{"id":522,"date":"2025-10-09T15:17:31","date_gmt":"2025-10-09T15:17:31","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=522"},"modified":"2025-10-24T12:59:47","modified_gmt":"2025-10-24T12:59:47","slug":"cross-product-and-applications-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/cross-product-and-applications-fresh-take\/","title":{"raw":"Cross Product and Applications: Fresh Take","rendered":"Cross Product and Applications: Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Calculate the cross product of two vectors and understand what it tells us<\/li>\r\n \t<li>Find a vector orthogonal to two given vectors<\/li>\r\n \t<li>Determine areas and volumes by using the cross product<\/li>\r\n \t<li>Calculate the torque of a given force and position vector<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>The Cross 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>cross product<\/strong> multiplies two vectors to produce a third vector that's perpendicular to both inputs. While the dot product gives a scalar, the cross product gives a vector.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Cross Product Formula<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">For [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} \\times \\mathbf{v} = \\langle u_2v_3 - u_3v_2, u_3v_1 - u_1v_3, u_1v_2 - u_2v_1 \\rangle[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">The result is orthogonal to both [latex]\\mathbf{u}[\/latex] and [latex]\\mathbf{v}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Direction: The Right-Hand Rule<\/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\">Point your right hand fingers along [latex]\\mathbf{u}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Curl them toward [latex]\\mathbf{v}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Your thumb points in the direction of [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex]<\/li>\r\n<\/ol>\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>Anticommutative<\/strong>: [latex]\\mathbf{u} \\times \\mathbf{v} = -(\\mathbf{v} \\times \\mathbf{u})[\/latex] (order matters!)<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Self cross product<\/strong>: [latex]\\mathbf{v} \\times \\mathbf{v} = \\mathbf{0}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Zero vector<\/strong>: [latex]\\mathbf{u} \\times \\mathbf{0} = \\mathbf{0}[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Unit Vector Cross Products<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Follow the cyclic pattern [latex]\\mathbf{i} \\to \\mathbf{j} \\to \\mathbf{k} \\to \\mathbf{i}[\/latex]:<\/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\">Forward: [latex]\\mathbf{i} \\times \\mathbf{j} = \\mathbf{k}[\/latex], [latex]\\mathbf{j} \\times \\mathbf{k} = \\mathbf{i}[\/latex], [latex]\\mathbf{k} \\times \\mathbf{i} = \\mathbf{j}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Backward (negative): [latex]\\mathbf{j} \\times \\mathbf{i} = -\\mathbf{k}[\/latex], etc.<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Magnitude<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]||\\mathbf{u} \\times \\mathbf{v}|| = ||\\mathbf{u}||\u00a0 ||\\mathbf{v}||\u00a0 \\sin \\theta[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Compare to the dot product (uses cosine). The cross product uses sine.<\/p>\r\n<p class=\"whitespace-normal break-words\">Cross products require three dimensions\u2014there's no direction simultaneously perpendicular to two 2D vectors without leaving the plane.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Find [latex]\\mathbf{p} \\times \\mathbf{q}[\/latex] for [latex]\\mathbf{p} =\\langle 5,1,2 \\rangle[\/latex] and [latex]\\mathbf{q} =\\langle -2,0,1 \\rangle[\/latex]. Express the answer using standard unit vectors.[reveal-answer q=\"125237156\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"125237156\"][latex]\\mathbf{i} -9\\mathbf{j}+2\\mathbf{k}[\/latex][\/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 example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=7753558&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=fEMzUqQTmoo&amp;video_target=tpm-plugin-aghj6e1q-fEMzUqQTmoo\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">You can view the transcript for <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP2.30_transcript.html\">\u201cCP 2.30\u201d here (opens in new window).<\/a><\/span><\/center><\/section><section class=\"textbox example\" aria-label=\"Example\">Suppose vectors [latex]\\textbf u[\/latex] and [latex]\\textbf v[\/latex] lie in the [latex]xy[\/latex]-plane (the [latex]z[\/latex]-component of each vector is zero). Now suppose the [latex]x[\/latex]- and [latex]y[\/latex]-components of [latex]\\textbf u[\/latex] and the [latex]y[\/latex]-component of [latex]\\textbf v[\/latex] are all positive, whereas the [latex]x[\/latex]-component of [latex]\\textbf v[\/latex] is negative. Assuming the coordinate axes are oriented in the usual positions, in which direction does [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex] point?[reveal-answer q=\"373093789\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"373093789\"]Up (the positive [latex]z[\/latex]-direction)[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Find [latex](\\mathbf{i} \\times \\mathbf{j}) \\times (\\mathbf{k} \\times \\mathbf{i})[\/latex].