{"id":397,"date":"2021-08-05T16:19:16","date_gmt":"2021-08-05T16:19:16","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=397"},"modified":"2022-10-20T23:57:14","modified_gmt":"2022-10-20T23:57:14","slug":"unit-vectors","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/unit-vectors\/","title":{"raw":"Unit Vectors","rendered":"Unit Vectors"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Express a vector in terms of unit vectors.<\/li>\r\n \t<li>Give two examples of vector quantities.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Unit Vectors<\/h2>\r\nA <strong>unit vector<\/strong> is a vector with magnitude [latex]1[\/latex]. For any nonzero vector [latex]{\\bf{v}}[\/latex], we can use scalar multiplication to find a unit vector [latex]{\\bf{u}}[\/latex] that has the same direction as [latex]{\\bf{v}}[\/latex]. To do this, we multiply the vector by the reciprocal of its magnitude:\r\n\r\n<center>[latex]{\\bf{u}} = \\frac{1}{||{\\bf{v}}||}{\\bf{v}}[\/latex].<\/center>Recall that when we defined scalar multiplication, we noted that [latex]||k{\\bf{v}}|| = |k| \\cdot ||{\\bf{v}}||[\/latex]. For [latex]{\\bf{u}} = \\frac{1}{||{\\bf{v}}||}{\\bf{v}}[\/latex], it follows that [latex]||{\\bf{u}}|| = \\frac{1}{||{\\bf{v}}||}(||{\\bf{v}}||) = 1[\/latex]. We say that [latex]{\\bf{u}}[\/latex] is the <em>unit vector in the direction of<\/em> [latex]{\\bf{v}}[\/latex] (Figure 18). The process of using scalar multiplication to find a unit vector with a given direction is called <strong>normalization<\/strong>.\r\n\r\n<center><\/center><center>[caption id=\"attachment_700\" align=\"alignnone\" width=\"285\"]<img class=\"size-full wp-image-700\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/17191802\/2-1-18.jpeg\" alt=\"This image has two figures. The first is a vector labeled \u201cv.\u201d The second figure is a vector in the same direction labeled \u201cu.\u201d This vector has a length of 1 unit.\" width=\"285\" height=\"97\" \/> Figure 1.\u00a0The vector [latex]{\\bf{v}}[\/latex] and associated unit vector\u00a0[latex]{\\bf{u}}=\\frac{1}{||{\\bf{v}}||}{\\bf{v}}[\/latex]. In this case, [latex]||{\\bf{v}}||&gt;1[\/latex].[\/caption]<\/center>\r\n<div id=\"fs-id1165693975960\" class=\"textbook exercises\">\r\n<h3>Example: Finding a Unit Vector<\/h3>\r\nLet [latex]{\\bf{v}} = \\langle 1,2 \\rangle[\/latex].\r\n<ul>\r\n \t<li style=\"list-style-type: none;\">\r\n<ol style=\"list-style-type: lower-alpha;\">a. Find a unit vector with the same direction as [latex]{\\bf{v}}[\/latex].<\/ol>\r\n<ol style=\"list-style-type: lower-alpha;\">b. Find a vector [latex]{\\bf{w}}[\/latex] with the same direction as [latex]{\\bf{v}}[\/latex] such that [latex]||{\\bf{w}}|| = 7[\/latex].<\/ol>\r\n<\/li>\r\n<\/ul>\r\n<div id=\"fs-id1167793478764\" class=\"exercise\">[reveal-answer q=\"fs-id1167794934754\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794934754\"]\r\n<ul>\r\n \t<li style=\"list-style-type: none;\">\r\n<ul>\r\n \t<li style=\"list-style-type: none;\">\r\n<ol style=\"list-style-type: lower-alpha;\">a. First, find the magnitude of [latex]{\\bf{v}}[\/latex], then divide the components of [latex]{\\bf{v}}[\/latex] by the magnitude:<\/ol>\r\n<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<center>[latex]||{\\bf{v}}|| = \\sqrt{1^2 + 2^2} = \\sqrt{1 + 4} = \\sqrt{5}[\/latex]<\/center>&nbsp;\r\n\r\n<center>[latex]{\\bf{u}} = \\frac{1}{||{\\bf{v}}||} {\\bf{v}} = \\frac{1}{\\sqrt{5}} \\langle 1,2 \\rangle = \\langle \\frac{1}{\\sqrt{5}}, \\frac{2}{\\sqrt{5}} \\rangle[\/latex].<\/center>\r\n<ul>\r\n \t<li style=\"list-style-type: none;\">\r\n<ul>\r\n \t<li style=\"list-style-type: none;\">\r\n<ol style=\"list-style-type: lower-alpha;\">b. The vector [latex]{\\bf{u}}[\/latex] is in the same direction as [latex]{\\bf{v}}[\/latex] and [latex]||{\\bf{u}}|| = 1[\/latex]. Use scalar multiplication to increase the length of [latex]{\\bf{u}}[\/latex] without changing direction:<\/ol>\r\n<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<center>[latex]{\\bf{w}} = 7{\\bf{u}} = 7 \\langle \\frac{1}{\\sqrt{5}}, \\frac{2}{\\sqrt{5}} \\rangle = \\langle \\frac{7}{\\sqrt{5}}, \\frac{14}{\\sqrt{5}} \\rangle.[\/latex]<\/center>\r\n<div class=\"exercise\">[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167233157091\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nLet [latex]{\\bf{v}} = \\langle 9,2 \\rangle[\/latex]. Find a vector with magnitude [latex]5[\/latex] in the opposite direction as [latex]{\\bf{v}}[\/latex].