{"id":915,"date":"2025-10-14T15:55:22","date_gmt":"2025-10-14T15:55:22","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=915"},"modified":"2025-10-22T14:29:29","modified_gmt":"2025-10-22T14:29:29","slug":"lines-and-planes-in-space-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/lines-and-planes-in-space-learn-it-2\/","title":{"raw":"Lines and Planes in Space: Learn It 2","rendered":"Lines and Planes in Space: Learn It 2"},"content":{"raw":"

Distance between a Point and a Line<\/h2>\r\n

You already know how to calculate the distance between two points in space. Now we expand this to find the distance between a point and an entire line. This calculation has many practical applications. Builders must consider \"setback\" requirements to ensure structures maintain proper distance from property lines. Airlines need to calculate distances between populated areas and proposed flight paths.<\/p>\r\n\r\n

\r\n

The distance from a point to a line<\/strong> is defined as the length of the shortest possible segment connecting the point to the line. This shortest segment is always perpendicular to the line.<\/p>\r\n

Let [latex]L[\/latex] be a line and [latex]M[\/latex] be any point not on the line. The distance [latex]d[\/latex] from [latex]M[\/latex] to [latex]L[\/latex] is the length of line segment [latex]\\overline{MP}[\/latex], where [latex]P[\/latex] is the point on [latex]L[\/latex] such that [latex]\\overline{MP}[\/latex] is perpendicular to [latex]L[\/latex].<\/p>\r\n\r\n<\/div>\r\n

[caption id=\"attachment_5160\" align=\"aligncenter\" width=\"382\"]\"This 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>\r\n
\r\n

In three dimensions, there's no obvious way to identify which point on the line creates the perpendicular segment. We solve this by selecting an arbitrary point on the line and using vector properties.<\/p>\r\n

Let [latex]P[\/latex] be any point on line [latex]L[\/latex] and let [latex]\\textbf{v}[\/latex] be a direction vector for [latex]L[\/latex]. For any point [latex]M[\/latex] not on the line, consider the vectors [latex]\\overrightarrow{PM}[\/latex] and [latex]\\textbf{v}[\/latex].<\/p>\r\n\r\n<\/div>\r\n

<\/div>\r\n
[caption id=\"attachment_5162\" align=\"aligncenter\" width=\"438\"]\"This 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>\r\n
\r\n

These vectors form two sides of a parallelogram. From the properties of cross products and geometry:<\/p>\r\n\r\n

    \r\n \t
  • The area of this parallelogram equals [latex]||\\overrightarrow{PM} \\times \\textbf{v}||[\/latex]<\/li>\r\n \t
  • The area also equals base \u00d7 height = [latex]||\\textbf{v}|| \\cdot d[\/latex]<\/li>\r\n<\/ul>\r\n

    Setting these equal gives us:<\/p>\r\n

    [latex]||\\overrightarrow{PM} \\times \\textbf{v}|| = ||\\textbf{v}|| \\cdot d[\/latex]<\/p>\r\n

    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

    \r\n

    distance from a point to a line<\/h3>\r\n

    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

    [latex]d=\\frac{||\\overrightarrow{PM}\\times{\\bf{v}}||}{||{\\bf{v}}||}[\/latex].<\/p>\r\n\r\n<\/section>

    Why This Works<\/strong>: The cross product [latex]\\overrightarrow{PM} \\times \\textbf{v}[\/latex] gives us the area of a parallelogram. Dividing by the base [latex]||\\textbf{v}||[\/latex] isolates the height, which is exactly the perpendicular distance we want.\r\n\r\n<\/section>
    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].[reveal-answer q=\"483752783\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"483752783\"]\r\n

    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

    [latex]\\overrightarrow{PM}=\\langle1-3,1-(-1),3-3\\rangle=\\langle-2,2,0\\rangle[\/latex].<\/p>\r\n

    To calculate the distance, we need to find [latex]\\overrightarrow{PM}\\times{\\bf{v}}[\/latex]:<\/p>\r\n

