{"id":532,"date":"2025-10-09T15:18:42","date_gmt":"2025-10-09T15:18:42","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=532"},"modified":"2025-10-24T13:06:03","modified_gmt":"2025-10-24T13:06:03","slug":"lines-and-planes-in-space-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/lines-and-planes-in-space-fresh-take\/","title":{"raw":"Lines and Planes in Space: Fresh Take","rendered":"Lines and Planes in Space: Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Write different types of equations (vector, parametric, symmetric) to describe lines in space<\/li>\r\n \t<li>Find the distance from a point to a given line<\/li>\r\n \t<li>Write equations for planes using vectors and scalars<\/li>\r\n \t<li>Find the angle between two planes<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2 data-type=\"title\">Equations for a Line in Space<\/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\">In 2D, you describe lines using slope. In 3D, you use a <strong>direction vector<\/strong> parallel to the line instead.<\/p>\r\n<p class=\"whitespace-normal break-words\">Two nonzero vectors [latex]\\mathbf{u}[\/latex] and [latex]\\mathbf{v}[\/latex] are <strong>parallel<\/strong> if [latex]\\mathbf{u} = k\\mathbf{v}[\/latex] for some scalar [latex]k[\/latex]. When [latex]k &gt; 0[\/latex], they point the same way; when [latex]k &lt; 0[\/latex], they point opposite ways.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Three Forms of Line Equations<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">A line through point [latex]P(x_0, y_0, z_0)[\/latex] with direction vector [latex]\\mathbf{v} = \\langle a, b, c \\rangle[\/latex]:<\/p>\r\n\r\n<ul>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Vector form<\/strong>: [latex]\\mathbf{r} = \\mathbf{r}_0 + t\\mathbf{v}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Parametric form<\/strong>: [latex]x = x_0 + ta, \\quad y = y_0 + tb, \\quad z = z_0 + tc[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Symmetric form<\/strong>: [latex]\\frac{x - x_0}{a} = \\frac{y - y_0}{b} = \\frac{z - z_0}{c}[\/latex] (when [latex]a, b, c \\neq 0[\/latex])<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Finding the Direction Vector<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Given two points on the line, form the direction vector by subtracting coordinates:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\mathbf{v} = \\langle x_1 - x_0, y_1 - y_0, z_1 - z_0 \\rangle[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Line Segments<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">To describe just the segment between points [latex]P[\/latex] and [latex]Q[\/latex], restrict the parameter: [latex]0 \\leq t \\leq 1[\/latex].<\/p>\r\n\r\n<ul>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Vector form<\/strong>: [latex]\\mathbf{r} = (1-t)\\mathbf{p} + t\\mathbf{q}, \\quad 0 \\leq t \\leq 1[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Parametric form<\/strong>: [latex]x = x_0 + t(x_1 - x_0)[\/latex], [latex]y = y_0 + t(y_1 - y_0)[\/latex], [latex]z = z_0 + t(z_1 - z_0)[\/latex], [latex]0 \\leq t \\leq 1[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">At [latex]t = 0[\/latex], you're at [latex]P[\/latex]; at [latex]t = 1[\/latex], you're at [latex]Q[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Line equations aren't unique\u2014different direction vectors or reference points produce different-looking equations for the same line.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Find parametric and symmetric equations of the line passing through points [latex](1, -3, 2)[\/latex] and [latex](5, -2 ,8)[\/latex].[reveal-answer q=\"586348756\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"586348756\"]\r\n<p id=\"fs-id1163723281622\">Possible set of parametric equations: [latex]x=1+4t[\/latex],\u00a0[latex]y=-3+t[\/latex], [latex]z=2+6t[\/latex];<\/p>\r\n<p id=\"fs-id1163723902128\">related set of symmetric equations: [latex]\\frac{x-1}4=y+3=\\frac{z-2}6[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Find parametric equations of the line segment between points [latex]P(-1, 3, 6)[\/latex] and\u00a0[latex]Q(-8, 2, 4)[\/latex].[reveal-answer q=\"745628734\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"745628734\"]\r\n\r\n[latex]x=-1-7t[\/latex],\u00a0[latex]y=3-t[\/latex],\u00a0[latex]z=6-2t[\/latex],\u00a0[latex]0\\leq t\\leq1[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<h2 data-type=\"title\">Distance between a Point and a Line<\/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\">Finding the distance between a point and a line in 3D has practical applications\u2014from construction setback requirements to airline flight path planning.<\/p>\r\n<p class=\"whitespace-normal break-words\">The <strong>distance from a point to a line<\/strong> is the length of the shortest segment connecting them. This shortest segment is always perpendicular to the line.