{"id":505,"date":"2025-10-09T15:16:02","date_gmt":"2025-10-09T15:16:02","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=505"},"modified":"2025-10-24T12:51:11","modified_gmt":"2025-10-24T12:51:11","slug":"vector-operations-in-space-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/vector-operations-in-space-fresh-take\/","title":{"raw":"Vector Operations in Space: Fresh Take","rendered":"Vector Operations in Space: Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Calculate the dot product of two given vectors<\/li>\r\n \t<li>Find angles between vectors and determine when vectors are perpendicular using dot products<\/li>\r\n \t<li>Explain what is meant by the vector projection of one vector onto another vector, and describe how to compute it<\/li>\r\n \t<li>Calculate the work done by a given force<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>The Dot Product<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">The <strong>dot product<\/strong> is a way to \"multiply\" two vectors, but the result is a scalar (a number), not another vector. It measures how much two vectors align with each other and is essential for calculating work, finding angles between vectors, and determining if vectors are perpendicular.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Calculating the Dot Product<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">For 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]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\mathbf{u} \\cdot \\mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Process<\/strong>: Multiply corresponding components, then add the results.<\/p>\r\n<p class=\"whitespace-normal break-words\">For 2D vectors [latex]\\mathbf{u} = \\langle u_1, u_2 \\rangle[\/latex] and [latex]\\mathbf{v} = \\langle v_1, v_2 \\rangle[\/latex], it's the same: [latex]\\mathbf{u} \\cdot \\mathbf{v} = u_1v_1 + u_2v_2[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Example<\/strong>: [latex]\\langle 3, 5, 2 \\rangle \\cdot \\langle -1, 3, 0 \\rangle = (3)(-1) + (5)(3) + (2)(0) = -3 + 15 + 0 = 12[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Key Properties<\/strong><\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Commutative<\/strong>: [latex]\\mathbf{u} \\cdot \\mathbf{v} = \\mathbf{v} \\cdot \\mathbf{u}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Distributive<\/strong>: [latex]\\mathbf{u} \\cdot (\\mathbf{v} + \\mathbf{w}) = \\mathbf{u} \\cdot \\mathbf{v} + \\mathbf{u} \\cdot \\mathbf{w}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Magnitude relationship<\/strong>: [latex]\\mathbf{v} \\cdot \\mathbf{v} = ||\\mathbf{v}||^2[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Important Observations<\/strong><\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\mathbf{0} \\cdot \\mathbf{v} = 0[\/latex] for any vector<\/li>\r\n \t<li class=\"whitespace-normal break-words\">A vector's dot product with itself equals its magnitude squared: [latex]\\mathbf{v} \\cdot \\mathbf{v} = ||\\mathbf{v}||^2[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Only the zero vector has a dot product of zero with itself<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">The dot product tells you how much two vectors point in the same direction. It's the foundation for calculating work (force in the direction of motion) and finding angles between vectors.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">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].[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<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=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\" data-mce-fragment=\"1\"><\/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><\/section><section class=\"textbox example\" aria-label=\"Example\">\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<\/section>\r\n<h2 data-type=\"title\">Using the Dot Product to Find the Angle between Two Vectors<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">The dot product has a geometric interpretation that connects it to angles between vectors. This makes it a powerful tool for finding angles and determining if vectors are perpendicular.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Geometric Dot Product Formula<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">For nonzero vectors [latex]\\mathbf{u}[\/latex] and [latex]\\mathbf{v}[\/latex] with angle [latex]\\theta[\/latex] between them:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\mathbf{u} \\cdot \\mathbf{v} = ||\\mathbf{u}|| , ||\\mathbf{v}|| \\cos \\theta[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Rearranging to solve for the angle:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\cos \\theta = \\frac{\\mathbf{u} \\cdot \\mathbf{v}}{||\\mathbf{u}|| , ||\\mathbf{v}||}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Then [latex]\\theta = \\arccos\\left(\\frac{\\mathbf{u} \\cdot \\mathbf{v}}{||\\mathbf{u}|| , ||\\mathbf{v}||}\\right)[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>What the Angle Tells You<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">The sign of [latex]\\cos \\theta[\/latex] reveals the relationship between vectors:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\cos \\theta &gt; 0[\/latex]: Acute angle (vectors point in similar directions)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\cos \\theta = 0[\/latex]: Right angle (vectors are perpendicular)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\cos \\theta &lt; 0[\/latex]: Obtuse angle (vectors point in opposite-ish directions)<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Orthogonal Vectors: The Perpendicular Test<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Nonzero vectors [latex]\\mathbf{u}[\/latex] and [latex]\\mathbf{v}[\/latex] are <strong>orthogonal<\/strong> (perpendicular) if and only if [latex]\\mathbf{u} \\cdot \\mathbf{v} = 0[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">To test if vectors are perpendicular, just compute their dot product. If it's zero, they're orthogonal.