{"id":918,"date":"2025-10-14T16:05:49","date_gmt":"2025-10-14T16:05:49","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=918"},"modified":"2025-10-22T14:56:21","modified_gmt":"2025-10-22T14:56:21","slug":"lines-and-planes-in-space-learn-it-5","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/lines-and-planes-in-space-learn-it-5\/","title":{"raw":"Lines and Planes in Space: Learn It 5","rendered":"Lines and Planes in Space: Learn It 5"},"content":{"raw":"

Parallel and Intersecting Planes<\/h2>\r\n

Unlike lines, which can be parallel, intersecting, or skew in 3D space, planes have only two possible relationships: they are either parallel<\/strong> or they intersect<\/strong>. There's no such thing as skew planes.<\/p>\r\n

Parallel planes<\/strong> have parallel normal vectors. If two planes have normal vectors [latex]\\mathbf{n}_1[\/latex] and [latex]\\mathbf{n}_2[\/latex], the planes are parallel when [latex]\\mathbf{n}_1 = k\\mathbf{n}_2[\/latex] for some scalar [latex]k[\/latex].<\/p>\r\n

Intersecting planes<\/strong> meet along a line. When two planes aren't parallel, they must intersect, and their intersection is always a straight line.<\/p>\r\n\r\n\r\n[caption id=\"attachment_5191\" align=\"aligncenter\" width=\"286\"]\"This Figure 5. The intersection of two nonparallel planes is always a line.[\/caption]\r\n\r\n

Why Always a Line?<\/strong>: Think about it geometrically. Two distinct planes can't intersect at just one point (that would require them to be identical), and they can't intersect at scattered points. The intersection must be a continuous set of points that forms a line.<\/section>\r\n

When two planes intersect, you can find parametric equations for their line of intersection by using both plane equations simultaneously.<\/p>\r\n\r\n

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).[caption id=\"attachment_5192\" align=\"aligncenter\" width=\"412\"]\"This 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

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

[latex]\\begin{align*}\r\nx&+y+z=0 \\\\\r\n2x&-y+z=0 \\\\\r\n\\hline \\\\\r\n3x&\\quad\\,\\,+2z=0.\r\n\\end{align*}[\/latex]<\/p>\r\n

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

[latex]\\begin{aligned}\r\nx+y+z&=0 \\\\\r\n-\\frac23z+y+z&=0 \\\\\r\ny+\\frac13z&=0 \\\\\r\ny&=-\\frac13z\r\n\\end{aligned}[\/latex].<\/p>\r\n

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<\/section>\r\n

Finding the Angle Between Planes<\/h3>\r\n

When planes intersect, you might need to find the angle between them. We can use normal vectors to calculate the angle between the two planes.<\/p>\r\n

The key insight: the angle between two planes equals the angle between their normal vectors.<\/p>\r\n\r\n\r\n[caption id=\"attachment_5195\" align=\"aligncenter\" width=\"412\"]\"This Figure 7. The angle between two planes has the same measure as the angle between the normal vectors for the planes.[\/caption]\r\n

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

[latex]\\large{\\cos\\theta=\\frac{|{\\bf{n}}_1\\cdot{\\bf{n}}_2|}{||{\\bf{n}}_1|| \\ ||{\\bf{n}}_2||}}[\/latex].<\/p>\r\n

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

\r\n

angle between planes<\/h3>\r\n

For planes with normal vectors [latex]\\mathbf{n}_1[\/latex] and [latex]\\mathbf{n}_2[\/latex], the angle [latex]\\theta[\/latex] between the planes is:<\/p>\r\n

[latex]\\cos \\theta = \\frac{|\\mathbf{n}_1 \\cdot \\mathbf{n}_2|}{|\\mathbf{n}_1| , |\\mathbf{n}_2|}[\/latex]<\/p>\r\n

We use absolute value to ensure [latex]0\u00b0 \\leq \\theta \\leq 90\u00b0[\/latex].<\/em><\/p>\r\n[latex]\\\\[\/latex]\r\n

Special Cases:<\/p>\r\n\r\n

    \r\n \t
  • Parallel planes<\/strong>: [latex]\\mathbf{n}_1 \\times \\mathbf{n}_2 = \\mathbf{0}[\/latex] (cross product is zero)<\/li>\r\n \t
  • Perpendicular planes<\/strong>: [latex]\\mathbf{n}_1 \\cdot \\mathbf{n}_2 = 0[\/latex] (dot product is zero)<\/li>\r\n<\/ul>\r\n<\/section>
    \r\n

