{"id":3287,"date":"2025-08-16T01:48:41","date_gmt":"2025-08-16T01:48:41","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=3287"},"modified":"2025-10-21T21:41:00","modified_gmt":"2025-10-21T21:41:00","slug":"vectors-apply-it","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/vectors-apply-it\/","title":{"raw":"Vectors: Apply It","rendered":"Vectors: Apply It"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>View vectors geometrically.<\/li>\r\n \t<li>Find magnitude and direction.<\/li>\r\n \t<li>Find the component form of a vector.<\/li>\r\n \t<li>Find the unit vector in the direction of v.<\/li>\r\n<\/ul>\r\n<\/section>\r\n<div data-test-render-count=\"1\">\r\n<div class=\"group relative pb-3\" data-is-streaming=\"false\">\r\n<div class=\"font-claude-response relative leading-[1.65rem] [&amp;_pre&gt;div]:bg-bg-000\/50 [&amp;_pre&gt;div]:border-0.5 [&amp;_pre&gt;div]:border-border-400 [&amp;_.ignore-pre-bg&gt;div]:bg-transparent [&amp;_.standard-markdown_:is(p,blockquote,h1,h2,h3,h4,h5,h6)]:pl-2 [&amp;_.standard-markdown_:is(p,blockquote,ul,ol,h1,h2,h3,h4,h5,h6)]:pr-8 [&amp;_.progressive-markdown_:is(p,blockquote,h1,h2,h3,h4,h5,h6)]:pl-2 [&amp;_.progressive-markdown_:is(p,blockquote,ul,ol,h1,h2,h3,h4,h5,h6)]:pr-8\">\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0 standard-markdown\">\r\n<p class=\"whitespace-normal break-words\">Vectors are essential tools for describing quantities that have both magnitude and direction, such as velocity, force, and displacement.<\/p>\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<p class=\"whitespace-normal break-words\">A delivery truck travels from a warehouse at point [latex]W(-3, 2)[\/latex] to a store at point [latex]S(4, 5)[\/latex], where coordinates represent city blocks. Find the position vector representing this displacement and graph both the original vector and position vector.<\/p>\r\n<p class=\"whitespace-normal break-words\">[reveal-answer q=\"233559\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"233559\"]To find the position vector, we subtract the initial point coordinates from the terminal point coordinates: [latex]\\begin{aligned} \\mathbf{v} &amp;= \\langle 4 - (-3), 5 - 2 \\rangle \\ &amp;= \\langle 7, 3 \\rangle \\end{aligned}[\/latex] The position vector [latex]\\mathbf{v} = \\langle 7, 3 \\rangle[\/latex] starts at the origin [latex](0, 0)[\/latex] and ends at the point [latex](7, 3)[\/latex]. This tells us the truck travels 7 blocks east and 3 blocks north.[\/hidden-answer]<\/p>\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">\r\n<p class=\"whitespace-normal break-words\">A drone flies from point [latex]A(-2, 1)[\/latex] to point [latex]B(3, 5)[\/latex], where coordinates are in meters. Find the position vector representing the drone's displacement.<\/p>\r\n<p class=\"whitespace-normal break-words\">[latex]\\mathbf{v} = \\langle[\/latex] [response area] [latex],[\/latex] [response area] [latex]\\rangle[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Correct answer:<\/strong> [latex]\\langle 5, 4 \\rangle[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Feedback for correct answer:<\/strong> Perfect! You correctly calculated [latex]3 - (-2) = 5[\/latex] for the horizontal component and [latex]5 - 1 = 4[\/latex] for the vertical component. The drone's displacement is 5 meters east and 4 meters north.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Feedback for incorrect answer:<\/strong> Remember to subtract the initial point from the terminal point. The horizontal component is [latex]3 - (-2)[\/latex] and the vertical component is [latex]5 - 1[\/latex]. Be careful with the signs when subtracting negative numbers!<\/p>\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<p class=\"whitespace-normal break-words\">A search and rescue helicopter needs to fly from its base at point [latex]P(-8, 1)[\/latex] to a hiker's location at point [latex]Q(-2, -5)[\/latex], where coordinates are in kilometers. Find the magnitude and direction of the displacement vector.