<\/p>\r\n<p class=\"whitespace-normal break-words\">In 3D, there's no obvious way to identify which point on the line creates the perpendicular segment. We solve this using the cross product.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The Formula<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Let [latex]L[\/latex] be a line through point [latex]P[\/latex] with direction vector [latex]\\mathbf{v}[\/latex]. For any point [latex]M[\/latex] not on the line:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]d = \\frac{||\\overrightarrow{PM} \\times \\mathbf{v}||}{||\\mathbf{v}||}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Why This Works<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Vectors [latex]\\overrightarrow{PM}[\/latex] and [latex]\\mathbf{v}[\/latex] form two sides of a parallelogram:<\/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>Area of parallelogram<\/strong>: [latex]||\\overrightarrow{PM} \\times \\mathbf{v}||[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Also area<\/strong>: base \u00d7 height = [latex]||\\mathbf{v}|| \\cdot d[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">Setting these equal and solving for [latex]d[\/latex] gives the formula. The cross product's magnitude captures the parallelogram area, and dividing by the base [latex]||\\mathbf{v}||[\/latex] isolates the height\u2014exactly the perpendicular distance we need.<\/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\">Choose any point [latex]P[\/latex] on the line<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Form vector [latex]\\overrightarrow{PM}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Find the direction vector [latex]\\mathbf{v}[\/latex] for the line<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Calculate [latex]\\overrightarrow{PM} \\times \\mathbf{v}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Apply the formula: [latex]d = \\frac{||\\overrightarrow{PM} \\times \\mathbf{v}||}{||\\mathbf{v}||}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Find the distance between point [latex](0, 3, 6)[\/latex] and the line with parametric equations [latex]x=1-t[\/latex],\u00a0[latex]y=1+2t[\/latex],\u00a0[latex]z=5+3t[\/latex].[reveal-answer q=\"367450281\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"367450281\"]\r\n\r\n[latex]\\sqrt{\\frac{10}7}[\/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=7753584&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=gvmJ46Cromg&amp;video_target=tpm-plugin-4lq369lh-gvmJ46Cromg\" 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\/CP2.45_transcript.html\">\u201cCP 2.45\u201d here (opens in new window).<\/a><\/center><\/section>\r\n<h2 data-type=\"title\">Relationships between Lines<\/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\">In 2D, two lines are either identical, parallel, or intersecting. Three dimensions adds a fourth possibility that can't exist in a plane.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Four Possible Relationships<\/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\"><strong>Equal lines<\/strong>: Same direction, same points<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Parallel lines<\/strong>: Same direction, never meet<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Intersecting lines<\/strong>: Cross at exactly one point<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Skew lines<\/strong>: Not parallel and don't intersect (unique to 3D!)<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>Skew lines<\/strong> have different directions but never meet\u2014like one line running east-west on a building's ground floor while another runs north-south on the second floor. They \"miss\" each other in space. In 2D, non-parallel lines must intersect. In 3D, they can miss each other entirely\u2014that's what makes skew lines possible.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Classification Strategy<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Answer two questions to determine the relationship:<\/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\"><strong>Are direction vectors parallel?<\/strong> (Check if [latex]\\mathbf{v}_1 = k\\mathbf{v}_2[\/latex] for some scalar [latex]k[\/latex])<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Do the lines share a point?