    [latex]\\begin{aligned}\r\n\\overrightarrow{PM}\\times{\\bf{v}}&=\\begin{vmatrix}{\\bf{i}}&{\\bf{j}}&{\\bf{k}} \\\\-2&2&0 \\\\ 4&2&1\\end{vmatrix} \\\\\r\n&=(2-1){\\bf{i}}-(-2-0){\\bf{j}}+(-4-8){\\bf{k}} \\\\\r\n&=2{\\bf{i}}+2{\\bf{j}}-12{\\bf{k}}.\r\n\\end{aligned}[\/latex]<\/p>\r\n

    Therefore, the distance between the point and the line is (Figure 4)<\/p>\r\n

    [latex]\\begin{aligned}\r\nd&=\\frac{||\\overrightarrow{PM}\\times{\\bf{v}}||}{||{\\bf{v}}||} \\\\\r\n&=\\frac{\\sqrt{2^2+2^2+12^2}}{\\sqrt{4^2+2^2+1^2}} \\\\\r\n&=\\frac{s\\sqrt{38}}{\\sqrt{21}}.\r\n\\end{aligned}[\/latex]<\/p>\r\n\r\n[caption id=\"attachment_5164\" align=\"aligncenter\" width=\"357\"]\"This 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]<\/section><\/div>","rendered":"

    Distance between a Point and a Line<\/h2>\n

    You already know how to calculate the distance between two points in space. Now we expand this to find the distance between a point and an entire line. This calculation has many practical applications. Builders must consider “setback” requirements to ensure structures maintain proper distance from property lines. Airlines need to calculate distances between populated areas and proposed flight paths.<\/p>\n

    \n

    The distance from a point to a line<\/strong> is defined as the length of the shortest possible segment connecting the point to the line. This shortest segment is always perpendicular to the line.<\/p>\n

    Let [latex]L[\/latex] be a line and [latex]M[\/latex] be any point not on the line. The distance [latex]d[\/latex] from [latex]M[\/latex] to [latex]L[\/latex] is the length of line segment [latex]\\overline{MP}[\/latex], where [latex]P[\/latex] is the point on [latex]L[\/latex] such that [latex]\\overline{MP}[\/latex] is perpendicular to [latex]L[\/latex].<\/p>\n<\/div>\n

    \n
    \"This
    Figure 2. The distance from point [latex]M[\/latex] to line [latex]L[\/latex] is the length of [latex]\\overline{MP}[\/latex].<\/figcaption><\/figure>\n<\/div>\n
    <\/div>\n
    \n

    In three dimensions, there’s no obvious way to identify which point on the line creates the perpendicular segment. We solve this by selecting an arbitrary point on the line and using vector properties.<\/p>\n

    Let [latex]P[\/latex] be any point on line [latex]L[\/latex] and let [latex]\\textbf{v}[\/latex] be a direction vector for [latex]L[\/latex]. For any point [latex]M[\/latex] not on the line, consider the vectors [latex]\\overrightarrow{PM}[\/latex] and [latex]\\textbf{v}[\/latex].<\/p>\n<\/div>\n

    <\/div>\n
    \n
    \"This
    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.<\/figcaption><\/figure>\n<\/div>\n
    <\/div>\n
    \n

    These vectors form two sides of a parallelogram. From the properties of cross products and geometry:<\/p>\n

      \n
    • The area of this parallelogram equals [latex]||\\overrightarrow{PM} \\times \\textbf{v}||[\/latex]<\/li>\n
    • The area also equals base \u00d7 height = [latex]||\\textbf{v}|| \\cdot d[\/latex]<\/li>\n<\/ul>\n

      Setting these equal gives us:<\/p>\n

      [latex]||\\overrightarrow{PM} \\times \\textbf{v}|| = ||\\textbf{v}|| \\cdot d[\/latex]<\/p>\n

      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

      \n

      distance from a point to a line<\/h3>\n

      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

      [latex]d=\\frac{||\\overrightarrow{PM}\\times{\\bf{v}}||}{||{\\bf{v}}||}[\/latex].<\/p>\n<\/section>\n

      Why This Works<\/strong>: The cross product [latex]\\overrightarrow{PM} \\times \\textbf{v}[\/latex] gives us the area of a parallelogram. Dividing by the base [latex]||\\textbf{v}||[\/latex] isolates the height, which is exactly the perpendicular distance we want.<\/p>\n<\/section>\n
      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