{"id":5447,"date":"2022-06-02T18:23:59","date_gmt":"2022-06-02T18:23:59","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=5447"},"modified":"2022-10-21T00:13:52","modified_gmt":"2022-10-21T00:13:52","slug":"the-dot-product","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/the-dot-product\/","title":{"raw":"The Dot Product","rendered":"The Dot Product"},"content":{"raw":"<div data-type=\"note\">\r\n<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul class=\"os-abstract\">\r\n \t<li><span class=\"os-abstract-content\">Calculate the dot product of two given vectors.<\/span><\/li>\r\n \t<li><span class=\"os-abstract-content\"> Determine whether two given vectors are perpendicular.<\/span><\/li>\r\n \t<li><span class=\"os-abstract-content\"> Find the direction cosines of a given vector.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<section id=\"fs-id1163723170959\" data-depth=\"1\">\r\n<h2 data-type=\"title\">The Dot Product and Its Properties<\/h2>\r\n<p id=\"fs-id1163724052669\">We have already learned how to add and subtract vectors. In this chapter, we investigate two types of vector multiplication. The first type of vector multiplication is called the dot product, based on the notation we use for it, and it is defined as follows:<\/p>\r\n\r\n<\/section><\/div>\r\n<div data-type=\"note\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">definition<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1163723772734\">The\u00a0<strong><span id=\"7110f9df-6aa8-4c5b-adf0-ee8ed66ad26f_term70\" data-type=\"term\">dot product<\/span><\/strong>\u00a0of vectors [latex]\\mathbf{u} = \\langle u_1, u_2, u_3 \\rangle[\/latex] and [latex]\\mathbf{v} =\\langle v_1, v_2, v_3 \\rangle[\/latex] is given by the sum of the products of the components<\/p>\r\n\r\n<center>[latex]\\mathbf{u \\cdot v}[\/latex][latex] = u_1v_1 + u_2v_2 + u_3v_3 [\/latex].<\/center><\/div>\r\n<p id=\"fs-id1163724101618\">Note that if [latex]\\mathbf u[\/latex] and [latex]\\mathbf v[\/latex] are two-dimensional vectors, we calculate the dot product in a similar fashion. Thus, if [latex]\\mathbf{u} = \\langle u_1, u_2\\rangle[\/latex] and [latex]\\mathbf{v} =\\langle v_1, v_2 \\rangle[\/latex], then<\/p>\r\n\r\n<center>[latex]\\mathbf{u \\cdot v}[\/latex][latex] = u_1v_1 + u_2v_2 [\/latex].<\/center>When two vectors are combined under addition or subtraction, the result is a vector. When two vectors are combined using the dot product, the result is a scalar. For this reason, the dot product is often called the\u00a0<em data-effect=\"italics\">scalar product<\/em>. It may also be called the\u00a0<em data-effect=\"italics\">inner product<\/em>.\r\n<div id=\"fs-id1163724036465\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: calculating dot products<\/h3>\r\n<ol id=\"fs-id1163724076122\" type=\"a\">\r\n \t<li>Find the dot product of [latex]\\mathbf{u} = \\langle 3,5,2 \\rangle[\/latex] and\u00a0[latex]\\mathbf{v} =\\langle -1,3,0 \\rangle[\/latex].<\/li>\r\n \t<li>Find the scalar product of [latex]\\mathbf{p} = 10\\mathbf{i} - 4\\mathbf{j} +7 \\mathbf{k}[\/latex] and\u00a0[latex]\\mathbf{q} = -2\\mathbf{i} + \\mathbf{j} + 6\\mathbf{k}[\/latex].<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"471238733\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"471238733\"]\r\n<ol id=\"fs-id1163723802236\" type=\"a\">\r\n \t<li>Substitute the vector components into the formula for the dot product:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1163723138109\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\"><center>\r\n[latex]\r\n\\begin{align*}\r\n\\mathbf{u \\cdot v} &amp;= u_1v_1 + u_2v_2 + u_3v_3 \\\\\r\n&amp;= 3(-1)+5(3)+2(0)= -3+15+0 = 12.\\\\\r\n\\end{align*}\r\n[\/latex]<\/center><\/div><\/li>\r\n \t<li>The calculation is the same if the vectors are written using standard unit vectors. We still have three components for each vector to substitute into the formula for the dot product:\r\n<center>\r\n[latex]\r\n\\begin{align*}\r\n\\mathbf{p \\cdot q} &amp;= u_1v_1 + u_2v_2 + u_3v_3 \\\\\r\n&amp;= 10(-2)+(-4)(1)+7(6)= -20-4+42 = 18.\\\\\r\n\\end{align*}\r\n[\/latex]<\/center><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nFind [latex]\\mathbf{u \\cdot v}[\/latex], where [latex]\\mathbf{u} = \\langle 2,9,-1 \\rangle[\/latex] and [latex]\\mathbf{v} =\\langle -3,1,-4 \\rangle[\/latex].\r\n\r\n[reveal-answer q=\"847271036\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"847271036\"]\r\n\r\n<span style=\"font-size: 1rem; text-align: initial;\">[latex]7[\/latex]<\/span>\r\n\r\n<span style=\"font-size: 1rem; text-align: initial;\">[\/hidden-answer]<\/span>\r\n\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=7753520&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=cKRUC3Ohi9U&amp;video_target=tpm-plugin-y8g1rfrt-cKRUC3Ohi9U\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center><center>You can view the transcript for <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP2.21_transcript.html\">\u201cCP 2.21\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]26041[\/ohm_question]\r\n\r\n<\/div>\r\n<p id=\"fs-id1163723670274\">Like vector addition and subtraction, the dot product has several algebraic properties. We prove three of these properties and leave the rest as exercises.<\/p>\r\n\r\n<div id=\"fs-id1163724074875\" class=\"theorem ui-has-child-title\" data-type=\"note\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">theorem: properties of the dot product<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1163723853886\">Let [latex]\\textbf u[\/latex], [latex]\\textbf v[\/latex], and [latex]\\textbf w[\/latex] be vectors, and let [latex]c[\/latex] be a scalar.<\/p>\r\n\\[ \\begin{array}{lrl}\r\n\\mbox{i.} &amp; \\mathbf{u \\cdot v} &amp;= \\mathbf{v \\cdot u} &amp; \\mbox{Commutative property} \\\\\r\n\\mbox{ii.} &amp; \\mathbf{u} \\cdot (\\mathbf{v} + \\textbf{w}) &amp;= \\mathbf{u \\cdot v} + \\mathbf{u \\cdot w} &amp; \\mbox{Distributive property} \\\\\r\n\\mbox{iii.} &amp; c(\\mathbf{u \\cdot v}) &amp;= (c\\mathbf{u}) \\cdot \\mathbf{v} = \\mathbf{u} \\cdot (c\\mathbf{v}) &amp; \\mbox{Associative property} \\\\\r\n\\mbox{iv.} &amp; \\mathbf{v} \\cdot \\mathbf{v} &amp;= ||\\mathbf{v}||^2 &amp; \\mbox{Property of magnitude} \\\\\r\n\\end{array}\\]\r\n\r\n<\/div>\r\n<h3 data-type=\"title\">Proof<\/h3>\r\n<p id=\"fs-id1163723709365\">Let [latex]\\mathbf{u} = \\langle u_1, u_2, u_3 \\rangle[\/latex] and [latex]\\mathbf{v} =\\langle v_1, v_2, v_3 \\rangle[\/latex]. Then<\/p>\r\n\r\n<center>\r\n[latex]\r\n\\begin{align*}\r\n\\mathbf{u \\cdot v} &amp;= \\langle u_1, u_2, u_3 \\rangle \\cdot \\langle v_1, v_2, v_3 \\rangle\\\\\r\n&amp;= u_1v_1 + u_2v_2 + u_3v_3\\\\\r\n&amp;= v_1u_1 + v_2u_2 + v_3u_3\\\\\r\n&amp;= \\langle v_1, v_2, v_3 \\rangle \\cdot \\langle u_1, u_2, u_3 \\rangle\\\\\r\n&amp;= \\mathbf{v \\cdot u}.\\\\\r\n\\end{align*}\r\n[\/latex]<\/center>\r\n<p id=\"fs-id1163723953410\">The associative property looks like the associative property for real-number multiplication, but pay close attention to the difference between scalar and vector objects:<\/p>\r\n\r\n<center>\r\n[latex]\r\n\\begin{align*}\r\nc (\\mathbf{u \\cdot v}) &amp;= c(u_1v_1 + u_2v_2 + u_3v_3)\\\\\r\n&amp;= c(u_1v_1) + c(u_2v_2) + c(u_3v_3)\\\\\r\n&amp;= (cu_1)v_1 + (cu_2)v_2 + (cu_3)v_3\\\\\r\n&amp;= \\langle cu_1, cu_2, cu_3 \\rangle \\cdot \\langle v_1, v_2, v_3 \\rangle\\\\\r\n&amp;= c\\langle u_1, u_2,u_3 \\rangle \\cdot \\langle v_1, v_2, v_3 \\rangle\\\\\r\n&amp;= (c\\mathbf{u}) \\cdot \\mathbf{v}.\\\\\r\n\\end{align*}\r\n[\/latex]<\/center>\r\n<p id=\"fs-id1163723796994\">The proof that [latex]c (\\mathbf{u \\cdot v}) = \\mathbf{u}\\cdot (c\\mathbf{v})[\/latex] is similar.<\/p>\r\n<p id=\"fs-id1163723299158\">The fourth property shows the relationship between the magnitude of a vector and its dot product with itself:<\/p>\r\n\r\n<center>\r\n[latex]\r\n\\begin{align*}\r\n\\mathbf{v} \\cdot \\mathbf{v} &amp;= \\langle v_1, v_2, v_3 \\rangle \\cdot \\langle v_1, v_2, v_3 \\rangle\\\\\r\n&amp;= (v_1)^2 + (v_2)^2 + (v_3)^2\\\\\r\n&amp;= [\\sqrt{(v_1)^2 + (v_2)^2 + (v_3)^2}]^2\\\\\r\n&amp;= ||\\mathbf{v}||^2.\\\\\r\n\\end{align*}\r\n[\/latex]<\/center>[latex]_\\blacksquare[\/latex]\r\n<p id=\"fs-id1163724047817\">Note that the definition of the dot product yields [latex]\\mathbf{0} \\cdot \\mathbf{v} = \\mathbf{0}[\/latex]. By property iv., if [latex]\\mathbf{v} \\cdot \\mathbf{v} = 0[\/latex], then [latex]\\mathbf{v} = \\mathbf{0}[\/latex].<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example: using properties of the dot product<\/h3>\r\n<p id=\"fs-id1163724075617\">Let [latex]\\mathbf{a} = \\langle 1,2,-3 \\rangle[\/latex], [latex]\\mathbf{b} = \\langle 0,2,4 \\rangle[\/latex], and [latex]\\mathbf{c} = \\langle 5,-1,3 \\rangle[\/latex]. Find each of the following products.