<\/strong> (Set parametric equations equal and solve)<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>Decision Table<\/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>Parallel directions + shared point<\/strong> \u2192 Equal lines<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Parallel directions + no shared point<\/strong> \u2192 Parallel but distinct<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Non-parallel directions + shared point<\/strong> \u2192 Intersecting lines<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Non-parallel directions + no shared point<\/strong> \u2192 Skew lines<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Testing for a Shared Point<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Set the parametric equations equal and solve the system:<\/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\">If you find consistent values for both parameters, the lines intersect at that point<\/li>\r\n \t<li class=\"whitespace-normal break-words\">If you get a contradiction, the lines are skew (or parallel if directions match)<\/li>\r\n<\/ul>\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<p id=\"fs-id1163724065582\">Describe the relationship between the lines with the following parametric equations:<\/p>\r\n<p style=\"text-align: center;\">[latex]x=1-4t[\/latex],\u00a0[latex]y=3+t[\/latex],\u00a0[latex]z=8-6t[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]x=2+3s[\/latex],\u00a0[latex]y=2s[\/latex],\u00a0[latex]z=-1-3s[\/latex].<\/p>\r\n[reveal-answer q=\"389456729\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"389456729\"]\r\n<div id=\"fs-id1163724065578-solution\" data-type=\"solution\">\r\n<div class=\"os-solution-container\">\r\n<p id=\"fs-id1163723110189\">These lines are skew because their direction vectors are not parallel and there is no point [latex](x, y, z)[\/latex] that lies on both lines.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<h2 data-type=\"title\">Equations for a Plane<\/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\">A plane is uniquely determined by three non-collinear points, but the most powerful way to describe a plane uses a <strong>normal vector<\/strong>\u2014a vector perpendicular to the plane.<\/p>\r\n<p class=\"whitespace-normal break-words\">A point [latex]Q[\/latex] lies on a plane if and only if the vector [latex]\\overrightarrow{PQ}[\/latex] (from a fixed point [latex]P[\/latex] on the plane to [latex]Q[\/latex]) is perpendicular to the normal vector [latex]\\mathbf{n}[\/latex]. Since orthogonal vectors have dot product zero: [latex]\\mathbf{n} \\cdot \\overrightarrow{PQ} = 0[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Three Forms of Plane Equations<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Given point [latex]P = (x_0, y_0, z_0)[\/latex] on the plane and normal vector [latex]\\mathbf{n} = \\langle a, b, c \\rangle[\/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\"><strong>Vector equation<\/strong>: [latex]\\mathbf{n} \\cdot \\overrightarrow{PQ} = 0[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Scalar equation<\/strong>: [latex]a(x - x_0) + b(y - y_0) + c(z - z_0) = 0[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>General form<\/strong>: [latex]ax + by + cz + d = 0[\/latex] where [latex]d = -ax_0 - by_0 - cz_0[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Finding a Plane Through Three Points<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Use the cross product to find a normal vector:<\/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 two vectors from the three points (e.g., [latex]\\overrightarrow{PQ}[\/latex] and [latex]\\overrightarrow{PR}[\/latex])<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Calculate [latex]\\mathbf{n} = \\overrightarrow{PQ} \\times \\overrightarrow{PR}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Use [latex]\\mathbf{n}[\/latex] and any of the three points to write the plane equation<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>Distance from a Point to a Plane<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">The shortest distance from point [latex]P[\/latex] to a plane is measured perpendicular to the plane. For a plane with normal vector [latex]\\mathbf{n}[\/latex] passing through point [latex]Q[\/latex]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]d = \\frac{|\\overrightarrow{QP} \\cdot \\mathbf{n}|}{||\\mathbf{n}||}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">This is the scalar projection of [latex]\\overrightarrow{QP}[\/latex] onto [latex]\\mathbf{n}[\/latex]\u2014the component of [latex]\\overrightarrow{QP}[\/latex] perpendicular to the plane.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<p id=\"fs-id1163724087484\">Find an equation of the plane containing the lines [latex]L_1[\/latex] and\u00a0[latex]L_2[\/latex]:<\/p>\r\n[latex]L_1[\/latex]:\u00a0[latex]x=-y=z[\/latex]\r\n\r\n[latex]L_2[\/latex]:\u00a0[latex]\\frac{x-3}2=y=z-2[\/latex]\r\n\r\n[reveal-answer q=\"498652076\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"498652076\"]\r\n\r\n[latex]-2(x-1)+(y+1)+3(z-1)=0[\/latex] or\u00a0[latex]-2x+y+3z=0[\/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=7753585&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=0dYLxxAAgV4&amp;video_target=tpm-plugin-79t9enhx-0dYLxxAAgV4\" 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\/CP2.47_transcript.html\">\u201cCP 2.47\u201d here (opens in new window).