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Direction Angles and Cosines<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">A vector's <strong>direction angles<\/strong> ([latex]\\alpha[\/latex], [latex]\\beta[\/latex], [latex]\\gamma[\/latex]) are the angles it makes with the positive [latex]x[\/latex]-, [latex]y[\/latex]-, and [latex]z[\/latex]-axes. Find them by taking dot products with [latex]\\mathbf{i}[\/latex], [latex]\\mathbf{j}[\/latex], and [latex]\\mathbf{k}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">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.[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<\/section><section class=\"textbox example\" aria-label=\"Example\">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]?[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<\/section><section class=\"textbox example\" aria-label=\"Example\">\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<\/section>\r\n<h2 data-type=\"title\">Projections<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Sometimes you need to know how much of one vector acts in the direction of another. <strong>Vector projection<\/strong> answers this question by finding the component of vector [latex]\\mathbf{v}[\/latex] that points in the direction of vector [latex]\\mathbf{u}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">The <strong>vector projection<\/strong> of [latex]\\mathbf{v}[\/latex] onto [latex]\\mathbf{u}[\/latex]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\text{proj}_{\\mathbf{u}}\\mathbf{v} = \\frac{\\mathbf{u} \\cdot \\mathbf{v}}{||\\mathbf{u}||^2}\\mathbf{u}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">The <strong>scalar projection<\/strong> (just the length, not the vector):<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\text{comp}_{\\mathbf{u}}\\mathbf{v} = \\frac{|\\mathbf{u} \\cdot \\mathbf{v}|}{||\\mathbf{u}||}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Think of projection as the \"shadow\" [latex]\\mathbf{v}[\/latex] casts onto [latex]\\mathbf{u}[\/latex] when light shines perpendicular to [latex]\\mathbf{u}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Decomposing Vectors<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Any vector [latex]\\mathbf{v}[\/latex] can be split into two orthogonal (perpendicular) parts:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\mathbf{v} = \\mathbf{p} + \\mathbf{q}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">where:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\mathbf{p} = \\text{proj}_{\\mathbf{u}}\\mathbf{v}[\/latex] (component parallel to [latex]\\mathbf{u}[\/latex])<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\mathbf{q} = \\mathbf{v} - \\mathbf{p}[\/latex] (component perpendicular to [latex]\\mathbf{u}[\/latex])<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">These components satisfy [latex]\\mathbf{p} \\cdot \\mathbf{q} = 0[\/latex], confirming they're orthogonal.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Process<\/strong>:<\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Calculate [latex]\\mathbf{p} = \\text{proj}_{\\mathbf{u}}\\mathbf{v}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Find [latex]\\mathbf{q} = \\mathbf{v} - \\mathbf{p}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Verify: Check that [latex]\\mathbf{p} \\cdot \\mathbf{q} = 0[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Express [latex]\\mathbf{v} = 5\\mathbf{i} - \\mathbf{j}[\/latex] as a sum of orthogonal vectors such that one of the vectors has the same direction as [latex]\\mathbf{u} = 4\\mathbf{i} +2\\mathbf{j}[\/latex].[reveal-answer q=\"114833389\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"114833389\"]\r\n\r\n[latex]\\mathbf{v} = \\mathbf{p} + \\mathbf{q}[\/latex], where\u00a0[latex]\\mathbf{p} = \\frac{18}{5}\\mathbf{i} + \\frac{19}{5}\\mathbf{j}[\/latex] and\u00a0[latex]\\mathbf{q} = \\frac{7}{5}\\mathbf{i} - \\frac{14}{5}\\mathbf{j}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above Try IT.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=7753522&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=p8Z4SXoXij8&amp;video_target=tpm-plugin-qxq1iaee-p8Z4SXoXij8\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/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\/CP2.27_transcript.html\">\u201cCP 2.27\u201d here (opens in new window).