<\/p>\r\n\r\n<ol id=\"fs-id1163724078047\" type=\"a\">\r\n \t<li>[latex](\\mathbf{a} \\cdot \\mathbf{b})\\mathbf{c} [\/latex]<\/li>\r\n \t<li>[latex]\\mathbf{a}\\cdot (2\\mathbf{c} )[\/latex]<\/li>\r\n \t<li>[latex]||\\mathbf{b}||^2[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"235187363\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"235187363\"]\r\n<ol id=\"fs-id1163723139665\" type=\"a\">\r\n \t<li>Note that this expression asks for the scalar multiple of\u00a0 [latex]\\mathbf{c}[\/latex]\u00a0by [latex]\\mathbf{a} \\cdot \\mathbf{b}[\/latex]:<center>\r\n[latex]\r\n\\begin{align*}\r\n(\\mathbf{a} \\cdot \\mathbf{b})\\mathbf{c} &amp;= (\\langle 1,2,-3 \\rangle \\cdot \\langle 0,2,4 \\rangle) \\langle 5,-1,3 \\rangle\\\\\r\n&amp;= (1(0)+2(2)+(-3)(4))\\langle 5,-1,3 \\rangle\\\\\r\n&amp;= -8\\langle 5,-1,3 \\rangle\\\\\r\n&amp;= \\langle -40,8,-24 \\rangle.\\\\\r\n\\end{align*}\r\n[\/latex]<\/center><\/li>\r\n \t<li>This expression is a dot product of vector\u00a0[latex]\\mathbf{a}[\/latex]\u00a0and scalar multiple 2[latex]\\mathbf{c}[\/latex]:\r\n<center>\r\n[latex]\r\n\\begin{align*}\r\n\\mathbf{a}\\cdot (2\\mathbf{c} ) &amp;= 2(\\mathbf{a}\\cdot \\mathbf{c})\\\\\r\n&amp;= 2(\\langle 1,2,-3 \\rangle \\cdot\\langle 5,-1,3 \\rangle)\\\\\r\n&amp;= 2(1(5) +2(-1)+(-3)(3))\\\\\r\n&amp;= 2(-6) = -12.\\\\\r\n\\end{align*}\r\n[\/latex]<\/center><\/li>\r\n \t<li>Simplifying this expression is a straightforward application of the dot product:\r\n<center>[latex]||\\mathbf{b}||^2 = \\mathbf{b} \\cdot \\mathbf{b} = \\langle 0,2,4 \\rangle\\cdot\\langle 0,2,4 \\rangle = 0^2+2^2 +4^2 = 0+4+16 = 20 [\/latex].<\/center><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<p id=\"fs-id1163723289025\">Find the following products for [latex]\\mathbf{p} = \\langle 7,0,2 \\rangle[\/latex], [latex]\\mathbf{q} = \\langle -2,2,-2 \\rangle[\/latex], and\u00a0[latex]\\mathbf{r} = \\langle 0,2,-3 \\rangle[\/latex].<\/p>\r\n\r\n<ol id=\"fs-id1163723199327\" type=\"a\">\r\n \t<li>[latex](\\mathbf{r} \\cdot \\mathbf{p})\\mathbf{q}[\/latex]<\/li>\r\n \t<li>[latex]||\\mathbf{p}||^2[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"645176472\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"645176472\"]\r\n\r\na.\u00a0[latex](\\mathbf{r} \\cdot \\mathbf{p})\\mathbf{q} = \\langle 12,-12,12 \\rangle[\/latex];\r\nb.\u00a0[latex]||\\mathbf{p}||^2 = 53[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2 data-type=\"title\">Using the Dot Product to Find the Angle between Two Vectors<\/h2>\r\n<p id=\"fs-id1163723246791\">When two nonzero vectors are placed in standard position, whether in two dimensions or three dimensions, they form an angle between them (Figure 1). The dot product provides a way to find the measure of this angle. This property is a result of the fact that we can express the dot product in terms of the cosine of the angle formed by two vectors.<\/p>\r\n\r\n[caption id=\"attachment_5071\" align=\"aligncenter\" width=\"277\"]<img class=\"size-full wp-image-5071\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/26211719\/Figure-2.44.jpg\" alt=\"This figure is two vectors with the same initial point. The first vector is labeled \u201cu,\u201d and the second vector is labeled \u201cv.\u201d The angle between the two vectors is labeled \u201ctheta.\u201d\" width=\"277\" height=\"157\" \/> Figure 1. Let [latex]\u03b8[\/latex] be the angle between two nonzero vectors [latex]{\\bf{u}}[\/latex] and [latex]{\\bf{v}}[\/latex] such that [latex]0\u2264\u03b8\u2264\u03c0[\/latex].[\/caption]\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">theorem: evaluating a dot product<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1163724054271\">The dot product of two vectors is the product of the magnitude of each vector and the cosine of the angle between them:<\/p>\r\n\r\n<center>[latex]\\mathbf{u} \\cdot \\mathbf{v} = ||\\mathbf{u} ||\\;||\\mathbf{v} ||\\cos{\\theta}[\/latex].<\/center><\/div>\r\n<h3 data-type=\"title\">Proof<\/h3>\r\nPlace vectors [latex]\\mathbf{u}[\/latex] and [latex]\\mathbf{v}[\/latex] in standard position and consider the vector [latex]\\mathbf{v} - \\mathbf{u}[\/latex] (Figure 2). These three vectors form a triangle with side lengths [latex]||\\mathbf{u}||[\/latex],\u00a0[latex]||\\mathbf{v}||[\/latex], and\u00a0[latex]||\\mathbf{v} - \\mathbf{u}||[\/latex].\r\n\r\n[caption id=\"attachment_5073\" align=\"aligncenter\" width=\"277\"]<img class=\"size-full wp-image-5073\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/26212632\/Figure-2.45.jpg\" alt=\"This figure is two vectors with the same initial point. The first vector is labeled \u201cu,\u201d and the second vector is labeled \u201cv.\u201d The angle between the two vectors is labeled \u201ctheta.\u201d There is also a third vector from the terminal point of vector u to the terminal point of vector v. It is labeled \u201cv \u2013 u.\u201d\" width=\"277\" height=\"158\" \/> Figure 2. The lengths of the sides of the triangle are given by the magnitudes of the vectors that form the triangle.[\/caption]\r\n<p id=\"fs-id1163724054158\">cosines describes the relationship among the side lengths of the triangle and the angle [latex]\\theta[\/latex]. Applying the law of cosines here gives<\/p>\r\n\r\n<center>[latex]||\\mathbf{v} - \\mathbf{u}||^2 = ||\\mathbf{u}||^2 + ||\\mathbf{v}||^2 - 2||\\mathbf{u}||\\;||\\mathbf{v}||\\cos{\\theta}[\/latex].<\/center>\r\n<p id=\"fs-id1163724085758\">The dot product provides a way to rewrite the left side of this equation:<\/p>\r\n\r\n<center>\r\n[latex]\r\n\\begin{align*}\r\n||\\mathbf{v} - \\mathbf{u}||^2 &amp;= (\\mathbf{v} - \\mathbf{u}) \\cdot (\\mathbf{v} - \\mathbf{u})\\\\\r\n&amp;= (\\mathbf{v} - \\mathbf{u}) \\cdot \\mathbf{v}-(\\mathbf{v} - \\mathbf{u}) \\cdot \\mathbf{u}\\\\\r\n&amp;= \\mathbf{v} \\cdot \\mathbf{v} - \\mathbf{u} \\cdot \\mathbf{v} - \\mathbf{v} \\cdot \\mathbf{u}+\\mathbf{u} \\cdot \\mathbf{u}\\\\\r\n&amp;= \\mathbf{v} \\cdot \\mathbf{v} - \\mathbf{u} \\cdot \\mathbf{v} - \\mathbf{u} \\cdot \\mathbf{v}+\\mathbf{u} \\cdot \\mathbf{u}\\\\\r\n&amp;= ||\\mathbf{v}||^2 - 2 \\mathbf{u} \\cdot \\mathbf{v} + ||\\mathbf{u}||^2.\\\\\r\n\\end{align*}\r\n[\/latex]<\/center>\r\n<p id=\"fs-id1163724104874\">Substituting into the law of cosines yields<\/p>\r\n\r\n<center>[latex]\r\n\\begin{align*}\r\n||\\mathbf{v} - \\mathbf{u}||^2 &amp;= ||\\mathbf{u}||^2 + ||\\mathbf{v}||^2 - 2||\\mathbf{u}||\\;||\\mathbf{v}||\\cos{\\theta}\\\\\r\n||\\mathbf{v}||^2 - 2 \\mathbf{u} \\cdot \\mathbf{v} + ||\\mathbf{u}||^2 &amp;= ||\\mathbf{u}||^2 + ||\\mathbf{v}||^2 - 2||\\mathbf{u}||\\;||\\mathbf{v}||\\cos{\\theta}\\\\\r\n- 2 \\mathbf{u} \\cdot \\mathbf{v} &amp;= - 2||\\mathbf{u}||\\;||\\mathbf{v}||\\cos{\\theta}\\\\\r\n\\mathbf{u} \\cdot \\mathbf{v} &amp;= ||\\mathbf{u}||\\;||\\mathbf{v}||\\cos{\\theta}.\\\\\r\n\\end{align*}\r\n[\/latex]<\/center>[latex]_\\blacksquare[\/latex]\r\n<p id=\"fs-id1163723122497\">We can use this form of the dot product to find the measure of the angle between two nonzero vectors. The following equation rearranges\u00a0the Dot Product\u00a0to solve for the cosine of the angle:<\/p>\r\n\r\n<center>[latex]\\cos{\\theta} = \\frac{\\mathbf{u} \\cdot \\mathbf{v}}{||\\mathbf{u}||\\;||\\mathbf{v}||}[\/latex].<\/center>\r\n<p id=\"fs-id1163723122563\">Using this equation, we can find the cosine of the angle between two nonzero vectors. Since we are considering the smallest angle between the vectors, we assume [latex]0^\\circ \\leq \\theta \\leq 180^\\circ[\/latex] (or [latex]0 \\leq \\theta \\leq \\pi[\/latex] if we are working in radians). The inverse cosine is unique over this range, so we are then able to determine the measure of the angle [latex]\\theta[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1163724080388\" class=\"ui-has-child-title\" data-type=\"example\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: finding the angle between two vectors<\/h3>\r\n<p id=\"fs-id1163724080398\">Find the measure of the angle between each pair of vectors.<\/p>\r\n\r\n<ol id=\"fs-id1163724080401\" type=\"a\">\r\n \t<li>[latex]\\mathbf{i} + \\mathbf{j} + \\mathbf{k}[\/latex] and\u00a0[latex]2\\mathbf{i} - \\mathbf{j} - 3\\mathbf{k}[\/latex]<\/li>\r\n \t<li>[latex]\\langle 2,5,6 \\rangle [\/latex] and\u00a0[latex]\\langle -2,-4,4 \\rangle [\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"723893215\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"723893215\"]\r\n<ol id=\"fs-id1163724080485\" type=\"a\">\r\n \t<li>To find the cosine of the angle formed by the two vectors, substitute the components of the vectors into\u00a0the rearranged Dot Product:<center>\r\n[latex]\r\n\\begin{align*}\r\n\\cos{\\theta} &amp;= \\frac{(\\mathbf{i} + \\mathbf{j} + \\mathbf{k}) \\cdot (2\\mathbf{i} - \\mathbf{j} - 3\\mathbf{k})}{||\\mathbf{i} + \\mathbf{j} + \\mathbf{k}|| \\cdot ||2\\mathbf{i} - \\mathbf{j} - 3\\mathbf{k}||}\\\\\r\n&amp;= \\frac{1(2)+(1)(-1)+(1)(-3)}{\\sqrt{1^2+1^2+1^2}\\sqrt{2^2+(-1)^2+(-3)^2}}\\\\\r\n&amp;= \\frac{-2}{\\sqrt{3}\\sqrt{14}} = \\frac{-2}{\\sqrt{42}}.\\\\\\end{align*}\r\n[\/latex]<\/center>\r\nTherefore, [latex]{\\theta} = \\arccos{\\frac{-2}{\\sqrt{42}}}[\/latex] rad.<\/li>\r\n \t<li>Start by finding the value of the cosine of the angle between the vectors:<span data-type=\"newline\">\r\n<\/span><center>\r\n[latex]\r\n\\begin{align*}\r\n\\cos{\\theta} &amp;= \\frac{\\langle 2,5,6 \\rangle \\cdot \\langle -2,-4,4 \\rangle}{||\\langle 2,5,6 \\rangle|| \\cdot ||\\langle -2,-4,4 \\rangle||}\\\\\r\n&amp;= \\frac{2(-2)+(5)(-4)+(6)(4)}{\\sqrt{2^2+5^2+6^2}\\sqrt{(-2)^2+(-4)^2+4^2}}\\\\\r\n&amp;= \\frac{0}{\\sqrt{65}\\sqrt{36}} = 0.