{"id":5457,"date":"2022-06-02T18:31:16","date_gmt":"2022-06-02T18:31:16","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=5457"},"modified":"2022-10-21T00:19:53","modified_gmt":"2022-10-21T00:19:53","slug":"equations-for-a-line-in-space","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/equations-for-a-line-in-space\/","title":{"raw":"Equations for a Line in Space","rendered":"Equations for a Line 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, parametric, and symmetric equations of a line through a given point in a given direction, and a line through two given points.<\/span><\/li>\r\n \t<li><span class=\"os-abstract-content\">Find the distance from a point to a given line.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<section id=\"fs-id1163724073036\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Equations for a Line in Space<\/h2>\r\n<p id=\"fs-id1163723200645\">Let\u2019s first explore what it means for two vectors to be parallel. Recall that parallel vectors must have the same or opposite directions. If two nonzero vectors, [latex]\\textbf u[\/latex] and [latex]\\textbf v[\/latex], are parallel, we claim there must be a scalar, [latex]k[\/latex], such that [latex]{\\bf{u}}=k{\\bf{v}}[\/latex]. If [latex]\\textbf u[\/latex] and [latex]\\textbf v[\/latex] have the same direction, simply choose [latex]k=\\frac{||{\\bf{u}}||}{||{\\bf{v}}||}[\/latex]. If [latex]\\textbf u[\/latex] and [latex]\\textbf v[\/latex] have opposite directions, choose [latex]k=-\\frac{||{\\bf{u}}||}{||{\\bf{v}}||}[\/latex]. Note that the converse holds as well. If [latex]{\\bf{u}}=k{\\bf{v}}[\/latex] for some scalar [latex]k[\/latex], then either [latex]\\textbf u[\/latex] and [latex]\\textbf v[\/latex] have the same direction [latex](k&gt;0)[\/latex] or opposite directions [latex](k&lt;0)[\/latex], so [latex]\\textbf u[\/latex] and [latex]\\textbf v[\/latex] are parallel. Therefore, two nonzero vectors [latex]\\textbf u[\/latex] and [latex]\\textbf v[\/latex] are parallel if and only if [latex]{\\bf{u}}=k{\\bf{v}}[\/latex] for some scalar [latex]k[\/latex]. By convention, the zero vector [latex]\\textbf 0[\/latex] is considered to be parallel to all vectors.<\/p>\r\n<p id=\"fs-id1163723994177\">As in two dimensions, we can describe a line in space using a point on the line and the direction of the line, or a parallel vector, which we call the\u00a0<strong><span id=\"60ce88e7-9116-48d3-bb30-013c89cddc64_term87\" data-type=\"term\">direction vector<\/span><\/strong>\u00a0(Figure 1). Let [latex]L[\/latex] be a line in space passing through point [latex]P(x_0, y_0, z_0)[\/latex]. Let [latex]{\\bf{v}}=\\langle a,b,c\\rangle[\/latex] be a vector parallel to [latex]L[\/latex]. Then, for any point on line [latex]Q(x, y, z)[\/latex], we know that [latex]\\overrightarrow{PQ}[\/latex] is parallel to [latex]\\textbf v[\/latex]. Thus, as we just discussed, there is a scalar, [latex]t[\/latex], such that [latex]\\overrightarrow{PQ}=t{\\bf{v}}[\/latex], which gives<\/p>\r\n\r\n<div id=\"fs-id1167793270756\" class=\"ui-has-child-title\" style=\"text-align: center;\" data-type=\"note\">[latex]\\begin{aligned}\r\n\\overrightarrow{PQ}&amp;=t{\\bf{v}} \\\\\r\n\\langle x-x_0,y-y_0,z-z_0\\rangle&amp;=t\\langle a,b,c\\rangle \\\\\r\n\\langle x-x_0,y-y_0,z-z_0\\rangle&amp;=\\langle ta,tb,tc\\rangle.\r\n\\end{aligned}[\/latex]<\/div>\r\n<\/section>[caption id=\"attachment_5157\" align=\"aligncenter\" width=\"433\"]<img class=\"size-full wp-image-5157\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27211934\/2.63.jpg\" alt=\"This figure is the first octant of the 3-dimensional coordinate system. There is a line segment passing through two points. The points are labeled \u201cP = (x sub 0, y sub 0, z sub 0)\u201d and \u201cQ = (x, y, z).\u201d There is also a vector in standard position drawn. The vector is labeled \u201cv = &lt;a, b, c&gt;.\u201d\" width=\"433\" height=\"390\" \/> Figure 1. Vector [latex]\\textbf v[\/latex] is the direction vector for [latex]\\overrightarrow{PQ}[\/latex].[\/caption]\r\n<p id=\"fs-id1163723876839\">Using vector operations, we can rewrite\u00a0this equation\u00a0as<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\langle x-x_0,y-y_0,z-z_0\\rangle&amp;=\\langle ta,tb,tc\\rangle \\\\\r\n\\langle x,y,z\\rangle -\\langle x_0,y_0,z_0\\rangle&amp;=t\\langle a,b,c\\rangle \\\\\r\n\\langle x,y,z\\rangle&amp;=\\langle x_0,y_0,z_0\\rangle+t\\langle a,b,c\\rangle\r\n\\end{aligned}[\/latex].<\/p>\r\nSetting [latex]{\\bf{r}}=\\langle x,y,z\\rangle[\/latex] and [latex]{\\bf{r}}_0=\\langle x_0,y_0,z_0\\rangle[\/latex], we now have the\u00a0<strong><span id=\"60ce88e7-9116-48d3-bb30-013c89cddc64_term88\" data-type=\"term\">vector equation of a line<\/span><\/strong>:\r\n<div id=\"fs-id1163723862928\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\"><\/div>\r\n<p style=\"text-align: center;\">[latex]{\\bf{r}}={\\bf{r}}_0+t{\\bf{v}}[\/latex].<\/p>\r\nEquating components,\u00a0the Vector Equation of a Line\u00a0shows that the following equations are simultaneously true: [latex]x-x_0=ta[\/latex], [latex]y-y_0=tb[\/latex], and [latex]z-z_0=tc[\/latex]. If we solve each of these equations for the component variables [latex]x[\/latex], [latex]y[\/latex], and [latex]z[\/latex], we get a set of equations in which each variable is defined in terms of the parameter [latex]t[\/latex] and that, together, describe the line. This set of three equations forms a set of\u00a0<strong><span id=\"60ce88e7-9116-48d3-bb30-013c89cddc64_term89\" data-type=\"term\">parametric equations of a line<\/span>:<\/strong>\r\n<div id=\"fs-id1163724080297\" data-type=\"equation\"><\/div>\r\n<p style=\"text-align: center;\">[latex]x=x_0+ta \\quad y=y_0+tb \\quad z=z_0+tc[\/latex].<\/p>\r\n<p id=\"fs-id1163723236850\">If we solve each of the equations for [latex]t[\/latex] assuming [latex]a[\/latex], [latex]b[\/latex], and [latex]c[\/latex] are nonzero, we get a different description of the same line:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\frac{x-x_0}a=t \\quad \\frac{y-y_o}b=t \\quad \\frac{z-z_0}c=t[\/latex].<\/p>\r\n<p id=\"fs-id1163723755477\">Because each expression equals [latex]t[\/latex], they all have the same value. We can set them equal to each other to create\u00a0<strong><span id=\"60ce88e7-9116-48d3-bb30-013c89cddc64_term90\" data-type=\"term\">symmetric equations of a line<\/span>:<\/strong><\/p>\r\n\r\n<div id=\"fs-id1167793270756\" class=\"ui-has-child-title\" style=\"text-align: center;\" data-type=\"note\">[latex]\\frac{x-x_0}a=\\frac{y-y_o}b=\\frac{z-z_0}c[\/latex].<\/div>\r\n<p id=\"fs-id1163724050718\">We summarize the results in the following theorem.<\/p>\r\n\r\n<div data-type=\"example\">\r\n<div class=\"textbox shaded\"><header>\r\n<h3 class=\"os-title\" style=\"text-align: center;\"><span class=\"os-title-label\">THEOREM: parametric and symmetric equations of a line<\/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-id1163723197352\">A line [latex]L[\/latex] parallel to vector [latex]{\\bf{v}}=\\langle a,b,c\\rangle[\/latex] and passing through point [latex]P(x_0, y_0, z_0)[\/latex] can be described by the following parametric equations:<\/p>\r\n<p style=\"text-align: center;\">[latex]x=x_0+ta, \\ y=y_0+tb, \\ z=z_0+tc[\/latex].