<\/a><\/center><\/section><section class=\"textbox example\" aria-label=\"Example\">Find the distance between point [latex]P=(5, -1, 0)[\/latex] and the plane given by [latex]4x+2y-z=3[\/latex].[reveal-answer q=\"348756277\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"348756277\"]\r\n\r\n[latex]\\frac{15}{\\sqrt{21}}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<h2 data-type=\"title\">Parallel and Intersecting Planes<\/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\">Unlike lines (which can be parallel, intersecting, or skew), planes have only two possible relationships: they're either <strong>parallel<\/strong> or they <strong>intersect<\/strong>. There's no such thing as skew planes.<\/p>\r\n<p class=\"whitespace-normal break-words\">Planes are parallel when their normal vectors are parallel: [latex]\\mathbf{n}_1 = k\\mathbf{n}_2[\/latex] for some scalar [latex]k[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">When planes aren't parallel, they must intersect along a line. The intersection is always a straight line, never just a point or scattered points.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Finding the Line of Intersection<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">To find parametric equations for the intersection line:<\/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\">Use both plane equations simultaneously<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Eliminate one variable by adding\/subtracting equations<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Express two variables in terms of the third<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Define the third variable as parameter [latex]t[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>Angle Between Planes<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">The angle between two planes equals the angle between their normal vectors:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\cos \\theta = \\frac{|\\mathbf{n}_1 \\cdot \\mathbf{n}_2|}{||\\mathbf{n}_1||\u00a0 ||\\mathbf{n}_2||}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">We use absolute value to ensure [latex]0\u00b0 \\leq \\theta \\leq 90\u00b0[\/latex].<\/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\"><strong>Parallel<\/strong>: [latex]\\mathbf{n}_1 \\times \\mathbf{n}_2 = \\mathbf{0}[\/latex] (cross product is zero)<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Perpendicular<\/strong>: [latex]\\mathbf{n}_1 \\cdot \\mathbf{n}_2 = 0[\/latex] (dot product is zero)<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Distance Between Parallel Planes<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">For point [latex]P(x_0, y_0, z_0)[\/latex] and plane [latex]ax + by + cz + k = 0[\/latex]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]d = \\frac{|ax_0 + by_0 + cz_0 + k|}{\\sqrt{a^2 + b^2 + c^2}}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">If planes are [latex]ax + by + cz + k_1 = 0[\/latex] and [latex]ax + by + cz + k_2 = 0[\/latex], the distance is [latex]\\frac{|k_1 - k_2|}{\\sqrt{a^2 + b^2 + c^2}}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Find parametric equations for the line formed by the intersection of planes [latex]x+y-z=3[\/latex] and\u00a0[latex]3x-y+3z=5[\/latex].[reveal-answer q=\"374865270\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"374865270\"]\r\n\r\n[latex]x=t[\/latex],\u00a0[latex]y=7-3t[\/latex],\u00a0[latex]z=4-2t[\/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=7753586&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=RjKvharjDJI&amp;video_target=tpm-plugin-e3oqvxiw-RjKvharjDJI\" 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\/CP2.49_transcript.html\"> \u201cCP 2.49\u201d here (opens in new window).<\/a><\/center><\/section><section class=\"textbox example\" aria-label=\"Example\">Find the measure of the angle between planes [latex]x+y-z=3[\/latex] and [latex]3x-y+3z=5[\/latex]. Give the answer in radians and round to two decimal places.[reveal-answer q=\"174365292\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"174365292\"]\r\n\r\n[latex]1.44[\/latex] rad\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Find the distance between parallel planes [latex]5x-2y+z=6[\/latex] and\u00a0[latex]5x-2y+z=-3[\/latex].[reveal-answer q=\"437562978\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"437562978\"]\r\n\r\n[latex]\\frac{9}{\\sqrt{30}}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Write different types of equations (vector, parametric, symmetric) to describe lines in space<\/li>\n<li>Find the distance from a point to a given line<\/li>\n<li>Write equations for planes using vectors and scalars<\/li>\n<li>Find the angle between two planes<\/li>\n<\/ul>\n<\/section>\n<h2 data-type=\"title\">Equations for a Line in Space<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">In 2D, you describe lines using slope. In 3D, you use a <strong>direction vector<\/strong> parallel to the line instead.