\\\\\\end{align*}\r\n[\/latex]<\/center>\r\nNow,[latex]\\cos{\\theta} = 0[\/latex] and\u00a0[latex]0 \\leq \\theta \\leq \\pi[\/latex], so\u00a0[latex]\\theta = \\frac{\\pi}{2}[\/latex].<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nFind the measure of the angle, in radians, formed by vectors [latex]\\mathbf{a} = \\langle 1,2,0 \\rangle [\/latex] and [latex]\\mathbf{b} = \\langle 2,4,1 \\rangle[\/latex]. Round to the nearest hundredth.\r\n\r\n[reveal-answer q=\"875123144\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"875123144\"]\r\n\r\n[latex]\\theta \\approx 0.22[\/latex] rad\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe angle between two vectors can be acute [latex](0 &lt; \\cos{\\theta} &lt; 1) [\/latex], obtuse [latex](-1 &lt; \\cos{\\theta} &lt; 0)[\/latex], or straight [latex](\\cos{\\theta} = -1)[\/latex]. If [latex]\\cos{\\theta} = 1[\/latex], then both vectors have the same direction. If [latex]\\cos{\\theta} = 0[\/latex], then the vectors, when placed in standard position, form a right angle (Figure 3). We can formalize this result into a theorem regarding orthogonal (perpendicular) vectors.\r\n\r\n[caption id=\"attachment_5075\" align=\"aligncenter\" width=\"671\"]<img class=\"size-full wp-image-5075\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/26215505\/Figure-2.46.jpg\" alt=\"This figure has 5 images. The first image has two vectors u and v. The angle between these two vectors is theta. Theta is an acute angle. The second image is has two vectors u and v. The angle between these vectors is theta. Theta is an obtuse angle. The third image is vectors u and v in opposite directions. The angle between u and v is a straight angle. The fourth image is u and v in the same direction. The fifth image is u and v with angle theta between them as a right angle.\" width=\"671\" height=\"524\" \/> Figure 3. (a) An acute angle has [latex]0&lt; \\cos{\\theta}&lt;1[\/latex]. (b) An obtuse angle has [latex]\u22121&lt;\\cos\\theta&lt;0[\/latex]. (c) A straight line has [latex]\\cos\\theta=\u22121[\/latex]. (d) If the vectors have the same direction, [latex]\\cos\\theta=1[\/latex]. (e) If the vectors are orthogonal (perpendicular), [latex]\\cos\\theta=0[\/latex][\/caption]\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">theorem: orthogonal vectors<\/h3>\r\nThe nonzero vectors [latex]\\mathbf{u}[\/latex] and [latex]\\mathbf{v}[\/latex] are\u00a0<strong>orthogonal vectors<\/strong>\u00a0if and only if [latex]\\mathbf{u} \\cdot \\mathbf{v} = 0[\/latex].\r\n\r\n<\/div>\r\n<h3 data-type=\"title\">Proof<\/h3>\r\n<p id=\"fs-id1163723295937\">Let [latex]\\mathbf{u}[\/latex] and [latex]\\mathbf{v}[\/latex] be nonzero vectors, and let [latex]\\theta[\/latex] denote the angle between them. First, assume [latex]\\mathbf{u} \\cdot \\mathbf{v} = 0[\/latex]. Then<\/p>\r\n\r\n<center>[latex]||\\mathbf{u}||\\;||\\mathbf{v}||\\cos{\\theta} = 0[\/latex].<\/center>\r\n<p id=\"fs-id1163723296024\">However, [latex]||\\mathbf{u}|| \\neq 0[\/latex] and [latex]||\\mathbf{v}|| \\neq 0[\/latex], so we must have [latex]\\cos{\\theta} = 0[\/latex]. Hence, [latex]\\theta = 90^\\circ[\/latex], and the vectors are orthogonal.<\/p>\r\n<p id=\"fs-id1163723193058\">Now assume [latex]\\mathbf{u}[\/latex] and [latex]\\mathbf{v}[\/latex] are orthogonal. Then [latex]\\theta = 90^\\circ[\/latex] and we have<\/p>\r\n\r\n<center>[latex]\\mathbf{u} \\cdot \\mathbf{v} = ||\\mathbf{u}||\\;||\\mathbf{v}||\\cos{\\theta} = ||\\mathbf{u}||\\;||\\mathbf{v}||\\cos{90^\\circ} = ||\\mathbf{u}||\\;||\\mathbf{v}||(0) = 0[\/latex].<\/center>[latex]_\\blacksquare[\/latex]\r\n\r\nThe terms\u00a0<span id=\"7110f9df-6aa8-4c5b-adf0-ee8ed66ad26f_term72\" class=\"no-emphasis\" data-type=\"term\"><em data-effect=\"italics\">orthogonal<\/em><\/span>,\u00a0<span id=\"7110f9df-6aa8-4c5b-adf0-ee8ed66ad26f_term73\" class=\"no-emphasis\" data-type=\"term\"><em data-effect=\"italics\">perpendicular<\/em><\/span>, and\u00a0<span id=\"7110f9df-6aa8-4c5b-adf0-ee8ed66ad26f_term74\" class=\"no-emphasis\" data-type=\"term\"><em data-effect=\"italics\">normal<\/em><\/span>\u00a0each indicate that mathematical objects are intersecting at right angles. The use of each term is determined mainly by its context. We say that vectors are orthogonal and lines are perpendicular. The term\u00a0<em data-effect=\"italics\">normal<\/em>\u00a0is used most often when measuring the angle made with a plane or other surface.\r\n<div id=\"fs-id1163723193086\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: identifying orthagonal vectors<\/h3>\r\nDetermine whether [latex]\\mathbf{p} = \\langle 1,0,5 \\rangle[\/latex] and [latex]\\mathbf{q} = \\langle 10,3,-2 \\rangle[\/latex] are orthogonal vectors.\r\n\r\n[reveal-answer q=\"934625143\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"934625143\"]\r\n<p id=\"fs-id1163723140417\">Using the definition, we need only check the dot product of the vectors:<\/p>\r\n\r\n<center>[latex]\\mathbf{p} \\cdot \\mathbf{q} = 1(10) +(0)(3) + (5)(-2) = 10+0-10 = 0[\/latex].<\/center>\r\n<p id=\"fs-id1163723140524\">Because [latex]\\mathbf{p} \\cdot \\mathbf{q} = 0[\/latex], the vectors are orthogonal (Figure 4).<\/p>\r\n\r\n[caption id=\"attachment_5076\" align=\"aligncenter\" width=\"500\"]<img class=\"size-full wp-image-5076\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/26215826\/Figure-2.47.jpg\" alt=\"This figure is the 3-dimensional coordinate system. There are two vectors in standard position. The vectors are labeled \u201cp\u201d and \u201cq.\u201d The angle between the vectors is a right angle.\" width=\"500\" height=\"465\" \/> Figure 4. Vectors [latex]{\\bf{p}}[\/latex] and [latex]{\\bf{q}}[\/latex] form a right angle when their initial points are aligned.[\/caption][\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nFor which value of [latex]x[\/latex] is [latex]\\mathbf{p} = \\langle 2,8,-1 \\rangle[\/latex] orthogonal to [latex]\\mathbf{q} = \\langle x,-1,2 \\rangle[\/latex]?\r\n\r\n[reveal-answer q=\"003746142\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"003746142\"]\r\n\r\n[latex]x=5[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: measuring the angle formed by two vectors<\/h3>\r\n<p id=\"fs-id1163723093514\">Let [latex]\\mathbf{v} = \\langle 2,3,3 \\rangle[\/latex]. Find the measures of the angles formed by the following vectors.<\/p>\r\n\r\n<ol id=\"fs-id1163723093549\" type=\"a\">\r\n \t<li>[latex]\\mathbf{v}[\/latex] and\u00a0[latex]\\mathbf{i}[\/latex]<\/li>\r\n \t<li>[latex]\\mathbf{v}[\/latex] and\u00a0[latex]\\mathbf{j}[\/latex]<\/li>\r\n \t<li>[latex]\\mathbf{v}[\/latex] and\u00a0[latex]\\mathbf{k}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"834690919\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"834690919\"]\r\n<ol id=\"fs-id1163723080236\" type=\"a\">\r\n \t<li>Let [latex]\\alpha[\/latex] be the angle formed by [latex]\\mathbf{v}[\/latex] and\u00a0[latex]\\mathbf{i}[\/latex]:\r\n<center>\r\n[latex]\r\n\\begin{align*}\r\n\\cos{\\alpha} &amp;= \\frac{\\mathbf{v} \\cdot \\mathbf{i} }{||\\mathbf{v}|| \\cdot ||\\mathbf{i}||}\\\\\r\n&amp;= \\frac{\\langle 2,3,3 \\rangle \\cdot \\langle 1,0,0 \\rangle}{\\sqrt{2^2+3^2+3^2}\\sqrt{1}}\\\\\r\n&amp;= \\frac{2}{\\sqrt{22}} =.\\\\\\end{align*}\r\n[\/latex]&nbsp;\r\n\r\n<\/center><center>\r\n[latex]\\alpha = \\arccos{\\frac{2}{\\sqrt{22}}} \\approx 1.130[\/latex] rad.<\/center><\/li>\r\n \t<li>Let [latex]\\beta[\/latex] represent the angle formed by [latex]\\mathbf{v}[\/latex] and\u00a0[latex]\\mathbf{j}[\/latex]:\r\n<center>\r\n[latex]\r\n\\begin{align*}\r\n\\cos{\\beta} &amp;= \\frac{\\mathbf{v} \\cdot \\mathbf{j} }{||\\mathbf{v}|| \\cdot ||\\mathbf{j}||}\\\\\r\n&amp;= \\frac{\\langle 2,3,3 \\rangle \\cdot \\langle 0,1,0 \\rangle}{\\sqrt{2^2+3^2+3^2}\\sqrt{1}}\\\\\r\n&amp;= \\frac{3}{\\sqrt{22}} =.\\\\\\end{align*}\r\n[\/latex]&nbsp;\r\n\r\n<\/center><center>\r\n[latex]\\beta = \\arccos{\\frac{3}{\\sqrt{22}}} \\approx 0.877[\/latex] rad.<\/center><\/li>\r\n \t<li>Let [latex]\\gamma[\/latex] represent the angle formed by [latex]\\mathbf{v}[\/latex] and\u00a0[latex]\\mathbf{k}[\/latex]:<center>\r\n[latex]\r\n\\begin{align*}\r\n\\cos{\\gamma} &amp;= \\frac{\\mathbf{v} \\cdot \\mathbf{k} }{||\\mathbf{v}|| \\cdot ||\\mathbf{k}||}\\\\\r\n&amp;= \\frac{\\langle 2,3,3 \\rangle \\cdot \\langle 0,0,1\\rangle}{\\sqrt{2^2+3^2+3^2}\\sqrt{1}}\\\\\r\n&amp;= \\frac{3}{\\sqrt{22}} .\\\\\\end{align*}\r\n[\/latex]&nbsp;\r\n\r\n<\/center><center>\r\n[latex]\\gamma = \\arccos{\\frac{3}{\\sqrt{22}}} \\approx 0.877[\/latex] rad.<\/center><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<p id=\"fs-id1163724081643\">Let [latex]\\mathbf{v} = \\langle 3,-5,1 \\rangle[\/latex]. Find the measure of the angles formed by each pair of vectors.