A plane can be determined by:<\/p>\r\n\r\n
- \r\n \t
- Three non-collinear points<\/strong> (the most basic definition)<\/li>\r\n \t
- Two distinct, intersecting lines<\/strong> (exactly one plane contains both)<\/li>\r\n \t
- One line and a point not on the line<\/strong> (the line and point determine the plane)<\/li>\r\n \t
- A point and a normal vector<\/strong><\/li>\r\n<\/ul>\r\n
The most powerful way to describe a plane uses a normal vector<\/strong>\u2014a vector perpendicular to the plane. Here's the key insight:<\/p>\r\n
Imagine a vector [latex]\\mathbf{n}[\/latex] pointing perpendicular to a plane, and a fixed point [latex]P[\/latex] on the plane. Any other point [latex]Q[\/latex] lies on the same plane if and only if the vector [latex]\\overrightarrow{PQ}[\/latex] is perpendicular to [latex]\\mathbf{n}[\/latex].<\/p>\r\n\r\n[caption id=\"attachment_5173\" align=\"aligncenter\" width=\"441\"]
![\"This](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27214230\/2.69.jpg\")
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]Since orthogonal vectors have dot product zero, this gives us our plane equation.\r\n\r\n\r\n plane equations<\/h3>\r\n
Given point [latex]P = (x_0, y_0, z_0)[\/latex] on the plane and normal vector [latex]\\mathbf{n} = \\langle a, b, c \\rangle[\/latex]:<\/p>\r\n\r\n
- \r\n \t
- Vector equation<\/strong>: [latex]\\mathbf{n} \\cdot \\overrightarrow{PQ} = 0[\/latex]<\/li>\r\n \t
- Scalar equation<\/strong>: [latex]a(x - x_0) + b(y - y_0) + c(z - z_0) = 0[\/latex]<\/li>\r\n \t
- General form<\/strong>: [latex]ax + by + cz + d = 0[\/latex] where [latex]d = -ax_0 - by_0 - cz_0[\/latex]<\/li>\r\n<\/ul>\r\n<\/section>\r\n
To see how these forms connect, let's expand the vector equation. If [latex]Q = (x,y,z)[\/latex] is any point on the plane, then [latex]\\overrightarrow{PQ} = \\langle x-x_0, y-y_0, z-z_0 \\rangle[\/latex]. The condition that this vector is orthogonal to [latex]\\mathbf{n}[\/latex] gives us:<\/p>\r\n
[latex] \\begin{aligned} \\mathbf{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>\r\n
Expanding this gives us [latex]ax + by + cz - ax_0 - by_0 - cz_0 = 0[\/latex], which we can write as [latex]ax + by + cz + d = 0[\/latex] where [latex]d = -ax_0 - by_0 - cz_0[\/latex].<\/p>\r\n\r\n
\r\n\r\nWhen given three points, you need to find a normal vector to write the plane equation. The cross product provides the solution.\r\n\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.[reveal-answer q=\"374652897\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"374652897\"]\r\n 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
[latex]\\begin{aligned}\r\n\\overrightarrow{PQ}&=\\langle0-1,2-1,1-(-2)\\rangle&=\\langle-1,1,3\\rangle \\\\\r\n\\overrightarrow{QR}&=\\langle-1-0,-1-2,0-1\\rangle&=\\langle-1,-3,-1\\rangle\r\n\\end{aligned}.[\/latex]<\/p>\r\n
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
[latex]\\begin{aligned}\r\n{\\bf{n}}&=\\overrightarrow{PQ}\\times\\overrightarrow{QR} \\\\\r\n&=\\begin{vmatrix}{\\bf{i}}&{\\bf{j}}&{\\bf{k}} \\\\ -1&1&3\\\\ -1&-3&-1\\end{vmatrix} \\\\\r\n&=(-1+9){\\bf{i}}-(1+3){\\bf{j}}+(3+1){\\bf{k}} \\\\\r\n&=8{\\bf{i}}-4{\\bf{j}}+4{\\bf{k}}.\r\n\\end{aligned}[\/latex]<\/p>\r\n
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
[latex]\\begin{aligned}\r\n8(x-1)-4(y-1)+4(z+2)&=0 \\\\\r\n8x-4y+4z+4&=0.\r\n\\end{aligned}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>
