{"id":812,"date":"2021-08-27T19:56:00","date_gmt":"2021-08-27T19:56:00","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=812"},"modified":"2024-10-09T17:12:35","modified_gmt":"2024-10-09T17:12:35","slug":"equations-for-a-plane-in-space","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/equations-for-a-plane-in-space\/","title":{"raw":"Equations for a Plane in Space","rendered":"Equations for a Plane in Space"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul class=\"os-abstract\">\r\n \t<li><span class=\"os-abstract-content\">Write the vector and scalar equations of a plane through a given point with a given normal.<\/span><\/li>\r\n \t<li><span class=\"os-abstract-content\">Find the distance from a point to a given plane.<\/span><\/li>\r\n \t<li><span class=\"os-abstract-content\">Find the angle between two planes.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<section id=\"fs-id1163723141280\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Equations for a Plane<\/h2>\r\n<p id=\"fs-id1163723141285\">We know that a line is determined by two points. In other words, for any two distinct points, there is exactly one line that passes through those points, whether in two dimensions or three. Similarly, given any three points that do not all lie on the same line, there is a unique plane that passes through these points. Just as a line is determined by two points, a plane is determined by three.<\/p>\r\n<p id=\"fs-id1163723141292\">This may be the simplest way to characterize a plane, but we can use other descriptions as well. For example, given two distinct, intersecting lines, there is exactly one plane containing both lines. A plane is also determined by a line and any point that does not lie on the line. These characterizations arise naturally from the idea that a plane is determined by three points. Perhaps the most surprising characterization of a plane is actually the most useful.<\/p>\r\n\r\n<\/section>\r\n<div data-type=\"example\">\r\n<p id=\"fs-id1163723141299\">Imagine a pair of orthogonal vectors that share an initial point. Visualize grabbing one of the vectors and twisting it. As you twist, the other vector spins around and sweeps out a plane. Here, we describe that concept mathematically. Let [latex]{\\bf{n}}=\\langle a,b,c\\rangle[\/latex] be a vector and [latex]P=(x_0, y_0, z_0)[\/latex] be a point. Then the set of all points [latex]Q=(x, y, z)[\/latex] such that [latex]\\overrightarrow{PQ}[\/latex] is orthogonal to [latex]\\textbf n[\/latex]\u00a0<span style=\"font-size: 1rem; text-align: initial;\">forms a plane (<\/span>Figure 1<span style=\"font-size: 1rem; text-align: initial;\">). We say that [latex]\\textbf n[\/latex] is a\u00a0<\/span><strong style=\"font-size: 1rem; text-align: initial;\"><span id=\"60ce88e7-9116-48d3-bb30-013c89cddc64_term92\" data-type=\"term\">normal vector<\/span>,<\/strong><span style=\"font-size: 1rem; text-align: initial;\"> or perpendicular to the plane. Remember, the dot product of orthogonal vectors is zero. This fact generates the\u00a0<\/span><strong style=\"font-size: 1rem; text-align: initial;\"><span id=\"60ce88e7-9116-48d3-bb30-013c89cddc64_term93\" data-type=\"term\">vector equation of a plane<\/span><\/strong><span style=\"font-size: 1rem; text-align: initial;\">: [latex]{\\bf{n}}\\cdot\\overrightarrow{PQ}=0[\/latex]. Rewriting this equation provides additional ways to describe the plane:<\/span><\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n{\\bf{n}}\\cdot\\overrightarrow{PQ}&amp;=0 \\\\\r\n\\langle a,b,c\\rangle\\cdot\\langle x-x_0,y-y_0,z-z_0\\rangle&amp;=0 \\\\\r\na(x-x_0)+b(y-y_0)+c(z-z_0)&amp;=0.\r\n\\end{aligned}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"os-figure\">[caption id=\"attachment_5173\" align=\"aligncenter\" width=\"441\"]<img class=\"size-full wp-image-5173\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27214230\/2.69.jpg\" alt=\"This figure is a parallelogram representing a plane. In the plane is a vector from point P to point Q. Perpendicular to the vector P Q is the vector n.\" width=\"441\" height=\"226\" \/> Figure 1. Given a point [latex]P[\/latex] and vector [latex]\\textbf n[\/latex], the set of all points [latex]Q[\/latex] with [latex]\\overrightarrow{PQ}[\/latex] orthogonal to [latex]\\textbf n[\/latex] forms a plane.[\/caption]<\/div>\r\n<div data-type=\"example\"><\/div>\r\n<div data-type=\"example\">\r\n<div class=\"textbox shaded\"><header>\r\n<h3 class=\"os-title\" style=\"text-align: center;\" data-type=\"title\"><span id=\"22\" class=\"os-title-label\" data-type=\"\">DEFINITION<\/span><\/h3>\r\n\r\n<hr \/>\r\n\r\n<\/header><section>\r\n<div class=\"os-note-body\">\r\n<p id=\"fs-id1163723211072\">Given a point [latex]P[\/latex] and vector [latex]\\textbf n[\/latex], the set of all points [latex]Q[\/latex] satisfying the equation [latex]{\\bf{n}}\\cdot\\overrightarrow{PQ}=0[\/latex] forms a plane. The equation<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\bf{n}}\\cdot\\overrightarrow{PQ}=0[\/latex].<\/p>\r\n<p id=\"fs-id1163724147360\">is known as the vector equation of a plane.<\/p>\r\n<p id=\"fs-id1163724062719\">The\u00a0<strong><span id=\"60ce88e7-9116-48d3-bb30-013c89cddc64_term94\" data-type=\"term\">scalar equation of a plane<\/span><\/strong>\u00a0containing point [latex]P=(x_0, y_0, z_0)[\/latex] with normal vector [latex]{\\bf{n}}=\\langle a,b,c\\rangle[\/latex] is<\/p>\r\n<p style=\"text-align: center;\">[latex]a(x-x_0)+b(y-y_0)+c(z-z_0)=0[\/latex].<\/p>\r\n<p id=\"fs-id1163723339633\">This equation can be expressed as [latex]ax+by+cz+d=0[\/latex], where [latex]d=-ax_0-by_0-cz_0[\/latex]. This form of the equation is sometimes called the<strong>\u00a0<span id=\"60ce88e7-9116-48d3-bb30-013c89cddc64_term95\" data-type=\"term\">general form of the equation of a plane<\/span><\/strong>.<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div>\r\nAs described earlier in this section, any three points that do not all lie on the same line determine a plane. Given three such points, we can find an equation for the plane containing these points.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: writing an equation of a plane given three points in the plane<\/h3>\r\nWrite an equation for the plane containing points [latex]P=(1,1,-2)[\/latex], [latex]Q=(0,2,1)[\/latex], and [latex]R=(-1,-1,0)[\/latex] in both standard and general forms.\r\n\r\n[reveal-answer q=\"374652897\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"374652897\"]\r\n<p id=\"fs-id1163723246705\">To write an equation for a plane, we must find a normal vector for the plane. We start by identifying two vectors in the plane:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\overrightarrow{PQ}&amp;=\\langle0-1,2-1,1-(-2)\\rangle&amp;=\\langle-1,1,3\\rangle \\\\\r\n\\overrightarrow{QR}&amp;=\\langle-1-0,-1-2,0-1\\rangle&amp;=\\langle-1,-3,-1\\rangle\r\n\\end{aligned}.[\/latex]<\/p>\r\n<p id=\"fs-id1163724184793\">The cross product [latex]\\overrightarrow{PQ}\\times\\overrightarrow{QR}[\/latex] is orthogonal to both [latex]\\overrightarrow{PQ}[\/latex] and [latex]\\overrightarrow{QR}[\/latex], so it is normal to the plane that contains these two vectors:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n{\\bf{n}}&amp;=\\overrightarrow{PQ}\\times\\overrightarrow{QR} \\\\\r\n&amp;=\\begin{vmatrix}{\\bf{i}}&amp;{\\bf{j}}&amp;{\\bf{k}} \\\\ -1&amp;1&amp;3\\\\ -1&amp;-3&amp;-1\\end{vmatrix} \\\\\r\n&amp;=(-1+9){\\bf{i}}-(1+3){\\bf{j}}+(3+1){\\bf{k}} \\\\\r\n&amp;=8{\\bf{i}}-4{\\bf{j}}+4{\\bf{k}}.\r\n\\end{aligned}[\/latex]<\/p>\r\n<p id=\"fs-id1163723128152\">Thus, [latex]{\\bf{n}}=\\langle8,-4,4\\rangle[\/latex], and we can choose any of the three given points to write an equation of the plane:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n8(x-1)-4(y-1)+4(z+2)&amp;=0 \\\\\r\n8x-4y+4z+4&amp;=0.\r\n\\end{aligned}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe scalar equations of a plane vary depending on the normal vector and point chosen.\r\n\r\n<\/div>\r\n<div data-type=\"example\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: writing an equation for a plane given a point and a line<\/h3>\r\nFind an equation of the plane that passes through point [latex](1, 4, 3)[\/latex] and contains the line given by [latex]x=\\frac{y-1}2=z+1[\/latex].