<\/p>\r\n\r\n<ol id=\"fs-id1163724081678\" type=\"a\">\r\n \t<li>[latex]\\mathbf{v} [\/latex] and\u00a0[latex]\\mathbf{i} [\/latex]<\/li>\r\n \t<li>[latex]\\mathbf{v} [\/latex] and\u00a0[latex]\\mathbf{j} [\/latex]<\/li>\r\n \t<li>[latex]\\mathbf{v} [\/latex] and\u00a0[latex]\\mathbf{k} [\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"552341772\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"552341772\"]\r\n\r\na. [latex]\\alpha \\approx 1.04[\/latex] rad; b. [latex]\\beta \\approx 2.58[\/latex] rad; c. [latex]\\gamma \\approx 1.40[\/latex] rad\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"fs-id1163724049105\">The angle a vector makes with each of the coordinate axes, called a direction angle, is very important in practical computations, especially in a field such as engineering. For example, in astronautical engineering, the angle at which a rocket is launched must be determined very precisely. A very small error in the angle can lead to the rocket going hundreds of miles off course. Direction angles are often calculated by using the dot product and the cosines of the angles, called the direction cosines. Therefore, we define both these angles and their cosines.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"ui-has-child-title\" data-type=\"note\">\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"os-title\" style=\"text-align: center;\" data-type=\"title\"><span id=\"20\" class=\"os-title-label\" data-type=\"\">DEFINITION<\/span><\/h3>\r\n\r\n<hr \/>\r\n\r\nThe angles formed by a nonzero vector and the coordinate axes are called the\u00a0<strong>direction angles<\/strong>\u00a0for the vector (Figure 5). The cosines for these angles are called the\u00a0<strong>direction cosines.<\/strong>\r\n\r\n<\/div>\r\n[caption id=\"attachment_5118\" align=\"aligncenter\" width=\"373\"]<img class=\"size-full wp-image-5118\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27152927\/Figure-2.48.jpg\" alt=\"This figure is the first octant of the 3-dimensional coordinate system. It has the standard unit vectors drawn on axes x, y, and z. There is also a vector drawn in the first octant labeled \u201cv.\u201d The angle between the x-axis and v is labeled \u201calpha.\u201d The angle between the y-axis and vector v is labeled \u201cbeta.\u201d The angle between the z-axis and vector v is labeled \u201cgamma.\u201d\" width=\"373\" height=\"340\" \/> Figure 5. Angle [latex]\\alpha[\/latex] is formed by vector [latex]{\\bf{v}}[\/latex] and unit vector [latex]{\\bf{i}}[\/latex]. Angle [latex]\\beta[\/latex] is formed by vector [latex]{\\bf{v}}[\/latex] and unit vector [latex]{\\bf{j}}[\/latex]. Angle [latex]\\gamma[\/latex] is formed by vector [latex]{\\bf{v}}[\/latex] and unit vector [latex]{\\bf{k}}[\/latex].[\/caption]\r\n<p id=\"fs-id1163724049202\">In\u00a0Example: Measuring the Angle Formed by Two Vectors, the direction cosines of [latex]\\mathbf{v} = \\langle 2,3,3 \\rangle[\/latex] are [latex]\\cos{\\alpha} = \\frac{2}{\\sqrt{22}}[\/latex], [latex]\\cos{\\beta} = \\frac{3}{\\sqrt{22}}[\/latex], and [latex]\\cos{\\gamma} = \\frac{3}{\\sqrt{22}}[\/latex]. The direction angles of [latex]\\mathbf{v}[\/latex] are [latex]\\alpha = 1.130[\/latex] rad,\u00a0[latex]\\beta = 0.877[\/latex] rad, and\u00a0[latex]\\gamma = 0.877[\/latex] rad.<\/p>\r\n<p id=\"fs-id1163724059755\">So far, we have focused mainly on vectors related to force, movement, and position in three-dimensional physical space. However, vectors are often used in more abstract ways. For example, suppose a fruit vendor sells apples, bananas, and oranges. On a given day, he sells 30 apples, 12 bananas, and 18 oranges. He might use a quantity vector, [latex]\\mathbf{q} = \\langle 30,12,18 \\rangle[\/latex], to represent the quantity of fruit he sold that day. Similarly, he might want to use a price vector, [latex]\\mathbf{p} = \\langle 0.50,0.25,1 \\rangle[\/latex], to indicate that he sells his apples for 50\u00a2 each, bananas for 25\u00a2 each, and oranges for $1 apiece. In this example, although we could still graph these vectors, we do not interpret them as literal representations of position in the physical world. We are simply using vectors to keep track of particular pieces of information about apples, bananas, and oranges.<\/p>\r\n<p id=\"fs-id1163724036513\">This idea might seem a little strange, but if we simply regard vectors as a way to order and store data, we find they can be quite a powerful tool. Going back to the fruit vendor, let\u2019s think about the dot product, [latex]\\mathbf{q} \\cdot \\mathbf{p}[\/latex]. We compute it by multiplying the number of apples sold (30) by the price per apple (50\u00a2), the number of bananas sold by the price per banana, and the number of oranges sold by the price per orange. We then add all these values together. So, in this example, the dot product tells us how much money the fruit vendor had in sales on that particular day.<\/p>\r\n<p id=\"fs-id1163724036549\">When we use vectors in this more general way, there is no reason to limit the number of components to three. What if the fruit vendor decides to start selling grapefruit? In that case, he would want to use four-dimensional quantity and price vectors to represent the number of apples, bananas, oranges, and grapefruit sold, and their unit prices. As you might expect, to calculate the dot product of four-dimensional vectors, we simply add the products of the components as before, but the sum has four terms instead of three.<\/p>\r\n\r\n<\/div>\r\n<div data-type=\"note\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: using vectors in an economic context<\/h3>\r\n<p id=\"fs-id1163724036566\">AAA Party Supply Store sells invitations, party favors, decorations, and food service items such as paper plates and napkins. When AAA buys its inventory, it pays 25\u00a2 per package for invitations and party favors. Decorations cost AAA 50\u00a2 each, and food service items cost 20\u00a2 per package. AAA sells invitations for $2.50 per package and party favors for $1.50 per package. Decorations sell for $4.50 each and food service items for $1.25 per package.<\/p>\r\n<p id=\"fs-id1163724036578\">During the month of May, AAA Party Supply Store sells 1258 invitations, 342 party favors, 2426 decorations, and 1354 food service items. Use vectors and dot products to calculate how much money AAA made in sales during the month of May. How much did the store make in profit?<\/p>\r\n[reveal-answer q=\"734614322\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"734614322\"]\r\n<p id=\"fs-id1163724036587\">The cost, price, and quantity vectors are<\/p>\r\n\r\n<center>\r\n[latex]\r\n\\begin{align*}\r\n\\mathbf{c} &amp;= \\langle 0.25,0.25,0.50,0.20 \\rangle\\\\\r\n\\mathbf{p}&amp;= \\langle 2.50,1.50,4.50,1.25 \\rangle\\\\\r\n\\mathbf{q}&amp;= \\langle 1258,342,2426,1354 \\rangle.\\\\\\end{align*}\r\n[\/latex]<\/center>\r\n<p id=\"fs-id1163724119482\">AAA sales for the month of May can be calculated using the dot product [latex]\\mathbf{p} \\cdot \\mathbf{q}[\/latex]. We have<\/p>\r\n\r\n<center>\r\n[latex]\r\n\\begin{align*}\r\n\\mathbf{p} \\cdot \\mathbf{q} &amp;= \\langle 2.50,1.50,4.50,1.25 \\rangle \\cdot \\langle 1258,342,2426,1354 \\rangle\\\\\r\n&amp;= 3145+513+10917+1692.5\\\\\r\n&amp;= 16267.5.\\\\\\end{align*}\r\n[\/latex]<\/center>\r\n<p id=\"fs-id1163723283999\">So, AAA took in $16,267.50 during the month of May.<\/p>\r\n<p id=\"fs-id1163723284003\">To calculate the profit, we must first calculate how much AAA paid for the items sold. We use the dot product [latex]\\mathbf{c} \\cdot \\mathbf{q}[\/latex] to get<\/p>\r\n\r\n<center>\r\n[latex]\r\n\\begin{align*}\r\n\\mathbf{c} \\cdot \\mathbf{q} &amp;= \\langle 0.25,0.25,0.50,0.20 \\rangle \\cdot \\langle 1258,342,2426,1354 \\rangle\\\\\r\n&amp;= 314.5+85.5+1213+270.8\\\\\r\n&amp;= 1883.8.\\\\\\end{align*}\r\n[\/latex]<\/center>\r\n<p id=\"fs-id1163723326660\">So, AAA paid $1,883.80 for the items they sold. Their profit, then, is given by<\/p>\r\n\r\n<center>\r\n[latex]\r\n\\begin{align*}\r\n\\mathbf{p} \\cdot \\mathbf{q} - \\mathbf{c} \\cdot \\mathbf{q} &amp;= 16267.5-1883.8\\\\\r\n&amp;= 14383.7.\\\\\\end{align*}\r\n[\/latex]<\/center>\r\n<p id=\"fs-id1163723326731\">Therefore, AAA Party Supply Store made $14,383.70 in May.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nOn June 1, AAA Party Supply Store decided to increase the price they charge for party favors to $2 per package. They also changed suppliers for their invitations, and are now able to purchase invitations for only 10\u00a2 per package. All their other costs and prices remain the same. If AAA sells 1408 invitations, 147 party favors, 2112 decorations, and 1894 food service items in the month of June, use vectors and dot products to calculate their total sales and profit for June.\r\n\r\n[reveal-answer q=\"000034625\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"000034625\"]\r\n\r\nSales = [latex]$15,685.50[\/latex]; profit = [latex]$14,073.15[\/latex]\r\n\r\n[\/hidden-answer]\r\n\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=7753521&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=FzU_D0eXcYs&amp;video_target=tpm-plugin-9rztup8m-FzU_D0eXcYs\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center><center>You can view the transcript for <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP2.26_transcript.html\">\u201cCP 2.26\u201d here (opens in new window).<\/a><\/center><\/div>","rendered":"<div data-type=\"note\">\n<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul class=\"os-abstract\">\n<li><span class=\"os-abstract-content\">Calculate the dot product of two given vectors.