[reveal-answer q=\"092380347\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"092380347\"][latex]-\\mathbf{i}[\/latex][\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Use the cross product properties to calculate [latex](\\mathbf{i} \\times \\mathbf{k}) \\times (\\mathbf{k} \\times \\mathbf{j})[\/latex].[reveal-answer q=\"092367174\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"092367174\"][latex]-\\mathbf{k}[\/latex][\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Use the properties of the cross product\u00a0to find the magnitude of [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex], where [latex]\\mathbf{u} = \\langle -8,0,0\\rangle[\/latex] and\u00a0[latex]\\mathbf{v} = \\langle 0,2,0\\rangle[\/latex].[reveal-answer q=\"635451352\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"635451352\"][latex]16[\/latex][\/hidden-answer]\r\n\r\n<\/section>\r\n<h2 data-type=\"title\">Determinants and the Cross 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 cross product formula [latex]\\mathbf{u} \\times \\mathbf{v} = \\langle u_2v_3 - u_3v_2, u_3v_1 - u_1v_3, u_1v_2 - u_2v_1 \\rangle[\/latex] is hard to remember. <strong>Determinant notation<\/strong> provides an organized, memorable alternative.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Quick Determinant Review<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>2\u00d72 determinant<\/strong>: [latex]\\begin{vmatrix} a_1 &amp; a_2 \\\\ b_1 &amp; b_2 \\end{vmatrix} = a_1b_2 - b_1a_2[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>3\u00d73 determinant<\/strong> (expand along first row with alternating signs): [latex]\\begin{vmatrix} a_1 &amp; a_2 &amp; a_3 \\\\ b_1 &amp; b_2 &amp; b_3 \\\\ c_1 &amp; c_2 &amp; c_3\\end{vmatrix} = a_1\\begin{vmatrix} b_2 &amp; b_3 \\\\ c_2 &amp; c_3 \\end{vmatrix} - a_2 \\begin{vmatrix} b_1 &amp; b_3 \\\\ c_1 &amp; c_3 \\end{vmatrix} + a_3\\begin{vmatrix} b_1 &amp; b_2 \\\\ c_1 &amp; c_2 \\end{vmatrix}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Cross Product Using Determinants<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">For [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} \\times \\mathbf{v} = \\begin{vmatrix} \\mathbf{i} &amp; \\mathbf{j} &amp; \\mathbf{k} \\ u_1 &amp; u_2 &amp; u_3 \\\\ v_1 &amp; v_2 &amp; v_3\\end{vmatrix}[\/latex]<\/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\">First row: [latex]\\mathbf{i}[\/latex], [latex]\\mathbf{j}[\/latex], [latex]\\mathbf{k}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Second row: [latex]\\mathbf{u}[\/latex] components<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Third row: [latex]\\mathbf{v}[\/latex] components<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Expand along first row with alternating signs ([latex]+[\/latex], [latex]-[\/latex], [latex]+[\/latex])<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\mathbf{u} \\times \\mathbf{v} = \\mathbf{i}\\begin{vmatrix} u_2 &amp; u_3 \\\\ v_2 &amp; v_3 \\end{vmatrix} - \\mathbf{j}\\begin{vmatrix} u_1 &amp; u_3 \\\\ v_1 &amp; v_3 \\end{vmatrix} + \\mathbf{k}\\begin{vmatrix} u_1 &amp; u_2 \\\\ v_1 &amp; v_2 \\end{vmatrix}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Expanding the determinant gives exactly the same components as the original formula, but in an organized, memorable format.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Note<\/strong>: This is technically a <strong>mnemonic device<\/strong> (memory aid), not a true determinant, since determinants contain only numbers, not vectors.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Evaluate the determinant [latex]\\begin{vmatrix} 1 &amp; -2 &amp; -1 \\\\ 3 &amp; 2 &amp; -3 \\\\ 1 &amp; 5 &amp; 4\\end{vmatrix}[\/latex].[reveal-answer q=\"178463041\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"178463041\"][latex]40[\/latex][\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Use determinant notation to find [latex]\\mathbf{a} \\times \\mathbf{b}[\/latex], where [latex]\\mathbf{a} = \\langle 8,2,3\\rangle[\/latex] and\u00a0[latex]\\mathbf{b} = \\langle -1,0,4\\rangle[\/latex].[reveal-answer q=\"457345863\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"457345863\"][latex]8\\mathbf{i} -35\\mathbf{j} +2\\mathbf{k}[\/latex].[\/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 example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=7809775&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=umfZKb2xRMw&amp;video_target=tpm-plugin-ma5mr889-umfZKb2xRMw\" 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\/CP2.36_transcript.html\">\u201cCP 2.36\u201d here (opens in new window).<\/a><\/center><\/section>\r\n<h2 data-type=\"title\">Using the Cross 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 cross product solves key geometric problems in 3D: finding perpendicular vectors, calculating areas of parallelograms and triangles, and more.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Finding Orthogonal Vectors<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Since [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex] is always perpendicular to both [latex]\\mathbf{u}[\/latex] and [latex]\\mathbf{v}[\/latex], it's the go-to method for finding orthogonal vectors.