<\/p>\n<p class=\"whitespace-normal break-words\">Two nonzero vectors [latex]\\mathbf{u}[\/latex] and [latex]\\mathbf{v}[\/latex] are <strong>parallel<\/strong> if [latex]\\mathbf{u} = k\\mathbf{v}[\/latex] for some scalar [latex]k[\/latex]. When [latex]k > 0[\/latex], they point the same way; when [latex]k < 0[\/latex], they point opposite ways.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Three Forms of Line Equations<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">A line through point [latex]P(x_0, y_0, z_0)[\/latex] with direction vector [latex]\\mathbf{v} = \\langle a, b, c \\rangle[\/latex]:<\/p>\n<ul>\n<li class=\"whitespace-normal break-words\"><strong>Vector form<\/strong>: [latex]\\mathbf{r} = \\mathbf{r}_0 + t\\mathbf{v}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Parametric form<\/strong>: [latex]x = x_0 + ta, \\quad y = y_0 + tb, \\quad z = z_0 + tc[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Symmetric form<\/strong>: [latex]\\frac{x - x_0}{a} = \\frac{y - y_0}{b} = \\frac{z - z_0}{c}[\/latex] (when [latex]a, b, c \\neq 0[\/latex])<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Finding the Direction Vector<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Given two points on the line, form the direction vector by subtracting coordinates:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\mathbf{v} = \\langle x_1 - x_0, y_1 - y_0, z_1 - z_0 \\rangle[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Line Segments<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">To describe just the segment between points [latex]P[\/latex] and [latex]Q[\/latex], restrict the parameter: [latex]0 \\leq t \\leq 1[\/latex].<\/p>\n<ul>\n<li class=\"whitespace-normal break-words\"><strong>Vector form<\/strong>: [latex]\\mathbf{r} = (1-t)\\mathbf{p} + t\\mathbf{q}, \\quad 0 \\leq t \\leq 1[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Parametric form<\/strong>: [latex]x = x_0 + t(x_1 - x_0)[\/latex], [latex]y = y_0 + t(y_1 - y_0)[\/latex], [latex]z = z_0 + t(z_1 - z_0)[\/latex], [latex]0 \\leq t \\leq 1[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">At [latex]t = 0[\/latex], you&#8217;re at [latex]P[\/latex]; at [latex]t = 1[\/latex], you&#8217;re at [latex]Q[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Line equations aren&#8217;t unique\u2014different direction vectors or reference points produce different-looking equations for the same line.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Find parametric and symmetric equations of the line passing through points [latex](1, -3, 2)[\/latex] and [latex](5, -2 ,8)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q586348756\">Show Solution<\/button><\/p>\n<div id=\"q586348756\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1163723281622\">Possible set of parametric equations: [latex]x=1+4t[\/latex],\u00a0[latex]y=-3+t[\/latex], [latex]z=2+6t[\/latex];<\/p>\n<p id=\"fs-id1163723902128\">related set of symmetric equations: [latex]\\frac{x-1}4=y+3=\\frac{z-2}6[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Find parametric equations of the line segment between points [latex]P(-1, 3, 6)[\/latex] and\u00a0[latex]Q(-8, 2, 4)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q745628734\">Show Solution<\/button><\/p>\n<div id=\"q745628734\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=-1-7t[\/latex],\u00a0[latex]y=3-t[\/latex],\u00a0[latex]z=6-2t[\/latex],\u00a0[latex]0\\leq t\\leq1[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<h2 data-type=\"title\">Distance between a Point and a Line<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Finding the distance between a point and a line in 3D has practical applications\u2014from construction setback requirements to airline flight path planning.<\/p>\n<p class=\"whitespace-normal break-words\">The <strong>distance from a point to a line<\/strong> is the length of the shortest segment connecting them. This shortest segment is always perpendicular to the line.<\/p>\n<p class=\"whitespace-normal break-words\">In 3D, there&#8217;s no obvious way to identify which point on the line creates the perpendicular segment. We solve this using the cross product.