<\/a><\/center><\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<p id=\"fs-id1163724087498\">A container ship leaves port traveling [latex]15^\\circ[\/latex] north of east. Its engine generates a speed of [latex]20[\/latex] knots along that path (see the following figure). In addition, the ocean current moves the ship southeast at a speed of [latex]2[\/latex] knots. Considering both the engine and the current, how fast is the ship moving in the direction [latex]15^\\circ[\/latex] north of east? Round the answer to two decimal places.<\/p>\r\n\r\n[caption id=\"attachment_5129\" align=\"aligncenter\" width=\"604\"]<img class=\"size-full wp-image-5129\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/08154708\/Checkpoint-2.28.jpg\" alt=\"This figure is an image of a ship. The ship is at the origin of two perpendicular axes. The horizontal axis is labeled \u201ceast.\u201d The second axis is vertical and labeled \u201cnorth.\u201d From the ship there are two vectors. The first is labeled \u201cv\u201d and has an angle of 15 degrees between the East axis and the vector v. The second vector is labeled \u201cw\u201d and has an angle of 45 degrees between the East axis and the vector w. Vector w is below the East axis in the fourth quadrant.\" width=\"604\" height=\"448\" \/> Figure 4. Find how fast the ship is moving in the direction [latex]45^{\\circ}[\/latex] south of east.[\/caption][reveal-answer q=\"185672416\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"185672416\"][latex]21[\/latex] knots.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<h2 data-type=\"title\">Work<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">The most important real-world application of the dot product is calculating <strong>work<\/strong>\u2014the energy required to move an object with force.<\/p>\r\n<p class=\"whitespace-normal break-words\">When force and motion point in the same direction, work is simple: [latex]W = F \\cdot d[\/latex] (force times distance). But when force acts at an angle to the motion\u2014like pulling a wagon handle at 55\u00b0\u2014only part of the force actually moves the object. The dot product automatically accounts for this angle.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Work Formula<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">When constant force [latex]\\mathbf{F}[\/latex] moves an object from point [latex]P[\/latex] to point [latex]Q[\/latex]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]W = \\mathbf{F} \\cdot \\overrightarrow{PQ} = ||\\mathbf{F}|| , ||\\overrightarrow{PQ}|| \\cos \\theta[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">where [latex]\\theta[\/latex] is the angle between the force and displacement vectors.<\/p>\r\n<p class=\"whitespace-normal break-words\">The term [latex]||\\mathbf{F}|| \\cos \\theta[\/latex] is the component of force in the direction of motion\u2014exactly what you get from projection. Work equals this component times the distance moved.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Special Cases<\/strong><\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\theta = 0\u00b0[\/latex] (same direction): Maximum work, [latex]W = ||\\mathbf{F}|| , ||\\overrightarrow{PQ}||[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\theta = 90\u00b0[\/latex] (perpendicular): No work, [latex]W = 0[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\theta = 180\u00b0[\/latex] (opposite direction): Negative work, [latex]W = -||\\mathbf{F}|| , ||\\overrightarrow{PQ}||[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Units<\/strong><\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>U.S. Standard<\/strong>: Force (pounds) \u00d7 Distance (feet) = Work (foot-pounds or ft\u22c5lb)<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Metric<\/strong>: Force (newtons) \u00d7 Distance (meters) = Work (joules or J)<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">One joule equals one newton-meter.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">A constant force of [latex]30[\/latex] lb is applied at an angle of [latex]60^\\circ[\/latex] to pull a handcart [latex]10[\/latex] ft across the ground (Figure 6). What is the work done by this force?[caption id=\"attachment_5133\" align=\"aligncenter\" width=\"396\"]<img class=\"size-full wp-image-5133\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27154741\/Figure-2.52.jpg\" alt=\"This figure is an image of a hand cart with a crate. The vertical handle of the hand cart has two vectors. The first is horizontal to the handle and labeled \u201cs.\u201d The second is from the handle and labeled \u201cF.\u201d The angle between the two vectors is 60 degrees.\" width=\"396\" height=\"413\" \/> Figure 6. Find the force [latex]{\\bf{F}}[\/latex].[\/caption][reveal-answer q=\"793301735\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"793301735\"][latex]150[\/latex] ft-lb\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Calculate the dot product of two given vectors<\/li>\n<li>Find angles between vectors and determine when vectors are perpendicular using dot products<\/li>\n<li>Explain what is meant by the vector projection of one vector onto another vector, and describe how to compute it<\/li>\n<li>Calculate the work done by a given force<\/li>\n<\/ul>\n<\/section>\n<h2>The Dot Product<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The <strong>dot product<\/strong> is a way to &#8220;multiply&#8221; two vectors, but the result is a scalar (a number), not another vector. It measures how much two vectors align with each other and is essential for calculating work, finding angles between vectors, and determining if vectors are perpendicular.