\r\n\r\n[reveal-answer q=\"289161783\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"289161783\"]\r\n<p id=\"fs-id1163723216073\">Symmetric equations describe the line that passes through point [latex](0, 1, -1)[\/latex] parallel to vector [latex]{\\bf{v}}_1=\\langle1,2,1\\rangle[\/latex] (see the following figure). Use this point and the given point, [latex](1, 4, 3)[\/latex], to identify a second vector parallel to the plane:<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\bf{v}}_2=\\langle1-0,4-1,3-(-1)\\rangle=\\langle1,3,4\\rangle[\/latex].<\/p>\r\n<p id=\"fs-id1163723199146\">Use the cross product of these vectors to identify a normal vector for the plane:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n{\\bf{n}}&amp;={\\bf{v}}_1\\times{\\bf{v}}_2 \\\\\r\n&amp;=\\begin{vmatrix}{\\bf{i}}&amp;{\\bf{j}}&amp;{\\bf{k}} \\\\1&amp;2&amp;1 \\\\1&amp;3&amp;4\\end{vmatrix} \\\\\r\n&amp;=(8-3){\\bf{i}}-(4-1){\\bf{j}}+(3-2){\\bf{k}} \\\\\r\n&amp;=5{\\bf{i}}-3{\\bf{j}}+{\\bf{k}}\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1163723277829\">The scalar equations for the plane are [latex]5x-3(y-1)+(z+1)=0[\/latex] and\u00a0[latex]5x-3y+z+4=0[\/latex].<\/p>\r\n\r\n[caption id=\"attachment_5180\" align=\"aligncenter\" width=\"387\"]<img class=\"size-full wp-image-5180\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31200056\/Example-2.50.jpg\" alt=\"This figure is the 3-dimensional coordinate system. There is a plane sketched. It is vertical, but skew to the z-axis.\" width=\"387\" height=\"478\" \/> Figure 2. Plane represented by the scalar equation [latex]5x-3y+z+4=0[\/latex].[\/caption][\/hidden-answer]<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\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<\/div>\r\nWatch the following video to see the worked solution to the above Try IT.\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=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\"><\/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>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]26045[\/ohm_question]\r\n\r\n<\/div>\r\nNow that we can write an equation for a plane, we can use the equation to find the distance [latex]d[\/latex] between a point [latex]P[\/latex] and the plane. It is defined as the shortest possible distance from [latex]P[\/latex] to a point on the plane.\r\n\r\n[caption id=\"attachment_5183\" align=\"aligncenter\" width=\"456\"]<img class=\"size-full wp-image-5183\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31200627\/2.70.jpg\" alt=\"This figure is the sketch of a parallelogram representing a plane. In the plane are points Q and R. there is a broken line from Q to R on the plane. There is a vector n out of the plane at point Q. Also, there is a vector labeled \u201cR P\u201d from point R to point P which is above the plane. This vector is perpendicular to the plane.\" width=\"456\" height=\"299\" \/> Figure 3. We want to find the shortest distance from point [latex]P[\/latex] to the plane. Let point [latex]R[\/latex] be the point in the plane such that, for any other point in the plane [latex]Q[\/latex], [latex]||\\overrightarrow{RP}||&lt;||\\overrightarrow{QP}||[\/latex].[\/caption]Just as we find the two-dimensional distance between a point and a line by calculating the length of a line segment perpendicular to the line, we find the three-dimensional distance between a point and a plane by calculating the length of a line segment perpendicular to the plane. Let [latex]R[\/latex] be the point in the plane such that [latex]\\overrightarrow{RP}[\/latex] is orthogonal to the plane, and let [latex]Q[\/latex] be an arbitrary point in the plane. Then the projection of vector [latex]\\overrightarrow{QP}[\/latex] onto the normal vector describes vector [latex]\\overrightarrow{RP}[\/latex], as shown in\u00a0Figure 3.\r\n<div class=\"textbox shaded\"><header>\r\n<h3 class=\"os-title\" style=\"text-align: center;\"><span class=\"os-title-label\">THEOREM: the distance between a plane and a point<\/span><\/h3>\r\n\r\n<hr \/>\r\n\r\n<\/header><section>\r\n<div class=\"os-note-body\">\r\n<p id=\"fs-id1163723334713\">Suppose a plane with normal vector [latex]\\textbf n[\/latex] passes through point [latex]Q[\/latex]. The distance [latex]d[\/latex] from the plane to a point [latex]P[\/latex] not in the plane is given by<\/p>\r\n<p style=\"text-align: center;\">[latex]d=\\left|\\left|\\text{proj}_{\\bf{n}}\\overrightarrow{QP}\\right|\\right|=\\left|\\text{comp}_{\\bf{n}}\\overrightarrow{QP}\\right|=\\frac{|\\overrightarrow{QP}\\cdot{\\bf{n}}|}{||{\\bf{n}}||}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: distance between a point and a plane<\/h3>\r\n<p id=\"fs-id1163724065502\">Find the distance between point [latex]P=(3, 1, 2)[\/latex] and the plane given by [latex]x-2y+z=5[\/latex] (see the following figure).<\/p>\r\n\r\n[caption id=\"attachment_5188\" align=\"aligncenter\" width=\"386\"]<img class=\"size-full wp-image-5188\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31201059\/Example-2.51.jpg\" alt=\"This figure is the 3-dimensional coordinate system. There is a point drawn at (3, 1, 2). The point is labeled \u201cP(3, 1, 2).\u201d There is a plane drawn. There is a perpendicular line from the plane to point P(3, 1, 2).\" width=\"386\" height=\"392\" \/> Figure 4. The distance between the plane given by [latex]x-2y+z=5[\/latex] and the point [latex]P=(3, 1, 2)[\/latex][\/caption]\r\n[reveal-answer q=\"947365278\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"947365278\"]\r\n<p id=\"fs-id1163723349521\">The coefficients of the plane\u2019s equation provide a normal vector for the plane: [latex]{\\bf{n}}\\langle1,-2,1\\rangle[\/latex]. To find vector [latex]\\overrightarrow{QP}[\/latex], we need a point in the plane. Any point will work, so set [latex]y=z=0[\/latex] to see that point [latex]Q=(5, 0, 0)[\/latex] lies in the plane. Find the component form of the vector from [latex]Q[\/latex] to\u00a0[latex]P[\/latex]:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\overrightarrow{QP}=\\langle3-5,1-0,2-0\\rangle=\\langle-2,1,2\\rangle[\/latex].<\/p>\r\n<p id=\"fs-id1163724120375\">Apply the distance formula from\u00a0The Distance between a Plane and a Point:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nd&amp;=\\frac{|\\overrightarrow{QP}\\cdot{\\bf{n}}|}{||{\\bf{n}}||} \\\\\r\n&amp;=\\frac{|\\langle-2,1,2\\rangle\\cdot\\langle1,-2,1\\rangle|}{\\sqrt{1^2+(-2)^2+1^2}} \\\\\r\n&amp;=\\frac{|-2-2+2|}{\\sqrt6} \\\\\r\n&amp;=\\frac{2}{\\sqrt6 .}\r\n\\end{aligned}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nFind the distance between point [latex]P=(5, -1, 0)[\/latex] and the plane given by [latex]4x+2y-z=3[\/latex].\r\n\r\n[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<\/div>\r\n<h2 data-type=\"title\">Parallel and Intersecting Planes<\/h2>\r\n<p id=\"fs-id1163724138356\">We have discussed the various possible relationships between two lines in two dimensions and three dimensions. When we describe the relationship between two planes in space, we have only two possibilities: the two distinct planes are parallel or they intersect. When two planes are parallel, their normal vectors are parallel. When two planes intersect, the intersection is a line (Figure 5).<\/p>\r\n\r\n\r\n[caption id=\"attachment_5191\" align=\"aligncenter\" width=\"286\"]<img class=\"size-full wp-image-5191\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31201453\/2.71.jpg\" alt=\"This figure is two planes that are intersecting. The intersection forms a line segment.\" width=\"286\" height=\"368\" \/> Figure 5. The intersection of two nonparallel planes is always a line.[\/caption]\r\n\r\nWe can use the equations of the two planes to find parametric equations for the line of intersection.\r\n\r\n<\/div>\r\n<div class=\"os-figure\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: finding the line of intersection for two planes<\/h3>\r\nFind parametric and symmetric equations for the line formed by the intersection of the planes given by [latex]x+y+z=0[\/latex] and [latex]2x-y+z=0[\/latex] (see the following figure).\r\n\r\n[caption id=\"attachment_5192\" align=\"aligncenter\" width=\"412\"]<img class=\"size-full wp-image-5192\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31201604\/Example-2.52.jpg\" alt=\"This figure is two planes intersecting in the 3-dimensional coordinate system.\" width=\"412\" height=\"453\" \/> Figure 6. The line formed by the intersection of the planes given by [latex]x+y+z=0[\/latex] and [latex]2x-y+z=0[\/latex][\/caption][reveal-answer q=\"038942561\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"038942561\"]\r\n<p id=\"fs-id1163723326629\">Note that the two planes have nonparallel normals, so the planes intersect. Further, the origin satisfies each equation, so we know the line of intersection passes through the origin. Add the plane equations so we can eliminate the one of the variables, in this case, [latex]y[\/latex]:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align*}\r\nx&amp;+y+z=0 \\\\\r\n2x&amp;-y+z=0 \\\\\r\n\\hline \\\\\r\n3x&amp;\\quad\\,\\,+2z=0.