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The Formula<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Let [latex]L[\/latex] be a line through point [latex]P[\/latex] with direction vector [latex]\\mathbf{v}[\/latex]. For any point [latex]M[\/latex] not on the line:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]d = \\frac{||\\overrightarrow{PM} \\times \\mathbf{v}||}{||\\mathbf{v}||}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Why This Works<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Vectors [latex]\\overrightarrow{PM}[\/latex] and [latex]\\mathbf{v}[\/latex] form two sides of a parallelogram:<\/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>Area of parallelogram<\/strong>: [latex]||\\overrightarrow{PM} \\times \\mathbf{v}||[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Also area<\/strong>: base \u00d7 height = [latex]||\\mathbf{v}|| \\cdot d[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">Setting these equal and solving for [latex]d[\/latex] gives the formula. The cross product&#8217;s magnitude captures the parallelogram area, and dividing by the base [latex]||\\mathbf{v}||[\/latex] isolates the height\u2014exactly the perpendicular distance we need.<\/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\">Choose any point [latex]P[\/latex] on the line<\/li>\n<li class=\"whitespace-normal break-words\">Form vector [latex]\\overrightarrow{PM}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Find the direction vector [latex]\\mathbf{v}[\/latex] for the line<\/li>\n<li class=\"whitespace-normal break-words\">Calculate [latex]\\overrightarrow{PM} \\times \\mathbf{v}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Apply the formula: [latex]d = \\frac{||\\overrightarrow{PM} \\times \\mathbf{v}||}{||\\mathbf{v}||}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Find the distance between point [latex](0, 3, 6)[\/latex] and the line with parametric equations [latex]x=1-t[\/latex],\u00a0[latex]y=1+2t[\/latex],\u00a0[latex]z=5+3t[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q367450281\">Show Solution<\/button><\/p>\n<div id=\"q367450281\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\sqrt{\\frac{10}7}[\/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=7753584&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=gvmJ46Cromg&amp;video_target=tpm-plugin-4lq369lh-gvmJ46Cromg\" 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\/CP2.45_transcript.html\">\u201cCP 2.45\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<h2 data-type=\"title\">Relationships between Lines<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">In 2D, two lines are either identical, parallel, or intersecting. Three dimensions adds a fourth possibility that can&#8217;t exist in a plane.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Four Possible Relationships<\/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\"><strong>Equal lines<\/strong>: Same direction, same points<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Parallel lines<\/strong>: Same direction, never meet<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Intersecting lines<\/strong>: Cross at exactly one point<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Skew lines<\/strong>: Not parallel and don&#8217;t intersect (unique to 3D!)<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>Skew lines<\/strong> have different directions but never meet\u2014like one line running east-west on a building&#8217;s ground floor while another runs north-south on the second floor. They &#8220;miss&#8221; each other in space. In 2D, non-parallel lines must intersect. In 3D, they can miss each other entirely\u2014that&#8217;s what makes skew lines possible.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Classification Strategy<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Answer two questions to determine the relationship:<\/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\"><strong>Are direction vectors parallel?<\/strong> (Check if [latex]\\mathbf{v}_1 = k\\mathbf{v}_2[\/latex] for some scalar [latex]k[\/latex])<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Do the lines share a point?<\/strong> (Set parametric equations equal and solve)<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>Decision Table<\/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>Parallel directions + shared point<\/strong> \u2192 Equal lines<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Parallel directions + no shared point<\/strong> \u2192 Parallel but distinct<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Non-parallel directions + shared point<\/strong> \u2192 Intersecting lines<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Non-parallel directions + no shared point<\/strong> \u2192 Skew lines<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Testing for a Shared Point<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Set the parametric equations equal and solve the system:<\/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\">If you find consistent values for both parameters, the lines intersect at that point<\/li>\n<li class=\"whitespace-normal break-words\">If you get a contradiction, the lines are