<\/p>\r\n<p id=\"fs-id1163723635234\">If the constants [latex]a[\/latex], [latex]b[\/latex], and [latex]c[\/latex] are all nonzero, then [latex]L[\/latex] can be described by the symmetric equation of the line:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\frac{x-x_0}a=\\frac{y-y_o}b=\\frac{z-z_0}c[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<p id=\"fs-id1163723757318\">The parametric equations of a line are not unique. Using a different parallel vector or a different point on the line leads to a different, equivalent representation. Each set of parametric equations leads to a related set of symmetric equations, so it follows that a symmetric equation of a line is not unique either.<\/p>\r\n\r\n<div id=\"fs-id1163723137804\" class=\"ui-has-child-title\" data-type=\"example\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: equations of a line in space<\/h3>\r\nFind parametric and symmetric equations of the line passing through points [latex](1, 4, -2)[\/latex] and\u00a0[latex](-3, 5, 0)[\/latex].\r\n\r\n[reveal-answer q=\"389475782\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"389475782\"]\r\n<p id=\"fs-id1163723796963\">First, identify a vector parallel to the line:<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\bf{v}}=\\langle-3-1,5-4,0-(-2)\\rangle=\\langle-4,1,2\\rangle[\/latex].<\/p>\r\n<p id=\"fs-id1163724059770\">Use either of the given points on the line to complete the parametric equations:<\/p>\r\n<p style=\"text-align: center;\">[latex]x=1-4t[\/latex],\u00a0[latex]y=4+t[\/latex], and\u00a0[latex]z=-2+2t[\/latex].<\/p>\r\n<p id=\"fs-id1163723281292\">Solve each equation for [latex]t[\/latex] to create the symmetric equation of the line:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\frac{x-1}{-4}=y-4=\\frac{z+2}2[\/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 and symmetric equations of the line passing through points [latex](1, -3, 2)[\/latex] and [latex](5, -2 ,8)[\/latex].\r\n\r\n[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<\/div>\r\n<p id=\"fs-id1163723200547\">Sometimes we don\u2019t want the equation of a whole line, just a line segment. In this case, we limit the values of our parameter [latex]t[\/latex]. For example, let [latex]P(x_0, y_0, z_0)[\/latex] and [latex]Q(x_1, y_1, z_1)[\/latex] be points on a line, and let [latex]{\\bf{p}}=\\langle x_0,y_0,z_0\\rangle[\/latex] and [latex]{\\bf{q}}=\\langle x_1,y_1,z_1\\rangle[\/latex] be the associated position vectors. In addition, let [latex]{\\bf{r}}=\\langle x,y,z\\rangle[\/latex]. We want to find a vector equation for the line segment between [latex]P[\/latex] and [latex]Q[\/latex]. Using [latex]P[\/latex] as our known point on the line, and [latex]\\overrightarrow{PQ}=\\langle x_1-x_0,y_1-y_0,z_1-z_0\\rangle[\/latex] as the direction vector equation,\u00a0the Vector Equation of a Line\u00a0gives<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\bf{r}}={\\bf{p}}+t \\left( \\overrightarrow{PQ}\\right)[\/latex].<\/p>\r\n<p id=\"fs-id1163723341582\">Using properties of vectors, then<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n{\\bf{r}}&amp;={\\bf{p}}+ t \\left(\\overrightarrow{PQ}\\right) \\\\\r\n&amp;=\\langle x_0,y_0,z_0\\rangle+t\\langle x_1-x_0,y_1-y_0,z_1-z_0\\rangle \\\\\r\n&amp;=\\langle x_0,y_0,z_0\\rangle+t(\\langle x_1,y_1,z_1\\rangle-\\langle x_0,y_0,z_0\\rangle) \\\\\r\n&amp;=\\langle x_0,y_0,z_0\\rangle+t\\langle x_1,y_1,z_1\\rangle-t\\langle x_0,y_0,z_0\\rangle \\\\\r\n&amp;=(1-t)\\langle x_0,y_0,z_0\\rangle+t\\langle x_1,y_1,z_1\\rangle \\\\\r\n&amp;=(1-t){\\bf{p}}+t{\\bf{q}}\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1163724050947\">Thus, the vector equation of the line passing through [latex]P[\/latex] and [latex]Q[\/latex] is<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\bf{r}}=(1-t){\\bf{p}}+t{\\bf{q}}[\/latex].<\/p>\r\n<p id=\"fs-id1163723363447\">Remember that we didn\u2019t want the equation of the whole line, just the line segment between [latex]P[\/latex] and [latex]Q[\/latex]. Notice that when [latex]t=0[\/latex], we have [latex]{\\bf{r}}={\\bf{p}}[\/latex], and when [latex]t=1[\/latex], we have [latex]{\\bf{r}}={\\bf{q}}[\/latex]. Therefore, the vector equation of the line segment between [latex]P[\/latex] and [latex]Q[\/latex] is<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\bf{r}}=(1-t){\\bf{p}}+t{\\bf{q}}[\/latex],\u00a0[latex]0\\leq t\\leq1[\/latex].<\/p>\r\n<p id=\"fs-id1163724067505\">Going back to\u00a0the Vector Equation of a Line, we can also find parametric equations for this line segment. We have<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n{\\bf{r}}&amp;={\\bf{p}}+\\left(\\overrightarrow{PQ}\\right) \\\\\r\n\\langle x,y,z\\rangle &amp;=\\langle x_0,y_0,z_0\\rangle+t\\langle x_1-x_0,y_1-y_0,z_1-z_0\\rangle \\\\\r\n&amp;=\\langle x_0+t(x_1-x_0),y_0+t(y_1-y_0),z_0+t(z_1-z_0)\\rangle\r\n\\end{aligned}[\/latex].<\/p>\r\nThen, the parametric equations are\r\n\r\n[latex]x=x_0+t(x_1-x_0)[\/latex],\u00a0[latex]y=y_0+t(y_1-y_0)[\/latex],\u00a0[latex]z=z_0+t(z_1-z_0)[\/latex],\u00a0[latex]0\\leq t\\leq1[\/latex].\r\n\r\n<\/div>\r\n<\/div>\r\n<div data-type=\"example\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: parametric equations of a line segment<\/h3>\r\nFind parametric equations of the line segment between the points [latex]P(2, 1, 4)[\/latex] and\u00a0[latex]Q(3, -1 ,3)[\/latex].\r\n\r\n[reveal-answer q=\"112936453\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"112936453\"]\r\n<p id=\"fs-id1163724069731\">By\u00a0the Parametric Equations of a Line, we have<\/p>\r\n<p style=\"text-align: center;\">[latex]x=x_0+t(x_1-x_0)[\/latex],\u00a0[latex]y=y_0+t(y_1-y_0)[\/latex],\u00a0[latex]z=z_0+t(z_1-z_0)[\/latex],\u00a0[latex]0\\leq t\\leq1[\/latex].<\/p>\r\n<p id=\"fs-id1163724054258\">Working with each component separately, we get<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nx&amp;=x_0+t(x_1-x_0) \\\\\r\n&amp;=2+t(3-2) \\\\\r\n&amp;=2+t,\r\n\\end{aligned}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\ny&amp;=y_0+t(y_1-y_0) \\\\\r\n&amp;=1+t(-1-1) \\\\\r\n&amp;=1-2t,\r\n\\end{aligned}[\/latex]<\/p>\r\n<p id=\"fs-id1163723280727\">and<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nz&amp;=z_0+t(z_1-z_0) \\\\\r\n&amp;=4+t(3-4) \\\\\r\n&amp;=4-t.\r\n\\end{aligned}[\/latex]<\/p>\r\n<p id=\"fs-id1163723200089\">Therefore, the parametric equations for the line segment are<\/p>\r\n<p style=\"text-align: center;\">[latex]x=2+t,y=1-2t,z=4-t, \\ 0\\leq t\\leq1[\/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 of the line segment between points [latex]P(-1, 3, 6)[\/latex] and\u00a0[latex]Q(-8, 2, 4)[\/latex].\r\n\r\n[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<\/div>\r\n<h2 data-type=\"title\">Distance between a Point and a Line<\/h2>\r\n<p id=\"fs-id1163724043792\">We already know how to calculate the distance between two points in space. We now expand this definition to describe the distance between a point and a line in space. Several real-world contexts exist when it is important to be able to calculate these distances. When building a home, for example, builders must consider \u201csetback\u201d requirements, when structures or fixtures have to be a certain distance from the property line. Air travel offers another example. Airlines are concerned about the distances between populated areas and proposed flight paths.