{"id":3291,"date":"2025-08-16T01:49:33","date_gmt":"2025-08-16T01:49:33","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=3291"},"modified":"2026-06-11T19:48:24","modified_gmt":"2026-06-11T19:48:24","slug":"operations-with-vectors-apply-it","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/operations-with-vectors-apply-it\/","title":{"raw":"Operations with Vectors: Apply It","rendered":"Operations with Vectors: Apply It"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Perform vector addition and scalar multiplication.<\/li>\r\n \t<li>Perform operations with vectors in terms of i and j .<\/li>\r\n \t<li>Find the dot product of two vectors.<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Vectors and Bearing<\/h2>\r\n<section class=\"textbox recall\" aria-label=\"Recall\">\r\n<p class=\"whitespace-normal break-words\">When adding vectors, we add corresponding components. The sum [latex]\\mathbf{u} + \\mathbf{v} = \\langle u_1, u_2 \\rangle + \\langle v_1, v_2 \\rangle = \\langle u_1 + v_1, u_2 + v_2 \\rangle[\/latex] represents the combined effect of both vectors.<\/p>\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">An airplane is flying at an airspeed of 200 miles per hour headed on a SE bearing of 140\u00b0. A north wind (from north to south) is blowing at 16.2 miles per hour. What are the ground speed and actual bearing of the plane?<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181212\/CNX_Precalc_Figure_08_08_0152.jpg\" alt=\"Image of a plan flying SE at 140 degrees and the north wind blowing.\" width=\"487\" height=\"462\" \/>[reveal-answer q=\"612904\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"612904\"]The ground speed is represented by [latex]x[\/latex] in the diagram, and we need to find the angle [latex]\\alpha [\/latex] in order to calculate the adjusted bearing, which will be [latex]140^\\circ +\\alpha [\/latex].\r\n\r\nNotice in Figure 19, that angle [latex]BCO[\/latex] must be equal to angle [latex]AOC[\/latex] by the rule of alternating interior angles, so angle [latex]BCO[\/latex] is 140\u00b0. We can find [latex]x[\/latex] by the Law of Cosines:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}{x}^{2}&amp;={\\left(16.2\\right)}^{2}+{\\left(200\\right)}^{2}-2\\left(16.2\\right)\\left(200\\right)\\cos \\left(140^\\circ \\right) \\\\ &amp;=45,226.41 \\\\ x&amp;=\\sqrt{45,226.41} \\\\ &amp;=212.7 \\end{align}[\/latex]<\/p>\r\nThe ground speed is approximately 213 miles per hour. Now we can calculate the bearing using the Law of Sines.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{\\sin \\alpha }{16.2}&amp;=\\frac{\\sin \\left(140^\\circ \\right)}{212.7} \\\\ \\sin \\alpha &amp;=\\frac{16.2\\sin \\left(140^\\circ \\right)}{212.7} \\\\ &amp;=0.04896 \\\\ {\\sin }^{-1}&amp;\\left(0.04896\\right)=2.8^\\circ \\end{align}[\/latex]<\/p>\r\nTherefore, the plane has a SE bearing of 140\u00b0+2.8\u00b0=142.8\u00b0. The ground speed is 212.7 miles per hour.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox proTip\" aria-label=\"Pro Tip\">In navigation, bearings are measured clockwise from north. To convert to standard position (counterclockwise from east), use the relationship: east component = [latex]|\\mathbf{v}|\\sin(\\text{bearing})[\/latex] and north component = [latex]|\\mathbf{v}|\\cos(\\text{bearing})[\/latex].<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]326263[\/ohm_question]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]326264[\/ohm_question]<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Perform vector addition and scalar multiplication.<\/li>\n<li>Perform operations with vectors in terms of i and j .<\/li>\n<li>Find the dot product of two vectors.<\/li>\n<\/ul>\n<\/section>\n<h2>Vectors and Bearing<\/h2>\n<section class=\"textbox recall\" aria-label=\"Recall\">\n<p class=\"whitespace-normal break-words\">When adding vectors, we add corresponding components. The sum [latex]\\mathbf{u} + \\mathbf{v} = \\langle u_1, u_2 \\rangle + \\langle v_1, v_2 \\rangle = \\langle u_1 + v_1, u_2 + v_2 \\rangle[\/latex] represents the combined effect of both vectors.<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">An airplane is flying at an airspeed of 200 miles per hour headed on a SE bearing of 140\u00b0. A north wind (from north to south) is blowing at 16.2 miles per hour. What are the ground speed and actual bearing of the plane?<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181212\/CNX_Precalc_Figure_08_08_0152.jpg\" alt=\"Image of a plan flying SE at 140 degrees and the north wind blowing.\" width=\"487\" height=\"462\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q612904\">Show Solution<\/button><\/p>\n<div id=\"q612904\" class=\"hidden-answer\" style=\"display: none\">The ground speed is represented by [latex]x[\/latex] in the diagram, and we need to find the angle [latex]\\alpha[\/latex] in order to calculate the adjusted bearing, which will be [latex]140^\\circ +\\alpha[\/latex].<\/p>\n<p>Notice in Figure 19, that angle [latex]BCO[\/latex] must be equal to angle [latex]AOC[\/latex] by the rule of alternating interior angles, so angle [latex]BCO[\/latex] is 140\u00b0. We can find [latex]x[\/latex] by the Law of Cosines:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}{x}^{2}&={\\left(16.2\\right)}^{2}+{\\left(200\\right)}^{2}-2\\left(16.2\\right)\\left(200\\right)\\cos \\left(140^\\circ \\right) \\\\ &=45,226.41 \\\\ x&=\\sqrt{45,226.41} \\\\ &=212.7 \\end{align}[\/latex]<\/p>\n<p>The ground speed is approximately 213 miles per hour. Now we can calculate the bearing using the Law of Sines.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{\\sin \\alpha }{16.2}&=\\frac{\\sin \\left(140^\\circ \\right)}{212.7} \\\\ \\sin \\alpha &=\\frac{16.2\\sin \\left(140^\\circ \\right)}{212.7} \\\\ &=0.04896 \\\\ {\\sin }^{-1}&\\left(0.04896\\right)=2.8^\\circ \\end{align}[\/latex]<\/p>\n<p>Therefore, the plane has a SE bearing of 140\u00b0+2.8\u00b0=142.8\u00b0. The ground speed is 212.7 miles per hour.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">In navigation, bearings are measured clockwise from north. To convert to standard position (counterclockwise from east), use the relationship: east component = [latex]|\\mathbf{v}|\\sin(\\text{bearing})[\/latex] and north component = [latex]|\\mathbf{v}|\\cos(\\text{bearing})[\/latex].<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm326263\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=326263&theme=lumen&iframe_resize_id=ohm326263&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm326264\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=326264&theme=lumen&iframe_resize_id=ohm326264&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":67,"menu_order":23,"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\/3291"}],"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\/3291\/revisions"}],"predecessor-version":[{"id":6250,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/3291\/revisions\/6250"}],"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\/3291\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=3291"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=3291"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=3291"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=3291"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}