<\/span><\/li>\n<li><span class=\"os-abstract-content\"> Determine whether two given vectors are perpendicular.<\/span><\/li>\n<li><span class=\"os-abstract-content\"> Find the direction cosines of a given vector.<\/span><\/li>\n<\/ul>\n<\/div>\n<section id=\"fs-id1163723170959\" data-depth=\"1\">\n<h2 data-type=\"title\">The Dot Product and Its Properties<\/h2>\n<p id=\"fs-id1163724052669\">We have already learned how to add and subtract vectors. In this chapter, we investigate two types of vector multiplication. The first type of vector multiplication is called the dot product, based on the notation we use for it, and it is defined as follows:<\/p>\n<\/section>\n<\/div>\n<div data-type=\"note\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">definition<\/h3>\n<hr \/>\n<p id=\"fs-id1163723772734\">The\u00a0<strong><span id=\"7110f9df-6aa8-4c5b-adf0-ee8ed66ad26f_term70\" data-type=\"term\">dot product<\/span><\/strong>\u00a0of vectors [latex]\\mathbf{u} = \\langle u_1, u_2, u_3 \\rangle[\/latex] and [latex]\\mathbf{v} =\\langle v_1, v_2, v_3 \\rangle[\/latex] is given by the sum of the products of the components<\/p>\n<div style=\"text-align: center;\">[latex]\\mathbf{u \\cdot v}[\/latex][latex]= u_1v_1 + u_2v_2 + u_3v_3[\/latex].<\/div>\n<\/div>\n<p id=\"fs-id1163724101618\">Note that if [latex]\\mathbf u[\/latex] and [latex]\\mathbf v[\/latex] are two-dimensional vectors, we calculate the dot product in a similar fashion. Thus, if [latex]\\mathbf{u} = \\langle u_1, u_2\\rangle[\/latex] and [latex]\\mathbf{v} =\\langle v_1, v_2 \\rangle[\/latex], then<\/p>\n<div style=\"text-align: center;\">[latex]\\mathbf{u \\cdot v}[\/latex][latex]= u_1v_1 + u_2v_2[\/latex].<\/div>\n<p>When two vectors are combined under addition or subtraction, the result is a vector. When two vectors are combined using the dot product, the result is a scalar. For this reason, the dot product is often called the\u00a0<em data-effect=\"italics\">scalar product<\/em>. It may also be called the\u00a0<em data-effect=\"italics\">inner product<\/em>.<\/p>\n<div id=\"fs-id1163724036465\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\n<div class=\"textbox exercises\">\n<h3>Example: calculating dot products<\/h3>\n<ol id=\"fs-id1163724076122\" type=\"a\">\n<li>Find the dot product of [latex]\\mathbf{u} = \\langle 3,5,2 \\rangle[\/latex] and\u00a0[latex]\\mathbf{v} =\\langle -1,3,0 \\rangle[\/latex].<\/li>\n<li>Find the scalar product of [latex]\\mathbf{p} = 10\\mathbf{i} - 4\\mathbf{j} +7 \\mathbf{k}[\/latex] and\u00a0[latex]\\mathbf{q} = -2\\mathbf{i} + \\mathbf{j} + 6\\mathbf{k}[\/latex].<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q471238733\">Show Solution<\/span><\/p>\n<div id=\"q471238733\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1163723802236\" type=\"a\">\n<li>Substitute the vector components into the formula for the dot product:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1163723138109\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  \\mathbf{u \\cdot v} &= u_1v_1 + u_2v_2 + u_3v_3 \\\\  &= 3(-1)+5(3)+2(0)= -3+15+0 = 12.\\\\  \\end{align*}[\/latex]<\/div>\n<\/div>\n<\/li>\n<li>The calculation is the same if the vectors are written using standard unit vectors. We still have three components for each vector to substitute into the formula for the dot product:\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  \\mathbf{p \\cdot q} &= u_1v_1 + u_2v_2 + u_3v_3 \\\\  &= 10(-2)+(-4)(1)+7(6)= -20-4+42 = 18.\\\\  \\end{align*}[\/latex]<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Find [latex]\\mathbf{u \\cdot v}[\/latex], where [latex]\\mathbf{u} = \\langle 2,9,-1 \\rangle[\/latex] and [latex]\\mathbf{v} =\\langle -3,1,-4 \\rangle[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q847271036\">Show Solution<\/span><\/p>\n<div id=\"q847271036\" class=\"hidden-answer\" style=\"display: none\">\n<p><span style=\"font-size: 1rem; text-align: initial;\">[latex]7[\/latex]<\/span><\/p>\n<p><span style=\"font-size: 1rem; text-align: initial;\"><\/div>\n<\/div>\n<p><\/span><\/p>\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=7753520&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=cKRUC3Ohi9U&amp;video_target=tpm-plugin-y8g1rfrt-cKRUC3Ohi9U\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the transcript for <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP2.21_transcript.html\">\u201cCP 2.21\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=\"ohm26041\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=26041&theme=oea&iframe_resize_id=ohm26041&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p id=\"fs-id1163723670274\">Like vector addition and subtraction, the dot product has several algebraic properties. We prove three of these properties and leave the rest as exercises.<\/p>\n<div id=\"fs-id1163724074875\" class=\"theorem ui-has-child-title\" data-type=\"note\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">theorem: properties of the dot product<\/h3>\n<hr \/>\n<p id=\"fs-id1163723853886\">Let [latex]\\textbf u[\/latex], [latex]\\textbf v[\/latex], and [latex]\\textbf w[\/latex] be vectors, and let [latex]c[\/latex] be a scalar.<\/p>\n<p>\\[ \\begin{array}{lrl}<br \/>\n\\mbox{i.} &amp; \\mathbf{u \\cdot v} &amp;= \\mathbf{v \\cdot u} &amp; \\mbox{Commutative property} \\\\<br \/>\n\\mbox{ii.} &amp; \\mathbf{u} \\cdot (\\mathbf{v} + \\textbf{w}) &amp;= \\mathbf{u \\cdot v} + \\mathbf{u \\cdot w} &amp; \\mbox{Distributive property} \\\\<br \/>\n\\mbox{iii.} &amp; c(\\mathbf{u \\cdot v}) &amp;= (c\\mathbf{u}) \\cdot \\mathbf{v} = \\mathbf{u} \\cdot (c\\mathbf{v}) &amp; \\mbox{Associative property} \\\\<br \/>\n\\mbox{iv.} &amp; \\mathbf{v} \\cdot \\mathbf{v} &amp;= ||\\mathbf{v}||^2 &amp; \\mbox{Property of magnitude} \\\\<br \/>\n\\end{array}\\]<\/p>\n<\/div>\n<h3 data-type=\"title\">Proof<\/h3>\n<p id=\"fs-id1163723709365\">Let [latex]\\mathbf{u} = \\langle u_1, u_2, u_3 \\rangle[\/latex] and [latex]\\mathbf{v} =\\langle v_1, v_2, v_3 \\rangle[\/latex]. Then<\/p>\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  \\mathbf{u \\cdot v} &= \\langle u_1, u_2, u_3 \\rangle \\cdot \\langle v_1, v_2, v_3 \\rangle\\\\  &= u_1v_1 + u_2v_2 + u_3v_3\\\\  &= v_1u_1 + v_2u_2 + v_3u_3\\\\  &= \\langle v_1, v_2, v_3 \\rangle \\cdot \\langle u_1, u_2, u_3 \\rangle\\\\  &= \\mathbf{v \\cdot u}.\\\\  \\end{align*}[\/latex]<\/div>\n<p id=\"fs-id1163723953410\">The associative property looks like the associative property for real-number multiplication, but pay close attention to the difference between scalar and vector objects:<\/p>\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  c (\\mathbf{u \\cdot v}) &= c(u_1v_1 + u_2v_2 + u_3v_3)\\\\  &= c(u_1v_1) + c(u_2v_2) + c(u_3v_3)\\\\  &= (cu_1)v_1 + (cu_2)v_2 + (cu_3)v_3\\\\  &= \\langle cu_1, cu_2, cu_3 \\rangle \\cdot \\langle v_1, v_2, v_3 \\rangle\\\\  &= c\\langle u_1, u_2,u_3 \\rangle \\cdot \\langle v_1, v_2, v_3 \\rangle\\\\  &= (c\\mathbf{u}) \\cdot \\mathbf{v}.\\\\  \\end{align*}[\/latex]<\/div>\n<p id=\"fs-id1163723796994\">The proof that [latex]c (\\mathbf{u \\cdot v}) = \\mathbf{u}\\cdot (c\\mathbf{v})[\/latex] is similar.<\/p>\n<p id=\"fs-id1163723299158\">The fourth property shows the relationship between the magnitude of a vector and its dot product with itself:<\/p>\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  \\mathbf{v} \\cdot \\mathbf{v} &= \\langle v_1, v_2, v_3 \\rangle \\cdot \\langle v_1, v_2, v_3 \\rangle\\\\  &= (v_1)^2 + (v_2)^2 + (v_3)^2\\\\  &= [\\sqrt{(v_1)^2 + (v_2)^2 + (v_3)^2}]^2\\\\  &= ||\\mathbf{v}||^2.\\\\  \\end{align*}[\/latex]<\/div>\n<p>[latex]_\\blacksquare[\/latex]<\/p>\n<p id=\"fs-id1163724047817\">Note that the definition of the dot product yields [latex]\\mathbf{0} \\cdot \\mathbf{v} = \\mathbf{0}[\/latex]. By property iv., if [latex]\\mathbf{v} \\cdot \\mathbf{v} = 0[\/latex], then [latex]\\mathbf{v} = \\mathbf{0}[\/latex].<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: using properties of the dot product<\/h3>\n<p id=\"fs-id1163724075617\">Let [latex]\\mathbf{a} = \\langle 1,2,-3 \\rangle[\/latex], [latex]\\mathbf{b} = \\langle 0,2,4 \\rangle[\/latex], and [latex]\\mathbf{c} = \\langle 5,-1,3 \\rangle[\/latex]. Find each of the following products.<\/p>\n<ol id=\"fs-id1163724078047\" type=\"a\">\n<li>[latex](\\mathbf{a} \\cdot \\mathbf{b})\\mathbf{c}[\/latex]<\/li>\n<li>[latex]\\mathbf{a}\\cdot (2\\mathbf{c} )[\/latex]<\/li>\n<li>[latex]||\\mathbf{b}||^2[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q235187363\">Show Solution<\/span><\/p>\n<div id=\"q235187363\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1163723139665\" type=\"a\">\n<li>Note that this expression asks for the scalar multiple of\u00a0 [latex]\\mathbf{c}[\/latex]\u00a0by [latex]\\mathbf{a} \\cdot \\mathbf{b}[\/latex]:\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  (\\mathbf{a} \\cdot \\mathbf{b})\\mathbf{c} &= (\\langle 1,2,-3 \\rangle \\cdot \\langle 0,2,4 \\rangle) \\langle 5,-1,3 \\rangle\\\\  &= (1(0)+2(2)+(-3)(4))\\langle 5,-1,3 \\rangle\\\\  &= -8\\langle 5,-1,3 \\rangle\\\\  &= \\langle -40,8,-24 \\rangle.