<\/p>\r\n<p id=\"fs-id1163723236930\">Let [latex]L[\/latex] be a line in the plane and let [latex]M[\/latex] be any point not on the line. Then, we define distance [latex]d[\/latex] from [latex]M[\/latex] to [latex]L[\/latex] as the length of line segment [latex]\\overline{MP}[\/latex], where [latex]P[\/latex] is a point on [latex]L[\/latex] such that [latex]\\overline{MP}[\/latex] is perpendicular to [latex]L[\/latex] (Figure 2).<\/p>\r\n\r\n<\/div>\r\n<div data-type=\"example\">[caption id=\"attachment_5160\" align=\"aligncenter\" width=\"382\"]<img class=\"size-full wp-image-5160\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27212154\/2.64.jpg\" alt=\"This figure has two line segments. The first line is labeled \u201cL\u201d and has point P on the segment. The second line segment is drawn from point P to point M and is perpendicular to line L. The second line segment is labeled \u201cd.\u201d\" width=\"382\" height=\"210\" \/> Figure 2. The distance from point [latex]M[\/latex] to line [latex]L[\/latex] is the length of [latex]\\overline{MP}[\/latex].[\/caption]<\/div>\r\n<div data-type=\"example\"><\/div>\r\n<div data-type=\"example\">When we\u2019re looking for the distance between a line and a point in space,\u00a0Figure 2\u00a0still applies. We still define the distance as the length of the perpendicular line segment connecting the point to the line. In space, however, there is no clear way to know which point on the line creates such a perpendicular line segment, so we select an arbitrary point on the line and use properties of vectors to calculate the distance. Therefore, let [latex]P[\/latex] be an arbitrary point on line [latex]L[\/latex] and let [latex]\\textbf v[\/latex] be a direction vector for [latex]L[\/latex] (Figure 3).<\/div>\r\n<div data-type=\"example\"><\/div>\r\n<div data-type=\"example\">[caption id=\"attachment_5162\" align=\"aligncenter\" width=\"438\"]<img class=\"size-full wp-image-5162\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27212515\/2.65.jpg\" alt=\"This figure has a line segment labeled \u201cL.\u201d On the line segment L there is point P. There is a vector drawn from point P to another point M. Also, from M there is a line segment drawn to line L. This segment is perpendicular to line L. There is also a vector labeled \u201cv\u201d on line segment L. A parallelogram has been formed with vector v, line segment P M, and two other segments back to line L.\" width=\"438\" height=\"234\" \/> Figure 3. Vectors [latex]\\overrightarrow{PM}[\/latex] and [latex]\\textbf v[\/latex] form two sides of a parallelogram with base [latex]||{\\bf{v}}||[\/latex] and height [latex]d[\/latex], which is the distance between a line and a point in space.[\/caption]<\/div>\r\n<div data-type=\"example\"><\/div>\r\n<div data-type=\"example\">\r\n<p id=\"fs-id1163724075400\">By\u00a0Theorem: Area of a Parallelogram, vectors [latex]\\overrightarrow{PM}[\/latex] and [latex]\\textbf v[\/latex] form two sides of a parallelogram with area [latex]||\\overrightarrow{PM}\\times{\\bf{v}}||[\/latex]. Using a formula from geometry, the area of this parallelogram can also be calculated as the product of its base and height:<\/p>\r\n<p style=\"text-align: center;\">[latex]||\\overrightarrow{PM}\\times{\\bf{v}}||=||{\\bf{v}}||d[\/latex].<\/p>\r\n<p id=\"fs-id1163724051243\">We can use this formula to find a general formula for the distance between a line in space and any point not on the line.<\/p>\r\n\r\n<\/div>\r\n<div data-type=\"example\">\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 line<\/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-id1163723987376\">Let [latex]L[\/latex] be a line in space passing through point [latex]P[\/latex] with direction vector [latex]\\textbf v[\/latex]. If [latex]M[\/latex] is any point not on [latex]L[\/latex], then the distance from [latex]M[\/latex] to [latex]L[\/latex] is<\/p>\r\n<p style=\"text-align: center;\">[latex]d=\\frac{||\\overrightarrow{PM}\\times{\\bf{v}}||}{||{\\bf{v}}||}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: calculating the distance from a point to a line<\/h3>\r\nFind the distance between t point [latex]M=(1, 1, 3)[\/latex] and line [latex]\\frac{x-3}4=\\frac{y+1}2=z-3[\/latex].\r\n\r\n[reveal-answer q=\"483752783\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"483752783\"]\r\n<p id=\"fs-id1163723304185\">From the symmetric equations of the line, we know that vector [latex]{\\bf{v}}\\langle4,2,1\\rangle[\/latex] is a direction vector for the line. Setting the symmetric equations of the line equal to zero, we see that point [latex]P(3, -1 , 3)[\/latex] lies on the line. Then,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\overrightarrow{PM}=\\langle1-3,1-(-1),3-3\\rangle=\\langle-2,2,0\\rangle[\/latex].<\/p>\r\n<p id=\"fs-id1163723203966\">To calculate the distance, we need to find [latex]\\overrightarrow{PM}\\times{\\bf{v}}[\/latex]:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\overrightarrow{PM}\\times{\\bf{v}}&amp;=\\begin{vmatrix}{\\bf{i}}&amp;{\\bf{j}}&amp;{\\bf{k}} \\\\-2&amp;2&amp;0 \\\\ 4&amp;2&amp;1\\end{vmatrix} \\\\\r\n&amp;=(2-1){\\bf{i}}-(-2-0){\\bf{j}}+(-4-8){\\bf{k}} \\\\\r\n&amp;=2{\\bf{i}}+2{\\bf{j}}-12{\\bf{k}}.\r\n\\end{aligned}[\/latex]<\/p>\r\n<p id=\"fs-id1163724039857\">Therefore, the distance between the point and the line is (Figure 4)<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nd&amp;=\\frac{||\\overrightarrow{PM}\\times{\\bf{v}}||}{||{\\bf{v}}||} \\\\\r\n&amp;=\\frac{\\sqrt{2^2+2^2+12^2}}{\\sqrt{4^2+2^2+1^2}} \\\\\r\n&amp;=\\frac{s\\sqrt{38}}{\\sqrt{21}}.\r\n\\end{aligned}[\/latex]<\/p>\r\n\r\n[caption id=\"attachment_5164\" align=\"aligncenter\" width=\"357\"]<img class=\"size-full wp-image-5164\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27212704\/2.66.jpg\" alt=\"This figure is the first octant of the 3-dimensional coordinate system. There is a 3-dimensional box drawn in the octant. There is a point labeled at (1, 1, 3). There is a line segment labeled \u201cL\u201d inside of the box. Also, there is a perpendicular line segment from the point to line L.\" width=\"357\" height=\"396\" \/> Figure 4. Point [latex](1, 1, 3)[\/latex] is approximately [latex]2.7[\/latex] units from the line with symmetric equations [latex]\\frac{x-3}4=\\frac{y+1}2=z-3[\/latex].[\/caption][\/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](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].\r\n\r\n[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<\/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=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\"><\/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>\r\n<h2 data-type=\"title\">Relationships between Lines<\/h2>\r\n<p id=\"fs-id1163724035950\">Given two lines in the two-dimensional plane, the lines are equal, they are parallel but not equal, or they intersect in a single point. In three dimensions, a fourth case is possible. If two lines in space are not parallel, but do not intersect, then the lines are said to be\u00a0<strong><span id=\"60ce88e7-9116-48d3-bb30-013c89cddc64_term91\" data-type=\"term\">skew lines<\/span><\/strong>\u00a0(Figure 5).