\r\n\\end{align*}[\/latex]<\/p>\r\n<p id=\"fs-id1163723321204\">This gives us [latex]x=-\\frac23z[\/latex]. We substitute this value into the first equation to express [latex]y[\/latex] in terms of [latex]z[\/latex]:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nx+y+z&amp;=0 \\\\\r\n-\\frac23z+y+z&amp;=0 \\\\\r\ny+\\frac13z&amp;=0 \\\\\r\ny&amp;=-\\frac13z\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1163723096825\">We now have the first two variables, [latex]x[\/latex] and [latex]y[\/latex], in terms of the third variable, [latex]z[\/latex]. Now we define [latex]z[\/latex] in terms of [latex]t[\/latex]. To eliminate the need for fractions, we choose to define the parameter [latex]t[\/latex] as [latex]t=-\\frac13z[\/latex]. Then, [latex]z=-3t[\/latex]. Substituting the parametric representation of [latex]z[\/latex] back into the other two equations, we see that the parametric equations for the line of intersection are [latex]x=2t[\/latex], [latex]y=t[\/latex], [latex]z=-3t[\/latex]. The symmetric equations for the line are [latex]\\frac{x}2=y=\\frac{z}{-3}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nFind 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].\r\n\r\n[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<\/div>\r\nWatch the following video to see the worked solution to the above Try IT.\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=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\"><\/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>In addition to finding the equation of the line of intersection between two planes, we may need to find the angle formed by the intersection of two planes. For example, builders constructing a house need to know the angle where different sections of the roof meet to know whether the roof will look good and drain properly. We can use normal vectors to calculate the angle between the two planes. We can do this because the angle between the normal vectors is the same as the angle between the planes.\u00a0Figure 7\u00a0shows why this is true.\r\n\r\n[caption id=\"attachment_5195\" align=\"aligncenter\" width=\"412\"]<img class=\"size-full wp-image-5195\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31201718\/2.72.jpg\" alt=\"This figure is two parallelograms representing planes. The planes intersect forming angle theta between them. The first plane as vector \u201cn sub 1\u201d normal to the plane. The second vector has vector \u201cn sub 2\u201d normal to the plane. The normal vectors intersect and form the angle theta.\" width=\"412\" height=\"316\" \/> Figure 7. The angle between two planes has the same measure as the angle between the normal vectors for the planes.[\/caption]\r\n<p id=\"fs-id1163723281415\">We can find the measure of the angle [latex]\\theta[\/latex] between two intersecting planes by first finding the cosine of the angle, using the following equation:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\cos\\theta=\\frac{|{\\bf{n}}_1\\cdot{\\bf{n}}_2|}{||{\\bf{n}}_1|| \\ ||{\\bf{n}}_2||}}[\/latex].<\/p>\r\n<p id=\"fs-id1163724078749\">We can then use the angle to determine whether two planes are parallel or orthogonal or if they intersect at some other angle.<\/p>\r\n\r\n<\/div>\r\n<div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: finding the angle between two planes<\/h3>\r\n<p id=\"fs-id1163724078763\">Determine whether each pair of planes is parallel, orthogonal, or neither. If the planes are intersecting, but not orthogonal, find the measure of the angle between them. Give the answer in radians and round to two decimal places.<\/p>\r\n\r\n<ol id=\"fs-id1163724078768\" type=\"a\">\r\n \t<li>[latex]x+2y-z=8[\/latex] and\u00a0[latex]2x+4y-2z=10[\/latex]<\/li>\r\n \t<li>[latex]2x-3y+2z=3[\/latex] and\u00a0[latex]6x+2y-3z=1[\/latex]<\/li>\r\n \t<li>[latex]x+y+z=4[\/latex] and\u00a0[latex]x-3y+5z=1[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"346528763\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"346528763\"]\r\n<ol id=\"fs-id1163723199442\" type=\"a\">\r\n \t<li>The normal vectors for these planes are [latex]{\\bf{n}}_1\\langle1,2,-1\\rangle[\/latex] and [latex]{\\bf{n}}_2=\\langle2,3,-2\\rangle[\/latex]. These two vectors are scalar multiples of each other. The normal vectors are parallel, so the planes are parallel.<\/li>\r\n \t<li>The normal vectors for these planes are [latex]{\\bf{n}}_1=\\langle2,-3,2\\rangle[\/latex] and [latex]{\\bf{n}}_2=\\langle6,2,-3\\rangle[\/latex]. Taking the dot product of these vectors, we have<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]{\\bf{n}}_1\\cdot{\\bf{n}}_2=\\langle2,-3,2\\rangle\\cdot\\langle6,2,-3\\rangle=2(6)-3(2)+2(-3)=0[\/latex].<\/p>\r\nThe normal vectors are orthogonal, so the corresponding planes are orthogonal as well.<\/li>\r\n \t<li>The normal vectors for these planes are [latex]{\\bf{n}}_1=\\langle1,1,1\\rangle[\/latex] and\u00a0[latex]{\\bf{n}}_2=\\langle1,-3,5\\rangle[\/latex]:<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\cos\\theta&amp;=\\frac{|{\\bf{n}}_1\\cdot{\\bf{n}}_2|}{||{\\bf{n}}_1|| \\ ||{\\bf{n}}_2||} \\\\\r\n&amp;=\\frac{|\\langle1,1,1\\rangle\\cdot\\langle1,-3,5\\rangle|}{\\sqrt{1^2+1^2+1^2}\\sqrt{1^2+(-3)^2+5^2}} \\\\\r\n&amp;=\\frac{3}{\\sqrt{105}}.\r\n\\end{aligned}[\/latex]<\/p>\r\nThe angle between the two planes is [latex]1.27[\/latex] rad, or approximately [latex]73^{\\small\\circ}[\/latex].<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nFind the measure of the angle 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.\r\n\r\n[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<\/div>\r\n<\/div>\r\n<div>\r\n<p id=\"fs-id1163724081830\">When we find that two planes are parallel, we may need to find the distance between them. To find this distance, we simply select a point in one of the planes. The distance from this point to the other plane is the distance between the planes.<\/p>\r\n<p id=\"fs-id1163724081835\">Previously, we introduced the formula for calculating this distance in\u00a0The Distance between a Plane and a Point:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{d=\\frac{\\overrightarrow{QP}\\cdot{\\bf{n}}}{||{\\bf{n}}||}}[\/latex],<\/p>\r\n<p id=\"fs-id1163724097578\">where [latex]Q[\/latex] is a point on the plane, [latex]P[\/latex] is a point not on the plane, and [latex]\\textbf n[\/latex]\u00a0<span style=\"font-size: 1rem; text-align: initial;\">is the normal vector that passes through point [latex]Q[\/latex]. Consider the distance from point [latex](x_0, y_0, z_0)[\/latex] to plane [latex]ax+by+cz+k=0[\/latex]. Let [latex](x_1, y_1, z_1)[\/latex] be any point in the plane. Substituting into the formula yields<\/span><\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\begin{aligned}\r\nd&amp;=\\frac{|a(x_0-x_1)+b(y_0-y_1)+c(z_0-z_1)|}{\\sqrt{a^2+b^2+c^2}} \\\\\r\n&amp;=\\frac{|ax_0+by_0+cz_0+k|}{\\sqrt{a^2+b^2+c^2}}\r\n\\end{aligned}}[\/latex].<\/p>\r\nWe state this result formally in the following theorem.\r\n<div id=\"fs-id1163724097697\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\r\n<div class=\"textbox shaded\"><header>\r\n<h3 class=\"os-title\" style=\"text-align: center;\"><span class=\"os-title-label\">THEOREM: distance from a point to a plane<\/span><\/h3>\r\n\r\n<hr \/>\r\n\r\n<\/header><section>\r\n<div class=\"os-note-body\">\r\n<p id=\"fs-id1163723268198\">Let [latex]P(x_0, y_0, z_0)[\/latex] be a point. The distance from [latex]P[\/latex] to plane [latex]ax+by+cz+k=0[\/latex] is given by<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{d=\\frac{|ax_0+by_0+cz_0+k|}{\\sqrt{a^2+b^2+c^2}}}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: finding the distance between parallel planes<\/h3>\r\nFind the distance between the two parallel planes given by [latex]2x+y-z=2[\/latex] and\u00a0[latex]2x+y-z=8[\/latex].\r\n\r\n[reveal-answer q=\"943725601\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"943725601\"]\r\n<p id=\"fs-id1163723334407\">Point [latex](1, 0, 0)[\/latex] lies in the first plane. The desired distance, then, is<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nd&amp;=\\frac{|ax_0+by_0+cz_0+k|}{\\sqrt{a^2+b^2+c^2}} \\\\\r\n&amp;=\\frac{|2(1)+1(0)+(-1)(0)+(-8)|}{\\sqrt{2^2+1^2+(-1)^2}} \\\\\r\n&amp;=\\frac{6}{\\sqrt6}=\\sqrt6\r\n\\end{aligned}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nFind the distance between parallel planes [latex]5x-2y+z=6[\/latex] and\u00a0[latex]5x-2y+z=-3[\/latex].