\r\n<div id=\"fs-id1167894444237\" class=\"exercise\">[reveal-answer q=\"fs-id1967793921284\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1967793921284\"]\r\n[latex] \\langle -\\frac{45}{\\sqrt{85}}, -\\frac{10}{\\sqrt{85}}\\rangle[\/latex]\r\n[\/hidden-answer]<\/div>\r\n<\/div>\r\nWe have seen how convenient it can be to write a vector in component form. Sometimes, though, it is more convenient to write a vector as a sum of a horizontal vector and a vertical vector. To make this easier, let\u2019s look at standard unit vectors. The <strong>standard unit vectors<\/strong> are the vectors [latex]{\\bf{i}} = \\langle 1,0 \\rangle[\/latex] and [latex]{\\bf{j}} = \\langle 0,1 \\rangle[\/latex] (Figure 2.18).\r\n\r\n[caption id=\"attachment_733\" align=\"aligncenter\" width=\"365\"]<img class=\"wp-image-733 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/24154043\/2-1-19.jpeg\" alt=\"This figure has the x and y axes of a coordinate system in the first quadrant. On the x-axis there is a vector labeled \u201ci,\u201d which equals &lt;1,0&gt;. The second vector is on the y-axis and is labeled \u201cj\u201d which equals &lt;0,1&gt;.\" width=\"365\" height=\"202\" \/> Figure 2. The standard unit vectors [latex]{\\bf{i}}[\/latex] and [latex]{\\bf{j}}[\/latex].[\/caption]By applying the properties of vectors, it is possible to express any vector in terms of [latex]{\\bf{i}}[\/latex] and [latex]{\\bf{j}}[\/latex] in what we call a <em>linear combination<\/em>:\r\n\r\n<center>[latex]||{\\bf{v}}|| = \\langle x,y \\rangle = \\langle x,0 \\rangle + \\langle 0,y \\rangle = x\\langle 1,0 \\rangle + y\\langle 0,1 \\rangle = x{\\bf{i}} + y{\\bf{j}}[\/latex].<\/center><center><\/center>Thus, [latex]{\\bf{v}}[\/latex] is the sum of a horizontal vector with magnitude [latex]x[\/latex], and a vertical vector with magnitude [latex]y[\/latex], as in the following figure.\r\n\r\n[caption id=\"attachment_735\" align=\"aligncenter\" width=\"174\"]<img class=\"wp-image-735 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/24154229\/2-1-20.jpeg\" alt=\"This figure is a right triangle. The horizontal side is labeled \u201cxi.\u201d The vertical side is labeled \u201cyj.\u201d The hypotenuse is a vector labeled \u201cv.\u201d\" width=\"174\" height=\"122\" \/> Figure 3. The vector [latex]{\\bf{v}}[\/latex] is the sum of [latex]x{\\bf{i}}[\/latex] and [latex]y{\\bf{j}}[\/latex].[\/caption]\r\n<div id=\"fs-id1165696275960\" class=\"textbook exercises\">\r\n<h3>Example: Using Standard Unit Vectors<\/h3>\r\n<ul>\r\n \t<li style=\"list-style-type: none;\">\r\n<ol style=\"list-style-type: lower-alpha;\">a. Express the vector [latex]{\\bf{w}} = \\langle 3,-4 \\rangle [\/latex] in terms of standard unit vectors.<\/ol>\r\n<ol style=\"list-style-type: lower-alpha;\">b. Vector [latex]{\\bf{u}}[\/latex] is a unit vector that forms an angle of [latex]60\u00b0[\/latex] with the positive [latex]x[\/latex]-axis. Use standard unit vectors to describe [latex]{\\bf{u}}[\/latex].<\/ol>\r\n<\/li>\r\n<\/ul>\r\n<div id=\"fs-id1167793478764\" class=\"exercise\">[reveal-answer q=\"fs-id1167798634754\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167798634754\"]\r\n<ul>\r\n \t<li style=\"list-style-type: none;\">\r\n<ul>\r\n \t<li style=\"list-style-type: none;\">\r\n<ol style=\"list-style-type: lower-alpha;\">a. Resolve vector [latex]{\\bf{w}}[\/latex] into a vector with a zero [latex]y[\/latex]-component and a vector with a zero [latex]x[\/latex]-component:<\/ol>\r\n<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<center>[latex]||{\\bf{w}}|| = \\langle 3,-4 \\rangle = 3{\\bf{i}} - 4{\\bf{j}}[\/latex]<\/center>\r\n<ul>\r\n \t<li style=\"list-style-type: none;\">\r\n<ul>\r\n \t<li style=\"list-style-type: none;\">\r\n<ol style=\"list-style-type: lower-alpha;\">b. Because [latex]{\\bf{u}}[\/latex] is a unit vector, the terminal point lies on the unit circle when the vector is placed in standard position (Figure 2.20).<\/ol>\r\n<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<div id=\"fs-id1167794064134\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {\\bf{u}} &amp; =\\hfill &amp; \\langle \\cos{60\u00b0}, \\sin{60\u00b0} \\rangle \\hfill \\\\ \\hfill &amp; =\\hfill &amp; \\langle \\frac{1}{2}, \\frac{\\sqrt{3}}{2} \\rangle \\hfill \\\\ \\hfill &amp; =\\hfill &amp; \\frac{1}{2}{\\bf{i}} - \\frac{\\sqrt{3}}{2}{\\bf{j}}\\end{array}[\/latex].<\/div>\r\n<div data-type=\"equation\" data-label=\"\"><\/div>\r\n[caption id=\"attachment_736\" align=\"aligncenter\" width=\"342\"]<img class=\"size-full wp-image-736\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/24154420\/2-1-21.jpeg\" alt=\"This figure is a unit circle. It is a circle centered at the origin. It has a vector with initial point at the origin and terminal point on the circle. The terminal point is labeled (cos(theta), sin(theta)). The length of the vector is 1 unit. There is also a right triangle formed with the vector as the hypotenuse. The horizontal side is labeled \u201ccos(theta)\u201d and the vertical side is labeled \u201csin(theta).