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Process<\/strong>: Calculate [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex], then normalize if you need a unit vector: [latex]\\frac{\\mathbf{u} \\times \\mathbf{v}}{||\\mathbf{u} \\times \\mathbf{v}||}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Both [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex] and [latex]\\mathbf{v} \\times \\mathbf{u}[\/latex] are orthogonal, but point in opposite directions.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Calculating Areas<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">The magnitude of the cross product gives areas directly:<\/p>\r\n\r\n<ul>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Parallelogram<\/strong>: If [latex]\\mathbf{u}[\/latex] and [latex]\\mathbf{v}[\/latex] form adjacent sides: [latex]\\text{Area} = ||\\mathbf{u} \\times \\mathbf{v}||[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Triangle<\/strong>: Half the parallelogram area: [latex]\\text{Area} = \\frac{1}{2}||\\mathbf{u} \\times \\mathbf{v}||[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Example<\/strong>: To find a triangle's area from three vertices [latex]P[\/latex], [latex]Q[\/latex], [latex]R[\/latex]:<\/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\">Form vectors [latex]\\overrightarrow{PQ}[\/latex] and [latex]\\overrightarrow{PR}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Calculate [latex]\\overrightarrow{PQ} \\times \\overrightarrow{PR}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Area = [latex]\\frac{1}{2}||\\overrightarrow{PQ} \\times \\overrightarrow{PR}||[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Find a unit vector orthogonal to both [latex]\\textbf a[\/latex] and [latex]\\textbf b[\/latex], where [latex]\\mathbf{a} = \\langle 4,0,3\\rangle[\/latex] and\u00a0[latex]\\mathbf{b} = \\langle 1,1,4\\rangle[\/latex].[reveal-answer q=\"748528349\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"748528349\"][latex]\\langle \\frac{-3}{\\sqrt{194}},\\frac{-13}{\\sqrt{194}},\\frac{4}{\\sqrt{194}}\\rangle[\/latex][\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Find the area of the parallelogram [latex]PQRS[\/latex] with vertices\u00a0[latex]P(1, 1, 0)[\/latex],\u00a0[latex]Q(7, 1, 0)[\/latex],\u00a0[latex]R(9, 4, 2)[\/latex], and\u00a0[latex]S(3, 4, 2)[\/latex].[reveal-answer q=\"387493274\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"387493274\"][latex]6\\sqrt{13}[\/latex][\/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 example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=7753559&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=L2XRUTcifoM&amp;video_target=tpm-plugin-mi2dpg43-L2XRUTcifoM\" 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\/CP2.38_transcript.html\">\u201cCP 2.38\u201d here (opens in new window).<\/a><\/center><\/section>\r\n<h2 data-type=\"title\">The Triple Scalar 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>triple scalar product<\/strong> combines a dot product with a cross product: [latex]\\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w})[\/latex]. The result is a scalar with important geometric meaning.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Determinant Formula<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Instead of computing [latex]\\mathbf{v} \\times \\mathbf{w}[\/latex] first, then taking the dot product, use a determinant:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w}) = \\begin{vmatrix}u_1 &amp; u_2 &amp; u_3 \\\\ v_1 &amp; v_2 &amp; v_3 \\\\ w_1 &amp; w_2 &amp; w_3 \\end{vmatrix}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">This is much faster\u2014just set up the determinant with vector components as rows and evaluate.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Order Matters<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Swapping vectors changes the determinant's sign (like swapping rows):<\/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{u} \\cdot (\\mathbf{v} \\times \\mathbf{w}) = -\\mathbf{u} \\cdot (\\mathbf{w} \\times \\mathbf{v})[\/latex] (switching cross product order)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w}) = \\mathbf{v} \\cdot (\\mathbf{w} \\times \\mathbf{u}) = \\mathbf{w} \\cdot (\\mathbf{u} \\times \\mathbf{v})[\/latex] (cyclic permutations give same value)<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Volume of a Parallelepiped<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">A <strong>parallelepiped<\/strong> is a 3D prism where all faces are parallelograms. When vectors [latex]\\mathbf{u}[\/latex], [latex]\\mathbf{v}[\/latex], and [latex]\\mathbf{w}[\/latex] form its edges:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]V = |\\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w})|[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Use absolute value because volume is always positive.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Quick 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\">Set up determinant with vector components<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Evaluate the determinant<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Take absolute value for volume<\/li>\r\n<\/ol>\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Calculate the triple scalar product [latex]{\\bf{a}}\\cdot({\\bf{b}}\\times{\\bf{c}})[\/latex], where [latex]{\\bf{a}}=\\langle2,-4,1\\rangle[\/latex],\u00a0[latex]{\\bf{b}}=\\langle0,3,-1\\rangle[\/latex], and\u00a0[latex]{\\bf{c}}=\\langle5,-3,3\\rangle[\/latex].