<\/p>\r\n[reveal-answer q=\"966819\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"966819\"]First, find the position vector: [latex]\\begin{aligned} \\mathbf{u} &amp;= \\langle -2 - (-8), -5 - 1 \\rangle \\ &amp;= \\langle 6, -6 \\rangle \\end{aligned}[\/latex] Next, calculate the magnitude using the Pythagorean Theorem: [latex]\\begin{aligned} |\\mathbf{u}| &amp;= \\sqrt{6^2 + (-6)^2} \\ &amp;= \\sqrt{36 + 36} \\ &amp;= \\sqrt{72} \\ &amp;= 6\\sqrt{2} \\ &amp;\\approx 8.49 \\text{ km} \\end{aligned}[\/latex] Now find the direction angle: [latex]\\begin{aligned} \\tan \\theta &amp;= \\frac{-6}{6} = -1 \\ \\theta &amp;= \\tan^{-1}(-1) = -45\u00b0 \\end{aligned}[\/latex] Since the vector [latex]\\langle 6, -6 \\rangle[\/latex] has a positive x-component and negative y-component, it terminates in the fourth quadrant. To express this as a positive angle measured counterclockwise from the positive x-axis: [latex]\\theta = -45\u00b0 + 360\u00b0 = 315\u00b0[\/latex] The helicopter needs to fly approximately [latex]8.49[\/latex] km at a bearing of [latex]315\u00b0[\/latex] (or equivalently, [latex]45\u00b0[\/latex] south of due east).[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">\r\n<p class=\"whitespace-normal break-words\">A sailboat travels from dock [latex]D(1, 4)[\/latex] to buoy [latex]B(7, 12)[\/latex], where coordinates are in nautical miles. Find the magnitude of the displacement vector (round to two decimal places).<\/p>\r\n<p class=\"whitespace-normal break-words\">[latex]|\\mathbf{v}| =[\/latex] [response area] nautical miles<\/p>\r\n[reveal-answer q=\"778611\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"778611\"]Correct answer: [latex]10.00[\/latex] (or [latex]10[\/latex]) Feedback for correct answer: Excellent! You found the position vector [latex]\\langle 6, 8 \\rangle[\/latex] and correctly calculated [latex]\\sqrt{6^2 + 8^2} = \\sqrt{36 + 64} = \\sqrt{100} = 10[\/latex]. This is actually a 3-4-5 right triangle scaled by 2! Feedback for incorrect answer: Start by finding the position vector: [latex]\\langle 7-1, 12-4 \\rangle = \\langle 6, 8 \\rangle[\/latex]. Then use the magnitude formula [latex]|\\mathbf{v}| = \\sqrt{6^2 + 8^2}[\/latex]. Don't forget to take the square root of your sum![\/hidden-answer]\r\n\r\n<\/section><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>View vectors geometrically.<\/li>\n<li>Find magnitude and direction.<\/li>\n<li>Find the component form of a vector.<\/li>\n<li>Find the unit vector in the direction of v.<\/li>\n<\/ul>\n<\/section>\n<div data-test-render-count=\"1\">\n<div class=\"group relative pb-3\" data-is-streaming=\"false\">\n<div class=\"font-claude-response relative leading-[1.65rem] [&amp;_pre&gt;div]:bg-bg-000\/50 [&amp;_pre&gt;div]:border-0.5 [&amp;_pre&gt;div]:border-border-400 [&amp;_.ignore-pre-bg&gt;div]:bg-transparent [&amp;_.standard-markdown_:is(p,blockquote,h1,h2,h3,h4,h5,h6)]:pl-2 [&amp;_.standard-markdown_:is(p,blockquote,ul,ol,h1,h2,h3,h4,h5,h6)]:pr-8 [&amp;_.progressive-markdown_:is(p,blockquote,h1,h2,h3,h4,h5,h6)]:pl-2 [&amp;_.progressive-markdown_:is(p,blockquote,ul,ol,h1,h2,h3,h4,h5,h6)]:pr-8\">\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0 standard-markdown\">\n<p class=\"whitespace-normal break-words\">Vectors are essential tools for describing quantities that have both magnitude and direction, such as velocity, force, and displacement.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-normal break-words\">A delivery truck travels from a warehouse at point [latex]W(-3, 2)[\/latex] to a store at point [latex]S(4, 5)[\/latex], where coordinates represent city blocks. Find the position vector representing this displacement and graph both the original vector and position vector.<\/p>\n<p class=\"whitespace-normal break-words\">\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q233559\">Show Solution<\/button><\/p>\n<div id=\"q233559\" class=\"hidden-answer\" style=\"display: none\">To find the position vector, we subtract the initial point coordinates from the terminal point coordinates: [latex]\\begin{aligned} \\mathbf{v} &= \\langle 4 - (-3), 5 - 2 \\rangle \\ &= \\langle 7, 3 \\rangle \\end{aligned}[\/latex] The position vector [latex]\\mathbf{v} = \\langle 7, 3 \\rangle[\/latex] starts at the origin [latex](0, 0)[\/latex] and ends at the point [latex](7, 3)[\/latex]. This tells us the truck travels 7 blocks east and 3 blocks north.<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\">\n<p class=\"whitespace-normal break-words\">A drone flies from point [latex]A(-2, 1)[\/latex] to point [latex]B(3, 5)[\/latex], where coordinates are in meters. Find the position vector representing the drone&#8217;s displacement.