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Calculating the Dot Product<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">For 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]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\mathbf{u} \\cdot \\mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Process<\/strong>: Multiply corresponding components, then add the results.<\/p>\n<p class=\"whitespace-normal break-words\">For 2D vectors [latex]\\mathbf{u} = \\langle u_1, u_2 \\rangle[\/latex] and [latex]\\mathbf{v} = \\langle v_1, v_2 \\rangle[\/latex], it&#8217;s the same: [latex]\\mathbf{u} \\cdot \\mathbf{v} = u_1v_1 + u_2v_2[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Example<\/strong>: [latex]\\langle 3, 5, 2 \\rangle \\cdot \\langle -1, 3, 0 \\rangle = (3)(-1) + (5)(3) + (2)(0) = -3 + 15 + 0 = 12[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Key Properties<\/strong><\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>Commutative<\/strong>: [latex]\\mathbf{u} \\cdot \\mathbf{v} = \\mathbf{v} \\cdot \\mathbf{u}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Distributive<\/strong>: [latex]\\mathbf{u} \\cdot (\\mathbf{v} + \\mathbf{w}) = \\mathbf{u} \\cdot \\mathbf{v} + \\mathbf{u} \\cdot \\mathbf{w}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Magnitude relationship<\/strong>: [latex]\\mathbf{v} \\cdot \\mathbf{v} = ||\\mathbf{v}||^2[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Important Observations<\/strong><\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">[latex]\\mathbf{0} \\cdot \\mathbf{v} = 0[\/latex] for any vector<\/li>\n<li class=\"whitespace-normal break-words\">A vector&#8217;s dot product with itself equals its magnitude squared: [latex]\\mathbf{v} \\cdot \\mathbf{v} = ||\\mathbf{v}||^2[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Only the zero vector has a dot product of zero with itself<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">The dot product tells you how much two vectors point in the same direction. It&#8217;s the foundation for calculating work (force in the direction of motion) and finding angles between vectors.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q847271036\">Show Solution<\/button><\/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<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=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\" data-mce-fragment=\"1\"><\/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<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q645176472\">Show Solution<\/button><\/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<\/section>\n<h2 data-type=\"title\">Using the Dot Product to Find the Angle between Two Vectors<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The dot product has a geometric interpretation that connects it to angles between vectors. This makes it a powerful tool for finding angles and determining if vectors are perpendicular.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Geometric Dot Product Formula<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">For nonzero vectors [latex]\\mathbf{u}[\/latex] and [latex]\\mathbf{v}[\/latex] with angle [latex]\\theta[\/latex] between them:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\mathbf{u} \\cdot \\mathbf{v} = ||\\mathbf{u}|| , ||\\mathbf{v}|| \\cos \\theta[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Rearranging to solve for the angle:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\cos \\theta = \\frac{\\mathbf{u} \\cdot \\mathbf{v}}{||\\mathbf{u}|| , ||\\mathbf{v}||}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Then [latex]\\theta = \\arccos\\left(\\frac{\\mathbf{u} \\cdot \\mathbf{v}}{||\\mathbf{u}|| , ||\\mathbf{v}||}\\right)[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>What the Angle Tells You<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The sign of [latex]\\cos \\theta[\/latex] reveals the relationship between vectors:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">[latex]\\cos \\theta > 0[\/latex]: Acute angle (vectors point in similar directions)<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\cos \\theta = 0[\/latex]: Right angle (vectors are perpendicular)<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\cos \\theta < 0[\/latex]: Obtuse angle (vectors point in opposite-ish directions)<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Orthogonal Vectors: The Perpendicular Test<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Nonzero vectors [latex]\\mathbf{u}[\/latex] and [latex]\\mathbf{v}[\/latex] are <strong>orthogonal<\/strong> (perpendicular) if and only if [latex]\\mathbf{u} \\cdot \\mathbf{v} = 0[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">To test if vectors are perpendicular, just compute their dot product. If it&#8217;s zero, they&#8217;re orthogonal.