[reveal-answer q=\"849527436\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"849527436\"][latex]17[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Find the volume of the parallelepiped formed by the vectors\u00a0[latex]{\\bf{a}}=3{\\bf{i}}+4{\\bf{j}}-{\\bf{k}}[\/latex],\u00a0[latex]{\\bf{b}}=2{\\bf{i}}-{\\bf{j}}-{\\bf{k}}[\/latex], and\u00a0[latex]{\\bf{c}}=3{\\bf{j}}+{\\bf{k}}[\/latex].[reveal-answer q=\"983459723\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"983459723\"][latex]8[\/latex] units<sup>3<\/sup>.[\/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=7753560&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=4Jv4U5G3qxA&amp;video_target=tpm-plugin-wayy3i76-4Jv4U5G3qxA\" 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\/CP2.40_transcript.html\">\u201cCP 2.40\u201d here (opens in new window).<\/a><\/center><\/section>\r\n<h2 data-type=\"title\">Applications of the Cross 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\">Cross products solve practical problems in physics and engineering, from testing if vectors lie in the same plane to calculating rotational forces.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Testing for Coplanar Vectors<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Three vectors are <strong>coplanar<\/strong> if they all lie in the same plane. The test: calculate the triple scalar product.<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w}) = 0[\/latex] if and only if the vectors are coplanar.<\/p>\r\n<p class=\"whitespace-normal break-words\">If the parallelepiped volume is zero, one dimension must be zero\u2014the vectors lie flat in a plane.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Finding Vectors Perpendicular to Planes<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">To find a vector perpendicular to a plane, take the cross product of any two non-parallel vectors in that plane.<\/p>\r\n<p class=\"whitespace-normal break-words\">Given three points [latex]P[\/latex], [latex]Q[\/latex], [latex]R[\/latex] in a plane, form vectors [latex]\\overrightarrow{PQ}[\/latex] and [latex]\\overrightarrow{QR}[\/latex], then [latex]\\overrightarrow{PQ} \\times \\overrightarrow{QR}[\/latex] is perpendicular to the plane.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Torque: Measuring Rotational Force<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Torque<\/strong> ([latex]\\tau[\/latex]) measures how effectively a force causes rotation. When you turn a wrench or tighten a screw, you're applying torque:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\tau = \\mathbf{r} \\times \\mathbf{F}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">where [latex]\\mathbf{r}[\/latex] is the position vector from the rotation axis to where force [latex]\\mathbf{F}[\/latex] is applied.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Key Facts<\/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>Magnitude<\/strong>: [latex]|\\tau| = |\\mathbf{r}| \\cdot |\\mathbf{F}| \\sin \\theta[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Direction<\/strong>: Points along the rotation axis (right-hand rule)<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Maximum torque<\/strong>: When force is perpendicular to [latex]\\mathbf{r}[\/latex] ([latex]\\theta = 90\u00b0[\/latex])<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Zero torque<\/strong>: When force points toward\/away from axis ([latex]\\theta = 0\u00b0[\/latex] or [latex]180\u00b0[\/latex])<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Units<\/strong>: Newton-meters (N\u22c5m) or foot-pounds (ft\u22c5lb)<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Example<\/strong>: Longer wrench = greater torque. That's why you can tighten a bolt more easily with a longer wrench\u2014you're increasing [latex]|\\mathbf{r}|[\/latex].<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Are the vectors [latex]{\\bf{a}}={\\bf{i}}+{\\bf{j}}-{\\bf{k}}[\/latex], [latex]{\\bf{b}}={\\bf{i}}-{\\bf{j}}+{\\bf{k}}[\/latex], and [latex]{\\bf{c}}={\\bf{i}}+{\\bf{j}}+{\\bf{k}}[\/latex] coplanar?[reveal-answer q=\"308475728\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"308475728\"]No, the triple scalar product is [latex]-4\\ne0[\/latex], so the three vectors form the adjacent edges of a parallelepiped. They are not coplanar.[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Calculate the force required to produce [latex]15\\text{N}\\cdot\\text{m}[\/latex] torque at an angle of [latex]30^{\\small\\circ}[\/latex] from a [latex]150[\/latex]-cm rod.[reveal-answer q=\"473650281\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"473650281\"][latex]20[\/latex] 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=7753561&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=VgSpno5eD9k&amp;video_target=tpm-plugin-ydgl3n50-VgSpno5eD9k\" 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\/CP2.42_transcript.html\">\u201cCP 2.42\u201d here (opens in new window).<\/a><\/center><\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Calculate the cross product of two vectors and understand what it tells us<\/li>\n<li>Find a vector orthogonal to two given vectors<\/li>\n<li>Determine areas and volumes by using the cross product<\/li>\n<li>Calculate the torque of a given force and position vector<\/li>\n<\/ul>\n<\/section>\n<h2>The Cross 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>cross product<\/strong> multiplies two vectors to produce a third vector that&#8217;s perpendicular to both inputs. While the dot product gives a scalar, the cross product gives a vector.