point at the origin and terminal point at (6, 8).\" width=\"267\" height=\"347\" \/> Figure 6. In component form, [latex]{\\bf{v}}=\\langle{6,8}\\rangle[\/latex].[\/caption]<\/center>\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li style=\"list-style-type: none;\">\r\n<ol style=\"list-style-type: lower-alpha;\">b. To find [latex]\\bf{v + w}[\/latex], add the [latex]x[\/latex]-components and the [latex]y[\/latex]-components separately:<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<center>[latex]\\bf{v + w}[\/latex][latex]=\\langle6,8\\rangle + \\langle\u22122,4\\rangle = \\langle4,12\\rangle[\/latex].<\/center>\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li style=\"list-style-type: none;\">\r\n<ol style=\"list-style-type: lower-alpha;\">c. To find [latex]3[\/latex][latex]\\bf{v}[\/latex], multiply [latex]\\bf{v}[\/latex] by the scalar [latex]k=3[\/latex]:<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<center>[latex]3[\/latex][latex]\\bf{v}[\/latex][latex]=3\\cdot\\langle6,8\\rangle = \\langle3\\cdot6,3\\cdot8\\rangle = \\langle18,24\\rangle[\/latex].<\/center>\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li style=\"list-style-type: none;\">\r\n<ol style=\"list-style-type: lower-alpha;\">d. To find [latex]\\bf{v}[\/latex][latex]\u22122[\/latex][latex]\\bf{w}[\/latex], find [latex]\u22122[\/latex][latex]\\bf{w}[\/latex] and add it to [latex]\\bf{v}[\/latex]:<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<center>[latex]\\bf{v}[\/latex][latex]\u22122[\/latex][latex]\\bf{w}[\/latex][latex]=\\langle6,8\\rangle \u2212 2 \\cdot \\langle\u22122,4\\rangle = \\langle6,8\\rangle + \\langle4,\u22128\\rangle = \\langle10,0\\rangle[\/latex].<\/center>[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167893957091\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nLet [latex]\\bf{a}[\/latex][latex]=\\langle7,1\\rangle[\/latex] and let [latex]\\bf{b}[\/latex] be the vector with initial point [latex](3,2)[\/latex] and terminal point [latex](\u22121,\u22121)[\/latex].\r\n<ol style=\"list-style-type: lower-alpha;\">a. Find [latex]\\bf{||a||}[\/latex].<\/ol>\r\n<ol style=\"list-style-type: lower-alpha;\">b. Express [latex]\\bf{b}[\/latex] in component form.<\/ol>\r\n<ol style=\"list-style-type: lower-alpha;\">c. Find [latex]3[\/latex][latex]\\bf{a}[\/latex][latex]\u22124[\/latex][latex]\\bf{b}[\/latex].<\/ol>\r\n<div id=\"fs-id1167893940237\" class=\"exercise\">[reveal-answer q=\"fs-id1967793933114\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1967793933114\"]\r\n<ol style=\"list-style-type: lower-alpha;\">a. [latex]\\bf{||a||}[\/latex][latex] = 5\\sqrt{2}[\/latex]<\/ol>\r\n<ol style=\"list-style-type: lower-alpha;\">b. [latex]\\bf{b}[\/latex][latex] =\\langle-4,-3\\rangle[\/latex]<\/ol>\r\n<ol style=\"list-style-type: lower-alpha;\">c. [latex]3[\/latex][latex]\\bf{a}[\/latex][latex]\u22124[\/latex][latex]\\bf{b}[\/latex][latex] =\\langle37,15\\rangle[\/latex]<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try IT.\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=7713415&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=sA0145-1P3Q&amp;video_target=tpm-plugin-7hhonzao-sA0145-1P3Q\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP2.5_transcript.html\">transcript for \u201cCP 2.5\u201d here (opens in new window)<\/a><\/center>Now that we have established the basic rules of vector arithmetic, we can state the properties of vector operations. We will prove two of these properties. The others can be proved in a similar manner.\r\n<div id=\"fs-id1167793372221\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Properties of Vector Operations Theorem<\/h3>\r\n\r\n<hr \/>\r\n\r\nLet [latex]\\bf{u}[\/latex],[latex]\\bf{v}[\/latex], and [latex]\\bf{w}[\/latex] be vectors in a plane. Let [latex]r[\/latex] and [latex]s[\/latex] be scalars.\r\n\r\n\\[ \\begin{array}{lrl}\r\n\\mbox{i.} &amp; \\textbf{u} + \\textbf{v} &amp;= \\textbf{v} +\\textbf{u} &amp; \\mbox{Commutative property} \\\\\r\n\\mbox{ii.} &amp; (\\textbf{u} + \\textbf{v}) + \\textbf{w} &amp;= \\textbf{u} + (\\textbf{v} + \\textbf{w}) &amp; \\mbox{Associative property} \\\\\r\n\\mbox{iii.} &amp; \\textbf{u} + 0 &amp;= \\textbf{u} &amp; \\mbox{Additive identity property} \\\\\r\n\\mbox{iv.} &amp; \\textbf{u} + (-\\textbf{u}) &amp;= 0 &amp; \\mbox{Additive inverse property} \\\\\r\n\\mbox{v.} &amp; r(s\\textbf{u}) &amp;= (rs)\\textbf{u} &amp; \\mbox{Associativity of scalar multiplication} \\\\\r\n\\mbox{vi.} &amp; (r+s)\\textbf{u} &amp;= r\\textbf{u} +s\\textbf{u} &amp; \\mbox{Distributive property} \\\\\r\n\\mbox{vii.