\r\n\r\n[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<\/div>\r\n<div data-type=\"title\">\r\n<div class=\"textbox tryit\">\r\n<h3>Activity: distance between two skew lines<\/h3>\r\n[caption id=\"attachment_5197\" align=\"aligncenter\" width=\"582\"]<img class=\"size-full wp-image-5197\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31201814\/2.73.jpg\" alt=\"This figure shows a system of pipes running in different directions in an industrial plant. Two skew pipes are highlighted in red.\" width=\"582\" height=\"571\" \/> Figure 8. Industrial pipe installations often feature pipes running in different directions. How can we find the distance between two skew pipes?[\/caption]\r\n<p id=\"fs-id1163723210042\">Finding the distance from a point to a line or from a line to a plane seems like a pretty abstract procedure. But, if the lines represent pipes in a chemical plant or tubes in an oil refinery or roads at an intersection of highways, confirming that the distance between them meets specifications can be both important and awkward to measure. One way is to model the two pipes as lines, using the techniques in this chapter, and then calculate the distance between them. The calculation involves forming vectors along the directions of the lines and using both the cross product and the dot product.<\/p>\r\n<p id=\"fs-id1163723210050\">The symmetric forms of two lines, [latex]L_1[\/latex] and [latex]L_2[\/latex], are<\/p>\r\n<p style=\"text-align: center;\">[latex]L_1[\/latex]:\u00a0[latex]\\frac{x-x_1}{a_1}=\\frac{y-y_1}{b_1}=\\frac{z-z_1}{c_1}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]L_2[\/latex]:\u00a0[latex]\\frac{x-x_2}{a_2}=\\frac{y-y_2}{b_2}=\\frac{z-z_2}{c_2}[\/latex].<\/p>\r\n<p id=\"fs-id1163723200693\">You are to develop a formula for the distance [latex]d[\/latex] between these two lines, in terms of the values [latex]a_1[\/latex], [latex]b_1[\/latex], [latex]c_1[\/latex]; [latex]a_2[\/latex], [latex]b_2[\/latex], [latex]c_2[\/latex]; [latex]x_1[\/latex], [latex]y_1[\/latex], [latex]z_1[\/latex]; and [latex]x_2[\/latex], [latex]y_2[\/latex], [latex]z_2[\/latex]. The distance between two lines is usually taken to mean the minimum distance, so this is the length of a line segment or the length of a vector that is perpendicular to both lines and intersects both lines.<\/p>\r\n\r\n<ol id=\"fs-id1163723200803\" type=\"1\">\r\n \t<li>First, write down two vectors, [latex]{\\bf{v}}_1[\/latex] and [latex]{\\bf{v}}_2[\/latex], that lie along [latex]L_1[\/latex] and [latex]L_2[\/latex], respectively.<\/li>\r\n \t<li>Find the cross product of these two vectors and call it [latex]\\textbf N[\/latex]. This vector is perpendicular to [latex]{\\bf{v}}_1[\/latex] and [latex]{\\bf{v}}_2[\/latex], and hence is perpendicular to both lines.<\/li>\r\n \t<li>From vector [latex]\\textbf N[\/latex], form a unit vector [latex]\\textbf n[\/latex] in the same direction.<\/li>\r\n \t<li>Use symmetric equations to find a convenient vector [latex]{\\bf{v}}_{12}[\/latex] that lies between any two points, one on each line. Again, this can be done directly from the symmetric equations.<\/li>\r\n \t<li>The dot product of two vectors is the magnitude of the projection of one vector onto the other\u2014that is, [latex]{\\bf{A}}\\cdot{\\bf{B}}=||{\\bf{A}}|| \\ ||{\\bf{B}}||\\cos\\theta[\/latex], where [latex]\\theta[\/latex] is the angle between the vectors. Using the dot product, find the projection of vector [latex]{\\bf{v}}_{12}[\/latex] found in step 4 onto unit vector [latex]\\textbf n[\/latex] found in step 3. This projection is perpendicular to both lines, and hence its length must be the perpendicular distance [latex]d[\/latex] between them. Note that the value of [latex]d[\/latex] may be negative, depending on your choice of vector [latex]{\\bf{v}}_{12}[\/latex] or the order of the cross product, so use absolute value signs around the numerator.<\/li>\r\n \t<li>Check that your formula gives the correct distance of [latex]|-25|\/\\sqrt{198}\\approx1.78[\/latex] between the following two lines:<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nL_1&amp;:\\frac{x-5}2=\\frac{y-3}4=\\frac{z-1}3 \\\\\r\nL_2&amp;:\\frac{x-6}3=\\frac{y-1}5=\\frac{z}7.\r\n\\end{aligned}[\/latex]<\/p>\r\n<\/li>\r\n \t<li>Is your general expression valid when the lines are parallel? If not, why not? (<em data-effect=\"italics\">Hint:<\/em>\u00a0What do you know about the value of the cross product of two parallel vectors? Where would that result show up in your expression for [latex]d[\/latex]?)<\/li>\r\n \t<li>Demonstrate that your expression for the distance is zero when the lines intersect. Recall that two lines intersect if they are not parallel and they are in the same plane. Hence, consider the direction of [latex]\\textbf n[\/latex] and [latex]{\\bf{v}}_{12}[\/latex]. What is the result of their dot product?<\/li>\r\n \t<li>Consider the following application. Engineers at a refinery have determined they need to install support struts between many of the gas pipes to reduce damaging vibrations. To minimize cost, they plan to install these struts at the closest points between adjacent skewed pipes. Because they have detailed schematics of the structure, they are able to determine the correct lengths of the struts needed, and hence manufacture and distribute them to the installation crews without spending valuable time making measurements.<span data-type=\"newline\">\r\n<\/span>The rectangular frame structure has the dimensions [latex]4.0\\times15.0\\times10.0\\text{ m}[\/latex] (height, width, and depth). One sector has a pipe entering the lower corner of the standard frame unit and exiting at the diametrically opposed corner (the one farthest away at the top); call this [latex]L_1[\/latex]. A second pipe enters and exits at the two different opposite lower corners; call this [latex]L_2[\/latex] (Figure 9).<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"CNX_Calc_Figure_12_05_017\" class=\"os-figure\">\r\n<div class=\"os-caption-container\">\r\n\r\n[caption id=\"attachment_5199\" align=\"aligncenter\" width=\"397\"]<img class=\"size-full wp-image-5199\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31201941\/2.74.jpg\" alt=\"This figure is a three-dimensional box in an x y z coordinate system. The box has dimensions x = 10 m, y = 15 m, and z = 4 m. Line L1 passes through a main diagonal of the box from the origin to the far corner. Line L2 passes through a diagonal in the base of the box with x-intercept 10 and y-intercept 15.\" width=\"397\" height=\"394\" \/> Figure 9. Two pipes cross through a standard frame unit.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>Write down the vectors along the lines representing those pipes, find the cross product between them from which to create the unit vector [latex]\\textbf n[\/latex], define a vector that spans two points on each line, and finally determine the minimum distance between the lines. (Take the origin to be at the lower corner of the first pipe.) Similarly, you may also develop the symmetric equations for each line and substitute directly into your formula.<\/li>\r\n<\/ol>\r\n<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<div data-type=\"note\">\r\n<div id=\"fs-id1163724036465\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\r\n<div id=\"fs-id1163724074875\" class=\"theorem ui-has-child-title\" data-type=\"note\">\r\n<div id=\"fs-id1163724080388\" class=\"ui-has-child-title\" data-type=\"example\">\r\n<div data-type=\"example\"><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul class=\"os-abstract\">\n<li><span class=\"os-abstract-content\">Write the vector and scalar equations of a plane through a given point with a given normal.<\/span><\/li>\n<li><span class=\"os-abstract-content\">Find the distance from a point to a given plane.<\/span><\/li>\n<li><span class=\"os-abstract-content\">Find the angle between two planes.<\/span><\/li>\n<\/ul>\n<\/div>\n<section id=\"fs-id1163723141280\" data-depth=\"1\">\n<h2 data-type=\"title\">Equations for a Plane<\/h2>\n<p id=\"fs-id1163723141285\">We know that a line is determined by two points. In other words, for any two distinct points, there is exactly one line that passes through those points, whether in two dimensions or three. Similarly, given any three points that do not all lie on the same line, there is a unique plane that passes through these points. Just as a line is determined by two points, a plane is determined by three.