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Cross Product Formula<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">For [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} \\times \\mathbf{v} = \\langle u_2v_3 - u_3v_2, u_3v_1 - u_1v_3, u_1v_2 - u_2v_1 \\rangle[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">The result is orthogonal to both [latex]\\mathbf{u}[\/latex] and [latex]\\mathbf{v}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Direction: The Right-Hand Rule<\/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\">Point your right hand fingers along [latex]\\mathbf{u}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Curl them toward [latex]\\mathbf{v}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Your thumb points in the direction of [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex]<\/li>\n<\/ol>\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>Anticommutative<\/strong>: [latex]\\mathbf{u} \\times \\mathbf{v} = -(\\mathbf{v} \\times \\mathbf{u})[\/latex] (order matters!)<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Self cross product<\/strong>: [latex]\\mathbf{v} \\times \\mathbf{v} = \\mathbf{0}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Zero vector<\/strong>: [latex]\\mathbf{u} \\times \\mathbf{0} = \\mathbf{0}[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Unit Vector Cross Products<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Follow the cyclic pattern [latex]\\mathbf{i} \\to \\mathbf{j} \\to \\mathbf{k} \\to \\mathbf{i}[\/latex]:<\/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\">Forward: [latex]\\mathbf{i} \\times \\mathbf{j} = \\mathbf{k}[\/latex], [latex]\\mathbf{j} \\times \\mathbf{k} = \\mathbf{i}[\/latex], [latex]\\mathbf{k} \\times \\mathbf{i} = \\mathbf{j}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Backward (negative): [latex]\\mathbf{j} \\times \\mathbf{i} = -\\mathbf{k}[\/latex], etc.<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Magnitude<\/strong><\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]||\\mathbf{u} \\times \\mathbf{v}|| = ||\\mathbf{u}||\u00a0 ||\\mathbf{v}||\u00a0 \\sin \\theta[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Compare to the dot product (uses cosine). The cross product uses sine.<\/p>\n<p class=\"whitespace-normal break-words\">Cross products require three dimensions\u2014there&#8217;s no direction simultaneously perpendicular to two 2D vectors without leaving the plane.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Find [latex]\\mathbf{p} \\times \\mathbf{q}[\/latex] for [latex]\\mathbf{p} =\\langle 5,1,2 \\rangle[\/latex] and [latex]\\mathbf{q} =\\langle -2,0,1 \\rangle[\/latex]. Express the answer using standard unit vectors.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q125237156\">Show Solution<\/button><\/p>\n<div id=\"q125237156\" class=\"hidden-answer\" style=\"display: none\">[latex]\\mathbf{i} -9\\mathbf{j}+2\\mathbf{k}[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=7753558&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=fEMzUqQTmoo&amp;video_target=tpm-plugin-aghj6e1q-fEMzUqQTmoo\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<div style=\"text-align: center;\"><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">You can view the transcript for <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP2.30_transcript.html\">\u201cCP 2.30\u201d here (opens in new window).<\/a><\/span><\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Suppose vectors [latex]\\textbf u[\/latex] and [latex]\\textbf v[\/latex] lie in the [latex]xy[\/latex]-plane (the [latex]z[\/latex]-component of each vector is zero). Now suppose the [latex]x[\/latex]&#8211; and [latex]y[\/latex]-components of [latex]\\textbf u[\/latex] and the [latex]y[\/latex]-component of [latex]\\textbf v[\/latex] are all positive, whereas the [latex]x[\/latex]-component of [latex]\\textbf v[\/latex] is negative. Assuming the coordinate axes are oriented in the usual positions, in which direction does [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex] point?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q373093789\">Show Solution<\/button><\/p>\n<div id=\"q373093789\" class=\"hidden-answer\" style=\"display: none\">Up (the positive [latex]z[\/latex]-direction)<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Find [latex](\\mathbf{i} \\times \\mathbf{j}) \\times (\\mathbf{k} \\times \\mathbf{i})[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q092380347\">Show Solution<\/button><\/p>\n<div id=\"q092380347\" class=\"hidden-answer\" style=\"display: none\">[latex]-\\mathbf{i}[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Use the cross product properties to calculate [latex](\\mathbf{i} \\times \\mathbf{k}) \\times (\\mathbf{k} \\times \\mathbf{j})[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q092367174\">Show Solution<\/button><\/p>\n<div id=\"q092367174\" class=\"hidden-answer\" style=\"display: none\">[latex]-\\mathbf{k}[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Use the properties of the cross product\u00a0to find the magnitude of [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex], where [latex]\\mathbf{u} = \\langle -8,0,0\\rangle[\/latex] and\u00a0[latex]\\mathbf{v} = \\langle 0,2,0\\rangle[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q635451352\">Show Solution<\/button><\/p>\n<div id=\"q635451352\" class=\"hidden-answer\" style=\"display: none\">[latex]16[\/latex]<\/div>\n<\/div>\n<\/section>\n<h2 data-type=\"title\">Determinants and the Cross Product<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The cross product formula [latex]\\mathbf{u} \\times \\mathbf{v} = \\langle u_2v_3 - u_3v_2, u_3v_1 - u_1v_3, u_1v_2 - u_2v_1 \\rangle[\/latex] is hard to remember. <strong>Determinant notation<\/strong> provides an organized, memorable alternative.