\u201d\" width=\"342\" height=\"342\" \/> Figure 4.\u00a0The terminal point of [latex]{\\bf{u}}[\/latex] lies on the unit circle [latex](\\cos{\\theta},\\sin{\\theta})[\/latex].[\/caption]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id11672665157091\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nLet [latex]{\\bf{a}} = \\langle 16,-11 \\rangle [\/latex] and let [latex]{\\bf{b}}[\/latex] be a unit vector that forms an angle of [latex]225\u00b0[\/latex] with the positive [latex]x[\/latex]-axis. Express [latex]{\\bf{a}}[\/latex] and [latex]{\\bf{b}}[\/latex] in terms of the standard unit vectors.\r\n<div id=\"fs-id1167895644237\" class=\"exercise\">[reveal-answer q=\"fs-id1968393921284\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1968393921284\"]\r\n[latex]{\\bf{a}} = 16{\\bf{i}} - 11{\\bf{j}}[\/latex]\r\n[latex]{\\bf{b}} = -\\frac{\\sqrt{2}}{2}{\\bf{i}} - \\frac{\\sqrt{2}}{2}{\\bf{j}}[\/latex]\r\n[\/hidden-answer]<\/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=7713416&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=azDVrpxgUko&amp;video_target=tpm-plugin-ihc7z8ph-azDVrpxgUko\" 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\/CP2.9_transcript.html\">transcript for \u201cCP 2.9\u201d here (opens in new window)<\/a><\/center>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]199015[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Applications of Vectors<\/h2>\r\nBecause vectors have both direction and magnitude, they are valuable tools for solving problems involving such applications as motion and force. Recall the boat example and the quarterback example we described earlier. Here we look at two other examples in detail.\r\n<div id=\"fs-id1165696175960\" class=\"textbook exercises\">\r\n<h3>Example: Finding Resultant Force<\/h3>\r\nJane\u2019s car is stuck in the mud. Lisa and Jed come along in a truck to help pull her out. They attach one end of a tow strap to the front of the car and the other end to the truck\u2019s trailer hitch, and the truck starts to pull. Meanwhile, Jane and Jed get behind the car and push. The truck generates a horizontal force of [latex]300[\/latex] lb on the car. Jane and Jed are pushing at a slight upward angle and generate a force of [latex]150[\/latex] lb on the car. These forces can be represented by vectors, as shown in Figure 22. The angle between these vectors is [latex]15\u00b0[\/latex]. Find the resultant force (the vector sum) and give its magnitude to the nearest tenth of a pound and its direction angle from the positive [latex]x[\/latex]-axis.\r\n\r\n[caption id=\"attachment_738\" align=\"aligncenter\" width=\"671\"]<img class=\"wp-image-738 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/24154738\/2-1-22.jpeg\" alt=\"This image is the side view of an automobile. From the front of the automobile there is a horizontal vector labeled \u201c300 pounds.\u201d Also, from the front of the automobile there is another vector labeled \u201c150 pounds.\u201d The angle between the two vectors is 15 degrees.\" width=\"671\" height=\"167\" \/> Figure 5. Two forces acting on a car in different directions.[\/caption]\r\n\r\n<div><\/div>\r\n<div id=\"fs-id1167793476464\" class=\"exercise\">\r\n\r\n[reveal-answer q=\"fs-id1167798676754\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"fs-id1167798676754\"]To find the effect of combining the two forces, add their representative vectors. First, express each vector in component form or in terms of the standard unit vectors. For this purpose, it is easiest if we align one of the vectors with the positive [latex]x[\/latex]-axis. The horizontal vector, then, has initial point [latex](0,0)[\/latex] and terminal point [latex](300,0)[\/latex]. It can be expressed as [latex]\\langle 300, 0\\rangle [\/latex] or [latex]300{\\bf{i}} [\/latex].The second vector has magnitude [latex]150[\/latex] and makes an angle of [latex]15\u00b0[\/latex] with the first, so we can express it as [latex]\\langle 150\\cos({15\u00b0}), 150\\sin({15\u00b0}) \\rangle [\/latex], or [latex]150\\cos({15\u00b0}){\\bf{i}} + 150\\sin({15\u00b0}){\\bf{j}}[\/latex]. Then, the sum of the vectors, or resultant vector, is [latex]{\\bf{r}} = \\langle 300,0 \\rangle + \\langle 150\\cos({15\u00b0}), 150\\sin({15\u00b0}) \\rangle[\/latex] , and we have\r\n\r\n<center>[latex]\\begin{array}{ccc}\\hfill {\\bf{r}} &amp; =\\hfill \\sqrt{(300 + 150\\cos({15\u00b0}))^2 + (150\\sin({15\u00b0}))^2} \\hfill \\\\ \\hfill &amp; \\approx 446.6 \\end{array}[\/latex]<\/center>The angle [latex]\u03b8[\/latex] made by [latex]{\\bf{r}}[\/latex] and the positive [latex]x[\/latex]-axis has [latex]\\tan{\u03b8} = \\frac{150\\sin({15\u00b0})}{(300 + 150\\cos({15\u00b0}))} \\approx 0.09[\/latex], so [latex]\u03b8 \\approx \\tan^{-1}{0.09} \\approx 5\u00b0[\/latex], which means the resultant force [latex]{\\bf{r}}[\/latex] has an angle of [latex]5\u00b0[\/latex] above the horizontal axis.