\\\\  \\end{align*}[\/latex]<\/div>\n<\/li>\n<li>This expression is a dot product of vector\u00a0[latex]\\mathbf{a}[\/latex]\u00a0and scalar multiple 2[latex]\\mathbf{c}[\/latex]:\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  \\mathbf{a}\\cdot (2\\mathbf{c} ) &= 2(\\mathbf{a}\\cdot \\mathbf{c})\\\\  &= 2(\\langle 1,2,-3 \\rangle \\cdot\\langle 5,-1,3 \\rangle)\\\\  &= 2(1(5) +2(-1)+(-3)(3))\\\\  &= 2(-6) = -12.\\\\  \\end{align*}[\/latex]<\/div>\n<\/li>\n<li>Simplifying this expression is a straightforward application of the dot product:\n<div style=\"text-align: center;\">[latex]||\\mathbf{b}||^2 = \\mathbf{b} \\cdot \\mathbf{b} = \\langle 0,2,4 \\rangle\\cdot\\langle 0,2,4 \\rangle = 0^2+2^2 +4^2 = 0+4+16 = 20[\/latex].<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p id=\"fs-id1163723289025\">Find the following products for [latex]\\mathbf{p} = \\langle 7,0,2 \\rangle[\/latex], [latex]\\mathbf{q} = \\langle -2,2,-2 \\rangle[\/latex], and\u00a0[latex]\\mathbf{r} = \\langle 0,2,-3 \\rangle[\/latex].<\/p>\n<ol id=\"fs-id1163723199327\" type=\"a\">\n<li>[latex](\\mathbf{r} \\cdot \\mathbf{p})\\mathbf{q}[\/latex]<\/li>\n<li>[latex]||\\mathbf{p}||^2[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q645176472\">Show Solution<\/span><\/p>\n<div id=\"q645176472\" class=\"hidden-answer\" style=\"display: none\">\n<p>a.\u00a0[latex](\\mathbf{r} \\cdot \\mathbf{p})\\mathbf{q} = \\langle 12,-12,12 \\rangle[\/latex];<br \/>\nb.\u00a0[latex]||\\mathbf{p}||^2 = 53[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2 data-type=\"title\">Using the Dot Product to Find the Angle between Two Vectors<\/h2>\n<p id=\"fs-id1163723246791\">When two nonzero vectors are placed in standard position, whether in two dimensions or three dimensions, they form an angle between them (Figure 1). The dot product provides a way to find the measure of this angle. This property is a result of the fact that we can express the dot product in terms of the cosine of the angle formed by two vectors.<\/p>\n<div id=\"attachment_5071\" style=\"width: 287px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5071\" class=\"size-full wp-image-5071\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/26211719\/Figure-2.44.jpg\" alt=\"This figure is two vectors with the same initial point. The first vector is labeled \u201cu,\u201d and the second vector is labeled \u201cv.\u201d The angle between the two vectors is labeled \u201ctheta.\u201d\" width=\"277\" height=\"157\" \/><\/p>\n<p id=\"caption-attachment-5071\" class=\"wp-caption-text\">Figure 1. Let [latex]\u03b8[\/latex] be the angle between two nonzero vectors [latex]{\\bf{u}}[\/latex] and [latex]{\\bf{v}}[\/latex] such that [latex]0\u2264\u03b8\u2264\u03c0[\/latex].<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">theorem: evaluating a dot product<\/h3>\n<hr \/>\n<p id=\"fs-id1163724054271\">The dot product of two vectors is the product of the magnitude of each vector and the cosine of the angle between them:<\/p>\n<div style=\"text-align: center;\">[latex]\\mathbf{u} \\cdot \\mathbf{v} = ||\\mathbf{u} ||\\;||\\mathbf{v} ||\\cos{\\theta}[\/latex].<\/div>\n<\/div>\n<h3 data-type=\"title\">Proof<\/h3>\n<p>Place vectors [latex]\\mathbf{u}[\/latex] and [latex]\\mathbf{v}[\/latex] in standard position and consider the vector [latex]\\mathbf{v} - \\mathbf{u}[\/latex] (Figure 2). These three vectors form a triangle with side lengths [latex]||\\mathbf{u}||[\/latex],\u00a0[latex]||\\mathbf{v}||[\/latex], and\u00a0[latex]||\\mathbf{v} - \\mathbf{u}||[\/latex].<\/p>\n<div id=\"attachment_5073\" style=\"width: 287px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5073\" class=\"size-full wp-image-5073\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/26212632\/Figure-2.45.jpg\" alt=\"This figure is two vectors with the same initial point. The first vector is labeled \u201cu,\u201d and the second vector is labeled \u201cv.\u201d The angle between the two vectors is labeled \u201ctheta.\u201d There is also a third vector from the terminal point of vector u to the terminal point of vector v. It is labeled \u201cv \u2013 u.\u201d\" width=\"277\" height=\"158\" \/><\/p>\n<p id=\"caption-attachment-5073\" class=\"wp-caption-text\">Figure 2. The lengths of the sides of the triangle are given by the magnitudes of the vectors that form the triangle.<\/p>\n<\/div>\n<p id=\"fs-id1163724054158\">cosines describes the relationship among the side lengths of the triangle and the angle [latex]\\theta[\/latex]. Applying the law of cosines here gives<\/p>\n<div style=\"text-align: center;\">[latex]||\\mathbf{v} - \\mathbf{u}||^2 = ||\\mathbf{u}||^2 + ||\\mathbf{v}||^2 - 2||\\mathbf{u}||\\;||\\mathbf{v}||\\cos{\\theta}[\/latex].<\/div>\n<p id=\"fs-id1163724085758\">The dot product provides a way to rewrite the left side of this equation:<\/p>\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  ||\\mathbf{v} - \\mathbf{u}||^2 &= (\\mathbf{v} - \\mathbf{u}) \\cdot (\\mathbf{v} - \\mathbf{u})\\\\  &= (\\mathbf{v} - \\mathbf{u}) \\cdot \\mathbf{v}-(\\mathbf{v} - \\mathbf{u}) \\cdot \\mathbf{u}\\\\  &= \\mathbf{v} \\cdot \\mathbf{v} - \\mathbf{u} \\cdot \\mathbf{v} - \\mathbf{v} \\cdot \\mathbf{u}+\\mathbf{u} \\cdot \\mathbf{u}\\\\  &= \\mathbf{v} \\cdot \\mathbf{v} - \\mathbf{u} \\cdot \\mathbf{v} - \\mathbf{u} \\cdot \\mathbf{v}+\\mathbf{u} \\cdot \\mathbf{u}\\\\  &= ||\\mathbf{v}||^2 - 2 \\mathbf{u} \\cdot \\mathbf{v} + ||\\mathbf{u}||^2.\\\\  \\end{align*}[\/latex]<\/div>\n<p id=\"fs-id1163724104874\">Substituting into the law of cosines yields<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align*}  ||\\mathbf{v} - \\mathbf{u}||^2 &= ||\\mathbf{u}||^2 + ||\\mathbf{v}||^2 - 2||\\mathbf{u}||\\;||\\mathbf{v}||\\cos{\\theta}\\\\  ||\\mathbf{v}||^2 - 2 \\mathbf{u} \\cdot \\mathbf{v} + ||\\mathbf{u}||^2 &= ||\\mathbf{u}||^2 + ||\\mathbf{v}||^2 - 2||\\mathbf{u}||\\;||\\mathbf{v}||\\cos{\\theta}\\\\  - 2 \\mathbf{u} \\cdot \\mathbf{v} &= - 2||\\mathbf{u}||\\;||\\mathbf{v}||\\cos{\\theta}\\\\  \\mathbf{u} \\cdot \\mathbf{v} &= ||\\mathbf{u}||\\;||\\mathbf{v}||\\cos{\\theta}.\\\\  \\end{align*}[\/latex]<\/div>\n<p>[latex]_\\blacksquare[\/latex]<\/p>\n<p id=\"fs-id1163723122497\">We can use this form of the dot product to find the measure of the angle between two nonzero vectors. The following equation rearranges\u00a0the Dot Product\u00a0to solve for the cosine of the angle:<\/p>\n<div style=\"text-align: center;\">[latex]\\cos{\\theta} = \\frac{\\mathbf{u} \\cdot \\mathbf{v}}{||\\mathbf{u}||\\;||\\mathbf{v}||}[\/latex].<\/div>\n<p id=\"fs-id1163723122563\">Using this equation, we can find the cosine of the angle between two nonzero vectors. Since we are considering the smallest angle between the vectors, we assume [latex]0^\\circ \\leq \\theta \\leq 180^\\circ[\/latex] (or [latex]0 \\leq \\theta \\leq \\pi[\/latex] if we are working in radians). The inverse cosine is unique over this range, so we are then able to determine the measure of the angle [latex]\\theta[\/latex].<\/p>\n<div id=\"fs-id1163724080388\" class=\"ui-has-child-title\" data-type=\"example\">\n<div class=\"textbox exercises\">\n<h3>Example: finding the angle between two vectors<\/h3>\n<p id=\"fs-id1163724080398\">Find the measure of the angle between each pair of vectors.<\/p>\n<ol id=\"fs-id1163724080401\" type=\"a\">\n<li>[latex]\\mathbf{i} + \\mathbf{j} + \\mathbf{k}[\/latex] and\u00a0[latex]2\\mathbf{i} - \\mathbf{j} - 3\\mathbf{k}[\/latex]<\/li>\n<li>[latex]\\langle 2,5,6 \\rangle[\/latex] and\u00a0[latex]\\langle -2,-4,4 \\rangle[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q723893215\">Show Solution<\/span><\/p>\n<div id=\"q723893215\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1163724080485\" type=\"a\">\n<li>To find the cosine of the angle formed by the two vectors, substitute the components of the vectors into\u00a0the rearranged Dot Product:\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  \\cos{\\theta} &= \\frac{(\\mathbf{i} + \\mathbf{j} + \\mathbf{k}) \\cdot (2\\mathbf{i} - \\mathbf{j} - 3\\mathbf{k})}{||\\mathbf{i} + \\mathbf{j} + \\mathbf{k}|| \\cdot ||2\\mathbf{i} - \\mathbf{j} - 3\\mathbf{k}||}\\\\  &= \\frac{1(2)+(1)(-1)+(1)(-3)}{\\sqrt{1^2+1^2+1^2}\\sqrt{2^2+(-1)^2+(-3)^2}}\\\\  &= \\frac{-2}{\\sqrt{3}\\sqrt{14}} = \\frac{-2}{\\sqrt{42}}.\\\\\\end{align*}[\/latex]<\/div>\n<p>Therefore, [latex]{\\theta} = \\arccos{\\frac{-2}{\\sqrt{42}}}[\/latex] rad.<\/li>\n<li>Start by finding the value of the cosine of the angle between the vectors:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  \\cos{\\theta} &= \\frac{\\langle 2,5,6 \\rangle \\cdot \\langle -2,-4,4 \\rangle}{||\\langle 2,5,6 \\rangle|| \\cdot ||\\langle -2,-4,4 \\rangle||}\\\\  &= \\frac{2(-2)+(5)(-4)+(6)(4)}{\\sqrt{2^2+5^2+6^2}\\sqrt{(-2)^2+(-4)^2+4^2}}\\\\  &= \\frac{0}{\\sqrt{65}\\sqrt{36}} = 0.\\\\\\end{align*}[\/latex]<\/div>\n<p>Now,[latex]\\cos{\\theta} = 0[\/latex] and\u00a0[latex]0 \\leq \\theta \\leq \\pi[\/latex], so\u00a0[latex]\\theta = \\frac{\\pi}{2}[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Find the measure of the angle, in radians, formed by vectors [latex]\\mathbf{a} = \\langle 1,2,0 \\rangle[\/latex] and [latex]\\mathbf{b} = \\langle 2,4,1 \\rangle[\/latex]. Round to the nearest hundredth.