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Direction Angles and Cosines<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">A vector&#8217;s <strong>direction angles<\/strong> ([latex]\\alpha[\/latex], [latex]\\beta[\/latex], [latex]\\gamma[\/latex]) are the angles it makes with the positive [latex]x[\/latex]-, [latex]y[\/latex]-, and [latex]z[\/latex]-axes. Find them by taking dot products with [latex]\\mathbf{i}[\/latex], [latex]\\mathbf{j}[\/latex], and [latex]\\mathbf{k}[\/latex].<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q875123144\">Show Solution<\/button><\/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<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q003746142\">Show Solution<\/button><\/p>\n<div id=\"q003746142\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=5[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q552341772\">Show Solution<\/button><\/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<\/section>\n<h2 data-type=\"title\">Projections<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Sometimes you need to know how much of one vector acts in the direction of another. <strong>Vector projection<\/strong> answers this question by finding the component of vector [latex]\\mathbf{v}[\/latex] that points in the direction of vector [latex]\\mathbf{u}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">The <strong>vector projection<\/strong> of [latex]\\mathbf{v}[\/latex] onto [latex]\\mathbf{u}[\/latex]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\text{proj}_{\\mathbf{u}}\\mathbf{v} = \\frac{\\mathbf{u} \\cdot \\mathbf{v}}{||\\mathbf{u}||^2}\\mathbf{u}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">The <strong>scalar projection<\/strong> (just the length, not the vector):<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\text{comp}_{\\mathbf{u}}\\mathbf{v} = \\frac{|\\mathbf{u} \\cdot \\mathbf{v}|}{||\\mathbf{u}||}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Think of projection as the &#8220;shadow&#8221; [latex]\\mathbf{v}[\/latex] casts onto [latex]\\mathbf{u}[\/latex] when light shines perpendicular to [latex]\\mathbf{u}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Decomposing Vectors<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Any vector [latex]\\mathbf{v}[\/latex] can be split into two orthogonal (perpendicular) parts:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\mathbf{v} = \\mathbf{p} + \\mathbf{q}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">where:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">[latex]\\mathbf{p} = \\text{proj}_{\\mathbf{u}}\\mathbf{v}[\/latex] (component parallel to [latex]\\mathbf{u}[\/latex])<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\mathbf{q} = \\mathbf{v} - \\mathbf{p}[\/latex] (component perpendicular to [latex]\\mathbf{u}[\/latex])<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">These components satisfy [latex]\\mathbf{p} \\cdot \\mathbf{q} = 0[\/latex], confirming they&#8217;re orthogonal.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Process<\/strong>:<\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Calculate [latex]\\mathbf{p} = \\text{proj}_{\\mathbf{u}}\\mathbf{v}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Find [latex]\\mathbf{q} = \\mathbf{v} - \\mathbf{p}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Verify: Check that [latex]\\mathbf{p} \\cdot \\mathbf{q} = 0[\/latex]<\/li>\n<\/ol>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Express [latex]\\mathbf{v} = 5\\mathbf{i} - \\mathbf{j}[\/latex] as a sum of orthogonal vectors such that one of the vectors has the same direction as [latex]\\mathbf{u} = 4\\mathbf{i} +2\\mathbf{j}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q114833389\">Show Solution<\/button><\/p>\n<div id=\"q114833389\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\mathbf{v} = \\mathbf{p} + \\mathbf{q}[\/latex], where\u00a0[latex]\\mathbf{p} = \\frac{18}{5}\\mathbf{i} + \\frac{19}{5}\\mathbf{j}[\/latex] and\u00a0[latex]\\mathbf{q} = \\frac{7}{5}\\mathbf{i} - \\frac{14}{5}\\mathbf{j}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above Try IT.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=7753522&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=p8Z4SXoXij8&amp;video_target=tpm-plugin-qxq1iaee-p8Z4SXoXij8\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the transcript for <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP2.27_transcript.html\">\u201cCP 2.27\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p id=\"fs-id1163724087498\">A container ship leaves port traveling [latex]15^\\circ[\/latex] north of east. Its engine generates a speed of [latex]20[\/latex] knots along that path (see the following figure). In addition, the ocean current moves the ship southeast at a speed of [latex]2[\/latex] knots. Considering both the engine and the current, how fast is the ship moving in the direction [latex]15^\\circ[\/latex] north of east? Round the answer to two decimal places.<\/p>\n<figure id=\"attachment_5129\" aria-describedby=\"caption-attachment-5129\" style=\"width: 604px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-5129\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/08154708\/Checkpoint-2.28.jpg\" alt=\"This figure is an image of a ship. The ship is at the origin of two perpendicular axes. The horizontal axis is labeled \u201ceast.