skew (or parallel if directions match)<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p id=\"fs-id1163724065582\">Describe the relationship between the lines with the following parametric equations:<\/p>\n<p style=\"text-align: center;\">[latex]x=1-4t[\/latex],\u00a0[latex]y=3+t[\/latex],\u00a0[latex]z=8-6t[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]x=2+3s[\/latex],\u00a0[latex]y=2s[\/latex],\u00a0[latex]z=-1-3s[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q389456729\">Show Solution<\/button><\/p>\n<div id=\"q389456729\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1163724065578-solution\" data-type=\"solution\">\n<div class=\"os-solution-container\">\n<p id=\"fs-id1163723110189\">These lines are skew because their direction vectors are not parallel and there is no point [latex](x, y, z)[\/latex] that lies on both lines.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<h2 data-type=\"title\">Equations for a Plane<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">A plane is uniquely determined by three non-collinear points, but the most powerful way to describe a plane uses a <strong>normal vector<\/strong>\u2014a vector perpendicular to the plane.<\/p>\n<p class=\"whitespace-normal break-words\">A point [latex]Q[\/latex] lies on a plane if and only if the vector [latex]\\overrightarrow{PQ}[\/latex] (from a fixed point [latex]P[\/latex] on the plane to [latex]Q[\/latex]) is perpendicular to the normal vector [latex]\\mathbf{n}[\/latex]. Since orthogonal vectors have dot product zero: [latex]\\mathbf{n} \\cdot \\overrightarrow{PQ} = 0[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Three Forms of Plane Equations<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Given point [latex]P = (x_0, y_0, z_0)[\/latex] on the plane and normal vector [latex]\\mathbf{n} = \\langle a, b, c \\rangle[\/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\"><strong>Vector equation<\/strong>: [latex]\\mathbf{n} \\cdot \\overrightarrow{PQ} = 0[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Scalar equation<\/strong>: [latex]a(x - x_0) + b(y - y_0) + c(z - z_0) = 0[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>General form<\/strong>: [latex]ax + by + cz + d = 0[\/latex] where [latex]d = -ax_0 - by_0 - cz_0[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Finding a Plane Through Three Points<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Use the cross product to find a normal vector:<\/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 two vectors from the three points (e.g., [latex]\\overrightarrow{PQ}[\/latex] and [latex]\\overrightarrow{PR}[\/latex])<\/li>\n<li class=\"whitespace-normal break-words\">Calculate [latex]\\mathbf{n} = \\overrightarrow{PQ} \\times \\overrightarrow{PR}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Use [latex]\\mathbf{n}[\/latex] and any of the three points to write the plane equation<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>Distance from a Point to a Plane<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The shortest distance from point [latex]P[\/latex] to a plane is measured perpendicular to the plane. For a plane with normal vector [latex]\\mathbf{n}[\/latex] passing through point [latex]Q[\/latex]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]d = \\frac{|\\overrightarrow{QP} \\cdot \\mathbf{n}|}{||\\mathbf{n}||}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">This is the scalar projection of [latex]\\overrightarrow{QP}[\/latex] onto [latex]\\mathbf{n}[\/latex]\u2014the component of [latex]\\overrightarrow{QP}[\/latex] perpendicular to the plane.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p id=\"fs-id1163724087484\">Find an equation of the plane containing the lines [latex]L_1[\/latex] and\u00a0[latex]L_2[\/latex]:<\/p>\n<p>[latex]L_1[\/latex]:\u00a0[latex]x=-y=z[\/latex]<\/p>\n<p>[latex]L_2[\/latex]:\u00a0[latex]\\frac{x-3}2=y=z-2[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q498652076\">Show Solution<\/button><\/p>\n<div id=\"q498652076\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]-2(x-1)+(y+1)+3(z-1)=0[\/latex] or\u00a0[latex]-2x+y+3z=0[\/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=7753585&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=0dYLxxAAgV4&amp;video_target=tpm-plugin-79t9enhx-0dYLxxAAgV4\" 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\/CP2.47_transcript.html\">\u201cCP 2.47\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Find the distance between point [latex]P=(5, -1, 0)[\/latex] and the plane given by [latex]4x+2y-z=3[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q348756277\">Show Solution<\/button><\/p>\n<div id=\"q348756277\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{15}{\\sqrt{21}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<h2 data-type=\"title\">Parallel and Intersecting Planes<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Unlike lines (which can be parallel, intersecting, or skew), planes have only two possible relationships: they&#8217;re either <strong>parallel<\/strong> or they <strong>intersect<\/strong>. There&#8217;s no such thing as skew planes.