The same plane can be described by different equations depending on which point and normal vector you choose. Scalar multiples of a normal vector still point in the same direction, so [latex]\\langle 8,-4,4 \\rangle[\/latex], [latex]\\langle 2,-1,1 \\rangle[\/latex], and [latex]\\langle -4,2,-2 \\rangle[\/latex] all describe the same plane orientation. [\/Tip Box]<\/section> \r\n Connection to Cross Products<\/strong>: Finding a plane through three points always involves:<\/p>\r\n\r\n
- \r\n \t
- Form two vectors from the three points<\/li>\r\n \t
- Take their cross product to get a normal vector<\/li>\r\n \t
- Use the normal vector and any of the three points to write the plane equation<\/li>\r\n<\/ol>\r\n<\/section>
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].[reveal-answer q=\"289161783\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"289161783\"]\r\n 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
[latex]{\\bf{v}}_2=\\langle1-0,4-1,3-(-1)\\rangle=\\langle1,3,4\\rangle[\/latex].<\/p>\r\n
Use the cross product of these vectors to identify a normal vector for the plane:<\/p>\r\n
[latex]\\begin{aligned}\r\n{\\bf{n}}&={\\bf{v}}_1\\times{\\bf{v}}_2 \\\\\r\n&=\\begin{vmatrix}{\\bf{i}}&{\\bf{j}}&{\\bf{k}} \\\\1&2&1 \\\\1&3&4\\end{vmatrix} \\\\\r\n&=(8-3){\\bf{i}}-(4-1){\\bf{j}}+(3-2){\\bf{k}} \\\\\r\n&=5{\\bf{i}}-3{\\bf{j}}+{\\bf{k}}\r\n\\end{aligned}[\/latex].<\/p>\r\n
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\"]
![\"This](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31200056\/Example-2.50.jpg\")
Figure 2. Plane represented by the scalar equation [latex]5x-3y+z+4=0[\/latex].[\/caption][\/hidden-answer]<\/section>[ohm_question hide_question_numbers=1]325840[\/ohm_question]<\/section>\r\n Distance from a Point to a Plane<\/h3>\r\nOnce you have a plane equation, you can find the shortest distance from any point to the plane. This distance is always measured along a line perpendicular to the plane.\r\n\r\n[caption id=\"attachment_5183\" align=\"aligncenter\" width=\"456\"]
![\"This](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31200627\/2.70.jpg\")
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}||<||\\overrightarrow{QP}||[\/latex].[\/caption]Just as you find the 2D distance from a point to a line by measuring perpendicular to the line, you find the 3D distance from a point to a plane by measuring perpendicular to the plane.<\/div>\r\nIf [latex]R[\/latex] is the closest point on the plane to [latex]P[\/latex], then [latex]\\overrightarrow{RP}[\/latex] must be parallel to the normal vector [latex]\\mathbf{n}[\/latex]. For any other point [latex]Q[\/latex] on the plane, the vector [latex]\\overrightarrow{QP}[\/latex] can be projected onto [latex]\\mathbf{n}[\/latex] to find [latex]\\overrightarrow{RP}[\/latex].\r\n\r\n\r\n the distance between a plane and a point<\/h3>\r\n
For a plane with normal vector [latex]\\mathbf{n}[\/latex] passing through point [latex]Q[\/latex], the distance [latex]d[\/latex] from the plane to point [latex]P[\/latex] (not on the plane) is:<\/p>\r\n
[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<\/section>
Why the Absolute Value? <\/strong>Distance is always positive, but [latex]\\overrightarrow{QP} \\cdot \\mathbf{n}[\/latex] can be positive or negative depending on which side of the plane point [latex]P[\/latex] lies. The absolute value ensures we get a positive distance<\/section> This formula is exactly the scalar projection of [latex]\\overrightarrow{QP}[\/latex] onto the normal vector [latex]\\mathbf{n}[\/latex]. The projection gives the component of [latex]\\overrightarrow{QP}[\/latex] in the direction perpendicular to the plane\u2014which is precisely the shortest distance.<\/section> \r\n 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\"]