} &amp; r(\\textbf{u} + \\textbf{v}) &amp;= r\\textbf{u} + r\\textbf{v} &amp; \\mbox{Distributive property} \\\\\r\n\\mbox{viii.} &amp; 1\\textbf{u} &amp;= \\textbf{u} , 0\\textbf{u} = 0 &amp; \\mbox{Identity and zero properties}\\end{array}\\]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<h4 data-type=\"title\">Proof of Cummutative Property<\/h4>\r\nLet [latex]{\\bf{u}}=\\langle x_1, y_1\\rangle[\/latex] and [latex]{\\bf{v}}=\\langle x_2, y_2\\rangle[\/latex]. Apply the commutative property for real numbers:\r\n\r\n<center>[latex]{\\bf{u}} + {\\bf{v}} = \\langle x_1 + x_2, y_1 + y_2\\rangle = \\langle x_2 + x_1, y_2 + y_1\\rangle = {\\bf{u}} + {\\bf{v}}[\/latex].<\/center>[latex]_\\blacksquare[\/latex]\r\n<h4 data-type=\"title\">Proof of Distrubutive Property<\/h4>\r\nApply the distributive property for real numbers:\r\n<div id=\"fs-id1167794064134\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\r\n\\hfill r({\\bf{u}} + {\\bf{v}}) &amp; =\\hfill &amp; r \\cdot \\langle x_1 + x_2,y_1 + y_2\\rangle \\hfill \\\\\r\n\\hfill &amp; =\\hfill &amp; \\langle r(x_1 + x_2), r(y_1 + y_2)\\rangle \\hfill \\\\\r\n\\hfill &amp; =\\hfill &amp; \\langle rx_1 + rx_2, ry_1 + ry_2\\rangle \\hfill \\\\\r\n\\hfill &amp; =\\hfill &amp; \\langle rx_1,ry_1\\rangle \\langle rx_2, ry_2\\rangle \\hfill \\\\\r\n\\hfill &amp; =\\hfill &amp; r{\\bf{u}} + r{\\bf{v}}\\end{array}[\/latex]<\/div>\r\n[latex]_\\blacksquare[\/latex]\r\n<div id=\"fs-id1167893967091\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nProve the additive inverse property.\r\n<div id=\"fs-id1167897940237\" class=\"exercise\">[reveal-answer q=\"fs-id1967793973114\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1967793973114\"]\r\nLet [latex]\\mathbf{u} = \\left \\langle x_1 , y_1 \\right \\rangle[\/latex].\r\nThen [latex]-\\mathbf{u} = \\left \\langle -x_1 , -y_1 \\right \\rangle[\/latex] and\r\n[latex]\r\n\\begin{align*}\r\n\\mathbf{u} + \\left ( -\\mathbf{u} \\right ) &amp;= \\left \\langle x_1 , y_1 \\right \\rangle + \\left \\langle -x_1 , -y_1 \\right \\rangle \\\\\r\n&amp;= \\left \\langle x_1 + \\left ( -x_1 \\right ) , y_1 + \\left ( -y_1 \\right ) \\right \\rangle \\\\\r\n&amp;= \\left \\langle x_1 - x_1 , y_1 - y_1 \\right \\rangle \\\\\r\n&amp;= \\left \\langle 0 , 0 \\right \\rangle \\\\\r\n&amp;= \\mathbf{0}\r\n\\end{align*}\r\n.[\/hidden-answer]<\/div>\r\n<\/div>\r\nWe have found the components of a vector given its initial and terminal points. In some cases, we may only have the magnitude and direction of a vector, not the points. For these vectors, we can identify the horizontal and vertical components using trigonometry (Figure 16).\r\n\r\n[caption id=\"attachment_697\" align=\"aligncenter\" width=\"262\"]<img class=\"size-full wp-image-697\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/17191244\/2-1-16.jpeg\" alt=\"This figure is a right triangle. There is an angle labeled theta. The two sides are labeled \u201cmagnitude of v times cosine theta\u201d and \u201cmagnitude of v times sine theta.\u201d The hypotenuse is labeled \u201cmagnitude of v.\u201d\" width=\"262\" height=\"258\" \/> Figure 7. The components of a vector form the legs of a right triangle, with the vector as the hypotenuse.[\/caption]\r\n\r\nConsider the angle [latex]\\theta[\/latex] formed by the vector [latex]{\\bf{v}}[\/latex] and the positive [latex]x[\/latex]-axis. We can see from the triangle that the components of vector [latex]{\\bf{v}}[\/latex] are [latex]\\langle ||{\\bf{v}}||\\cos{\\theta},||{\\bf{v}}||\\sin{\\theta} \\rangle[\/latex]. Therefore, given an angle and the magnitude of a vector, we can use the cosine and sine of the angle to find the components of the vector.\r\n<div id=\"fs-id1167793975960\" class=\"textbook exercises\">\r\n<h3>Example: Finding the Component Form of a Vector Using Trigonometry<\/h3>\r\nFind the component form of a vector with magnitude [latex]4[\/latex] that forms an angle of [latex]\u221245\u00b0[\/latex] with the [latex]x[\/latex]-axis.\r\n<div id=\"fs-id1167793469764\" class=\"exercise\">[reveal-answer q=\"fs-id1167794955754\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794955754\"]\r\nLet [latex]x[\/latex] and [latex]y[\/latex] represent the components of the vector (Figure 2.16). Then [latex]x = 4\\cos({\u221245\u00b0}) = 2\\sqrt{2}[\/latex] and [latex]y = 4\\sin({\u221245\u00b0}) = \u22122\\sqrt{2}[\/latex]. The component form of the vector is [latex]\\langle 2\\sqrt{2},\u22122\\sqrt{2} \\rangle[\/latex].[caption id=\"attachment_698\" align=\"aligncenter\" width=\"186\"]<img class=\"wp-image-698 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/17191328\/2-1-17.jpeg\" alt=\"This figure is a right triangle. The two sides are labeled \u201cx\u201d and \u201cy.