\r\n<div class=\"exercise\">[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<div><\/div>\r\n<div id=\"fs-id1235696175960\" class=\"textbook exercises\">\r\n<h3>Example: Finding Resultant Velocity<\/h3>\r\nAn airplane flies due west at an airspeed of [latex]425[\/latex] mph. The wind is blowing from the northeast at [latex]40[\/latex] mph. What is the ground speed of the airplane? What is the bearing of the airplane?\r\n<div id=\"fs-id1163193476464\" class=\"exercise\">[reveal-answer q=\"fs-id2167798676754\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id2167798676754\"]Let\u2019s start by sketching the situation described (Figure 23).[caption id=\"attachment_740\" align=\"aligncenter\" width=\"487\"]<img class=\"size-full wp-image-740\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/24155011\/2-1-23.jpeg\" alt=\"This figure is the image of an airplane. Coming out of the front of the airplane are two vectors. The first vector is labeled \u201c425\u201d and the second vector is labeled \u201c40.\u201d The angle between the vectors is 45 degrees.\" width=\"487\" height=\"312\" \/> Figure 5.\u00a0Initially, the plane travels due west. The wind is from the northeast, so it is blowing to the southwest. The angle between the plane\u2019s course and the wind is [latex]45^{\\circ}[\/latex] (Figure not drawn to scale.)[\/caption]Set up a sketch so that the initial points of the vectors lie at the origin. Then, the plane\u2019s velocity vector is [latex]{\\bf{p}} = -425{\\bf{i}}[\/latex]. The vector describing the wind makes an angle of [latex]225\u00b0[\/latex] with the positive [latex]x[\/latex]-axis:<center>[latex]{\\bf{w}} = \\langle 40\\cos({225\u00b0}), 40\\sin({225\u00b0}) \\rangle = \\langle -\\frac{40}{\\sqrt{2}} - \\frac{40}{\\sqrt{2}} \\rangle = - \\frac{40}{\\sqrt{2}}{\\bf{i}} - \\frac{40}{\\sqrt{2}}{\\bf{j}} [\/latex].<\/center>When the airspeed and the wind act together on the plane, we can add their vectors to find the resultant force:\r\n\r\n<center>[latex]{\\bf{p}} + {\\bf{w}} = -425{\\bf{i}} + (- \\frac{40}{\\sqrt{2}}{\\bf{i}} - \\frac{40}{\\sqrt{2}}{\\bf{j}}) = (-425 - \\frac{40}{\\sqrt{2}}){\\bf{i}} - \\frac{40}{\\sqrt{2}}{\\bf{j}} [\/latex].<\/center>The magnitude of the resultant vector shows the effect of the wind on the ground speed of the airplane:\r\n\r\n<center>[latex]||{\\bf{p}} + {\\bf{w}}|| = \\sqrt{(-425 - \\frac{40}{\\sqrt{2}})^2 + (-\\frac{40}{\\sqrt{2}})^2} \\approx 454.17 [\/latex]mph<\/center>As a result of the wind, the plane is traveling at approximately 454 mph relative to the ground.To determine the bearing of the airplane, we want to find the direction of the vector [latex]{\\bf{p}} + {\\bf{w}}[\/latex]:\r\n\r\n<center>[latex]\\begin{array}{ccc}\\hfill \\tan{\u03b8} &amp; =\\hfill &amp; \\frac{-\\frac{40}{\\sqrt{2}}}{(-425 - \\frac{40}{\\sqrt{2}})} \\approx 0.06 \\hfill \\\\ \\hfill \u03b8 &amp; \\approx \\hfill &amp;3.57\u00b0 \\end{array}[\/latex]<\/center>The overall direction of the plane is [latex]3.57\u00b0[\/latex] south of west.\r\n<div class=\"exercise\">[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id11672654357091\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nAn airplane flies due north at an airspeed of [latex]550[\/latex] mph. The wind is blowing from the northwest at [latex]50[\/latex] mph. What is the ground speed of the airplane?\r\n<div id=\"fs-id1167895634237\" class=\"exercise\">[reveal-answer q=\"fs-id1968393921227\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1968393921227\"]\r\nApproximately [latex]516[\/latex] mph\r\n[\/hidden-answer]<\/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=7713417&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=qTfx88_PwnU&amp;video_target=tpm-plugin-68pwohq8-qTfx88_PwnU\" 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\/CP2.10_transcript.html\">transcript for \u201cCP 2.10\u201d here (opens in new window)<\/a><\/center>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Express a vector in terms of unit vectors.