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Quick Determinant Review<\/strong><\/p>\n<p class=\"whitespace-normal break-words\"><strong>2\u00d72 determinant<\/strong>: [latex]\\begin{vmatrix} a_1 & a_2 \\\\ b_1 & b_2 \\end{vmatrix} = a_1b_2 - b_1a_2[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>3\u00d73 determinant<\/strong> (expand along first row with alternating signs): [latex]\\begin{vmatrix} a_1 & a_2 & a_3 \\\\ b_1 & b_2 & b_3 \\\\ c_1 & c_2 & c_3\\end{vmatrix} = a_1\\begin{vmatrix} b_2 & b_3 \\\\ c_2 & c_3 \\end{vmatrix} - a_2 \\begin{vmatrix} b_1 & b_3 \\\\ c_1 & c_3 \\end{vmatrix} + a_3\\begin{vmatrix} b_1 & b_2 \\\\ c_1 & c_2 \\end{vmatrix}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Cross Product Using Determinants<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">For [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} \\times \\mathbf{v} = \\begin{vmatrix} \\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\ u_1 & u_2 & u_3 \\\\ v_1 & v_2 & v_3\\end{vmatrix}[\/latex]<\/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\">First row: [latex]\\mathbf{i}[\/latex], [latex]\\mathbf{j}[\/latex], [latex]\\mathbf{k}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Second row: [latex]\\mathbf{u}[\/latex] components<\/li>\n<li class=\"whitespace-normal break-words\">Third row: [latex]\\mathbf{v}[\/latex] components<\/li>\n<li class=\"whitespace-normal break-words\">Expand along first row with alternating signs ([latex]+[\/latex], [latex]-[\/latex], [latex]+[\/latex])<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\mathbf{u} \\times \\mathbf{v} = \\mathbf{i}\\begin{vmatrix} u_2 & u_3 \\\\ v_2 & v_3 \\end{vmatrix} - \\mathbf{j}\\begin{vmatrix} u_1 & u_3 \\\\ v_1 & v_3 \\end{vmatrix} + \\mathbf{k}\\begin{vmatrix} u_1 & u_2 \\\\ v_1 & v_2 \\end{vmatrix}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Expanding the determinant gives exactly the same components as the original formula, but in an organized, memorable format.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Note<\/strong>: This is technically a <strong>mnemonic device<\/strong> (memory aid), not a true determinant, since determinants contain only numbers, not vectors.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Evaluate the determinant [latex]\\begin{vmatrix} 1 & -2 & -1 \\\\ 3 & 2 & -3 \\\\ 1 & 5 & 4\\end{vmatrix}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q178463041\">Show Solution<\/button><\/p>\n<div id=\"q178463041\" class=\"hidden-answer\" style=\"display: none\">[latex]40[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Use determinant notation to find [latex]\\mathbf{a} \\times \\mathbf{b}[\/latex], where [latex]\\mathbf{a} = \\langle 8,2,3\\rangle[\/latex] and\u00a0[latex]\\mathbf{b} = \\langle -1,0,4\\rangle[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q457345863\">Show Solution<\/button><\/p>\n<div id=\"q457345863\" class=\"hidden-answer\" style=\"display: none\">[latex]8\\mathbf{i} -35\\mathbf{j} +2\\mathbf{k}[\/latex].<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=7809775&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=umfZKb2xRMw&amp;video_target=tpm-plugin-ma5mr889-umfZKb2xRMw\" 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\/CP2.36_transcript.html\">\u201cCP 2.36\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<h2 data-type=\"title\">Using the Cross Product<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The cross product solves key geometric problems in 3D: finding perpendicular vectors, calculating areas of parallelograms and triangles, and more.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Finding Orthogonal Vectors<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Since [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex] is always perpendicular to both [latex]\\mathbf{u}[\/latex] and [latex]\\mathbf{v}[\/latex], it&#8217;s the go-to method for finding orthogonal vectors.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Process<\/strong>: Calculate [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex], then normalize if you need a unit vector: [latex]\\frac{\\mathbf{u} \\times \\mathbf{v}}{||\\mathbf{u} \\times \\mathbf{v}||}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Both [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex] and [latex]\\mathbf{v} \\times \\mathbf{u}[\/latex] are orthogonal, but point in opposite directions.