<\/p>\n<p id=\"fs-id1163723141292\">This may be the simplest way to characterize a plane, but we can use other descriptions as well. For example, given two distinct, intersecting lines, there is exactly one plane containing both lines. A plane is also determined by a line and any point that does not lie on the line. These characterizations arise naturally from the idea that a plane is determined by three points. Perhaps the most surprising characterization of a plane is actually the most useful.<\/p>\n<\/section>\n<div data-type=\"example\">\n<p id=\"fs-id1163723141299\">Imagine a pair of orthogonal vectors that share an initial point. Visualize grabbing one of the vectors and twisting it. As you twist, the other vector spins around and sweeps out a plane. Here, we describe that concept mathematically. Let [latex]{\\bf{n}}=\\langle a,b,c\\rangle[\/latex] be a vector and [latex]P=(x_0, y_0, z_0)[\/latex] be a point. Then the set of all points [latex]Q=(x, y, z)[\/latex] such that [latex]\\overrightarrow{PQ}[\/latex] is orthogonal to [latex]\\textbf n[\/latex]\u00a0<span style=\"font-size: 1rem; text-align: initial;\">forms a plane (<\/span>Figure 1<span style=\"font-size: 1rem; text-align: initial;\">). We say that [latex]\\textbf n[\/latex] is a\u00a0<\/span><strong style=\"font-size: 1rem; text-align: initial;\"><span id=\"60ce88e7-9116-48d3-bb30-013c89cddc64_term92\" data-type=\"term\">normal vector<\/span>,<\/strong><span style=\"font-size: 1rem; text-align: initial;\"> or perpendicular to the plane. Remember, the dot product of orthogonal vectors is zero. This fact generates the\u00a0<\/span><strong style=\"font-size: 1rem; text-align: initial;\"><span id=\"60ce88e7-9116-48d3-bb30-013c89cddc64_term93\" data-type=\"term\">vector equation of a plane<\/span><\/strong><span style=\"font-size: 1rem; text-align: initial;\">: [latex]{\\bf{n}}\\cdot\\overrightarrow{PQ}=0[\/latex]. Rewriting this equation provides additional ways to describe the plane:<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {\\bf{n}}\\cdot\\overrightarrow{PQ}&=0 \\\\  \\langle a,b,c\\rangle\\cdot\\langle x-x_0,y-y_0,z-z_0\\rangle&=0 \\\\  a(x-x_0)+b(y-y_0)+c(z-z_0)&=0.  \\end{aligned}[\/latex]<\/p>\n<\/div>\n<div class=\"os-figure\">\n<div id=\"attachment_5173\" style=\"width: 451px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5173\" class=\"size-full wp-image-5173\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27214230\/2.69.jpg\" alt=\"This figure is a parallelogram representing a plane. In the plane is a vector from point P to point Q. Perpendicular to the vector P Q is the vector n.\" width=\"441\" height=\"226\" \/><\/p>\n<p id=\"caption-attachment-5173\" class=\"wp-caption-text\">Figure 1. Given a point [latex]P[\/latex] and vector [latex]\\textbf n[\/latex], the set of all points [latex]Q[\/latex] with [latex]\\overrightarrow{PQ}[\/latex] orthogonal to [latex]\\textbf n[\/latex] forms a plane.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"example\"><\/div>\n<div data-type=\"example\">\n<div class=\"textbox shaded\">\n<header>\n<h3 class=\"os-title\" style=\"text-align: center;\" data-type=\"title\"><span id=\"22\" class=\"os-title-label\" data-type=\"\">DEFINITION<\/span><\/h3>\n<hr \/>\n<\/header>\n<section>\n<div class=\"os-note-body\">\n<p id=\"fs-id1163723211072\">Given a point [latex]P[\/latex] and vector [latex]\\textbf n[\/latex], the set of all points [latex]Q[\/latex] satisfying the equation [latex]{\\bf{n}}\\cdot\\overrightarrow{PQ}=0[\/latex] forms a plane. The equation<\/p>\n<p style=\"text-align: center;\">[latex]{\\bf{n}}\\cdot\\overrightarrow{PQ}=0[\/latex].<\/p>\n<p id=\"fs-id1163724147360\">is known as the vector equation of a plane.<\/p>\n<p id=\"fs-id1163724062719\">The\u00a0<strong><span id=\"60ce88e7-9116-48d3-bb30-013c89cddc64_term94\" data-type=\"term\">scalar equation of a plane<\/span><\/strong>\u00a0containing point [latex]P=(x_0, y_0, z_0)[\/latex] with normal vector [latex]{\\bf{n}}=\\langle a,b,c\\rangle[\/latex] is<\/p>\n<p style=\"text-align: center;\">[latex]a(x-x_0)+b(y-y_0)+c(z-z_0)=0[\/latex].<\/p>\n<p id=\"fs-id1163723339633\">This equation can be expressed as [latex]ax+by+cz+d=0[\/latex], where [latex]d=-ax_0-by_0-cz_0[\/latex]. This form of the equation is sometimes called the<strong>\u00a0<span id=\"60ce88e7-9116-48d3-bb30-013c89cddc64_term95\" data-type=\"term\">general form of the equation of a plane<\/span><\/strong>.<\/p>\n<\/div>\n<\/section>\n<\/div>\n<p>As described earlier in this section, any three points that do not all lie on the same line determine a plane. Given three such points, we can find an equation for the plane containing these points.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: writing an equation of a plane given three points in the plane<\/h3>\n<p>Write an equation for the plane containing points [latex]P=(1,1,-2)[\/latex], [latex]Q=(0,2,1)[\/latex], and [latex]R=(-1,-1,0)[\/latex] in both standard and general forms.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q374652897\">Show Solution<\/span><\/p>\n<div id=\"q374652897\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1163723246705\">To write an equation for a plane, we must find a normal vector for the plane. We start by identifying two vectors in the plane:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\overrightarrow{PQ}&=\\langle0-1,2-1,1-(-2)\\rangle&=\\langle-1,1,3\\rangle \\\\  \\overrightarrow{QR}&=\\langle-1-0,-1-2,0-1\\rangle&=\\langle-1,-3,-1\\rangle  \\end{aligned}.[\/latex]<\/p>\n<p id=\"fs-id1163724184793\">The cross product [latex]\\overrightarrow{PQ}\\times\\overrightarrow{QR}[\/latex] is orthogonal to both [latex]\\overrightarrow{PQ}[\/latex] and [latex]\\overrightarrow{QR}[\/latex], so it is normal to the plane that contains these two vectors:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {\\bf{n}}&=\\overrightarrow{PQ}\\times\\overrightarrow{QR} \\\\  &=\\begin{vmatrix}{\\bf{i}}&{\\bf{j}}&{\\bf{k}} \\\\ -1&1&3\\\\ -1&-3&-1\\end{vmatrix} \\\\  &=(-1+9){\\bf{i}}-(1+3){\\bf{j}}+(3+1){\\bf{k}} \\\\  &=8{\\bf{i}}-4{\\bf{j}}+4{\\bf{k}}.  \\end{aligned}[\/latex]<\/p>\n<p id=\"fs-id1163723128152\">Thus, [latex]{\\bf{n}}=\\langle8,-4,4\\rangle[\/latex], and we can choose any of the three given points to write an equation of the plane:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  8(x-1)-4(y-1)+4(z+2)&=0 \\\\  8x-4y+4z+4&=0.  \\end{aligned}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The scalar equations of a plane vary depending on the normal vector and point chosen.<\/p>\n<\/div>\n<div data-type=\"example\">\n<div class=\"textbox exercises\">\n<h3>Example: writing an equation for a plane given a point and a line<\/h3>\n<p>Find an equation of the plane that passes through point [latex](1, 4, 3)[\/latex] and contains the line given by [latex]x=\\frac{y-1}2=z+1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q289161783\">Show Solution<\/span><\/p>\n<div id=\"q289161783\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1163723216073\">Symmetric equations describe the line that passes through point [latex](0, 1, -1)[\/latex] parallel to vector [latex]{\\bf{v}}_1=\\langle1,2,1\\rangle[\/latex] (see the following figure). Use this point and the given point, [latex](1, 4, 3)[\/latex], to identify a second vector parallel to the plane:<\/p>\n<p style=\"text-align: center;\">[latex]{\\bf{v}}_2=\\langle1-0,4-1,3-(-1)\\rangle=\\langle1,3,4\\rangle[\/latex].<\/p>\n<p id=\"fs-id1163723199146\">Use the cross product of these vectors to identify a normal vector for the plane:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {\\bf{n}}&={\\bf{v}}_1\\times{\\bf{v}}_2 \\\\  &=\\begin{vmatrix}{\\bf{i}}&{\\bf{j}}&{\\bf{k}} \\\\1&2&1 \\\\1&3&4\\end{vmatrix} \\\\  &=(8-3){\\bf{i}}-(4-1){\\bf{j}}+(3-2){\\bf{k}} \\\\  &=5{\\bf{i}}-3{\\bf{j}}+{\\bf{k}}  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1163723277829\">The scalar equations for the plane are [latex]5x-3(y-1)+(z+1)=0[\/latex] and\u00a0[latex]5x-3y+z+4=0[\/latex].<\/p>\n<div id=\"attachment_5180\" style=\"width: 397px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5180\" class=\"size-full wp-image-5180\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31200056\/Example-2.50.jpg\" alt=\"This figure is the 3-dimensional coordinate system. There is a plane sketched. It is vertical, but skew to the z-axis.