<\/li>\n<li>Give two examples of vector quantities.<\/li>\n<\/ul>\n<\/div>\n<h2>Unit Vectors<\/h2>\n<p>A <strong>unit vector<\/strong> is a vector with magnitude [latex]1[\/latex]. For any nonzero vector [latex]{\\bf{v}}[\/latex], we can use scalar multiplication to find a unit vector [latex]{\\bf{u}}[\/latex] that has the same direction as [latex]{\\bf{v}}[\/latex]. To do this, we multiply the vector by the reciprocal of its magnitude:<\/p>\n<div style=\"text-align: center;\">[latex]{\\bf{u}} = \\frac{1}{||{\\bf{v}}||}{\\bf{v}}[\/latex].<\/div>\n<p>Recall that when we defined scalar multiplication, we noted that [latex]||k{\\bf{v}}|| = |k| \\cdot ||{\\bf{v}}||[\/latex]. For [latex]{\\bf{u}} = \\frac{1}{||{\\bf{v}}||}{\\bf{v}}[\/latex], it follows that [latex]||{\\bf{u}}|| = \\frac{1}{||{\\bf{v}}||}(||{\\bf{v}}||) = 1[\/latex]. We say that [latex]{\\bf{u}}[\/latex] is the <em>unit vector in the direction of<\/em> [latex]{\\bf{v}}[\/latex] (Figure 18). The process of using scalar multiplication to find a unit vector with a given direction is called <strong>normalization<\/strong>.<\/p>\n<div style=\"text-align: center;\"><\/div>\n<div style=\"text-align: center;\">\n<div id=\"attachment_700\" style=\"width: 295px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-700\" class=\"size-full wp-image-700\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/17191802\/2-1-18.jpeg\" alt=\"This image has two figures. The first is a vector labeled \u201cv.\u201d The second figure is a vector in the same direction labeled \u201cu.\u201d This vector has a length of 1 unit.\" width=\"285\" height=\"97\" \/><\/p>\n<p id=\"caption-attachment-700\" class=\"wp-caption-text\">Figure 1.\u00a0The vector [latex]{\\bf{v}}[\/latex] and associated unit vector\u00a0[latex]{\\bf{u}}=\\frac{1}{||{\\bf{v}}||}{\\bf{v}}[\/latex]. In this case, [latex]||{\\bf{v}}||&gt;1[\/latex].<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165693975960\" class=\"textbook exercises\">\n<h3>Example: Finding a Unit Vector<\/h3>\n<p>Let [latex]{\\bf{v}} = \\langle 1,2 \\rangle[\/latex].<\/p>\n<ul>\n<li style=\"list-style-type: none;\">\n<ol style=\"list-style-type: lower-alpha;\">          <\/ol>\n<ol style=\"list-style-type: lower-alpha;\">               <\/ol>\n<\/li>\n<\/ul>\n<div id=\"fs-id1167793478764\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167794934754\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167794934754\" class=\"hidden-answer\" style=\"display: none\">\n<ul>\n<li style=\"list-style-type: none;\">\n<ul>\n<li style=\"list-style-type: none;\">\n<ol style=\"list-style-type: lower-alpha;\">               <\/ol>\n<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<div style=\"text-align: center;\">[latex]||{\\bf{v}}|| = \\sqrt{1^2 + 2^2} = \\sqrt{1 + 4} = \\sqrt{5}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\">[latex]{\\bf{u}} = \\frac{1}{||{\\bf{v}}||} {\\bf{v}} = \\frac{1}{\\sqrt{5}} \\langle 1,2 \\rangle = \\langle \\frac{1}{\\sqrt{5}}, \\frac{2}{\\sqrt{5}} \\rangle[\/latex].<\/div>\n<ul>\n<li style=\"list-style-type: none;\">\n<ul>\n<li style=\"list-style-type: none;\">\n<ol style=\"list-style-type: lower-alpha;\">                          <\/ol>\n<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<div style=\"text-align: center;\">[latex]{\\bf{w}} = 7{\\bf{u}} = 7 \\langle \\frac{1}{\\sqrt{5}}, \\frac{2}{\\sqrt{5}} \\rangle = \\langle \\frac{7}{\\sqrt{5}}, \\frac{14}{\\sqrt{5}} \\rangle.[\/latex]<\/div>\n<div class=\"exercise\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167233157091\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Let [latex]{\\bf{v}} = \\langle 9,2 \\rangle[\/latex]. Find a vector with magnitude [latex]5[\/latex] in the opposite direction as [latex]{\\bf{v}}[\/latex].<\/p>\n<div id=\"fs-id1167894444237\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1967793921284\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1967793921284\" class=\"hidden-answer\" style=\"display: none\">\n[latex]\\langle -\\frac{45}{\\sqrt{85}}, -\\frac{10}{\\sqrt{85}}\\rangle[\/latex]\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>We have seen how convenient it can be to write a vector in component form. Sometimes, though, it is more convenient to write a vector as a sum of a horizontal vector and a vertical vector. To make this easier, let\u2019s look at standard unit vectors. The <strong>standard unit vectors<\/strong> are the vectors [latex]{\\bf{i}} = \\langle 1,0 \\rangle[\/latex] and [latex]{\\bf{j}} = \\langle 0,1 \\rangle[\/latex] (Figure 2.18).