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Calculating Areas<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The magnitude of the cross product gives areas directly:<\/p>\n<ul>\n<li class=\"whitespace-normal break-words\"><strong>Parallelogram<\/strong>: If [latex]\\mathbf{u}[\/latex] and [latex]\\mathbf{v}[\/latex] form adjacent sides: [latex]\\text{Area} = ||\\mathbf{u} \\times \\mathbf{v}||[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Triangle<\/strong>: Half the parallelogram area: [latex]\\text{Area} = \\frac{1}{2}||\\mathbf{u} \\times \\mathbf{v}||[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Example<\/strong>: To find a triangle&#8217;s area from three vertices [latex]P[\/latex], [latex]Q[\/latex], [latex]R[\/latex]:<\/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\">Form vectors [latex]\\overrightarrow{PQ}[\/latex] and [latex]\\overrightarrow{PR}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Calculate [latex]\\overrightarrow{PQ} \\times \\overrightarrow{PR}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Area = [latex]\\frac{1}{2}||\\overrightarrow{PQ} \\times \\overrightarrow{PR}||[\/latex]<\/li>\n<\/ol>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Find a unit vector orthogonal to both [latex]\\textbf a[\/latex] and [latex]\\textbf b[\/latex], where [latex]\\mathbf{a} = \\langle 4,0,3\\rangle[\/latex] and\u00a0[latex]\\mathbf{b} = \\langle 1,1,4\\rangle[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q748528349\">Show Solution<\/button><\/p>\n<div id=\"q748528349\" class=\"hidden-answer\" style=\"display: none\">[latex]\\langle \\frac{-3}{\\sqrt{194}},\\frac{-13}{\\sqrt{194}},\\frac{4}{\\sqrt{194}}\\rangle[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Find the area of the parallelogram [latex]PQRS[\/latex] with vertices\u00a0[latex]P(1, 1, 0)[\/latex],\u00a0[latex]Q(7, 1, 0)[\/latex],\u00a0[latex]R(9, 4, 2)[\/latex], and\u00a0[latex]S(3, 4, 2)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q387493274\">Show Solution<\/button><\/p>\n<div id=\"q387493274\" class=\"hidden-answer\" style=\"display: none\">[latex]6\\sqrt{13}[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=7753559&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=L2XRUTcifoM&amp;video_target=tpm-plugin-mi2dpg43-L2XRUTcifoM\" 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\/CP2.38_transcript.html\">\u201cCP 2.38\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<h2 data-type=\"title\">The Triple Scalar 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>triple scalar product<\/strong> combines a dot product with a cross product: [latex]\\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w})[\/latex]. The result is a scalar with important geometric meaning.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Determinant Formula<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Instead of computing [latex]\\mathbf{v} \\times \\mathbf{w}[\/latex] first, then taking the dot product, use a determinant:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w}) = \\begin{vmatrix}u_1 & u_2 & u_3 \\\\ v_1 & v_2 & v_3 \\\\ w_1 & w_2 & w_3 \\end{vmatrix}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">This is much faster\u2014just set up the determinant with vector components as rows and evaluate.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Order Matters<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Swapping vectors changes the determinant&#8217;s sign (like swapping rows):<\/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{u} \\cdot (\\mathbf{v} \\times \\mathbf{w}) = -\\mathbf{u} \\cdot (\\mathbf{w} \\times \\mathbf{v})[\/latex] (switching cross product order)<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w}) = \\mathbf{v} \\cdot (\\mathbf{w} \\times \\mathbf{u}) = \\mathbf{w} \\cdot (\\mathbf{u} \\times \\mathbf{v})[\/latex] (cyclic permutations give same value)<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Volume of a Parallelepiped<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">A <strong>parallelepiped<\/strong> is a 3D prism where all faces are parallelograms. When vectors [latex]\\mathbf{u}[\/latex], [latex]\\mathbf{v}[\/latex], and [latex]\\mathbf{w}[\/latex] form its edges:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]V = |\\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w})|[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Use absolute value because volume is always positive.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Quick 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\">Set up determinant with vector components<\/li>\n<li class=\"whitespace-normal break-words\">Evaluate the determinant<\/li>\n<li class=\"whitespace-normal break-words\">Take absolute value for volume<\/li>\n<\/ol>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Calculate the triple scalar product [latex]{\\bf{a}}\\cdot({\\bf{b}}\\times{\\bf{c}})[\/latex], where [latex]{\\bf{a}}=\\langle2,-4,1\\rangle[\/latex],\u00a0[latex]{\\bf{b}}=\\langle0,3,-1\\rangle[\/latex], and\u00a0[latex]{\\bf{c}}=\\langle5,-3,3\\rangle[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q849527436\">Show Solution<\/button><\/p>\n<div id=\"q849527436\" class=\"hidden-answer\" style=\"display: none\">[latex]17[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Find the volume of the parallelepiped formed by the vectors\u00a0[latex]{\\bf{a}}=3{\\bf{i}}+4{\\bf{j}}-{\\bf{k}}[\/latex],\u00a0[latex]{\\bf{b}}=2{\\bf{i}}-{\\bf{j}}-{\\bf{k}}[\/latex], and\u00a0[latex]{\\bf{c}}=3{\\bf{j}}+{\\bf{k}}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q983459723\">Show Solution<\/button><\/p>\n<div id=\"q983459723\" class=\"hidden-answer\" style=\"display: none\">[latex]8[\/latex] units<sup>3<\/sup>.