<\/p>\n<p class=\"whitespace-normal break-words\">[latex]\\mathbf{v} = \\langle[\/latex] [response area] [latex],[\/latex] [response area] [latex]\\rangle[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Correct answer:<\/strong> [latex]\\langle 5, 4 \\rangle[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Feedback for correct answer:<\/strong> Perfect! You correctly calculated [latex]3 - (-2) = 5[\/latex] for the horizontal component and [latex]5 - 1 = 4[\/latex] for the vertical component. The drone&#8217;s displacement is 5 meters east and 4 meters north.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Feedback for incorrect answer:<\/strong> Remember to subtract the initial point from the terminal point. The horizontal component is [latex]3 - (-2)[\/latex] and the vertical component is [latex]5 - 1[\/latex]. Be careful with the signs when subtracting negative numbers!<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-normal break-words\">A search and rescue helicopter needs to fly from its base at point [latex]P(-8, 1)[\/latex] to a hiker&#8217;s location at point [latex]Q(-2, -5)[\/latex], where coordinates are in kilometers. Find the magnitude and direction of the displacement vector.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q966819\">Show Solution<\/button><\/p>\n<div id=\"q966819\" class=\"hidden-answer\" style=\"display: none\">First, find the position vector: [latex]\\begin{aligned} \\mathbf{u} &= \\langle -2 - (-8), -5 - 1 \\rangle \\ &= \\langle 6, -6 \\rangle \\end{aligned}[\/latex] Next, calculate the magnitude using the Pythagorean Theorem: [latex]\\begin{aligned} |\\mathbf{u}| &= \\sqrt{6^2 + (-6)^2} \\ &= \\sqrt{36 + 36} \\ &= \\sqrt{72} \\ &= 6\\sqrt{2} \\ &\\approx 8.49 \\text{ km} \\end{aligned}[\/latex] Now find the direction angle: [latex]\\begin{aligned} \\tan \\theta &= \\frac{-6}{6} = -1 \\ \\theta &= \\tan^{-1}(-1) = -45\u00b0 \\end{aligned}[\/latex] Since the vector [latex]\\langle 6, -6 \\rangle[\/latex] has a positive x-component and negative y-component, it terminates in the fourth quadrant. To express this as a positive angle measured counterclockwise from the positive x-axis: [latex]\\theta = -45\u00b0 + 360\u00b0 = 315\u00b0[\/latex] The helicopter needs to fly approximately [latex]8.49[\/latex] km at a bearing of [latex]315\u00b0[\/latex] (or equivalently, [latex]45\u00b0[\/latex] south of due east).<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\">\n<p class=\"whitespace-normal break-words\">A sailboat travels from dock [latex]D(1, 4)[\/latex] to buoy [latex]B(7, 12)[\/latex], where coordinates are in nautical miles. Find the magnitude of the displacement vector (round to two decimal places).<\/p>\n<p class=\"whitespace-normal break-words\">[latex]|\\mathbf{v}| =[\/latex] [response area] nautical miles<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q778611\">Show Solution<\/button><\/p>\n<div id=\"q778611\" class=\"hidden-answer\" style=\"display: none\">Correct answer: [latex]10.00[\/latex] (or [latex]10[\/latex]) Feedback for correct answer: Excellent! You found the position vector [latex]\\langle 6, 8 \\rangle[\/latex] and correctly calculated [latex]\\sqrt{6^2 + 8^2} = \\sqrt{36 + 64} = \\sqrt{100} = 10[\/latex]. This is actually a 3-4-5 right triangle scaled by 2! Feedback for incorrect answer: Start by finding the position vector: [latex]\\langle 7-1, 12-4 \\rangle = \\langle 6, 8 \\rangle[\/latex]. Then use the magnitude formula [latex]|\\mathbf{v}| = \\sqrt{6^2 + 8^2}[\/latex]. Don&#8217;t forget to take the square root of your sum!<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"author":67,"menu_order":18,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":520,"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\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/3287"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/67"}],"version-history":[{"count":6,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/3287\/revisions"}],"predecessor-version":[{"id":4807,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/3287\/revisions\/4807"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/520"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/3287\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=3287"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=3287"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=3287"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=3287"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}