\u201d The hypotenuse is labeled \u201c4.\u201d There is also an angle labeled \u201c45 degrees.\u201d The hypotenuse is represented as a vector.\" width=\"186\" height=\"190\" \/> Figure 8. Use trigonometric ratios, [latex]x=||{\\bf{v}}||\\cos{\\theta}[\/latex] and[latex]y=||{\\bf{v}}||\\sin{\\theta}[\/latex], to identify the components of the vector.[\/caption]<\/div>\r\n<div class=\"exercise\">[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167893157091\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFind the component form of vector [latex]{\\bf{v}}[\/latex] with magnitude [latex]10[\/latex] that forms an angle of [latex]120\u00b0[\/latex] with the positive [latex]x[\/latex]-axis.\r\n<div id=\"fs-id1167893944237\" class=\"exercise\">[reveal-answer q=\"fs-id1967793933284\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1967793933284\"]\r\n[latex]{\\bf{v}} = \\langle \u22125,5\\sqrt{3}\\rangle[\/latex]\r\n[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]26031[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Express a vector in component form.<\/li>\n<li>Explain the formula for the magnitude of a vector.<\/li>\n<\/ul>\n<\/div>\n<p>Working with vectors in a plane is easier when we are working in a coordinate system. When the initial points and terminal points of vectors are given in Cartesian coordinates, computations become straightforward.<\/p>\n<div id=\"fs-id1167793900968\" class=\"textbook exercises\">\n<h3>Example: Comparing Vectors<\/h3>\n<p>Are [latex]\\bf{v}[\/latex] and [latex]\\bf{w}[\/latex] equivalent vectors?<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li style=\"list-style-type: none;\">\n<ol style=\"list-style-type: lower-alpha;\">         <\/ol>\n<\/li>\n<\/ol>\n<ol style=\"list-style-type: lower-alpha;\">        <\/ol>\n<ol style=\"list-style-type: lower-alpha;\">\n<li style=\"list-style-type: none;\">\n<ol style=\"list-style-type: lower-alpha;\">          <\/ol>\n<\/li>\n<\/ol>\n<ol style=\"list-style-type: lower-alpha;\">        <\/ol>\n<div id=\"fs-id1167793461764\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167795055165\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167795055165\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">                       <\/ol>\n<div id=\"attachment_690\" style=\"width: 352px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-690\" class=\"wp-image-690 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/17190445\/2-1-10.jpeg\" alt=\"This figure is a Cartesian coordinate system with two vectors. The first vector labeled \u201cv\u201d has initial point at (3, 2) and terminal point (7, 2). It is parallel to the x-axis. The second vector is labeled \u201cw\u201d and has initial point (1, -4) and terminal point (1, 0). It is parallel to the y-axis.\" width=\"342\" height=\"384\" \/><\/p>\n<p id=\"caption-attachment-690\" class=\"wp-caption-text\">Figure 1. These vectors are not equivalent.<\/p>\n<\/div>\n<ol style=\"list-style-type: lower-alpha;\">                             <\/ol>\n<div id=\"attachment_691\" style=\"width: 277px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-691\" class=\"wp-image-691 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/17190539\/2-1-11.jpeg\" alt=\"This figure is a Cartesian coordinate system with two vectors. The first vector labeled \u201cv\u201d has initial point at (0, 0) and terminal point (1, 1). The second vector is labeled \u201cw\u201d and has initial point (-2, 2) and terminal point (-1, 3).\" width=\"267\" height=\"272\" \/><\/p>\n<p id=\"caption-attachment-691\" class=\"wp-caption-text\">Figure 2. These vectors are equivalent.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167793958097\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Which of the following vectors are equivalent?<\/p>\n<div id=\"attachment_692\" style=\"width: 652px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-692\" class=\"wp-image-692 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/17190648\/2-1-12.jpeg\" alt=\"alt=&quot;This figure is a coordinate system with 6 vectors, each labeled a through f. Three of the vectors, \u201ca,\u201d \u201cb,\u201d and \u201ce\u201d have the same length and are pointing in the same direction.&quot;\" width=\"642\" height=\"384\" \/><\/p>\n<p id=\"caption-attachment-692\" class=\"wp-caption-text\">Figure 3. Determine which vectors are equivalent.