<\/p>\n<div id=\"attachment_733\" style=\"width: 375px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-733\" class=\"wp-image-733 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/24154043\/2-1-19.jpeg\" alt=\"This figure has the x and y axes of a coordinate system in the first quadrant. On the x-axis there is a vector labeled \u201ci,\u201d which equals &lt;1,0&gt;. The second vector is on the y-axis and is labeled \u201cj\u201d which equals &lt;0,1&gt;.\" width=\"365\" height=\"202\" \/><\/p>\n<p id=\"caption-attachment-733\" class=\"wp-caption-text\">Figure 2. The standard unit vectors [latex]{\\bf{i}}[\/latex] and [latex]{\\bf{j}}[\/latex].<\/p>\n<\/div>\n<p>By applying the properties of vectors, it is possible to express any vector in terms of [latex]{\\bf{i}}[\/latex] and [latex]{\\bf{j}}[\/latex] in what we call a <em>linear combination<\/em>:<\/p>\n<div style=\"text-align: center;\">[latex]||{\\bf{v}}|| = \\langle x,y \\rangle = \\langle x,0 \\rangle + \\langle 0,y \\rangle = x\\langle 1,0 \\rangle + y\\langle 0,1 \\rangle = x{\\bf{i}} + y{\\bf{j}}[\/latex].<\/div>\n<div style=\"text-align: center;\"><\/div>\n<p>Thus, [latex]{\\bf{v}}[\/latex] is the sum of a horizontal vector with magnitude [latex]x[\/latex], and a vertical vector with magnitude [latex]y[\/latex], as in the following figure.<\/p>\n<div id=\"attachment_735\" style=\"width: 184px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-735\" class=\"wp-image-735 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/24154229\/2-1-20.jpeg\" alt=\"This figure is a right triangle. The horizontal side is labeled \u201cxi.\u201d The vertical side is labeled \u201cyj.\u201d The hypotenuse is a vector labeled \u201cv.\u201d\" width=\"174\" height=\"122\" \/><\/p>\n<p id=\"caption-attachment-735\" class=\"wp-caption-text\">Figure 3. The vector [latex]{\\bf{v}}[\/latex] is the sum of [latex]x{\\bf{i}}[\/latex] and [latex]y{\\bf{j}}[\/latex].<\/p>\n<\/div>\n<div id=\"fs-id1165696275960\" class=\"textbook exercises\">\n<h3>Example: Using Standard Unit Vectors<\/h3>\n<ul>\n<li style=\"list-style-type: none;\">\n<ol style=\"list-style-type: lower-alpha;\">               <\/ol>\n<ol style=\"list-style-type: lower-alpha;\">                       <\/ol>\n<\/li>\n<\/ul>\n<div id=\"fs-id1167793478764\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167798634754\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167798634754\" class=\"hidden-answer\" style=\"display: none\">\n<ul>\n<li style=\"list-style-type: none;\">\n<ul>\n<li style=\"list-style-type: none;\">\n<ol style=\"list-style-type: lower-alpha;\">                 <\/ol>\n<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<div style=\"text-align: center;\">[latex]||{\\bf{w}}|| = \\langle 3,-4 \\rangle = 3{\\bf{i}} - 4{\\bf{j}}[\/latex]<\/div>\n<ul>\n<li style=\"list-style-type: none;\">\n<ul>\n<li style=\"list-style-type: none;\">\n<ol style=\"list-style-type: lower-alpha;\">                        <\/ol>\n<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<div id=\"fs-id1167794064134\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {\\bf{u}} & =\\hfill & \\langle \\cos{60\u00b0}, \\sin{60\u00b0} \\rangle \\hfill \\\\ \\hfill & =\\hfill & \\langle \\frac{1}{2}, \\frac{\\sqrt{3}}{2} \\rangle \\hfill \\\\ \\hfill & =\\hfill & \\frac{1}{2}{\\bf{i}} - \\frac{\\sqrt{3}}{2}{\\bf{j}}\\end{array}[\/latex].<\/div>\n<div data-type=\"equation\" data-label=\"\"><\/div>\n<div id=\"attachment_736\" style=\"width: 352px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-736\" class=\"size-full wp-image-736\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/24154420\/2-1-21.jpeg\" alt=\"This figure is a unit circle. It is a circle centered at the origin. It has a vector with initial point at the origin and terminal point on the circle. The terminal point is labeled (cos(theta), sin(theta)). The length of the vector is 1 unit. There is also a right triangle formed with the vector as the hypotenuse. The horizontal side is labeled \u201ccos(theta)\u201d and the vertical side is labeled \u201csin(theta).\u201d\" width=\"342\" height=\"342\" \/><\/p>\n<p id=\"caption-attachment-736\" class=\"wp-caption-text\">Figure 4.\u00a0The terminal point of [latex]{\\bf{u}}[\/latex] lies on the unit circle [latex](\\cos{\\theta},\\sin{\\theta})[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id11672665157091\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Let [latex]{\\bf{a}} = \\langle 16,-11 \\rangle[\/latex] and let [latex]{\\bf{b}}[\/latex] be a unit vector that forms an angle of [latex]225\u00b0[\/latex] with the positive [latex]x[\/latex]-axis. Express [latex]{\\bf{a}}[\/latex] and [latex]{\\bf{b}}[\/latex] in terms of the standard unit vectors.