<\/p>\r\n\r\n\r\n[caption id=\"attachment_5166\" align=\"aligncenter\" width=\"418\"]<img class=\"size-full wp-image-5166\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27212930\/2.67.jpg\" alt=\"This figure has two line segments. They are 3-dimensional, are not parallel, and do not intersect. The directions are different and one is above the other.\" width=\"418\" height=\"117\" \/> Figure 5. In three dimensions, it is possible that two lines do not cross, even when they have different directions.[\/caption]\r\n\r\n<\/div>\r\n<div data-type=\"example\">To classify lines as parallel but not equal, equal, intersecting, or skew, we need to know two things: whether the direction vectors are parallel and whether the lines share a point (Figure 6).<\/div>\r\n<div data-type=\"example\"><\/div>\r\n<div data-type=\"example\">\r\n\r\n[caption id=\"attachment_5167\" align=\"aligncenter\" width=\"610\"]<img class=\"size-full wp-image-5167\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27213057\/2.68.jpg\" alt=\"This figure is a table with two rows and two columns. Above the columns is the question \u201cLines share a common point?\u201d The first column is labeled \u201cyes,\u201d and the second column is labeled \u201cno.\u201d To the left of the rows is the question \u201cDirection vectors are parallel?\u201d The first row is labeled \u201cyes,\u201d and the second row is labeled \u201cno.\u201d The entries of the first row are \u201cequal\u201d and \u201cparallel but not equal.\u201d The entries in the second row are \u201cintersecting\u201d and \u201cskew.\u201d\" width=\"610\" height=\"334\" \/> Figure 6. Determine the relationship between two lines based on whether their direction vectors are parallel and whether they share a point.[\/caption]\r\n\r\n<\/div>\r\n<div data-type=\"example\"><\/div>\r\n<div data-type=\"example\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: classifying lines in space<\/h3>\r\n<p id=\"fs-id1163723283999\">For each pair of lines, determine whether the lines are equal, parallel but not equal, skew, or intersecting.<\/p>\r\n\r\n<ol id=\"fs-id1163723284004\" type=\"a\">\r\n \t<li><span data-type=\"newline\">[latex]L_1[\/latex]:\u00a0[latex]x=2s-1[\/latex],\u00a0[latex]y=s-1[\/latex],\u00a0[latex]z=s-4[\/latex]\r\n<\/span>[latex]L_2[\/latex]:\u00a0[latex]x=t-3[\/latex],\u00a0[latex]y=3t+8[\/latex],\u00a0[latex]z=5-2t[\/latex]<\/li>\r\n \t<li><span data-type=\"newline\">[latex]L_1[\/latex]:\u00a0[latex]x=-y=z[\/latex]\r\n<\/span>[latex]L_2[\/latex]:\u00a0[latex]\\frac{x-3}2=y=z-2[\/latex]<\/li>\r\n \t<li><span data-type=\"newline\">[latex]L_1[\/latex]:\u00a0[latex]x=6s-1[\/latex],\u00a0[latex]y=-2s[\/latex], [latex]z=3s+1[\/latex]\r\n<\/span>[latex]L_2[\/latex]:\u00a0[latex]\\frac{x-4}6=\\frac{y+3}{-2}=\\frac{z-1}3[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"849520983\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"849520983\"]\r\n<ol id=\"fs-id1163724052863\" type=\"a\">\r\n \t<li>Line [latex]L_1[\/latex] has direction vector [latex]{\\bf{v_1}}=\\langle2,1,1\\rangle[\/latex]; line [latex]L_2[\/latex] has direction vector [latex]{\\bf{v_2}}=\\langle1,3,-2\\rangle[\/latex]. Because the direction vectors are not parallel vectors, the lines are either intersecting or skew. To determine whether the lines intersect, we see if there is a point, [latex](x, y, z)[\/latex], that lies on both lines. To find this point, we use the parametric equations to create a system of equalities:<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]2s-1=t-3; \\quad s-1=3t+8; \\quad s-4=5-2t[\/latex].<\/p>\r\n<span data-type=\"newline\">\r\n<\/span>By the first equation, [latex]t=2s+2[\/latex]. Substituting into the second equation yields<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\ns-1&amp;=3(2s+2)+8 \\\\\r\ns-1&amp;=6s+6+8 \\\\\r\n5s&amp;=-15 \\\\\r\ns&amp;=-3.\r\n\\end{aligned}[\/latex]<\/p>\r\n<span data-type=\"newline\">\r\n<\/span>Substitution into the third equation, however, yields a contradiction:<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\ns-4&amp;=5-2(2s+2) \\\\\r\ns-4&amp;=5-4s-4 \\\\\r\n5s&amp;=5 \\\\\r\ns&amp;=1.\r\n\\end{aligned}[\/latex]<\/p>\r\n<span data-type=\"newline\">\r\n<\/span>There is no single point that satisfies the parametric equations for [latex]L_1[\/latex] and [latex]L_2[\/latex] simultaneously. These lines do not intersect, so they are skew (see the following figure).<span data-type=\"newline\">\r\n<\/span>\r\n\r\n[caption id=\"attachment_5168\" align=\"aligncenter\" width=\"376\"]<img class=\"size-full wp-image-5168\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27213249\/Example-2.48.jpg\" alt=\"This figure is the 3-dimensional coordinate system. There are two skew lines drawn. They do not intersect and are not parallel.\" width=\"376\" height=\"380\" \/> Figure 7. Lines [latex]L_1[\/latex] and [latex]L_2[\/latex] are skew.[\/caption]<\/li>\r\n \t<li>Line [latex]L_1[\/latex] has direction vector [latex]{\\bf{v_1}}=\\langle1,-1,1\\rangle[\/latex] and passes through the origin, [latex](0, 0, 0)[\/latex]. Line [latex]L_2[\/latex] has a different direction vector, [latex]{\\bf{v_2}}=\\langle2,1,1\\rangle[\/latex], so these lines are not parallel or equal. Let [latex]r[\/latex] represent the parameter for line [latex]L_1[\/latex] and let [latex]s[\/latex] represent the parameter for [latex]L_2[\/latex]:<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nx&amp;=r &amp;\\quad x&amp;=2s+3 \\\\\r\ny&amp;=-r &amp;\\quad y&amp;=s \\\\\r\nz&amp;=r &amp;\\quad z&amp;=s+2\r\n\\end{aligned}[\/latex].<\/p>\r\n<span data-type=\"newline\">\r\n<\/span>Solve the system of equations to find [latex]r=1[\/latex] and [latex]s=-1[\/latex]. If we need to find the point of intersection, we can substitute these parameters into the original equations to get [latex](1, -1, 1)[\/latex] (see the following figure).<span data-type=\"newline\">\r\n<\/span>\r\n\r\n[caption id=\"attachment_5169\" align=\"aligncenter\" width=\"514\"]<img class=\"size-full wp-image-5169\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27213452\/Example-2.48b.jpg\" alt=\"This figure is the 3-dimensional coordinate system. There are two lines drawn. They intersect at the point (1,-1,1).\" width=\"514\" height=\"501\" \/> Figure 8. Lines [latex]L_1[\/latex] and [latex]L_2[\/latex] intersect at [latex](1, -1, 1)[\/latex].[\/caption]<\/li>\r\n \t<li>Lines [latex]L_1[\/latex] and [latex]L_2[\/latex] have equivalent direction vectors: [latex]{\\bf{v}}=\\langle6,-2,3\\rangle[\/latex]. These two lines are parallel (see the following figure).<span data-type=\"newline\">\r\n<\/span>[caption id=\"attachment_5170\" align=\"aligncenter\" width=\"397\"]<img class=\"size-full wp-image-5170\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27213531\/Example-2.48c.jpg\" alt=\"This figure is the 3-dimensional coordinate system. There are two lines drawn. They do not intersect and are parallel.\" width=\"397\" height=\"378\" \/> Figure 9. Lines [latex]L_1[\/latex] and [latex]L_2[\/latex] are parallel.[\/caption]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<p id=\"fs-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<\/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, parametric, and symmetric equations of a line through a given point in a given direction, and a line through two given points.<\/span><\/li>\n<li><span class=\"os-abstract-content\">Find the distance from a point to a given line.