<\/p>\n<\/div>\n<div id=\"fs-id1167793940339\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167794933124\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167794933124\" class=\"hidden-answer\" style=\"display: none\">\nVectors [latex]\\bf{a}[\/latex], [latex]\\bf{b}[\/latex], and [latex]\\bf{e}[\/latex] are equivalent.\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>We have seen how to plot a vector when we are given an initial point and a terminal point. However, because a vector can be placed anywhere in a plane, it may be easier to perform calculations with a vector when its initial point coincides with the origin. We call a vector with its initial point at the origin a <strong>standard-position vector<\/strong>. Because the initial point of any vector in standard position is known to be [latex](0,0)[\/latex], we can describe the vector by looking at the coordinates of its terminal point. Thus, if vector [latex]\\bf{v}[\/latex] has its initial point at the origin and its terminal point at [latex](x,y)[\/latex], we write the vector in component form as<\/p>\n<div style=\"text-align: center;\">[latex]\\bf{v}[\/latex] [latex]= \\langle x,y \\rangle[\/latex].<\/div>\n<p>When a vector is written in component form like this, the scalars [latex]x[\/latex] and [latex]y[\/latex] are called the <strong>components <\/strong>of [latex]\\bf{v}[\/latex].<\/p>\n<div id=\"fs-id1167793372221\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\n<hr \/>\n<p>The vector with initial point [latex](0,0)[\/latex] and terminal point [latex](x,y)[\/latex] can be written in component form as<\/p>\n<div style=\"text-align: center;\">[latex]\\bf{v}[\/latex] [latex]= \\langle x,y \\rangle[\/latex].<\/div>\n<p>The scalars [latex]x[\/latex] and [latex]y[\/latex] are called the components of [latex]\\bf{v}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Recall that vectors are named with lowercase letters in bold type or by drawing an arrow over their name. We have also learned that we can name a vector by its component form, with the coordinates of its terminal point in angle brackets. However, when writing the component form of a vector, it is important to distinguish between [latex]\\langle x,y \\rangle[\/latex] and [latex](x,y)[\/latex]. The first ordered pair uses angle brackets to describe a vector, whereas the second uses parentheses to describe a point in a plane. The initial point of [latex]\\langle x,y \\rangle[\/latex] is [latex](0,0)[\/latex]; the terminal point of [latex]\\langle x,y \\rangle[\/latex] is [latex](x,y)[\/latex].<\/p>\n<p>When we have a vector not already in standard position, we can determine its component form in one of two ways. We can use a geometric approach, in which we sketch the vector in the coordinate plane, and then sketch an equivalent standard-position vector. Alternatively, we can find it algebraically, using the coordinates of the initial point and the terminal point. To find it algebraically, we subtract the [latex]x[\/latex]-coordinate of the initial point from the [latex]x[\/latex]-coordinate of the terminal point to get the [latex]x[\/latex] component, and we subtract the [latex]y[\/latex]-coordinate of the initial point from the [latex]y[\/latex]-coordinate of the terminal point to get the [latex]y[\/latex] component.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Component Form of a Vector<\/h3>\n<hr \/>\n<p>Let [latex]\\bf{v}[\/latex] be a vector with initial point [latex](x_i,y_i)[\/latex] and terminal point [latex](x_t,y_t)[\/latex]. Then we can express [latex]\\bf{v}[\/latex] in component form as<br \/>\n[latex]\\bf{v}[\/latex] [latex]=\\langle x_t \u2212 x_i,y_t \u2212 y_i\\rangle[\/latex].<\/p>\n<\/div>\n<div id=\"fs-id1167793900968\" class=\"textbook exercises\">\n<h3>Example: Expressing Vectors in Component Form<\/h3>\n<p>Express vector [latex]\\bf{v}[\/latex] with initial point [latex](\u22123,4)[\/latex] and terminal point [latex](1,2)[\/latex] in component form.<\/p>\n<div id=\"fs-id1167893461764\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167895055165\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167895055165\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li style=\"list-style-type: none;\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li style=\"list-style-type: none;\">\n<ol style=\"list-style-type: decimal;\">         <\/ol>\n<ol style=\"list-style-type: decimal;\">                 <\/ol>\n<ol style=\"list-style-type: decimal;\">                 <\/ol>\n<ol style=\"list-style-type: decimal;\">             <\/ol>\n<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<div style=\"text-align: center;\">[latex]\\bf{v}[\/latex] [latex]= \\langle4,\u22122\\rangle.