<\/p>\n<div id=\"fs-id1167895644237\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1968393921284\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1968393921284\" class=\"hidden-answer\" style=\"display: none\">\n[latex]{\\bf{a}} = 16{\\bf{i}} - 11{\\bf{j}}[\/latex]<br \/>\n[latex]{\\bf{b}} = -\\frac{\\sqrt{2}}{2}{\\bf{i}} - \\frac{\\sqrt{2}}{2}{\\bf{j}}[\/latex]\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=7713416&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=azDVrpxgUko&amp;video_target=tpm-plugin-ihc7z8ph-azDVrpxgUko\" 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\/CP2.9_transcript.html\">transcript for \u201cCP 2.9\u201d here (opens in new window)<\/a><\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm199015\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=199015&theme=oea&iframe_resize_id=ohm199015&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Applications of Vectors<\/h2>\n<p>Because vectors have both direction and magnitude, they are valuable tools for solving problems involving such applications as motion and force. Recall the boat example and the quarterback example we described earlier. Here we look at two other examples in detail.<\/p>\n<div id=\"fs-id1165696175960\" class=\"textbook exercises\">\n<h3>Example: Finding Resultant Force<\/h3>\n<p>Jane\u2019s car is stuck in the mud. Lisa and Jed come along in a truck to help pull her out. They attach one end of a tow strap to the front of the car and the other end to the truck\u2019s trailer hitch, and the truck starts to pull. Meanwhile, Jane and Jed get behind the car and push. The truck generates a horizontal force of [latex]300[\/latex] lb on the car. Jane and Jed are pushing at a slight upward angle and generate a force of [latex]150[\/latex] lb on the car. These forces can be represented by vectors, as shown in Figure 22. The angle between these vectors is [latex]15\u00b0[\/latex]. Find the resultant force (the vector sum) and give its magnitude to the nearest tenth of a pound and its direction angle from the positive [latex]x[\/latex]-axis.<\/p>\n<div id=\"attachment_738\" style=\"width: 681px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-738\" class=\"wp-image-738 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/24154738\/2-1-22.jpeg\" alt=\"This image is the side view of an automobile. From the front of the automobile there is a horizontal vector labeled \u201c300 pounds.\u201d Also, from the front of the automobile there is another vector labeled \u201c150 pounds.\u201d The angle between the two vectors is 15 degrees.\" width=\"671\" height=\"167\" \/><\/p>\n<p id=\"caption-attachment-738\" class=\"wp-caption-text\">Figure 5. Two forces acting on a car in different directions.<\/p>\n<\/div>\n<div><\/div>\n<div id=\"fs-id1167793476464\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167798676754\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167798676754\" class=\"hidden-answer\" style=\"display: none\">To find the effect of combining the two forces, add their representative vectors. First, express each vector in component form or in terms of the standard unit vectors. For this purpose, it is easiest if we align one of the vectors with the positive [latex]x[\/latex]-axis. The horizontal vector, then, has initial point [latex](0,0)[\/latex] and terminal point [latex](300,0)[\/latex]. It can be expressed as [latex]\\langle 300, 0\\rangle[\/latex] or [latex]300{\\bf{i}}[\/latex].The second vector has magnitude [latex]150[\/latex] and makes an angle of [latex]15\u00b0[\/latex] with the first, so we can express it as [latex]\\langle 150\\cos({15\u00b0}), 150\\sin({15\u00b0}) \\rangle[\/latex], or [latex]150\\cos({15\u00b0}){\\bf{i}} + 150\\sin({15\u00b0}){\\bf{j}}[\/latex]. Then, the sum of the vectors, or resultant vector, is [latex]{\\bf{r}} = \\langle 300,0 \\rangle + \\langle 150\\cos({15\u00b0}), 150\\sin({15\u00b0}) \\rangle[\/latex] , and we have<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\bf{r}} & =\\hfill \\sqrt{(300 + 150\\cos({15\u00b0}))^2 + (150\\sin({15\u00b0}))^2} \\hfill \\\\ \\hfill & \\approx 446.6 \\end{array}[\/latex]<\/div>\n<p>The angle [latex]\u03b8[\/latex] made by [latex]{\\bf{r}}[\/latex] and the positive [latex]x[\/latex]-axis has [latex]\\tan{\u03b8} = \\frac{150\\sin({15\u00b0})}{(300 + 150\\cos({15\u00b0}))} \\approx 0.09[\/latex], so [latex]\u03b8 \\approx \\tan^{-1}{0.09} \\approx 5\u00b0[\/latex], which means the resultant force [latex]{\\bf{r}}[\/latex] has an angle of [latex]5\u00b0[\/latex] above the horizontal axis.