<\/p>\n<p class=\"whitespace-normal break-words\">Planes are parallel when their normal vectors are parallel: [latex]\\mathbf{n}_1 = k\\mathbf{n}_2[\/latex] for some scalar [latex]k[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">When planes aren&#8217;t parallel, they must intersect along a line. The intersection is always a straight line, never just a point or scattered points.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Finding the Line of Intersection<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">To find parametric equations for the intersection line:<\/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\">Use both plane equations simultaneously<\/li>\n<li class=\"whitespace-normal break-words\">Eliminate one variable by adding\/subtracting equations<\/li>\n<li class=\"whitespace-normal break-words\">Express two variables in terms of the third<\/li>\n<li class=\"whitespace-normal break-words\">Define the third variable as parameter [latex]t[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>Angle Between Planes<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The angle between two planes equals the angle between their normal vectors:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\cos \\theta = \\frac{|\\mathbf{n}_1 \\cdot \\mathbf{n}_2|}{||\\mathbf{n}_1||\u00a0 ||\\mathbf{n}_2||}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">We use absolute value to ensure [latex]0\u00b0 \\leq \\theta \\leq 90\u00b0[\/latex].<\/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\"><strong>Parallel<\/strong>: [latex]\\mathbf{n}_1 \\times \\mathbf{n}_2 = \\mathbf{0}[\/latex] (cross product is zero)<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Perpendicular<\/strong>: [latex]\\mathbf{n}_1 \\cdot \\mathbf{n}_2 = 0[\/latex] (dot product is zero)<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Distance Between Parallel Planes<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">For point [latex]P(x_0, y_0, z_0)[\/latex] and plane [latex]ax + by + cz + k = 0[\/latex]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]d = \\frac{|ax_0 + by_0 + cz_0 + k|}{\\sqrt{a^2 + b^2 + c^2}}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">If planes are [latex]ax + by + cz + k_1 = 0[\/latex] and [latex]ax + by + cz + k_2 = 0[\/latex], the distance is [latex]\\frac{|k_1 - k_2|}{\\sqrt{a^2 + b^2 + c^2}}[\/latex].<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Find parametric equations for the line formed by the intersection of planes [latex]x+y-z=3[\/latex] and\u00a0[latex]3x-y+3z=5[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q374865270\">Show Solution<\/button><\/p>\n<div id=\"q374865270\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=t[\/latex],\u00a0[latex]y=7-3t[\/latex],\u00a0[latex]z=4-2t[\/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=7753586&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=RjKvharjDJI&amp;video_target=tpm-plugin-e3oqvxiw-RjKvharjDJI\" 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\/CP2.49_transcript.html\"> \u201cCP 2.49\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Find the measure of the angle between planes [latex]x+y-z=3[\/latex] and [latex]3x-y+3z=5[\/latex]. Give the answer in radians and round to two decimal places.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q174365292\">Show Solution<\/button><\/p>\n<div id=\"q174365292\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]1.44[\/latex] rad<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Find the distance between parallel planes [latex]5x-2y+z=6[\/latex] and\u00a0[latex]5x-2y+z=-3[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q437562978\">Show Solution<\/button><\/p>\n<div id=\"q437562978\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{9}{\\sqrt{30}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":17,"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-532","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\/532","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":4,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/532\/revisions"}],"predecessor-version":[{"id":1750,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/532\/revisions\/1750"}],"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\/532\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=532"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=532"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=532"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=532"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}