    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

      \r\n \t
    1. [latex]x+2y-z=8[\/latex] and\u00a0[latex]2x+4y-2z=10[\/latex]<\/li>\r\n \t
    2. [latex]2x-3y+2z=3[\/latex] and\u00a0[latex]6x+2y-3z=1[\/latex]<\/li>\r\n \t
    3. [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
        \r\n \t
      1. 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
      2. 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\r\n<\/span>\r\n

        [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

      3. 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]:\r\n<\/span>\r\n

        [latex]\\begin{aligned}\r\n\\cos\\theta&=\\frac{|{\\bf{n}}_1\\cdot{\\bf{n}}_2|}{||{\\bf{n}}_1|| \\ ||{\\bf{n}}_2||} \\\\\r\n&=\\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&=\\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<\/section>\r\n

        Distance Between Parallel Planes<\/h3>\r\n

        When two planes are parallel, you can find the distance between them by selecting any point on one plane and calculating its distance to the other plane.<\/p>\r\n\r\n

        \r\n

        distance from point to plane<\/h3>\r\n

        The distance from point [latex]P(x_0, y_0, z_0)[\/latex] to plane [latex]ax + by + cz + k = 0[\/latex] is:<\/p>\r\n

        [latex]d = \\frac{|ax_0 + by_0 + cz_0 + k|}{\\sqrt{a^2 + b^2 + c^2}}[\/latex]<\/p>\r\n\r\n<\/section>

        \r\n

        How This Works<\/strong><\/p>\r\n

        If [latex](x_1, y_1, z_1)[\/latex] is any point on the plane [latex]ax + by + cz + k = 0[\/latex], then [latex]ax_1 + by_1 + cz_1 + k = 0[\/latex], so [latex]k = -ax_1 - by_1 - cz_1[\/latex].<\/p>\r\n

        Using our general distance formula:<\/p>\r\n

        [latex]d = \\frac{|a(x_0-x_1) + b(y_0-y_1) + c(z_0-z_1)|}{\\sqrt{a^2+b^2+c^2}}[\/latex]<\/p>\r\n

        Expanding the numerator:<\/p>\r\n

        [latex]d = \\frac{|ax_0 + by_0 + cz_0 - ax_1 - by_1 - cz_1|}{\\sqrt{a^2+b^2+c^2}} = \\frac{|ax_0 + by_0 + cz_0 + k|}{\\sqrt{a^2+b^2+c^2}}[\/latex]<\/p>\r\n\r\n<\/section>

        For parallel planes [latex]ax + by + cz + k_1 = 0[\/latex] and [latex]ax + by + cz + k_2 = 0[\/latex], the distance between them is [latex]\\frac{|k_1 - k_2|}{\\sqrt{a^2 + b^2 + c^2}}[\/latex]. Just pick the origin as your test point!<\/section>
        Find the distance between the two parallel planes given by [latex]2x+y-z=2[\/latex] and\u00a0[latex]2x+y-z=8[\/latex].[reveal-answer q=\"943725601\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"943725601\"]\r\n

        Point [latex](1, 0, 0)[\/latex] lies in the first plane. The desired distance, then, is<\/p>\r\n

        [latex]\\begin{aligned}\r\nd&=\\frac{|ax_0+by_0+cz_0+k|}{\\sqrt{a^2+b^2+c^2}} \\\\\r\n&=\\frac{|2(1)+1(0)+(-1)(0)+(-8)|}{\\sqrt{2^2+1^2+(-1)^2}} \\\\\r\n&=\\frac{6}{\\sqrt6}=\\sqrt6\r\n\\end{aligned}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"

        Parallel and Intersecting Planes<\/h2>\n

        Unlike lines, which can be parallel, intersecting, or skew in 3D space, planes have only two possible relationships: they are either parallel<\/strong> or they intersect<\/strong>. There’s no such thing as skew planes.<\/p>\n

        Parallel planes<\/strong> have parallel normal vectors. If two planes have normal vectors [latex]\\mathbf{n}_1[\/latex] and [latex]\\mathbf{n}_2[\/latex], the planes are parallel when [latex]\\mathbf{n}_1 = k\\mathbf{n}_2[\/latex] for some scalar [latex]k[\/latex].<\/p>\n

        Intersecting planes<\/strong> meet along a line. When two planes aren’t parallel, they must intersect, and their intersection is always a straight line.<\/p>\n

        \"This
        Figure 5. The intersection of two nonparallel planes is always a line.<\/figcaption><\/figure>\n
        Why Always a Line?<\/strong>: Think about it geometrically. Two distinct planes can’t intersect at just one point (that would require them to be identical), and they can’t intersect at scattered points. The intersection must be a continuous set of points that forms a line.<\/section>\n

        When two planes intersect, you can find parametric equations for their line of intersection by using both plane equations simultaneously.<\/p>\n

        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
        \"This
        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]<\/figcaption><\/figure>\n