<\/p>\n<div class=\"exercise\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div><\/div>\n<div id=\"fs-id1235696175960\" class=\"textbook exercises\">\n<h3>Example: Finding Resultant Velocity<\/h3>\n<p>An airplane flies due west at an airspeed of [latex]425[\/latex] mph. The wind is blowing from the northeast at [latex]40[\/latex] mph. What is the ground speed of the airplane? What is the bearing of the airplane?<\/p>\n<div id=\"fs-id1163193476464\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id2167798676754\">Show Solution<\/span><\/p>\n<div id=\"qfs-id2167798676754\" class=\"hidden-answer\" style=\"display: none\">Let\u2019s start by sketching the situation described (Figure 23).<\/p>\n<div id=\"attachment_740\" style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-740\" class=\"size-full wp-image-740\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/24155011\/2-1-23.jpeg\" alt=\"This figure is the image of an airplane. Coming out of the front of the airplane are two vectors. The first vector is labeled \u201c425\u201d and the second vector is labeled \u201c40.\u201d The angle between the vectors is 45 degrees.\" width=\"487\" height=\"312\" \/><\/p>\n<p id=\"caption-attachment-740\" class=\"wp-caption-text\">Figure 5.\u00a0Initially, the plane travels due west. The wind is from the northeast, so it is blowing to the southwest. The angle between the plane\u2019s course and the wind is [latex]45^{\\circ}[\/latex] (Figure not drawn to scale.)<\/p>\n<\/div>\n<p>Set up a sketch so that the initial points of the vectors lie at the origin. Then, the plane\u2019s velocity vector is [latex]{\\bf{p}} = -425{\\bf{i}}[\/latex]. The vector describing the wind makes an angle of [latex]225\u00b0[\/latex] with the positive [latex]x[\/latex]-axis:<\/p>\n<div style=\"text-align: center;\">[latex]{\\bf{w}} = \\langle 40\\cos({225\u00b0}), 40\\sin({225\u00b0}) \\rangle = \\langle -\\frac{40}{\\sqrt{2}} - \\frac{40}{\\sqrt{2}} \\rangle = - \\frac{40}{\\sqrt{2}}{\\bf{i}} - \\frac{40}{\\sqrt{2}}{\\bf{j}}[\/latex].<\/div>\n<p>When the airspeed and the wind act together on the plane, we can add their vectors to find the resultant force:<\/p>\n<div style=\"text-align: center;\">[latex]{\\bf{p}} + {\\bf{w}} = -425{\\bf{i}} + (- \\frac{40}{\\sqrt{2}}{\\bf{i}} - \\frac{40}{\\sqrt{2}}{\\bf{j}}) = (-425 - \\frac{40}{\\sqrt{2}}){\\bf{i}} - \\frac{40}{\\sqrt{2}}{\\bf{j}}[\/latex].<\/div>\n<p>The magnitude of the resultant vector shows the effect of the wind on the ground speed of the airplane:<\/p>\n<div style=\"text-align: center;\">[latex]||{\\bf{p}} + {\\bf{w}}|| = \\sqrt{(-425 - \\frac{40}{\\sqrt{2}})^2 + (-\\frac{40}{\\sqrt{2}})^2} \\approx 454.17[\/latex]mph<\/div>\n<p>As a result of the wind, the plane is traveling at approximately 454 mph relative to the ground.To determine the bearing of the airplane, we want to find the direction of the vector [latex]{\\bf{p}} + {\\bf{w}}[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\tan{\u03b8} & =\\hfill & \\frac{-\\frac{40}{\\sqrt{2}}}{(-425 - \\frac{40}{\\sqrt{2}})} \\approx 0.06 \\hfill \\\\ \\hfill \u03b8 & \\approx \\hfill &3.57\u00b0 \\end{array}[\/latex]<\/div>\n<p>The overall direction of the plane is [latex]3.57\u00b0[\/latex] south of west.<\/p>\n<div class=\"exercise\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id11672654357091\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>An airplane flies due north at an airspeed of [latex]550[\/latex] mph. The wind is blowing from the northwest at [latex]50[\/latex] mph. What is the ground speed of the airplane?<\/p>\n<div id=\"fs-id1167895634237\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1968393921227\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1968393921227\" class=\"hidden-answer\" style=\"display: none\">\nApproximately [latex]516[\/latex] mph\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=7713417&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=qTfx88_PwnU&amp;video_target=tpm-plugin-68pwohq8-qTfx88_PwnU\" 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\/CP2.10_transcript.html\">transcript for \u201cCP 2.10\u201d here (opens in new window)<\/a><\/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-397\">\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 2.9. <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><li>CP 2.10. <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":5,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 3\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"CP 2.9\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"CP 2.10\",\"author\":\"Ryan 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