[\/latex]<\/div>\n<div id=\"attachment_693\" style=\"width: 427px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-693\" class=\"size-full wp-image-693\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/17190801\/2-1-13.jpeg\" alt=\"This figure is a coordinate system. There are two vectors on the graph. The first vector has initial point at the origin and terminal point at (4, -2). The horizontal distance from the initial to the terminal point for the vector is labeled as \u201c4 units.\u201d The vertical distance from the initial to the terminal point is labeled as \u201c2 units.\u201d The second vector has initial point at (-3, 4) and terminal point at (1, 2). The horizontal distance from the initial to the terminal point for the vector is labeled as \u201c4 units.\u201d The vertical distance from the initial to the terminal point is labeled as \u201c2 units.\u201d\" width=\"417\" height=\"384\" \/><\/p>\n<p id=\"caption-attachment-693\" class=\"wp-caption-text\">Figure 4.\u00a0These vectors are equivalent.<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li style=\"list-style-type: none;\">\n<ol style=\"list-style-type: lower-alpha;\"> <\/ol>\n<\/li>\n<\/ol>\n<ol style=\"list-style-type: lower-alpha;\">\n<li style=\"list-style-type: none;\">\n<ol style=\"list-style-type: lower-alpha;\">                                  <\/ol>\n<\/li>\n<\/ol>\n<div style=\"text-align: center;\">[latex]x_t - x_i = 1 - (-3) = 4[\/latex]<\/div>\n<ol style=\"list-style-type: lower-alpha;\">\n<li style=\"list-style-type: none;\">\n<ol style=\"list-style-type: lower-alpha;\">            <\/ol>\n<\/li>\n<\/ol>\n<div style=\"text-align: center;\">[latex]y_t - y_i = 2 - 4 = -2[\/latex]<\/div>\n<ol style=\"list-style-type: lower-alpha;\">\n<li style=\"list-style-type: none;\">\n<ol style=\"list-style-type: lower-alpha;\">   <\/ol>\n<\/li>\n<\/ol>\n<div style=\"text-align: center;\">[latex]\\bf{v}[\/latex] [latex]=\\langle x_t \u2212 x_i,y_t \u2212 y_i\\rangle[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\">[latex]=\\langle 1 - (-3), 2 - 4\\rangle[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\">[latex]=\\langle 4, -2\\rangle[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167793958097\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Vector [latex]\\bf{w}[\/latex] has initial point [latex](\u22124,\u22125)[\/latex] and terminal point [latex](\u22121,2)[\/latex]. Express [latex]\\bf{w}[\/latex] in component form.<\/p>\n<div id=\"fs-id1167793940339\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1168794933124\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1168794933124\" class=\"hidden-answer\" style=\"display: none\">\n[latex]\\langle 3,7 \\rangle[\/latex]\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>To find the magnitude of a vector, we calculate the distance between its initial point and its terminal point. The magnitude of vector<br \/>\n[latex]\\bf{v}[\/latex][latex]=\\langle x,y\\rangle[\/latex] is denoted [latex]\\bf{||v||}[\/latex], or [latex]\\bf{|v|}[\/latex], and can be computed using the formula<\/p>\n<div style=\"text-align: center;\">[latex]\\bf{||v||}[\/latex] [latex]=\\sqrt{x^2 + y^2}[\/latex]<\/div>\n<p>.<\/p>\n<p>Note that because this vector is written in component form, it is equivalent to a vector in standard position, with its initial point at the origin and terminal point [latex](x,y)[\/latex]. Thus, it suffices to calculate the magnitude of the vector in standard position. Using the distance formula to calculate the distance between initial point [latex](0,0)[\/latex] and terminal point [latex](x,y)[\/latex], we have<\/p>\n<div style=\"text-align: center;\">[latex]\\bf{||v||}[\/latex][latex]=\\sqrt{(x - 0)^2 + (y - 0)^2}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\">[latex]=\\sqrt{x^2 + y^2}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Based on this formula, it is clear that for any vector [latex]\\bf{v}[\/latex], [latex]\\bf{||v||}[\/latex][latex]\\geq 0[\/latex], and [latex]\\bf{||v||}[\/latex][latex]= 0[\/latex] if and only if [latex]\\bf{v}[\/latex][latex]= 0[\/latex].<\/p>\n<p>The magnitude of a vector can also be derived using the Pythagorean theorem, as in the following figure.<\/p>\n<div id=\"attachment_694\" style=\"width: 292px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-694\" class=\"size-full wp-image-694\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/17190913\/2-1-14.jpeg\" alt=\"This figure is a right triangle. The two sides are labeled \u201cx\u201d and \u201cy.\u201d The hypotenuse is represented as a vector and is labeled \u201csquare root (x^2 + y^2).