<\/span><\/li>\n<\/ul>\n<\/div>\n<section id=\"fs-id1163724073036\" data-depth=\"1\">\n<h2 data-type=\"title\">Equations for a Line in Space<\/h2>\n<p id=\"fs-id1163723200645\">Let\u2019s first explore what it means for two vectors to be parallel. Recall that parallel vectors must have the same or opposite directions. If two nonzero vectors, [latex]\\textbf u[\/latex] and [latex]\\textbf v[\/latex], are parallel, we claim there must be a scalar, [latex]k[\/latex], such that [latex]{\\bf{u}}=k{\\bf{v}}[\/latex]. If [latex]\\textbf u[\/latex] and [latex]\\textbf v[\/latex] have the same direction, simply choose [latex]k=\\frac{||{\\bf{u}}||}{||{\\bf{v}}||}[\/latex]. If [latex]\\textbf u[\/latex] and [latex]\\textbf v[\/latex] have opposite directions, choose [latex]k=-\\frac{||{\\bf{u}}||}{||{\\bf{v}}||}[\/latex]. Note that the converse holds as well. If [latex]{\\bf{u}}=k{\\bf{v}}[\/latex] for some scalar [latex]k[\/latex], then either [latex]\\textbf u[\/latex] and [latex]\\textbf v[\/latex] have the same direction [latex](k>0)[\/latex] or opposite directions [latex](k<0)[\/latex], so [latex]\\textbf u[\/latex] and [latex]\\textbf v[\/latex] are parallel. Therefore, two nonzero vectors [latex]\\textbf u[\/latex] and [latex]\\textbf v[\/latex] are parallel if and only if [latex]{\\bf{u}}=k{\\bf{v}}[\/latex] for some scalar [latex]k[\/latex]. By convention, the zero vector [latex]\\textbf 0[\/latex] is considered to be parallel to all vectors.<\/p>\n<p id=\"fs-id1163723994177\">As in two dimensions, we can describe a line in space using a point on the line and the direction of the line, or a parallel vector, which we call the\u00a0<strong><span id=\"60ce88e7-9116-48d3-bb30-013c89cddc64_term87\" data-type=\"term\">direction vector<\/span><\/strong>\u00a0(Figure 1). Let [latex]L[\/latex] be a line in space passing through point [latex]P(x_0, y_0, z_0)[\/latex]. Let [latex]{\\bf{v}}=\\langle a,b,c\\rangle[\/latex] be a vector parallel to [latex]L[\/latex]. Then, for any point on line [latex]Q(x, y, z)[\/latex], we know that [latex]\\overrightarrow{PQ}[\/latex] is parallel to [latex]\\textbf v[\/latex]. Thus, as we just discussed, there is a scalar, [latex]t[\/latex], such that [latex]\\overrightarrow{PQ}=t{\\bf{v}}[\/latex], which gives<\/p>\n<div id=\"fs-id1167793270756\" class=\"ui-has-child-title\" style=\"text-align: center;\" data-type=\"note\">[latex]\\begin{aligned}  \\overrightarrow{PQ}&=t{\\bf{v}} \\\\  \\langle x-x_0,y-y_0,z-z_0\\rangle&=t\\langle a,b,c\\rangle \\\\  \\langle x-x_0,y-y_0,z-z_0\\rangle&=\\langle ta,tb,tc\\rangle.  \\end{aligned}[\/latex]<\/div>\n<\/section>\n<div id=\"attachment_5157\" style=\"width: 443px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5157\" class=\"size-full wp-image-5157\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27211934\/2.63.jpg\" alt=\"This figure is the first octant of the 3-dimensional coordinate system. There is a line segment passing through two points. The points are labeled \u201cP = (x sub 0, y sub 0, z sub 0)\u201d and \u201cQ = (x, y, z).\u201d There is also a vector in standard position drawn. The vector is labeled \u201cv = &lt;a, b, c&gt;.\u201d\" width=\"433\" height=\"390\" \/><\/p>\n<p id=\"caption-attachment-5157\" class=\"wp-caption-text\">Figure 1. Vector [latex]\\textbf v[\/latex] is the direction vector for [latex]\\overrightarrow{PQ}[\/latex].<\/p>\n<\/div>\n<p id=\"fs-id1163723876839\">Using vector operations, we can rewrite\u00a0this equation\u00a0as<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\langle x-x_0,y-y_0,z-z_0\\rangle&=\\langle ta,tb,tc\\rangle \\\\  \\langle x,y,z\\rangle -\\langle x_0,y_0,z_0\\rangle&=t\\langle a,b,c\\rangle \\\\  \\langle x,y,z\\rangle&=\\langle x_0,y_0,z_0\\rangle+t\\langle a,b,c\\rangle  \\end{aligned}[\/latex].<\/p>\n<p>Setting [latex]{\\bf{r}}=\\langle x,y,z\\rangle[\/latex] and [latex]{\\bf{r}}_0=\\langle x_0,y_0,z_0\\rangle[\/latex], we now have the\u00a0<strong><span id=\"60ce88e7-9116-48d3-bb30-013c89cddc64_term88\" data-type=\"term\">vector equation of a line<\/span><\/strong>:<\/p>\n<div id=\"fs-id1163723862928\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\"><\/div>\n<p style=\"text-align: center;\">[latex]{\\bf{r}}={\\bf{r}}_0+t{\\bf{v}}[\/latex].<\/p>\n<p>Equating components,\u00a0the Vector Equation of a Line\u00a0shows that the following equations are simultaneously true: [latex]x-x_0=ta[\/latex], [latex]y-y_0=tb[\/latex], and [latex]z-z_0=tc[\/latex]. If we solve each of these equations for the component variables [latex]x[\/latex], [latex]y[\/latex], and [latex]z[\/latex], we get a set of equations in which each variable is defined in terms of the parameter [latex]t[\/latex] and that, together, describe the line. This set of three equations forms a set of\u00a0<strong><span id=\"60ce88e7-9116-48d3-bb30-013c89cddc64_term89\" data-type=\"term\">parametric equations of a line<\/span>:<\/strong><\/p>\n<div id=\"fs-id1163724080297\" data-type=\"equation\"><\/div>\n<p style=\"text-align: center;\">[latex]x=x_0+ta \\quad y=y_0+tb \\quad z=z_0+tc[\/latex].<\/p>\n<p id=\"fs-id1163723236850\">If we solve each of the equations for [latex]t[\/latex] assuming [latex]a[\/latex], [latex]b[\/latex], and [latex]c[\/latex] are nonzero, we get a different description of the same line:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{x-x_0}a=t \\quad \\frac{y-y_o}b=t \\quad \\frac{z-z_0}c=t[\/latex].<\/p>\n<p id=\"fs-id1163723755477\">Because each expression equals [latex]t[\/latex], they all have the same value. We can set them equal to each other to create\u00a0<strong><span id=\"60ce88e7-9116-48d3-bb30-013c89cddc64_term90\" data-type=\"term\">symmetric equations of a line<\/span>:<\/strong><\/p>\n<div id=\"fs-id1167793270756\" class=\"ui-has-child-title\" style=\"text-align: center;\" data-type=\"note\">[latex]\\frac{x-x_0}a=\\frac{y-y_o}b=\\frac{z-z_0}c[\/latex].<\/div>\n<p id=\"fs-id1163724050718\">We summarize the results in the following theorem.<\/p>\n<div data-type=\"example\">\n<div class=\"textbox shaded\">\n<header>\n<h3 class=\"os-title\" style=\"text-align: center;\"><span class=\"os-title-label\">THEOREM: parametric and symmetric equations of a line<\/span><\/h3>\n<hr \/>\n<\/header>\n<section>\n<div class=\"os-note-body\">\n<p id=\"fs-id1163723197352\">A line [latex]L[\/latex] parallel to vector [latex]{\\bf{v}}=\\langle a,b,c\\rangle[\/latex] and passing through point [latex]P(x_0, y_0, z_0)[\/latex] can be described by the following parametric equations:<\/p>\n<p style=\"text-align: center;\">[latex]x=x_0+ta, \\ y=y_0+tb, \\ z=z_0+tc[\/latex].<\/p>\n<p id=\"fs-id1163723635234\">If the constants [latex]a[\/latex], [latex]b[\/latex], and [latex]c[\/latex] are all nonzero, then [latex]L[\/latex] can be described by the symmetric equation of the line:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{x-x_0}a=\\frac{y-y_o}b=\\frac{z-z_0}c[\/latex].<\/p>\n<\/div>\n<\/section>\n<\/div>\n<p id=\"fs-id1163723757318\">The parametric equations of a line are not unique. Using a different parallel vector or a different point on the line leads to a different, equivalent representation. Each set of parametric equations leads to a related set of symmetric equations, so it follows that a symmetric equation of a line is not unique either.