{"id":491,"date":"2025-10-09T15:14:52","date_gmt":"2025-10-09T15:14:52","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=491"},"modified":"2025-11-17T14:19:41","modified_gmt":"2025-11-17T14:19:41","slug":"vector-basics-apply-it","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/vector-basics-apply-it\/","title":{"raw":"Vector Basics: Apply It","rendered":"Vector Basics: Apply It"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Explain what vectors are and how to write them using proper notation, with real-world examples of vector quantities<\/li>\r\n \t<li>Perform basic vector operations (scalar multiplication, addition, subtraction)<\/li>\r\n \t<li>Write vectors using their components and find how long they are (their magnitude)<\/li>\r\n \t<li>Express a vector in terms of unit vectors<\/li>\r\n<\/ul>\r\n<\/section>\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\r\n<section class=\"textbox example\" aria-label=\"Example\">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 5. 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<\/section><section class=\"textbox example\" aria-label=\"Example\">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?\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 6).[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 6. Initially, 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:<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:<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 [latex]454[\/latex] 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]:<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<\/section><section class=\"textbox example\" aria-label=\"Example\">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?\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<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<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\" data-mce-fragment=\"1\"><\/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><\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Explain what vectors are and how to write them using proper notation, with real-world examples of vector quantities<\/li>\n<li>Perform basic vector operations (scalar multiplication, addition, subtraction)<\/li>\n<li>Write vectors using their components and find how long they are (their magnitude)<\/li>\n<li>Express a vector in terms of unit vectors<\/li>\n<\/ul>\n<\/section>\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<section class=\"textbox example\" aria-label=\"Example\">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 5. 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<figure id=\"attachment_738\" aria-describedby=\"caption-attachment-738\" style=\"width: 671px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" 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\" \/><figcaption id=\"caption-attachment-738\" class=\"wp-caption-text\">Figure 5. Two forces acting on a car in different directions.<\/figcaption><\/figure>\n<div><\/div>\n<div id=\"fs-id1167793476464\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167798676754\">Show Solution<\/button><\/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<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id2167798676754\">Show Solution<\/button><\/p>\n<div id=\"qfs-id2167798676754\" class=\"hidden-answer\" style=\"display: none\">Let\u2019s start by sketching the situation described (Figure 6).<\/p>\n<figure id=\"attachment_740\" aria-describedby=\"caption-attachment-740\" style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" 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\" \/><figcaption id=\"caption-attachment-740\" class=\"wp-caption-text\">Figure 6. Initially, 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.)<\/figcaption><\/figure>\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 [latex]454[\/latex] 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<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1968393921227\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1968393921227\" class=\"hidden-answer\" style=\"display: none\">\nApproximately [latex]516[\/latex] mph\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/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\" data-mce-fragment=\"1\"><\/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<\/section>\n","protected":false},"author":15,"menu_order":9,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-491","chapter","type-chapter","status-publish","hentry"],"part":420,"module-header":"apply_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/491","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":5,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/491\/revisions"}],"predecessor-version":[{"id":2273,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/491\/revisions\/2273"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/420"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/491\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=491"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=491"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=491"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=491"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}