\u201d\" width=\"282\" height=\"119\" \/><\/p>\n<p id=\"caption-attachment-694\" class=\"wp-caption-text\">Figure 5. If you use the components of a vector to define a right triangle, the magnitude of the vector is the length of the triangle\u2019s hypotenuse.<\/p>\n<\/div>\n<p>We have defined scalar multiplication and vector addition geometrically. Expressing vectors in component form allows us to perform these same operations algebraically.<\/p>\n<div id=\"fs-id1167793372221\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\n<hr \/>\n<p>Let [latex]\\bf{v}[\/latex][latex]= \\langle x_1, y_1\\rangle.[\/latex] and [latex]\\bf{w}[\/latex][latex]= \\langle x_2, y_2\\rangle[\/latex] be vectors, and let [latex]k[\/latex] be a scalar.<\/p>\n<p><strong>Scalar multiplication<\/strong>: [latex]k\\bf{v}[\/latex][latex]=\\langle kx_1, ky_1\\rangle[\/latex]<\/p>\n<p><strong>Vector addition<\/strong>: [latex]\\bf{v + w}[\/latex][latex]=\\langle x_1, y_1\\rangle + \\langle x_2, y_2\\rangle = \\langle x_1 + x_2, y_1 + y_2\\rangle[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167793905960\" class=\"textbook exercises\">\n<h3>Example: Performing Operations in Component Form<\/h3>\n<p>Let [latex]\\bf{v}[\/latex] be the vector with initial point [latex](2,5)[\/latex] and terminal point [latex](8,13)[\/latex], and let [latex]\\bf{w}[\/latex][latex]= \\langle\u22122,4\\rangle[\/latex].<\/p>\n<ol style=\"list-style-type: lower-alpha;\">            <\/ol>\n<ol style=\"list-style-type: lower-alpha;\">   <\/ol>\n<ol style=\"list-style-type: lower-alpha;\">  <\/ol>\n<ol style=\"list-style-type: lower-alpha;\"> <\/ol>\n<div id=\"fs-id1167793461764\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167794055754\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167794055754\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li style=\"list-style-type: none;\">\n<ol style=\"list-style-type: lower-alpha;\">                                              <\/ol>\n<\/li>\n<\/ol>\n<div style=\"text-align: center;\">[latex]\\bf{||v||}[\/latex] [latex]=\\sqrt{6^2 + 8^2} = \\sqrt{36+64} = \\sqrt{100} = 10[\/latex].<\/div>\n<div style=\"text-align: center;\"><\/div>\n<div style=\"text-align: center;\">\n<div id=\"attachment_696\" style=\"width: 277px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-696\" class=\"wp-image-696 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/17191050\/2-1-15.jpeg\" alt=\"This figure is the first quadrant of a coordinate system. It has two vectors. The first vector has initial point at (2, 5) and terminal point (8, 13). The second vector has initial point at the origin and terminal point at (6, 8).\" width=\"267\" height=\"347\" \/><\/p>\n<p id=\"caption-attachment-696\" class=\"wp-caption-text\">Figure 6. In component form, [latex]{\\bf{v}}=\\langle{6,8}\\rangle[\/latex].<\/p>\n<\/div>\n<\/div>\n<ol style=\"list-style-type: lower-alpha;\">\n<li style=\"list-style-type: none;\">\n<ol style=\"list-style-type: lower-alpha;\">            <\/ol>\n<\/li>\n<\/ol>\n<div style=\"text-align: center;\">[latex]\\bf{v + w}[\/latex][latex]=\\langle6,8\\rangle + \\langle\u22122,4\\rangle = \\langle4,12\\rangle[\/latex].<\/div>\n<ol style=\"list-style-type: lower-alpha;\">\n<li style=\"list-style-type: none;\">\n<ol style=\"list-style-type: lower-alpha;\">         <\/ol>\n<\/li>\n<\/ol>\n<div style=\"text-align: center;\">[latex]3[\/latex][latex]\\bf{v}[\/latex][latex]=3\\cdot\\langle6,8\\rangle = \\langle3\\cdot6,3\\cdot8\\rangle = \\langle18,24\\rangle[\/latex].<\/div>\n<ol style=\"list-style-type: lower-alpha;\">\n<li style=\"list-style-type: none;\">\n<ol style=\"list-style-type: lower-alpha;\">          <\/ol>\n<\/li>\n<\/ol>\n<div style=\"text-align: center;\">[latex]\\bf{v}[\/latex][latex]\u22122[\/latex][latex]\\bf{w}[\/latex][latex]=\\langle6,8\\rangle \u2212 2 \\cdot \\langle\u22122,4\\rangle = \\langle6,8\\rangle + \\langle4,\u22128\\rangle = \\langle10,0\\rangle[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167893957091\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Let [latex]\\bf{a}[\/latex][latex]=\\langle7,1\\rangle[\/latex] and let [latex]\\bf{b}[\/latex] be the vector with initial point [latex](3,2)[\/latex] and terminal point [latex](\u22121,\u22121)[\/latex].<\/p>\n<ol style=\"list-style-type: lower-alpha;\">  <\/ol>\n<ol style=\"list-style-type: lower-alpha;\">     <\/ol>\n<ol style=\"list-style-type: lower-alpha;\">  <\/ol>\n<div id=\"fs-id1167893940237\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1967793933114\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1967793933114\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">   <\/ol>\n<ol style=\"list-style-type: lower-alpha;\">  <\/ol>\n<ol style=\"list-style-type: lower-alpha;\">  <\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try IT.