<\/p>\n<div id=\"fs-id1163723137804\" class=\"ui-has-child-title\" data-type=\"example\">\n<div class=\"textbox exercises\">\n<h3>Example: equations of a line in space<\/h3>\n<p>Find parametric and symmetric equations of the line passing through points [latex](1, 4, -2)[\/latex] and\u00a0[latex](-3, 5, 0)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q389475782\">Show Solution<\/span><\/p>\n<div id=\"q389475782\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1163723796963\">First, identify a vector parallel to the line:<\/p>\n<p style=\"text-align: center;\">[latex]{\\bf{v}}=\\langle-3-1,5-4,0-(-2)\\rangle=\\langle-4,1,2\\rangle[\/latex].<\/p>\n<p id=\"fs-id1163724059770\">Use either of the given points on the line to complete the parametric equations:<\/p>\n<p style=\"text-align: center;\">[latex]x=1-4t[\/latex],\u00a0[latex]y=4+t[\/latex], and\u00a0[latex]z=-2+2t[\/latex].<\/p>\n<p id=\"fs-id1163723281292\">Solve each equation for [latex]t[\/latex] to create the symmetric equation of the line:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{x-1}{-4}=y-4=\\frac{z+2}2[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q586348756\">Show Solution<\/span><\/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<\/div>\n<p id=\"fs-id1163723200547\">Sometimes we don\u2019t want the equation of a whole line, just a line segment. In this case, we limit the values of our parameter [latex]t[\/latex]. For example, let [latex]P(x_0, y_0, z_0)[\/latex] and [latex]Q(x_1, y_1, z_1)[\/latex] be points on a line, and let [latex]{\\bf{p}}=\\langle x_0,y_0,z_0\\rangle[\/latex] and [latex]{\\bf{q}}=\\langle x_1,y_1,z_1\\rangle[\/latex] be the associated position vectors. In addition, let [latex]{\\bf{r}}=\\langle x,y,z\\rangle[\/latex]. We want to find a vector equation for the line segment between [latex]P[\/latex] and [latex]Q[\/latex]. Using [latex]P[\/latex] as our known point on the line, and [latex]\\overrightarrow{PQ}=\\langle x_1-x_0,y_1-y_0,z_1-z_0\\rangle[\/latex] as the direction vector equation,\u00a0the Vector Equation of a Line\u00a0gives<\/p>\n<p style=\"text-align: center;\">[latex]{\\bf{r}}={\\bf{p}}+t \\left( \\overrightarrow{PQ}\\right)[\/latex].<\/p>\n<p id=\"fs-id1163723341582\">Using properties of vectors, then<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {\\bf{r}}&={\\bf{p}}+ t \\left(\\overrightarrow{PQ}\\right) \\\\  &=\\langle x_0,y_0,z_0\\rangle+t\\langle x_1-x_0,y_1-y_0,z_1-z_0\\rangle \\\\  &=\\langle x_0,y_0,z_0\\rangle+t(\\langle x_1,y_1,z_1\\rangle-\\langle x_0,y_0,z_0\\rangle) \\\\  &=\\langle x_0,y_0,z_0\\rangle+t\\langle x_1,y_1,z_1\\rangle-t\\langle x_0,y_0,z_0\\rangle \\\\  &=(1-t)\\langle x_0,y_0,z_0\\rangle+t\\langle x_1,y_1,z_1\\rangle \\\\  &=(1-t){\\bf{p}}+t{\\bf{q}}  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1163724050947\">Thus, the vector equation of the line passing through [latex]P[\/latex] and [latex]Q[\/latex] is<\/p>\n<p style=\"text-align: center;\">[latex]{\\bf{r}}=(1-t){\\bf{p}}+t{\\bf{q}}[\/latex].<\/p>\n<p id=\"fs-id1163723363447\">Remember that we didn\u2019t want the equation of the whole line, just the line segment between [latex]P[\/latex] and [latex]Q[\/latex]. Notice that when [latex]t=0[\/latex], we have [latex]{\\bf{r}}={\\bf{p}}[\/latex], and when [latex]t=1[\/latex], we have [latex]{\\bf{r}}={\\bf{q}}[\/latex]. Therefore, the vector equation of the line segment between [latex]P[\/latex] and [latex]Q[\/latex] is<\/p>\n<p style=\"text-align: center;\">[latex]{\\bf{r}}=(1-t){\\bf{p}}+t{\\bf{q}}[\/latex],\u00a0[latex]0\\leq t\\leq1[\/latex].<\/p>\n<p id=\"fs-id1163724067505\">Going back to\u00a0the Vector Equation of a Line, we can also find parametric equations for this line segment. We have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {\\bf{r}}&={\\bf{p}}+\\left(\\overrightarrow{PQ}\\right) \\\\  \\langle x,y,z\\rangle &=\\langle x_0,y_0,z_0\\rangle+t\\langle x_1-x_0,y_1-y_0,z_1-z_0\\rangle \\\\  &=\\langle x_0+t(x_1-x_0),y_0+t(y_1-y_0),z_0+t(z_1-z_0)\\rangle  \\end{aligned}[\/latex].<\/p>\n<p>Then, the parametric equations are<\/p>\n<p>[latex]x=x_0+t(x_1-x_0)[\/latex],\u00a0[latex]y=y_0+t(y_1-y_0)[\/latex],\u00a0[latex]z=z_0+t(z_1-z_0)[\/latex],\u00a0[latex]0\\leq t\\leq1[\/latex].<\/p>\n<\/div>\n<\/div>\n<div data-type=\"example\">\n<div class=\"textbox exercises\">\n<h3>Example: parametric equations of a line segment<\/h3>\n<p>Find parametric equations of the line segment between the points [latex]P(2, 1, 4)[\/latex] and\u00a0[latex]Q(3, -1 ,3)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q112936453\">Show Solution<\/span><\/p>\n<div id=\"q112936453\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1163724069731\">By\u00a0the Parametric Equations of a Line, we have<\/p>\n<p style=\"text-align: center;\">[latex]x=x_0+t(x_1-x_0)[\/latex],\u00a0[latex]y=y_0+t(y_1-y_0)[\/latex],\u00a0[latex]z=z_0+t(z_1-z_0)[\/latex],\u00a0[latex]0\\leq t\\leq1[\/latex].<\/p>\n<p id=\"fs-id1163724054258\">Working with each component separately, we get<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  x&=x_0+t(x_1-x_0) \\\\  &=2+t(3-2) \\\\  &=2+t,  \\end{aligned}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  y&=y_0+t(y_1-y_0) \\\\  &=1+t(-1-1) \\\\  &=1-2t,  \\end{aligned}[\/latex]<\/p>\n<p id=\"fs-id1163723280727\">and<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  z&=z_0+t(z_1-z_0) \\\\  &=4+t(3-4) \\\\  &=4-t.  \\end{aligned}[\/latex]<\/p>\n<p id=\"fs-id1163723200089\">Therefore, the parametric equations for the line segment are<\/p>\n<p style=\"text-align: center;\">[latex]x=2+t,y=1-2t,z=4-t, \\ 0\\leq t\\leq1[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q745628734\">Show Solution<\/span><\/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<\/div>\n<h2 data-type=\"title\">Distance between a Point and a Line<\/h2>\n<p id=\"fs-id1163724043792\">We already know how to calculate the distance between two points in space. We now expand this definition to describe the distance between a point and a line in space. Several real-world contexts exist when it is important to be able to calculate these distances. When building a home, for example, builders must consider \u201csetback\u201d requirements, when structures or fixtures have to be a certain distance from the property line. Air travel offers another example. Airlines are concerned about the distances between populated areas and proposed flight paths.<\/p>\n<p id=\"fs-id1163723236930\">Let [latex]L[\/latex] be a line in the plane and let [latex]M[\/latex] be any point not on the line. Then, we define distance [latex]d[\/latex] from [latex]M[\/latex] to [latex]L[\/latex] as the length of line segment [latex]\\overline{MP}[\/latex], where [latex]P[\/latex] is a point on [latex]L[\/latex] such that [latex]\\overline{MP}[\/latex] is perpendicular to [latex]L[\/latex] (Figure 2).<\/p>\n<\/div>\n<div data-type=\"example\">\n<div id=\"attachment_5160\" style=\"width: 392px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5160\" class=\"size-full wp-image-5160\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27212154\/2.64.jpg\" alt=\"This figure has two line segments. The first line is labeled \u201cL\u201d and has point P on the segment. The second line segment is drawn from point P to point M and is perpendicular to line L. The second line segment is labeled \u201cd.