\u201d The second axis is vertical and labeled \u201cnorth.\u201d From the ship there are two vectors. The first is labeled \u201cv\u201d and has an angle of 15 degrees between the East axis and the vector v. The second vector is labeled \u201cw\u201d and has an angle of 45 degrees between the East axis and the vector w. Vector w is below the East axis in the fourth quadrant.\" width=\"604\" height=\"448\" \/><figcaption id=\"caption-attachment-5129\" class=\"wp-caption-text\">Figure 4. Find how fast the ship is moving in the direction [latex]45^{\\circ}[\/latex] south of east.<\/figcaption><\/figure>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q185672416\">Show Solution<\/button><\/p>\n<div id=\"q185672416\" class=\"hidden-answer\" style=\"display: none\">[latex]21[\/latex] knots.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<h2 data-type=\"title\">Work<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The most important real-world application of the dot product is calculating <strong>work<\/strong>\u2014the energy required to move an object with force.<\/p>\n<p class=\"whitespace-normal break-words\">When force and motion point in the same direction, work is simple: [latex]W = F \\cdot d[\/latex] (force times distance). But when force acts at an angle to the motion\u2014like pulling a wagon handle at 55\u00b0\u2014only part of the force actually moves the object. The dot product automatically accounts for this angle.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Work Formula<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">When constant force [latex]\\mathbf{F}[\/latex] moves an object from point [latex]P[\/latex] to point [latex]Q[\/latex]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]W = \\mathbf{F} \\cdot \\overrightarrow{PQ} = ||\\mathbf{F}|| , ||\\overrightarrow{PQ}|| \\cos \\theta[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">where [latex]\\theta[\/latex] is the angle between the force and displacement vectors.<\/p>\n<p class=\"whitespace-normal break-words\">The term [latex]||\\mathbf{F}|| \\cos \\theta[\/latex] is the component of force in the direction of motion\u2014exactly what you get from projection. Work equals this component times the distance moved.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Special Cases<\/strong><\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">[latex]\\theta = 0\u00b0[\/latex] (same direction): Maximum work, [latex]W = ||\\mathbf{F}|| , ||\\overrightarrow{PQ}||[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\theta = 90\u00b0[\/latex] (perpendicular): No work, [latex]W = 0[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\theta = 180\u00b0[\/latex] (opposite direction): Negative work, [latex]W = -||\\mathbf{F}|| , ||\\overrightarrow{PQ}||[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Units<\/strong><\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>U.S. Standard<\/strong>: Force (pounds) \u00d7 Distance (feet) = Work (foot-pounds or ft\u22c5lb)<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Metric<\/strong>: Force (newtons) \u00d7 Distance (meters) = Work (joules or J)<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">One joule equals one newton-meter.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">A constant force of [latex]30[\/latex] lb is applied at an angle of [latex]60^\\circ[\/latex] to pull a handcart [latex]10[\/latex] ft across the ground (Figure 6). What is the work done by this force?<\/p>\n<figure id=\"attachment_5133\" aria-describedby=\"caption-attachment-5133\" style=\"width: 396px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-5133\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27154741\/Figure-2.52.jpg\" alt=\"This figure is an image of a hand cart with a crate. The vertical handle of the hand cart has two vectors. The first is horizontal to the handle and labeled \u201cs.\u201d The second is from the handle and labeled \u201cF.\u201d The angle between the two vectors is 60 degrees.\" width=\"396\" height=\"413\" \/><figcaption id=\"caption-attachment-5133\" class=\"wp-caption-text\">Figure 6. Find the force [latex]{\\bf{F}}[\/latex].<\/figcaption><\/figure>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q793301735\">Show Solution<\/button><\/p>\n<div id=\"q793301735\" class=\"hidden-answer\" style=\"display: none\">[latex]150[\/latex] ft-lb<\/p>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":21,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-505","chapter","type-chapter","status-publish","hentry"],"part":420,"module-header":"fresh_take","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/505","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":5,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/505\/revisions"}],"predecessor-version":[{"id":891,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/505\/revisions\/891"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/420"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/505\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=505"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=505"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=505"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=505"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}