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=7713415&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=sA0145-1P3Q&amp;video_target=tpm-plugin-7hhonzao-sA0145-1P3Q\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP2.5_transcript.html\">transcript for \u201cCP 2.5\u201d here (opens in new window)<\/a><\/div>\n<p>Now that we have established the basic rules of vector arithmetic, we can state the properties of vector operations. We will prove two of these properties. The others can be proved in a similar manner.<\/p>\n<div id=\"fs-id1167793372221\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Properties of Vector Operations Theorem<\/h3>\n<hr \/>\n<p>Let [latex]\\bf{u}[\/latex],[latex]\\bf{v}[\/latex], and [latex]\\bf{w}[\/latex] be vectors in a plane. Let [latex]r[\/latex] and [latex]s[\/latex] be scalars.<\/p>\n<p>\\[ \\begin{array}{lrl}<br \/>\n\\mbox{i.} &amp; \\textbf{u} + \\textbf{v} &amp;= \\textbf{v} +\\textbf{u} &amp; \\mbox{Commutative property} \\\\<br \/>\n\\mbox{ii.} &amp; (\\textbf{u} + \\textbf{v}) + \\textbf{w} &amp;= \\textbf{u} + (\\textbf{v} + \\textbf{w}) &amp; \\mbox{Associative property} \\\\<br \/>\n\\mbox{iii.} &amp; \\textbf{u} + 0 &amp;= \\textbf{u} &amp; \\mbox{Additive identity property} \\\\<br \/>\n\\mbox{iv.} &amp; \\textbf{u} + (-\\textbf{u}) &amp;= 0 &amp; \\mbox{Additive inverse property} \\\\<br \/>\n\\mbox{v.} &amp; r(s\\textbf{u}) &amp;= (rs)\\textbf{u} &amp; \\mbox{Associativity of scalar multiplication} \\\\<br \/>\n\\mbox{vi.} &amp; (r+s)\\textbf{u} &amp;= r\\textbf{u} +s\\textbf{u} &amp; \\mbox{Distributive property} \\\\<br \/>\n\\mbox{vii.} &amp; r(\\textbf{u} + \\textbf{v}) &amp;= r\\textbf{u} + r\\textbf{v} &amp; \\mbox{Distributive property} \\\\<br \/>\n\\mbox{viii.} &amp; 1\\textbf{u} &amp;= \\textbf{u} , 0\\textbf{u} = 0 &amp; \\mbox{Identity and zero properties}\\end{array}\\]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h4 data-type=\"title\">Proof of Cummutative Property<\/h4>\n<p>Let [latex]{\\bf{u}}=\\langle x_1, y_1\\rangle[\/latex] and [latex]{\\bf{v}}=\\langle x_2, y_2\\rangle[\/latex]. Apply the commutative property for real numbers:<\/p>\n<div style=\"text-align: center;\">[latex]{\\bf{u}} + {\\bf{v}} = \\langle x_1 + x_2, y_1 + y_2\\rangle = \\langle x_2 + x_1, y_2 + y_1\\rangle = {\\bf{u}} + {\\bf{v}}[\/latex].<\/div>\n<p>[latex]_\\blacksquare[\/latex]<\/p>\n<h4 data-type=\"title\">Proof of Distrubutive Property<\/h4>\n<p>Apply the distributive property for real numbers:<\/p>\n<div id=\"fs-id1167794064134\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}  \\hfill r({\\bf{u}} + {\\bf{v}}) & =\\hfill & r \\cdot \\langle x_1 + x_2,y_1 + y_2\\rangle \\hfill \\\\  \\hfill & =\\hfill & \\langle r(x_1 + x_2), r(y_1 + y_2)\\rangle \\hfill \\\\  \\hfill & =\\hfill & \\langle rx_1 + rx_2, ry_1 + ry_2\\rangle \\hfill \\\\  \\hfill & =\\hfill & \\langle rx_1,ry_1\\rangle \\langle rx_2, ry_2\\rangle \\hfill \\\\  \\hfill & =\\hfill & r{\\bf{u}} + r{\\bf{v}}\\end{array}[\/latex]<\/div>\n<p>[latex]_\\blacksquare[\/latex]<\/p>\n<div id=\"fs-id1167893967091\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Prove the additive inverse property.<\/p>\n<div id=\"fs-id1167897940237\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1967793973114\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1967793973114\" class=\"hidden-answer\" style=\"display: none\">\nLet [latex]\\mathbf{u} = \\left \\langle x_1 , y_1 \\right \\rangle[\/latex].<br \/>\nThen [latex]-\\mathbf{u} = \\left \\langle -x_1 , -y_1 \\right \\rangle[\/latex] and<br \/>\n[latex][\/latex]<br \/>\n\\begin{align*}<br \/>\n\\mathbf{u} + \\left ( -\\mathbf{u} \\right ) &amp;= \\left \\langle x_1 , y_1 \\right \\rangle + \\left \\langle -x_1 , -y_1 \\right \\rangle \\\\<br \/>\n&amp;= \\left \\langle x_1 + \\left ( -x_1 \\right ) , y_1 + \\left ( -y_1 \\right ) \\right \\rangle \\\\<br \/>\n&amp;= \\left \\langle x_1 - x_1 , y_1 - y_1 \\right \\rangle \\\\<br \/>\n&amp;= \\left \\langle 0 , 0 \\right \\rangle \\\\<br \/>\n&amp;= \\mathbf{0}<br \/>\n\\end{align*}<br \/>\n.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>We have found the components of a vector given its initial and terminal points. In some cases, we may only have the magnitude and direction of a vector, not the points. For these vectors, we can identify the horizontal and vertical components using trigonometry (Figure 16).