\" width=\"387\" height=\"478\" \/><\/p>\n<p id=\"caption-attachment-5180\" class=\"wp-caption-text\">Figure 2. Plane represented by the scalar equation [latex]5x-3y+z+4=0[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q498652076\">Show Solution<\/span><\/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<\/div>\n<p>Watch the following video to see the worked solution to the above Try IT.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=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\"><\/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<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm26045\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=26045&theme=oea&iframe_resize_id=ohm26045&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>Now that we can write an equation for a plane, we can use the equation to find the distance [latex]d[\/latex] between a point [latex]P[\/latex] and the plane. It is defined as the shortest possible distance from [latex]P[\/latex] to a point on the plane.<\/p>\n<div id=\"attachment_5183\" style=\"width: 466px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5183\" class=\"size-full wp-image-5183\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31200627\/2.70.jpg\" alt=\"This figure is the sketch of a parallelogram representing a plane. In the plane are points Q and R. there is a broken line from Q to R on the plane. There is a vector n out of the plane at point Q. Also, there is a vector labeled \u201cR P\u201d from point R to point P which is above the plane. This vector is perpendicular to the plane.\" width=\"456\" height=\"299\" \/><\/p>\n<p id=\"caption-attachment-5183\" class=\"wp-caption-text\">Figure 3. We want to find the shortest distance from point [latex]P[\/latex] to the plane. Let point [latex]R[\/latex] be the point in the plane such that, for any other point in the plane [latex]Q[\/latex], [latex]||\\overrightarrow{RP}||&lt;||\\overrightarrow{QP}||[\/latex].<\/p>\n<\/div>\n<p>Just as we find the two-dimensional distance between a point and a line by calculating the length of a line segment perpendicular to the line, we find the three-dimensional distance between a point and a plane by calculating the length of a line segment perpendicular to the plane. Let [latex]R[\/latex] be the point in the plane such that [latex]\\overrightarrow{RP}[\/latex] is orthogonal to the plane, and let [latex]Q[\/latex] be an arbitrary point in the plane. Then the projection of vector [latex]\\overrightarrow{QP}[\/latex] onto the normal vector describes vector [latex]\\overrightarrow{RP}[\/latex], as shown in\u00a0Figure 3.<\/p>\n<div class=\"textbox shaded\">\n<header>\n<h3 class=\"os-title\" style=\"text-align: center;\"><span class=\"os-title-label\">THEOREM: the distance between a plane and a point<\/span><\/h3>\n<hr \/>\n<\/header>\n<section>\n<div class=\"os-note-body\">\n<p id=\"fs-id1163723334713\">Suppose a plane with normal vector [latex]\\textbf n[\/latex] passes through point [latex]Q[\/latex]. The distance [latex]d[\/latex] from the plane to a point [latex]P[\/latex] not in the plane is given by<\/p>\n<p style=\"text-align: center;\">[latex]d=\\left|\\left|\\text{proj}_{\\bf{n}}\\overrightarrow{QP}\\right|\\right|=\\left|\\text{comp}_{\\bf{n}}\\overrightarrow{QP}\\right|=\\frac{|\\overrightarrow{QP}\\cdot{\\bf{n}}|}{||{\\bf{n}}||}[\/latex].<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: distance between a point and a plane<\/h3>\n<p id=\"fs-id1163724065502\">Find the distance between point [latex]P=(3, 1, 2)[\/latex] and the plane given by [latex]x-2y+z=5[\/latex] (see the following figure).<\/p>\n<div id=\"attachment_5188\" style=\"width: 396px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5188\" class=\"size-full wp-image-5188\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31201059\/Example-2.51.jpg\" alt=\"This figure is the 3-dimensional coordinate system. There is a point drawn at (3, 1, 2). The point is labeled \u201cP(3, 1, 2).\u201d There is a plane drawn. There is a perpendicular line from the plane to point P(3, 1, 2).\" width=\"386\" height=\"392\" \/><\/p>\n<p id=\"caption-attachment-5188\" class=\"wp-caption-text\">Figure 4. The distance between the plane given by [latex]x-2y+z=5[\/latex] and the point [latex]P=(3, 1, 2)[\/latex]<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q947365278\">Show Solution<\/span><\/p>\n<div id=\"q947365278\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1163723349521\">The coefficients of the plane\u2019s equation provide a normal vector for the plane: [latex]{\\bf{n}}\\langle1,-2,1\\rangle[\/latex]. To find vector [latex]\\overrightarrow{QP}[\/latex], we need a point in the plane. Any point will work, so set [latex]y=z=0[\/latex] to see that point [latex]Q=(5, 0, 0)[\/latex] lies in the plane. Find the component form of the vector from [latex]Q[\/latex] to\u00a0[latex]P[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\overrightarrow{QP}=\\langle3-5,1-0,2-0\\rangle=\\langle-2,1,2\\rangle[\/latex].<\/p>\n<p id=\"fs-id1163724120375\">Apply the distance formula from\u00a0The Distance between a Plane and a Point:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  d&=\\frac{|\\overrightarrow{QP}\\cdot{\\bf{n}}|}{||{\\bf{n}}||} \\\\  &=\\frac{|\\langle-2,1,2\\rangle\\cdot\\langle1,-2,1\\rangle|}{\\sqrt{1^2+(-2)^2+1^2}} \\\\  &=\\frac{|-2-2+2|}{\\sqrt6} \\\\  &=\\frac{2}{\\sqrt6 .}  \\end{aligned}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q348756277\">Show Solution<\/span><\/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<\/div>\n<h2 data-type=\"title\">Parallel and Intersecting Planes<\/h2>\n<p id=\"fs-id1163724138356\">We have discussed the various possible relationships between two lines in two dimensions and three dimensions. When we describe the relationship between two planes in space, we have only two possibilities: the two distinct planes are parallel or they intersect. When two planes are parallel, their normal vectors are parallel. When two planes intersect, the intersection is a line (Figure 5).<\/p>\n<div id=\"attachment_5191\" style=\"width: 296px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5191\" class=\"size-full wp-image-5191\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31201453\/2.71.jpg\" alt=\"This figure is two planes that are intersecting. The intersection forms a line segment.\" width=\"286\" height=\"368\" \/><\/p>\n<p id=\"caption-attachment-5191\" class=\"wp-caption-text\">Figure 5. The intersection of two nonparallel planes is always a line.<\/p>\n<\/div>\n<p>We can use the equations of the two planes to find parametric equations for the line of intersection.<\/p>\n<\/div>\n<div class=\"os-figure\">\n<div class=\"textbox exercises\">\n<h3>Example: finding the line of intersection for two planes<\/h3>\n<p>Find parametric and symmetric equations for the line formed by the intersection of the planes given by [latex]x+y+z=0[\/latex] and [latex]2x-y+z=0[\/latex] (see the following figure).<\/p>\n<div id=\"attachment_5192\" style=\"width: 422px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5192\" class=\"size-full wp-image-5192\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31201604\/Example-2.52.jpg\" alt=\"This figure is two planes intersecting in the 3-dimensional coordinate system.\" width=\"412\" height=\"453\" \/><\/p>\n<p id=\"caption-attachment-5192\" class=\"wp-caption-text\">Figure 6. The line formed by the intersection of the planes given by [latex]x+y+z=0[\/latex] and [latex]2x-y+z=0[\/latex]<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q038942561\">Show Solution<\/span><\/p>\n<div id=\"q038942561\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1163723326629\">Note that the two planes have nonparallel normals, so the planes intersect. Further, the origin satisfies each equation, so we know the line of intersection passes through the origin. Add the plane equations so we can eliminate the one of the variables, in this case, [latex]y[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align*}  x&+y+z=0 \\\\  2x&-y+z=0 \\\\  \\hline \\\\  3x&\\quad\\,\\,+2z=0.  \\end{align*}[\/latex]<\/p>\n<p id=\"fs-id1163723321204\">This gives us [latex]x=-\\frac23z[\/latex]. We substitute this value into the first equation to express [latex]y[\/latex] in terms of [latex]z[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  x+y+z&=0 \\\\  -\\frac23z+y+z&=0 \\\\  y+\\frac13z&=0 \\\\  y&=-\\frac13z  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1163723096825\">We now have the first two variables, [latex]x[\/latex] and [latex]y[\/latex], in terms of the third variable, [latex]z[\/latex]. Now we define [latex]z[\/latex] in terms of [latex]t[\/latex]. To eliminate the need for fractions, we choose to define the parameter [latex]t[\/latex] as [latex]t=-\\frac13z[\/latex]. Then, [latex]z=-3t[\/latex]. Substituting the parametric representation of [latex]z[\/latex] back into the other two equations, we see that the parametric equations for the line of intersection are [latex]x=2t[\/latex], [latex]y=t[\/latex], [latex]z=-3t[\/latex]. The symmetric equations for the line are [latex]\\frac{x}2=y=\\frac{z}{-3}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q374865270\">Show Solution<\/span><\/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<\/div>\n<p>Watch the following video to see the worked solution to the above Try IT.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=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\"><\/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<p>In addition to finding the equation of the line of intersection between two planes, we may need to find the angle formed by the intersection of two planes. For example, builders constructing a house need to know the angle where different sections of the roof meet to know whether the roof will look good and drain properly. We can use normal vectors to calculate the angle between the two planes. We can do this because the angle between the normal vectors is the same as the angle between the planes.\u00a0Figure 7\u00a0shows why this is true.<\/p>\n<div id=\"attachment_5195\" style=\"width: 422px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5195\" class=\"size-full wp-image-5195\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31201718\/2.72.jpg\" alt=\"This figure is two parallelograms representing planes. The planes intersect forming angle theta between them. The first plane as vector \u201cn sub 1\u201d normal to the plane. The second vector has vector \u201cn sub 2\u201d normal to the plane. The normal vectors intersect and form the angle theta.\" width=\"412\" height=\"316\" \/><\/p>\n<p id=\"caption-attachment-5195\" class=\"wp-caption-text\">Figure 7. The angle between two planes has the same measure as the angle between the normal vectors for the planes.<\/p>\n<\/div>\n<p id=\"fs-id1163723281415\">We can find the measure of the angle [latex]\\theta[\/latex] between two intersecting planes by first finding the cosine of the angle, using the following equation:<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\cos\\theta=\\frac{|{\\bf{n}}_1\\cdot{\\bf{n}}_2|}{||{\\bf{n}}_1|| \\ ||{\\bf{n}}_2||}}[\/latex].<\/p>\n<p id=\"fs-id1163724078749\">We can then use the angle to determine whether two planes are parallel or orthogonal or if they intersect at some other angle.<\/p>\n<\/div>\n<div>\n<div class=\"textbox exercises\">\n<h3>Example: finding the angle between two planes<\/h3>\n<p id=\"fs-id1163724078763\">Determine whether each pair of planes is parallel, orthogonal, or neither. If the planes are intersecting, but not orthogonal, find the measure of the angle between them. Give the answer in radians and round to two decimal places.<\/p>\n<ol id=\"fs-id1163724078768\" type=\"a\">\n<li>[latex]x+2y-z=8[\/latex] and\u00a0[latex]2x+4y-2z=10[\/latex]<\/li>\n<li>[latex]2x-3y+2z=3[\/latex] and\u00a0[latex]6x+2y-3z=1[\/latex]<\/li>\n<li>[latex]x+y+z=4[\/latex] and\u00a0[latex]x-3y+5z=1[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q346528763\">Show Solution<\/span><\/p>\n<div id=\"q346528763\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1163723199442\" type=\"a\">\n<li>The normal vectors for these planes are [latex]{\\bf{n}}_1\\langle1,2,-1\\rangle[\/latex] and [latex]{\\bf{n}}_2=\\langle2,3,-2\\rangle[\/latex]. These two vectors are scalar multiples of each other. The normal vectors are parallel, so the planes are parallel.<\/li>\n<li>The normal vectors for these planes are [latex]{\\bf{n}}_1=\\langle2,-3,2\\rangle[\/latex] and [latex]{\\bf{n}}_2=\\langle6,2,-3\\rangle[\/latex]. Taking the dot product of these vectors, we have<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]{\\bf{n}}_1\\cdot{\\bf{n}}_2=\\langle2,-3,2\\rangle\\cdot\\langle6,2,-3\\rangle=2(6)-3(2)+2(-3)=0[\/latex].<\/p>\n<p>The normal vectors are orthogonal, so the corresponding planes are orthogonal as well.<\/li>\n<li>The normal vectors for these planes are [latex]{\\bf{n}}_1=\\langle1,1,1\\rangle[\/latex] and\u00a0[latex]{\\bf{n}}_2=\\langle1,-3,5\\rangle[\/latex]:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\cos\\theta&=\\frac{|{\\bf{n}}_1\\cdot{\\bf{n}}_2|}{||{\\bf{n}}_1|| \\ ||{\\bf{n}}_2||} \\\\  &=\\frac{|\\langle1,1,1\\rangle\\cdot\\langle1,-3,5\\rangle|}{\\sqrt{1^2+1^2+1^2}\\sqrt{1^2+(-3)^2+5^2}} \\\\  &=\\frac{3}{\\sqrt{105}}.  \\end{aligned}[\/latex]<\/p>\n<p>The angle between the two planes is [latex]1.27[\/latex] rad, or approximately [latex]73^{\\small\\circ}[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Find the measure of the angle 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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q174365292\">Show Solution<\/span><\/p>\n<div id=\"q174365292\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]1.44[\/latex] rad<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div>\n<p id=\"fs-id1163724081830\">When we find that two planes are parallel, we may need to find the distance between them. To find this distance, we simply select a point in one of the planes. The distance from this point to the other plane is the distance between the planes.<\/p>\n<p id=\"fs-id1163724081835\">Previously, we introduced the formula for calculating this distance in\u00a0The Distance between a Plane and a Point:<\/p>\n<p style=\"text-align: center;\">[latex]\\large{d=\\frac{\\overrightarrow{QP}\\cdot{\\bf{n}}}{||{\\bf{n}}||}}[\/latex],<\/p>\n<p id=\"fs-id1163724097578\">where [latex]Q[\/latex] is a point on the plane, [latex]P[\/latex] is a point not on the plane, and [latex]\\textbf n[\/latex]\u00a0<span style=\"font-size: 1rem; text-align: initial;\">is the normal vector that passes through point [latex]Q[\/latex]. Consider the distance from point [latex](x_0, y_0, z_0)[\/latex] to plane [latex]ax+by+cz+k=0[\/latex]. Let [latex](x_1, y_1, z_1)[\/latex] be any point in the plane. Substituting into the formula yields<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\begin{aligned}  d&=\\frac{|a(x_0-x_1)+b(y_0-y_1)+c(z_0-z_1)|}{\\sqrt{a^2+b^2+c^2}} \\\\  &=\\frac{|ax_0+by_0+cz_0+k|}{\\sqrt{a^2+b^2+c^2}}  \\end{aligned}}[\/latex].<\/p>\n<p>We state this result formally in the following theorem.<\/p>\n<div id=\"fs-id1163724097697\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\n<div class=\"textbox shaded\">\n<header>\n<h3 class=\"os-title\" style=\"text-align: center;\"><span class=\"os-title-label\">THEOREM: distance from a point to a plane<\/span><\/h3>\n<hr \/>\n<\/header>\n<section>\n<div class=\"os-note-body\">\n<p id=\"fs-id1163723268198\">Let [latex]P(x_0, y_0, z_0)[\/latex] be a point. The distance from [latex]P[\/latex] to plane [latex]ax+by+cz+k=0[\/latex] is given by<\/p>\n<p style=\"text-align: center;\">[latex]\\large{d=\\frac{|ax_0+by_0+cz_0+k|}{\\sqrt{a^2+b^2+c^2}}}[\/latex].<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: finding the distance between parallel planes<\/h3>\n<p>Find the distance between the two parallel planes given by [latex]2x+y-z=2[\/latex] and\u00a0[latex]2x+y-z=8[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q943725601\">Show Solution<\/span><\/p>\n<div id=\"q943725601\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1163723334407\">Point [latex](1, 0, 0)[\/latex] lies in the first plane. The desired distance, then, is<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  d&=\\frac{|ax_0+by_0+cz_0+k|}{\\sqrt{a^2+b^2+c^2}} \\\\  &=\\frac{|2(1)+1(0)+(-1)(0)+(-8)|}{\\sqrt{2^2+1^2+(-1)^2}} \\\\  &=\\frac{6}{\\sqrt6}=\\sqrt6  \\end{aligned}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q437562978\">Show Solution<\/span><\/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<\/div>\n<div data-type=\"title\">\n<div class=\"textbox tryit\">\n<h3>Activity: distance between two skew lines<\/h3>\n<div id=\"attachment_5197\" style=\"width: 592px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5197\" class=\"size-full wp-image-5197\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31201814\/2.73.jpg\" alt=\"This figure shows a system of pipes running in different directions in an industrial plant. Two skew pipes are highlighted in red.