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q875123144\">Show Solution<\/span><\/p>\n<div id=\"q875123144\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\theta \\approx 0.22[\/latex] rad<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The angle between two vectors can be acute [latex](0 < \\cos{\\theta} < 1)[\/latex], obtuse [latex](-1 < \\cos{\\theta} < 0)[\/latex], or straight [latex](\\cos{\\theta} = -1)[\/latex]. If [latex]\\cos{\\theta} = 1[\/latex], then both vectors have the same direction. If [latex]\\cos{\\theta} = 0[\/latex], then the vectors, when placed in standard position, form a right angle (Figure 3). We can formalize this result into a theorem regarding orthogonal (perpendicular) vectors.\n\n\n\n<div id=\"attachment_5075\" style=\"width: 681px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5075\" class=\"size-full wp-image-5075\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/26215505\/Figure-2.46.jpg\" alt=\"This figure has 5 images. The first image has two vectors u and v. The angle between these two vectors is theta. Theta is an acute angle. The second image is has two vectors u and v. The angle between these vectors is theta. Theta is an obtuse angle. The third image is vectors u and v in opposite directions. The angle between u and v is a straight angle. The fourth image is u and v in the same direction. The fifth image is u and v with angle theta between them as a right angle.\" width=\"671\" height=\"524\" \/><\/p>\n<p id=\"caption-attachment-5075\" class=\"wp-caption-text\">Figure 3. (a) An acute angle has [latex]0&lt; \\cos{\\theta}&lt;1[\/latex]. (b) An obtuse angle has [latex]\u22121&lt;\\cos\\theta&lt;0[\/latex]. (c) A straight line has [latex]\\cos\\theta=\u22121[\/latex]. (d) If the vectors have the same direction, [latex]\\cos\\theta=1[\/latex]. (e) If the vectors are orthogonal (perpendicular), [latex]\\cos\\theta=0[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">theorem: orthogonal vectors<\/h3>\n<p>The nonzero vectors [latex]\\mathbf{u}[\/latex] and [latex]\\mathbf{v}[\/latex] are\u00a0<strong>orthogonal vectors<\/strong>\u00a0if and only if [latex]\\mathbf{u} \\cdot \\mathbf{v} = 0[\/latex].<\/p>\n<\/div>\n<h3 data-type=\"title\">Proof<\/h3>\n<p id=\"fs-id1163723295937\">Let [latex]\\mathbf{u}[\/latex] and [latex]\\mathbf{v}[\/latex] be nonzero vectors, and let [latex]\\theta[\/latex] denote the angle between them. First, assume [latex]\\mathbf{u} \\cdot \\mathbf{v} = 0[\/latex]. Then<\/p>\n<div style=\"text-align: center;\">[latex]||\\mathbf{u}||\\;||\\mathbf{v}||\\cos{\\theta} = 0[\/latex].<\/div>\n<p id=\"fs-id1163723296024\">However, [latex]||\\mathbf{u}|| \\neq 0[\/latex] and [latex]||\\mathbf{v}|| \\neq 0[\/latex], so we must have [latex]\\cos{\\theta} = 0[\/latex]. Hence, [latex]\\theta = 90^\\circ[\/latex], and the vectors are orthogonal.<\/p>\n<p id=\"fs-id1163723193058\">Now assume [latex]\\mathbf{u}[\/latex] and [latex]\\mathbf{v}[\/latex] are orthogonal. Then [latex]\\theta = 90^\\circ[\/latex] and we have<\/p>\n<div style=\"text-align: center;\">[latex]\\mathbf{u} \\cdot \\mathbf{v} = ||\\mathbf{u}||\\;||\\mathbf{v}||\\cos{\\theta} = ||\\mathbf{u}||\\;||\\mathbf{v}||\\cos{90^\\circ} = ||\\mathbf{u}||\\;||\\mathbf{v}||(0) = 0[\/latex].<\/div>\n<p>[latex]_\\blacksquare[\/latex]<\/p>\n<p>The terms\u00a0<span id=\"7110f9df-6aa8-4c5b-adf0-ee8ed66ad26f_term72\" class=\"no-emphasis\" data-type=\"term\"><em data-effect=\"italics\">orthogonal<\/em><\/span>,\u00a0<span id=\"7110f9df-6aa8-4c5b-adf0-ee8ed66ad26f_term73\" class=\"no-emphasis\" data-type=\"term\"><em data-effect=\"italics\">perpendicular<\/em><\/span>, and\u00a0<span id=\"7110f9df-6aa8-4c5b-adf0-ee8ed66ad26f_term74\" class=\"no-emphasis\" data-type=\"term\"><em data-effect=\"italics\">normal<\/em><\/span>\u00a0each indicate that mathematical objects are intersecting at right angles. The use of each term is determined mainly by its context. We say that vectors are orthogonal and lines are perpendicular. The term\u00a0<em data-effect=\"italics\">normal<\/em>\u00a0is used most often when measuring the angle made with a plane or other surface.<\/p>\n<div id=\"fs-id1163723193086\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\n<div class=\"textbox exercises\">\n<h3>Example: identifying orthagonal vectors<\/h3>\n<p>Determine whether [latex]\\mathbf{p} = \\langle 1,0,5 \\rangle[\/latex] and [latex]\\mathbf{q} = \\langle 10,3,-2 \\rangle[\/latex] are orthogonal vectors.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q934625143\">Show Solution<\/span><\/p>\n<div id=\"q934625143\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1163723140417\">Using the definition, we need only check the dot product of the vectors:<\/p>\n<div style=\"text-align: center;\">[latex]\\mathbf{p} \\cdot \\mathbf{q} = 1(10) +(0)(3) + (5)(-2) = 10+0-10 = 0[\/latex].<\/div>\n<p id=\"fs-id1163723140524\">Because [latex]\\mathbf{p} \\cdot \\mathbf{q} = 0[\/latex], the vectors are orthogonal (Figure 4).<\/p>\n<div id=\"attachment_5076\" style=\"width: 510px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5076\" class=\"size-full wp-image-5076\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/26215826\/Figure-2.47.jpg\" alt=\"This figure is the 3-dimensional coordinate system. There are two vectors in standard position. The vectors are labeled \u201cp\u201d and \u201cq.\u201d The angle between the vectors is a right angle.\" width=\"500\" height=\"465\" \/><\/p>\n<p id=\"caption-attachment-5076\" class=\"wp-caption-text\">Figure 4. Vectors [latex]{\\bf{p}}[\/latex] and [latex]{\\bf{q}}[\/latex] form a right angle when their initial points are aligned.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>For which value of [latex]x[\/latex] is [latex]\\mathbf{p} = \\langle 2,8,-1 \\rangle[\/latex] orthogonal to [latex]\\mathbf{q} = \\langle x,-1,2 \\rangle[\/latex]?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q003746142\">Show Solution<\/span><\/p>\n<div id=\"q003746142\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=5[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: measuring the angle formed by two vectors<\/h3>\n<p id=\"fs-id1163723093514\">Let [latex]\\mathbf{v} = \\langle 2,3,3 \\rangle[\/latex]. Find the measures of the angles formed by the following vectors.<\/p>\n<ol id=\"fs-id1163723093549\" type=\"a\">\n<li>[latex]\\mathbf{v}[\/latex] and\u00a0[latex]\\mathbf{i}[\/latex]<\/li>\n<li>[latex]\\mathbf{v}[\/latex] and\u00a0[latex]\\mathbf{j}[\/latex]<\/li>\n<li>[latex]\\mathbf{v}[\/latex] and\u00a0[latex]\\mathbf{k}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q834690919\">Show Solution<\/span><\/p>\n<div id=\"q834690919\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1163723080236\" type=\"a\">\n<li>Let [latex]\\alpha[\/latex] be the angle formed by [latex]\\mathbf{v}[\/latex] and\u00a0[latex]\\mathbf{i}[\/latex]:\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  \\cos{\\alpha} &= \\frac{\\mathbf{v} \\cdot \\mathbf{i} }{||\\mathbf{v}|| \\cdot ||\\mathbf{i}||}\\\\  &= \\frac{\\langle 2,3,3 \\rangle \\cdot \\langle 1,0,0 \\rangle}{\\sqrt{2^2+3^2+3^2}\\sqrt{1}}\\\\  &= \\frac{2}{\\sqrt{22}} =.\\\\\\end{align*}[\/latex]&nbsp;<\/p>\n<\/div>\n<div style=\"text-align: center;\">\n[latex]\\alpha = \\arccos{\\frac{2}{\\sqrt{22}}} \\approx 1.130[\/latex] rad.<\/div>\n<\/li>\n<li>Let [latex]\\beta[\/latex] represent the angle formed by [latex]\\mathbf{v}[\/latex] and\u00a0[latex]\\mathbf{j}[\/latex]:\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  \\cos{\\beta} &= \\frac{\\mathbf{v} \\cdot \\mathbf{j} }{||\\mathbf{v}|| \\cdot ||\\mathbf{j}||}\\\\  &= \\frac{\\langle 2,3,3 \\rangle \\cdot \\langle 0,1,0 \\rangle}{\\sqrt{2^2+3^2+3^2}\\sqrt{1}}\\\\  &= \\frac{3}{\\sqrt{22}} =.\\\\\\end{align*}[\/latex]&nbsp;<\/p>\n<\/div>\n<div style=\"text-align: center;\">\n[latex]\\beta = \\arccos{\\frac{3}{\\sqrt{22}}} \\approx 0.877[\/latex] rad.<\/div>\n<\/li>\n<li>Let [latex]\\gamma[\/latex] represent the angle formed by [latex]\\mathbf{v}[\/latex] and\u00a0[latex]\\mathbf{k}[\/latex]:\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  \\cos{\\gamma} &= \\frac{\\mathbf{v} \\cdot \\mathbf{k} }{||\\mathbf{v}|| \\cdot ||\\mathbf{k}||}\\\\  &= \\frac{\\langle 2,3,3 \\rangle \\cdot \\langle 0,0,1\\rangle}{\\sqrt{2^2+3^2+3^2}\\sqrt{1}}\\\\  &= \\frac{3}{\\sqrt{22}} .\\\\\\end{align*}[\/latex]&nbsp;<\/p>\n<\/div>\n<div style=\"text-align: center;\">\n[latex]\\gamma = \\arccos{\\frac{3}{\\sqrt{22}}} \\approx 0.877[\/latex] rad.<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p id=\"fs-id1163724081643\">Let [latex]\\mathbf{v} = \\langle 3,-5,1 \\rangle[\/latex]. Find the measure of the angles formed by each pair of vectors.