<\/p>\n<div id=\"attachment_697\" style=\"width: 272px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-697\" class=\"size-full wp-image-697\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/17191244\/2-1-16.jpeg\" alt=\"This figure is a right triangle. There is an angle labeled theta. The two sides are labeled \u201cmagnitude of v times cosine theta\u201d and \u201cmagnitude of v times sine theta.\u201d The hypotenuse is labeled \u201cmagnitude of v.\u201d\" width=\"262\" height=\"258\" \/><\/p>\n<p id=\"caption-attachment-697\" class=\"wp-caption-text\">Figure 7. The components of a vector form the legs of a right triangle, with the vector as the hypotenuse.<\/p>\n<\/div>\n<p>Consider the angle [latex]\\theta[\/latex] formed by the vector [latex]{\\bf{v}}[\/latex] and the positive [latex]x[\/latex]-axis. We can see from the triangle that the components of vector [latex]{\\bf{v}}[\/latex] are [latex]\\langle ||{\\bf{v}}||\\cos{\\theta},||{\\bf{v}}||\\sin{\\theta} \\rangle[\/latex]. Therefore, given an angle and the magnitude of a vector, we can use the cosine and sine of the angle to find the components of the vector.<\/p>\n<div id=\"fs-id1167793975960\" class=\"textbook exercises\">\n<h3>Example: Finding the Component Form of a Vector Using Trigonometry<\/h3>\n<p>Find the component form of a vector with magnitude [latex]4[\/latex] that forms an angle of [latex]\u221245\u00b0[\/latex] with the [latex]x[\/latex]-axis.<\/p>\n<div id=\"fs-id1167793469764\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167794955754\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167794955754\" class=\"hidden-answer\" style=\"display: none\">\nLet [latex]x[\/latex] and [latex]y[\/latex] represent the components of the vector (Figure 2.16). Then [latex]x = 4\\cos({\u221245\u00b0}) = 2\\sqrt{2}[\/latex] and [latex]y = 4\\sin({\u221245\u00b0}) = \u22122\\sqrt{2}[\/latex]. The component form of the vector is [latex]\\langle 2\\sqrt{2},\u22122\\sqrt{2} \\rangle[\/latex].<\/p>\n<div id=\"attachment_698\" style=\"width: 196px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-698\" class=\"wp-image-698 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/17191328\/2-1-17.jpeg\" alt=\"This figure is a right triangle. The two sides are labeled \u201cx\u201d and \u201cy.\u201d The hypotenuse is labeled \u201c4.\u201d There is also an angle labeled \u201c45 degrees.\u201d The hypotenuse is represented as a vector.\" width=\"186\" height=\"190\" \/><\/p>\n<p id=\"caption-attachment-698\" class=\"wp-caption-text\">Figure 8. Use trigonometric ratios, [latex]x=||{\\bf{v}}||\\cos{\\theta}[\/latex] and[latex]y=||{\\bf{v}}||\\sin{\\theta}[\/latex], to identify the components of the vector.<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167893157091\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find the component form of vector [latex]{\\bf{v}}[\/latex] with magnitude [latex]10[\/latex] that forms an angle of [latex]120\u00b0[\/latex] with the positive [latex]x[\/latex]-axis.<\/p>\n<div id=\"fs-id1167893944237\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1967793933284\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1967793933284\" class=\"hidden-answer\" style=\"display: none\">\n[latex]{\\bf{v}} = \\langle \u22125,5\\sqrt{3}\\rangle[\/latex]\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm26031\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=26031&theme=oea&iframe_resize_id=ohm26031&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-5436\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>CP 2.5. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 3. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/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":349141,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 3\",\"author\":\"Gilbert Strang, Edwin (Jed) 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Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-5436","chapter","type-chapter","status-publish","hentry"],"part":20,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/5436","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/349141"}],"version-history":[{"count":9,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/5436\/revisions"}],"predecessor-version":[{"id":6429,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/5436\/revisions\/6429"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/20"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/5436\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=5436"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=5436"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=5436"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=5436"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}