\" width=\"582\" height=\"571\" \/><\/p>\n<p id=\"caption-attachment-5197\" class=\"wp-caption-text\">Figure 8. Industrial pipe installations often feature pipes running in different directions. How can we find the distance between two skew pipes?<\/p>\n<\/div>\n<p id=\"fs-id1163723210042\">Finding the distance from a point to a line or from a line to a plane seems like a pretty abstract procedure. But, if the lines represent pipes in a chemical plant or tubes in an oil refinery or roads at an intersection of highways, confirming that the distance between them meets specifications can be both important and awkward to measure. One way is to model the two pipes as lines, using the techniques in this chapter, and then calculate the distance between them. The calculation involves forming vectors along the directions of the lines and using both the cross product and the dot product.<\/p>\n<p id=\"fs-id1163723210050\">The symmetric forms of two lines, [latex]L_1[\/latex] and [latex]L_2[\/latex], are<\/p>\n<p style=\"text-align: center;\">[latex]L_1[\/latex]:\u00a0[latex]\\frac{x-x_1}{a_1}=\\frac{y-y_1}{b_1}=\\frac{z-z_1}{c_1}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]L_2[\/latex]:\u00a0[latex]\\frac{x-x_2}{a_2}=\\frac{y-y_2}{b_2}=\\frac{z-z_2}{c_2}[\/latex].<\/p>\n<p id=\"fs-id1163723200693\">You are to develop a formula for the distance [latex]d[\/latex] between these two lines, in terms of the values [latex]a_1[\/latex], [latex]b_1[\/latex], [latex]c_1[\/latex]; [latex]a_2[\/latex], [latex]b_2[\/latex], [latex]c_2[\/latex]; [latex]x_1[\/latex], [latex]y_1[\/latex], [latex]z_1[\/latex]; and [latex]x_2[\/latex], [latex]y_2[\/latex], [latex]z_2[\/latex]. The distance between two lines is usually taken to mean the minimum distance, so this is the length of a line segment or the length of a vector that is perpendicular to both lines and intersects both lines.<\/p>\n<ol id=\"fs-id1163723200803\" type=\"1\">\n<li>First, write down two vectors, [latex]{\\bf{v}}_1[\/latex] and [latex]{\\bf{v}}_2[\/latex], that lie along [latex]L_1[\/latex] and [latex]L_2[\/latex], respectively.<\/li>\n<li>Find the cross product of these two vectors and call it [latex]\\textbf N[\/latex]. This vector is perpendicular to [latex]{\\bf{v}}_1[\/latex] and [latex]{\\bf{v}}_2[\/latex], and hence is perpendicular to both lines.<\/li>\n<li>From vector [latex]\\textbf N[\/latex], form a unit vector [latex]\\textbf n[\/latex] in the same direction.<\/li>\n<li>Use symmetric equations to find a convenient vector [latex]{\\bf{v}}_{12}[\/latex] that lies between any two points, one on each line. Again, this can be done directly from the symmetric equations.<\/li>\n<li>The dot product of two vectors is the magnitude of the projection of one vector onto the other\u2014that is, [latex]{\\bf{A}}\\cdot{\\bf{B}}=||{\\bf{A}}|| \\ ||{\\bf{B}}||\\cos\\theta[\/latex], where [latex]\\theta[\/latex] is the angle between the vectors. Using the dot product, find the projection of vector [latex]{\\bf{v}}_{12}[\/latex] found in step 4 onto unit vector [latex]\\textbf n[\/latex] found in step 3. This projection is perpendicular to both lines, and hence its length must be the perpendicular distance [latex]d[\/latex] between them. Note that the value of [latex]d[\/latex] may be negative, depending on your choice of vector [latex]{\\bf{v}}_{12}[\/latex] or the order of the cross product, so use absolute value signs around the numerator.<\/li>\n<li>Check that your formula gives the correct distance of [latex]|-25|\/\\sqrt{198}\\approx1.78[\/latex] between the following two lines:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  L_1&:\\frac{x-5}2=\\frac{y-3}4=\\frac{z-1}3 \\\\  L_2&:\\frac{x-6}3=\\frac{y-1}5=\\frac{z}7.  \\end{aligned}[\/latex]<\/p>\n<\/li>\n<li>Is your general expression valid when the lines are parallel? If not, why not? (<em data-effect=\"italics\">Hint:<\/em>\u00a0What do you know about the value of the cross product of two parallel vectors? Where would that result show up in your expression for [latex]d[\/latex]?)<\/li>\n<li>Demonstrate that your expression for the distance is zero when the lines intersect. Recall that two lines intersect if they are not parallel and they are in the same plane. Hence, consider the direction of [latex]\\textbf n[\/latex] and [latex]{\\bf{v}}_{12}[\/latex]. What is the result of their dot product?<\/li>\n<li>Consider the following application. Engineers at a refinery have determined they need to install support struts between many of the gas pipes to reduce damaging vibrations. To minimize cost, they plan to install these struts at the closest points between adjacent skewed pipes. Because they have detailed schematics of the structure, they are able to determine the correct lengths of the struts needed, and hence manufacture and distribute them to the installation crews without spending valuable time making measurements.<span data-type=\"newline\"><br \/>\n<\/span>The rectangular frame structure has the dimensions [latex]4.0\\times15.0\\times10.0\\text{ m}[\/latex] (height, width, and depth). One sector has a pipe entering the lower corner of the standard frame unit and exiting at the diametrically opposed corner (the one farthest away at the top); call this [latex]L_1[\/latex]. A second pipe enters and exits at the two different opposite lower corners; call this [latex]L_2[\/latex] (Figure 9).<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"CNX_Calc_Figure_12_05_017\" class=\"os-figure\">\n<div class=\"os-caption-container\">\n<div id=\"attachment_5199\" style=\"width: 407px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5199\" class=\"size-full wp-image-5199\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31201941\/2.74.jpg\" alt=\"This figure is a three-dimensional box in an x y z coordinate system. The box has dimensions x = 10 m, y = 15 m, and z = 4 m. Line L1 passes through a main diagonal of the box from the origin to the far corner. Line L2 passes through a diagonal in the base of the box with x-intercept 10 and y-intercept 15.\" width=\"397\" height=\"394\" \/><\/p>\n<p id=\"caption-attachment-5199\" class=\"wp-caption-text\">Figure 9. Two pipes cross through a standard frame unit.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span>Write down the vectors along the lines representing those pipes, find the cross product between them from which to create the unit vector [latex]\\textbf n[\/latex], define a vector that spans two points on each line, and finally determine the minimum distance between the lines. (Take the origin to be at the lower corner of the first pipe.) Similarly, you may also develop the symmetric equations for each line and substitute directly into your formula.<\/li>\n<\/ol>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<div data-type=\"note\">\n<div id=\"fs-id1163724036465\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\n<div id=\"fs-id1163724074875\" class=\"theorem ui-has-child-title\" data-type=\"note\">\n<div id=\"fs-id1163724080388\" class=\"ui-has-child-title\" data-type=\"example\">\n<div data-type=\"example\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-812\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>CP 2.47. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>CP 2.49. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 3. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":349141,"menu_order":22,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 3\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"CP 2.47\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"CP 2.49\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-812","chapter","type-chapter","status-publish","hentry"],"part":20,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/812","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/349141"}],"version-history":[{"count":91,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/812\/revisions"}],"predecessor-version":[{"id":6505,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/812\/revisions\/6505"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/20"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/812\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=812"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=812"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=812"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=812"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}