<\/p>\n<ol id=\"fs-id1163724081678\" type=\"a\">\n<li>[latex]\\mathbf{v}[\/latex] and\u00a0[latex]\\mathbf{i}[\/latex]<\/li>\n<li>[latex]\\mathbf{v}[\/latex] and\u00a0[latex]\\mathbf{j}[\/latex]<\/li>\n<li>[latex]\\mathbf{v}[\/latex] and\u00a0[latex]\\mathbf{k}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q552341772\">Show Solution<\/span><\/p>\n<div id=\"q552341772\" class=\"hidden-answer\" style=\"display: none\">\n<p>a. [latex]\\alpha \\approx 1.04[\/latex] rad; b. [latex]\\beta \\approx 2.58[\/latex] rad; c. [latex]\\gamma \\approx 1.40[\/latex] rad<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1163724049105\">The angle a vector makes with each of the coordinate axes, called a direction angle, is very important in practical computations, especially in a field such as engineering. For example, in astronautical engineering, the angle at which a rocket is launched must be determined very precisely. A very small error in the angle can lead to the rocket going hundreds of miles off course. Direction angles are often calculated by using the dot product and the cosines of the angles, called the direction cosines. Therefore, we define both these angles and their cosines.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"ui-has-child-title\" data-type=\"note\">\n<div class=\"textbox shaded\">\n<h3 class=\"os-title\" style=\"text-align: center;\" data-type=\"title\"><span id=\"20\" class=\"os-title-label\" data-type=\"\">DEFINITION<\/span><\/h3>\n<hr \/>\n<p>The angles formed by a nonzero vector and the coordinate axes are called the\u00a0<strong>direction angles<\/strong>\u00a0for the vector (Figure 5). The cosines for these angles are called the\u00a0<strong>direction cosines.<\/strong><\/p>\n<\/div>\n<div id=\"attachment_5118\" style=\"width: 383px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5118\" class=\"size-full wp-image-5118\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27152927\/Figure-2.48.jpg\" alt=\"This figure is the first octant of the 3-dimensional coordinate system. It has the standard unit vectors drawn on axes x, y, and z. There is also a vector drawn in the first octant labeled \u201cv.\u201d The angle between the x-axis and v is labeled \u201calpha.\u201d The angle between the y-axis and vector v is labeled \u201cbeta.\u201d The angle between the z-axis and vector v is labeled \u201cgamma.\u201d\" width=\"373\" height=\"340\" \/><\/p>\n<p id=\"caption-attachment-5118\" class=\"wp-caption-text\">Figure 5. Angle [latex]\\alpha[\/latex] is formed by vector [latex]{\\bf{v}}[\/latex] and unit vector [latex]{\\bf{i}}[\/latex]. Angle [latex]\\beta[\/latex] is formed by vector [latex]{\\bf{v}}[\/latex] and unit vector [latex]{\\bf{j}}[\/latex]. Angle [latex]\\gamma[\/latex] is formed by vector [latex]{\\bf{v}}[\/latex] and unit vector [latex]{\\bf{k}}[\/latex].<\/p>\n<\/div>\n<p id=\"fs-id1163724049202\">In\u00a0Example: Measuring the Angle Formed by Two Vectors, the direction cosines of [latex]\\mathbf{v} = \\langle 2,3,3 \\rangle[\/latex] are [latex]\\cos{\\alpha} = \\frac{2}{\\sqrt{22}}[\/latex], [latex]\\cos{\\beta} = \\frac{3}{\\sqrt{22}}[\/latex], and [latex]\\cos{\\gamma} = \\frac{3}{\\sqrt{22}}[\/latex]. The direction angles of [latex]\\mathbf{v}[\/latex] are [latex]\\alpha = 1.130[\/latex] rad,\u00a0[latex]\\beta = 0.877[\/latex] rad, and\u00a0[latex]\\gamma = 0.877[\/latex] rad.<\/p>\n<p id=\"fs-id1163724059755\">So far, we have focused mainly on vectors related to force, movement, and position in three-dimensional physical space. However, vectors are often used in more abstract ways. For example, suppose a fruit vendor sells apples, bananas, and oranges. On a given day, he sells 30 apples, 12 bananas, and 18 oranges. He might use a quantity vector, [latex]\\mathbf{q} = \\langle 30,12,18 \\rangle[\/latex], to represent the quantity of fruit he sold that day. Similarly, he might want to use a price vector, [latex]\\mathbf{p} = \\langle 0.50,0.25,1 \\rangle[\/latex], to indicate that he sells his apples for 50\u00a2 each, bananas for 25\u00a2 each, and oranges for $1 apiece. In this example, although we could still graph these vectors, we do not interpret them as literal representations of position in the physical world. We are simply using vectors to keep track of particular pieces of information about apples, bananas, and oranges.<\/p>\n<p id=\"fs-id1163724036513\">This idea might seem a little strange, but if we simply regard vectors as a way to order and store data, we find they can be quite a powerful tool. Going back to the fruit vendor, let\u2019s think about the dot product, [latex]\\mathbf{q} \\cdot \\mathbf{p}[\/latex]. We compute it by multiplying the number of apples sold (30) by the price per apple (50\u00a2), the number of bananas sold by the price per banana, and the number of oranges sold by the price per orange. We then add all these values together. So, in this example, the dot product tells us how much money the fruit vendor had in sales on that particular day.<\/p>\n<p id=\"fs-id1163724036549\">When we use vectors in this more general way, there is no reason to limit the number of components to three. What if the fruit vendor decides to start selling grapefruit? In that case, he would want to use four-dimensional quantity and price vectors to represent the number of apples, bananas, oranges, and grapefruit sold, and their unit prices. As you might expect, to calculate the dot product of four-dimensional vectors, we simply add the products of the components as before, but the sum has four terms instead of three.<\/p>\n<\/div>\n<div data-type=\"note\">\n<div class=\"textbox exercises\">\n<h3>Example: using vectors in an economic context<\/h3>\n<p id=\"fs-id1163724036566\">AAA Party Supply Store sells invitations, party favors, decorations, and food service items such as paper plates and napkins. When AAA buys its inventory, it pays 25\u00a2 per package for invitations and party favors. Decorations cost AAA 50\u00a2 each, and food service items cost 20\u00a2 per package. AAA sells invitations for $2.50 per package and party favors for $1.50 per package. Decorations sell for $4.50 each and food service items for $1.25 per package.<\/p>\n<p id=\"fs-id1163724036578\">During the month of May, AAA Party Supply Store sells 1258 invitations, 342 party favors, 2426 decorations, and 1354 food service items. Use vectors and dot products to calculate how much money AAA made in sales during the month of May. How much did the store make in profit?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q734614322\">Show Solution<\/span><\/p>\n<div id=\"q734614322\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1163724036587\">The cost, price, and quantity vectors are<\/p>\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  \\mathbf{c} &= \\langle 0.25,0.25,0.50,0.20 \\rangle\\\\  \\mathbf{p}&= \\langle 2.50,1.50,4.50,1.25 \\rangle\\\\  \\mathbf{q}&= \\langle 1258,342,2426,1354 \\rangle.\\\\\\end{align*}[\/latex]<\/div>\n<p id=\"fs-id1163724119482\">AAA sales for the month of May can be calculated using the dot product [latex]\\mathbf{p} \\cdot \\mathbf{q}[\/latex]. We have<\/p>\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  \\mathbf{p} \\cdot \\mathbf{q} &= \\langle 2.50,1.50,4.50,1.25 \\rangle \\cdot \\langle 1258,342,2426,1354 \\rangle\\\\  &= 3145+513+10917+1692.5\\\\  &= 16267.5.\\\\\\end{align*}[\/latex]<\/div>\n<p id=\"fs-id1163723283999\">So, AAA took in $16,267.50 during the month of May.<\/p>\n<p id=\"fs-id1163723284003\">To calculate the profit, we must first calculate how much AAA paid for the items sold. We use the dot product [latex]\\mathbf{c} \\cdot \\mathbf{q}[\/latex] to get<\/p>\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  \\mathbf{c} \\cdot \\mathbf{q} &= \\langle 0.25,0.25,0.50,0.20 \\rangle \\cdot \\langle 1258,342,2426,1354 \\rangle\\\\  &= 314.5+85.5+1213+270.8\\\\  &= 1883.8.\\\\\\end{align*}[\/latex]<\/div>\n<p id=\"fs-id1163723326660\">So, AAA paid $1,883.80 for the items they sold. Their profit, then, is given by<\/p>\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  \\mathbf{p} \\cdot \\mathbf{q} - \\mathbf{c} \\cdot \\mathbf{q} &= 16267.5-1883.8\\\\  &= 14383.7.\\\\\\end{align*}[\/latex]<\/div>\n<p id=\"fs-id1163723326731\">Therefore, AAA Party Supply Store made $14,383.70 in May.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>On June 1, AAA Party Supply Store decided to increase the price they charge for party favors to $2 per package. They also changed suppliers for their invitations, and are now able to purchase invitations for only 10\u00a2 per package. All their other costs and prices remain the same. If AAA sells 1408 invitations, 147 party favors, 2112 decorations, and 1894 food service items in the month of June, use vectors and dot products to calculate their total sales and profit for June.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q000034625\">Show Solution<\/span><\/p>\n<div id=\"q000034625\" class=\"hidden-answer\" style=\"display: none\">\n<p>Sales = [latex]$15,685.50[\/latex]; profit = [latex]$14,073.15[\/latex]<\/p>\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=7753521&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=FzU_D0eXcYs&amp;video_target=tpm-plugin-9rztup8m-FzU_D0eXcYs\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the transcript for <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP2.26_transcript.html\">\u201cCP 2.26\u201d here (opens in new window).<\/a><\/div>\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-5447\">\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 2.21. <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 2.26. <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 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