\u201d\" width=\"382\" height=\"210\" \/><\/p>\n<p id=\"caption-attachment-5160\" class=\"wp-caption-text\">Figure 2. The distance from point [latex]M[\/latex] to line [latex]L[\/latex] is the length of [latex]\\overline{MP}[\/latex].<\/p>\n<\/div>\n<\/div>\n<div data-type=\"example\"><\/div>\n<div data-type=\"example\">When we\u2019re looking for the distance between a line and a point in space,\u00a0Figure 2\u00a0still applies. We still define the distance as the length of the perpendicular line segment connecting the point to the line. In space, however, there is no clear way to know which point on the line creates such a perpendicular line segment, so we select an arbitrary point on the line and use properties of vectors to calculate the distance. Therefore, let [latex]P[\/latex] be an arbitrary point on line [latex]L[\/latex] and let [latex]\\textbf v[\/latex] be a direction vector for [latex]L[\/latex] (Figure 3).<\/div>\n<div data-type=\"example\"><\/div>\n<div data-type=\"example\">\n<div id=\"attachment_5162\" style=\"width: 448px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5162\" class=\"size-full wp-image-5162\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27212515\/2.65.jpg\" alt=\"This figure has a line segment labeled \u201cL.\u201d On the line segment L there is point P. There is a vector drawn from point P to another point M. Also, from M there is a line segment drawn to line L. This segment is perpendicular to line L. There is also a vector labeled \u201cv\u201d on line segment L. A parallelogram has been formed with vector v, line segment P M, and two other segments back to line L.\" width=\"438\" height=\"234\" \/><\/p>\n<p id=\"caption-attachment-5162\" class=\"wp-caption-text\">Figure 3. Vectors [latex]\\overrightarrow{PM}[\/latex] and [latex]\\textbf v[\/latex] form two sides of a parallelogram with base [latex]||{\\bf{v}}||[\/latex] and height [latex]d[\/latex], which is the distance between a line and a point in space.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"example\"><\/div>\n<div data-type=\"example\">\n<p id=\"fs-id1163724075400\">By\u00a0Theorem: Area of a Parallelogram, vectors [latex]\\overrightarrow{PM}[\/latex] and [latex]\\textbf v[\/latex] form two sides of a parallelogram with area [latex]||\\overrightarrow{PM}\\times{\\bf{v}}||[\/latex]. Using a formula from geometry, the area of this parallelogram can also be calculated as the product of its base and height:<\/p>\n<p style=\"text-align: center;\">[latex]||\\overrightarrow{PM}\\times{\\bf{v}}||=||{\\bf{v}}||d[\/latex].<\/p>\n<p id=\"fs-id1163724051243\">We can use this formula to find a general formula for the distance between a line in space and any point not on the line.<\/p>\n<\/div>\n<div data-type=\"example\">\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 line<\/span><\/h3>\n<hr \/>\n<\/header>\n<section>\n<div class=\"os-note-body\">\n<p id=\"fs-id1163723987376\">Let [latex]L[\/latex] be a line in space passing through point [latex]P[\/latex] with direction vector [latex]\\textbf v[\/latex]. If [latex]M[\/latex] is any point not on [latex]L[\/latex], then the distance from [latex]M[\/latex] to [latex]L[\/latex] is<\/p>\n<p style=\"text-align: center;\">[latex]d=\\frac{||\\overrightarrow{PM}\\times{\\bf{v}}||}{||{\\bf{v}}||}[\/latex].<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: calculating the distance from a point to a line<\/h3>\n<p>Find the distance between t point [latex]M=(1, 1, 3)[\/latex] and line [latex]\\frac{x-3}4=\\frac{y+1}2=z-3[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q483752783\">Show Solution<\/span><\/p>\n<div id=\"q483752783\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1163723304185\">From the symmetric equations of the line, we know that vector [latex]{\\bf{v}}\\langle4,2,1\\rangle[\/latex] is a direction vector for the line. Setting the symmetric equations of the line equal to zero, we see that point [latex]P(3, -1 , 3)[\/latex] lies on the line. Then,<\/p>\n<p style=\"text-align: center;\">[latex]\\overrightarrow{PM}=\\langle1-3,1-(-1),3-3\\rangle=\\langle-2,2,0\\rangle[\/latex].<\/p>\n<p id=\"fs-id1163723203966\">To calculate the distance, we need to find [latex]\\overrightarrow{PM}\\times{\\bf{v}}[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\overrightarrow{PM}\\times{\\bf{v}}&=\\begin{vmatrix}{\\bf{i}}&{\\bf{j}}&{\\bf{k}} \\\\-2&2&0 \\\\ 4&2&1\\end{vmatrix} \\\\  &=(2-1){\\bf{i}}-(-2-0){\\bf{j}}+(-4-8){\\bf{k}} \\\\  &=2{\\bf{i}}+2{\\bf{j}}-12{\\bf{k}}.  \\end{aligned}[\/latex]<\/p>\n<p id=\"fs-id1163724039857\">Therefore, the distance between the point and the line is (Figure 4)<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  d&=\\frac{||\\overrightarrow{PM}\\times{\\bf{v}}||}{||{\\bf{v}}||} \\\\  &=\\frac{\\sqrt{2^2+2^2+12^2}}{\\sqrt{4^2+2^2+1^2}} \\\\  &=\\frac{s\\sqrt{38}}{\\sqrt{21}}.  \\end{aligned}[\/latex]<\/p>\n<div id=\"attachment_5164\" style=\"width: 367px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5164\" class=\"size-full wp-image-5164\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27212704\/2.66.jpg\" alt=\"This figure is the first octant of the 3-dimensional coordinate system. There is a 3-dimensional box drawn in the octant. There is a point labeled at (1, 1, 3). There is a line segment labeled \u201cL\u201d inside of the box. Also, there is a perpendicular line segment from the point to line L.\" width=\"357\" height=\"396\" \/><\/p>\n<p id=\"caption-attachment-5164\" class=\"wp-caption-text\">Figure 4. Point [latex](1, 1, 3)[\/latex] is approximately [latex]2.7[\/latex] units from the line with symmetric equations [latex]\\frac{x-3}4=\\frac{y+1}2=z-3[\/latex].<\/p>\n<\/div>\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](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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q367450281\">Show Solution<\/span><\/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<\/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=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\"><\/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<h2 data-type=\"title\">Relationships between Lines<\/h2>\n<p id=\"fs-id1163724035950\">Given two lines in the two-dimensional plane, the lines are equal, they are parallel but not equal, or they intersect in a single point. In three dimensions, a fourth case is possible. If two lines in space are not parallel, but do not intersect, then the lines are said to be\u00a0<strong><span id=\"60ce88e7-9116-48d3-bb30-013c89cddc64_term91\" data-type=\"term\">skew lines<\/span><\/strong>\u00a0(Figure 5).<\/p>\n<div id=\"attachment_5166\" style=\"width: 428px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5166\" class=\"size-full wp-image-5166\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27212930\/2.67.jpg\" alt=\"This figure has two line segments. They are 3-dimensional, are not parallel, and do not intersect. The directions are different and one is above the other.\" width=\"418\" height=\"117\" \/><\/p>\n<p id=\"caption-attachment-5166\" class=\"wp-caption-text\">Figure 5. In three dimensions, it is possible that two lines do not cross, even when they have different directions.