![\"This](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31201059\/Example-2.51.jpg\")
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\nThe 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
[latex]\\overrightarrow{QP}=\\langle3-5,1-0,2-0\\rangle=\\langle-2,1,2\\rangle[\/latex].<\/p>\r\n
Apply the distance formula from\u00a0The Distance between a Plane and a Point:<\/p>\r\n
[latex]\\begin{aligned}\r\nd&=\\frac{|\\overrightarrow{QP}\\cdot{\\bf{n}}|}{||{\\bf{n}}||} \\\\\r\n&=\\frac{|\\langle-2,1,2\\rangle\\cdot\\langle1,-2,1\\rangle|}{\\sqrt{1^2+(-2)^2+1^2}} \\\\\r\n&=\\frac{|-2-2+2|}{\\sqrt6} \\\\\r\n&=\\frac{2}{\\sqrt6 .}\r\n\\end{aligned}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"
Equations for a Plane<\/h2>\n
Just as a line is uniquely determined by two points, a plane <\/strong>is uniquely determined by three points that don’t all lie on the same line. There are several ways to characterize a plane mathematically.<\/p>\n
A plane can be determined by:<\/p>\n
- \n
- Three non-collinear points<\/strong> (the most basic definition)<\/li>\n
- Two distinct, intersecting lines<\/strong> (exactly one plane contains both)<\/li>\n
- One line and a point not on the line<\/strong> (the line and point determine the plane)<\/li>\n
- A point and a normal vector<\/strong><\/li>\n<\/ul>\n
The most powerful way to describe a plane uses a normal vector<\/strong>\u2014a vector perpendicular to the plane. Here’s the key insight:<\/p>\n
Imagine a vector [latex]\\mathbf{n}[\/latex] pointing perpendicular to a plane, and a fixed point [latex]P[\/latex] on the plane. Any other point [latex]Q[\/latex] lies on the same plane if and only if the vector [latex]\\overrightarrow{PQ}[\/latex] is perpendicular to [latex]\\mathbf{n}[\/latex].<\/p>\n
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.<\/figcaption><\/figure>\n Since orthogonal vectors have dot product zero, this gives us our plane equation.<\/p>\n
\n plane equations<\/h3>\n
Given point [latex]P = (x_0, y_0, z_0)[\/latex] on the plane and normal vector [latex]\\mathbf{n} = \\langle a, b, c \\rangle[\/latex]:<\/p>\n
- \n
- Vector equation<\/strong>: [latex]\\mathbf{n} \\cdot \\overrightarrow{PQ} = 0[\/latex]<\/li>\n
- Scalar equation<\/strong>: [latex]a(x - x_0) + b(y - y_0) + c(z - z_0) = 0[\/latex]<\/li>\n
- General form<\/strong>: [latex]ax + by + cz + d = 0[\/latex] where [latex]d = -ax_0 - by_0 - cz_0[\/latex]<\/li>\n<\/ul>\n<\/section>\n
To see how these forms connect, let’s expand the vector equation. If [latex]Q = (x,y,z)[\/latex] is any point on the plane, then [latex]\\overrightarrow{PQ} = \\langle x-x_0, y-y_0, z-z_0 \\rangle[\/latex]. The condition that this vector is orthogonal to [latex]\\mathbf{n}[\/latex] gives us:<\/p>\n
[latex]\\begin{aligned} \\mathbf{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
Expanding this gives us [latex]ax + by + cz - ax_0 - by_0 - cz_0 = 0[\/latex], which we can write as [latex]ax + by + cz + d = 0[\/latex] where [latex]d = -ax_0 - by_0 - cz_0[\/latex].<\/p>\n
\nWhen given three points, you need to find a normal vector to write the plane equation. The cross product provides the solution.<\/p>\n
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 - Scalar equation<\/strong>: [latex]a(x - x_0) + b(y - y_0) + c(z - z_0) = 0[\/latex]<\/li>\n
- Two distinct, intersecting lines<\/strong> (exactly one plane contains both)<\/li>\n
- Three non-collinear points<\/strong> (the most basic definition)<\/li>\n
- Scalar equation<\/strong>: [latex]a(x - x_0) + b(y - y_0) + c(z - z_0) = 0[\/latex]<\/li>\r\n \t
- Two distinct, intersecting lines<\/strong> (exactly one plane contains both)<\/li>\r\n \t