<\/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=7753560&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=4Jv4U5G3qxA&amp;video_target=tpm-plugin-wayy3i76-4Jv4U5G3qxA\" 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\/CP2.40_transcript.html\">\u201cCP 2.40\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<h2 data-type=\"title\">Applications of the Cross Product<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Cross products solve practical problems in physics and engineering, from testing if vectors lie in the same plane to calculating rotational forces.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Testing for Coplanar Vectors<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Three vectors are <strong>coplanar<\/strong> if they all lie in the same plane. The test: calculate the triple scalar product.<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w}) = 0[\/latex] if and only if the vectors are coplanar.<\/p>\n<p class=\"whitespace-normal break-words\">If the parallelepiped volume is zero, one dimension must be zero\u2014the vectors lie flat in a plane.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Finding Vectors Perpendicular to Planes<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">To find a vector perpendicular to a plane, take the cross product of any two non-parallel vectors in that plane.<\/p>\n<p class=\"whitespace-normal break-words\">Given three points [latex]P[\/latex], [latex]Q[\/latex], [latex]R[\/latex] in a plane, form vectors [latex]\\overrightarrow{PQ}[\/latex] and [latex]\\overrightarrow{QR}[\/latex], then [latex]\\overrightarrow{PQ} \\times \\overrightarrow{QR}[\/latex] is perpendicular to the plane.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Torque: Measuring Rotational Force<\/strong><\/p>\n<p class=\"whitespace-normal break-words\"><strong>Torque<\/strong> ([latex]\\tau[\/latex]) measures how effectively a force causes rotation. When you turn a wrench or tighten a screw, you&#8217;re applying torque:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\tau = \\mathbf{r} \\times \\mathbf{F}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">where [latex]\\mathbf{r}[\/latex] is the position vector from the rotation axis to where force [latex]\\mathbf{F}[\/latex] is applied.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Key Facts<\/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>Magnitude<\/strong>: [latex]|\\tau| = |\\mathbf{r}| \\cdot |\\mathbf{F}| \\sin \\theta[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Direction<\/strong>: Points along the rotation axis (right-hand rule)<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Maximum torque<\/strong>: When force is perpendicular to [latex]\\mathbf{r}[\/latex] ([latex]\\theta = 90\u00b0[\/latex])<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Zero torque<\/strong>: When force points toward\/away from axis ([latex]\\theta = 0\u00b0[\/latex] or [latex]180\u00b0[\/latex])<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Units<\/strong>: Newton-meters (N\u22c5m) or foot-pounds (ft\u22c5lb)<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Example<\/strong>: Longer wrench = greater torque. That&#8217;s why you can tighten a bolt more easily with a longer wrench\u2014you&#8217;re increasing [latex]|\\mathbf{r}|[\/latex].<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Are the vectors [latex]{\\bf{a}}={\\bf{i}}+{\\bf{j}}-{\\bf{k}}[\/latex], [latex]{\\bf{b}}={\\bf{i}}-{\\bf{j}}+{\\bf{k}}[\/latex], and [latex]{\\bf{c}}={\\bf{i}}+{\\bf{j}}+{\\bf{k}}[\/latex] coplanar?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q308475728\">Show Solution<\/button><\/p>\n<div id=\"q308475728\" class=\"hidden-answer\" style=\"display: none\">No, the triple scalar product is [latex]-4\\ne0[\/latex], so the three vectors form the adjacent edges of a parallelepiped. They are not coplanar.<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Calculate the force required to produce [latex]15\\text{N}\\cdot\\text{m}[\/latex] torque at an angle of [latex]30^{\\small\\circ}[\/latex] from a [latex]150[\/latex]-cm rod.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q473650281\">Show Solution<\/button><\/p>\n<div id=\"q473650281\" class=\"hidden-answer\" style=\"display: none\">[latex]20[\/latex] 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=7753561&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=VgSpno5eD9k&amp;video_target=tpm-plugin-ydgl3n50-VgSpno5eD9k\" 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\/CP2.42_transcript.html\">\u201cCP 2.42\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":10,"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-522","chapter","type-chapter","status-publish","hentry"],"part":422,"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\/522","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":6,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/522\/revisions"}],"predecessor-version":[{"id":1749,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/522\/revisions\/1749"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/422"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/522\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=522"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=522"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=522"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=522"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}