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"example\">To classify lines as parallel but not equal, equal, intersecting, or skew, we need to know two things: whether the direction vectors are parallel and whether the lines share a point (Figure 6).<\/div>\n<div data-type=\"example\"><\/div>\n<div data-type=\"example\">\n<div id=\"attachment_5167\" style=\"width: 620px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5167\" class=\"size-full wp-image-5167\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27213057\/2.68.jpg\" alt=\"This figure is a table with two rows and two columns. Above the columns is the question \u201cLines share a common point?\u201d The first column is labeled \u201cyes,\u201d and the second column is labeled \u201cno.\u201d To the left of the rows is the question \u201cDirection vectors are parallel?\u201d The first row is labeled \u201cyes,\u201d and the second row is labeled \u201cno.\u201d The entries of the first row are \u201cequal\u201d and \u201cparallel but not equal.\u201d The entries in the second row are \u201cintersecting\u201d and \u201cskew.\u201d\" width=\"610\" height=\"334\" \/><\/p>\n<p id=\"caption-attachment-5167\" class=\"wp-caption-text\">Figure 6. Determine the relationship between two lines based on whether their direction vectors are parallel and whether they share a point.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"example\"><\/div>\n<div data-type=\"example\">\n<div class=\"textbox exercises\">\n<h3>Example: classifying lines in space<\/h3>\n<p id=\"fs-id1163723283999\">For each pair of lines, determine whether the lines are equal, parallel but not equal, skew, or intersecting.<\/p>\n<ol id=\"fs-id1163723284004\" type=\"a\">\n<li><span data-type=\"newline\">[latex]L_1[\/latex]:\u00a0[latex]x=2s-1[\/latex],\u00a0[latex]y=s-1[\/latex],\u00a0[latex]z=s-4[\/latex]<br \/>\n<\/span>[latex]L_2[\/latex]:\u00a0[latex]x=t-3[\/latex],\u00a0[latex]y=3t+8[\/latex],\u00a0[latex]z=5-2t[\/latex]<\/li>\n<li><span data-type=\"newline\">[latex]L_1[\/latex]:\u00a0[latex]x=-y=z[\/latex]<br \/>\n<\/span>[latex]L_2[\/latex]:\u00a0[latex]\\frac{x-3}2=y=z-2[\/latex]<\/li>\n<li><span data-type=\"newline\">[latex]L_1[\/latex]:\u00a0[latex]x=6s-1[\/latex],\u00a0[latex]y=-2s[\/latex], [latex]z=3s+1[\/latex]<br \/>\n<\/span>[latex]L_2[\/latex]:\u00a0[latex]\\frac{x-4}6=\\frac{y+3}{-2}=\\frac{z-1}3[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q849520983\">Show Solution<\/span><\/p>\n<div id=\"q849520983\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1163724052863\" type=\"a\">\n<li>Line [latex]L_1[\/latex] has direction vector [latex]{\\bf{v_1}}=\\langle2,1,1\\rangle[\/latex]; line [latex]L_2[\/latex] has direction vector [latex]{\\bf{v_2}}=\\langle1,3,-2\\rangle[\/latex]. Because the direction vectors are not parallel vectors, the lines are either intersecting or skew. To determine whether the lines intersect, we see if there is a point, [latex](x, y, z)[\/latex], that lies on both lines. To find this point, we use the parametric equations to create a system of equalities:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]2s-1=t-3; \\quad s-1=3t+8; \\quad s-4=5-2t[\/latex].<\/p>\n<p><span data-type=\"newline\"><br \/>\n<\/span>By the first equation, [latex]t=2s+2[\/latex]. Substituting into the second equation yields<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  s-1&=3(2s+2)+8 \\\\  s-1&=6s+6+8 \\\\  5s&=-15 \\\\  s&=-3.  \\end{aligned}[\/latex]<\/p>\n<p><span data-type=\"newline\"><br \/>\n<\/span>Substitution into the third equation, however, yields a contradiction:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  s-4&=5-2(2s+2) \\\\  s-4&=5-4s-4 \\\\  5s&=5 \\\\  s&=1.  \\end{aligned}[\/latex]<\/p>\n<p><span data-type=\"newline\"><br \/>\n<\/span>There is no single point that satisfies the parametric equations for [latex]L_1[\/latex] and [latex]L_2[\/latex] simultaneously. These lines do not intersect, so they are skew (see the following figure).<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"attachment_5168\" style=\"width: 386px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5168\" class=\"size-full wp-image-5168\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27213249\/Example-2.48.jpg\" alt=\"This figure is the 3-dimensional coordinate system. There are two skew lines drawn. They do not intersect and are not parallel.\" width=\"376\" height=\"380\" \/><\/p>\n<p id=\"caption-attachment-5168\" class=\"wp-caption-text\">Figure 7. Lines [latex]L_1[\/latex] and [latex]L_2[\/latex] are skew.<\/p>\n<\/div>\n<\/li>\n<li>Line [latex]L_1[\/latex] has direction vector [latex]{\\bf{v_1}}=\\langle1,-1,1\\rangle[\/latex] and passes through the origin, [latex](0, 0, 0)[\/latex]. Line [latex]L_2[\/latex] has a different direction vector, [latex]{\\bf{v_2}}=\\langle2,1,1\\rangle[\/latex], so these lines are not parallel or equal. Let [latex]r[\/latex] represent the parameter for line [latex]L_1[\/latex] and let [latex]s[\/latex] represent the parameter for [latex]L_2[\/latex]:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  x&=r &\\quad x&=2s+3 \\\\  y&=-r &\\quad y&=s \\\\  z&=r &\\quad z&=s+2  \\end{aligned}[\/latex].<\/p>\n<p><span data-type=\"newline\"><br \/>\n<\/span>Solve the system of equations to find [latex]r=1[\/latex] and [latex]s=-1[\/latex]. If we need to find the point of intersection, we can substitute these parameters into the original equations to get [latex](1, -1, 1)[\/latex] (see the following figure).<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"attachment_5169\" style=\"width: 524px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5169\" class=\"size-full wp-image-5169\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27213452\/Example-2.48b.jpg\" alt=\"This figure is the 3-dimensional coordinate system. There are two lines drawn. They intersect at the point (1,-1,1).\" width=\"514\" height=\"501\" \/><\/p>\n<p id=\"caption-attachment-5169\" class=\"wp-caption-text\">Figure 8. Lines [latex]L_1[\/latex] and [latex]L_2[\/latex] intersect at [latex](1, -1, 1)[\/latex].<\/p>\n<\/div>\n<\/li>\n<li>Lines [latex]L_1[\/latex] and [latex]L_2[\/latex] have equivalent direction vectors: [latex]{\\bf{v}}=\\langle6,-2,3\\rangle[\/latex]. These two lines are parallel (see the following figure).<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"attachment_5170\" style=\"width: 407px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5170\" class=\"size-full wp-image-5170\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27213531\/Example-2.48c.jpg\" alt=\"This figure is the 3-dimensional coordinate system. There are two lines drawn. They do not intersect and are parallel.\" width=\"397\" height=\"378\" \/><\/p>\n<p id=\"caption-attachment-5170\" class=\"wp-caption-text\">Figure 9. Lines [latex]L_1[\/latex] and [latex]L_2[\/latex] are parallel.<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p id=\"fs-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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q389456729\">Show Solution<\/span><\/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<\/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-5457\">\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 2.45. <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":21,"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 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