4.8: Planes in Rⁿ
- Find the vector and scalar equations of a plane.
Much like the above discussion with lines, vectors can be used to determine planes in \(\mathbb{R}^n\). Given a vector \(\vec{n}\) in \(\mathbb{R}^n\) and a point \(P_0\), it is possible to find a unique plane which contains \(P_0\) and is perpendicular to the given vector.
Below is a video on graphing a plane using traces.
Let \(\vec{n}\) be a nonzero vector in \(\mathbb{R}^n\). Then \(\vec{n}\) is called a normal vector to a plane if and only if \[\vec{n} \bullet \vec{v} = 0\nonumber \] for every vector \(\vec{v}\) in the plane.
In other words, we say that \(\vec{n}\) is orthogonal (perpendicular) to every vector in the plane.
Consider now a plane with normal vector given by \(\vec{n}\), and containing a point \(P_0\). Notice that this plane is unique. If \(P\) is an arbitrary point on this plane, then by definition the normal vector is orthogonal to the vector between \(P_0\) and \(P\). Letting \(\overrightarrow{0P}\) and \(\overrightarrow{0P_0}\) be the position vectors of points \(P\) and \(P_0\) respectively, it follows that \[\vec{n} \bullet (\overrightarrow{0P} - \overrightarrow{0P_0}) = 0\nonumber \] or \[\vec{n} \bullet \overrightarrow{P_0P} = 0\nonumber \]
The first of these equations gives the vector equation of the plane.
Let \(\vec{n}\) be the normal vector for a plane which contains a point \(P_0\). If \(P\) is an arbitrary point on this plane, then the vector equation of the plane is given by \[\vec{n} \bullet (\overrightarrow{0P} - \overrightarrow{0P_0}) = 0\nonumber \]
Below is a video on determining the equation of a plane using a normal vector.
Notice that this equation can be used to determine if a point \(P\) is contained in a certain plane.
Let \(\vec{n} = \left[ \begin{array}{r} 1 \\ 2 \\ 3 \end{array} \right]\) be the normal vector for a plane which contains the point \(P_0 = \left( 2, 1, 4 \right)\). Determine if the point \(P = \left( 5, 4, 1 \right)\) is contained in this plane.
Solution
By Definition \(\PageIndex{2}\) , \(P\) is a point in the plane if it satisfies the equation \[\vec{n} \bullet (\overrightarrow{0P} - \overrightarrow{0P_0}) = 0\nonumber \]
Given the above \(\vec{n}\), \(P_0\), and \(P\), this equation becomes \[\begin{aligned} \left[ \begin{array}{r} 1 \\ 2 \\ 3 \end{array} \right] \bullet \left( \left[ \begin{array}{r} 5 \\ 4 \\ 1 \end{array} \right] - \left[ \begin{array}{r} 2 \\ 1 \\ 4 \end{array} \right] \right) &= \left[ \begin{array}{r} 1 \\ 2 \\ 3 \end{array} \right] \bullet \left( \left[ \begin{array}{r} 3 \\ 3 \\ -3 \end{array} \right] \right) \\ &= 3 + 6 - 9 = 0\end{aligned}\]
Therefore \(P = ( 5, 4, 1)\) is contained in the plane.
Suppose \(\vec{n} = \left[ \begin{array}{c} a \\ b \\ c \end{array} \right]\), \(P = \left( x,y,z\right)\) and \(P_0 = (x_0, y_0, z_0 )\).
Then \[\begin{aligned} \vec{n} \bullet (\overrightarrow{0P} - \overrightarrow{0P_0}) &= 0 \\ \left[ \begin{array}{c} a \\ b \\ c \end{array} \right] \bullet \left( \left[ \begin{array}{c} x \\ y \\ z \end{array} \right] - \left[ \begin{array}{c} x_0 \\ y_0 \\ z_0 \end{array} \right] \right) &= 0 \\ \left[ \begin{array}{c} a \\ b \\ c \end{array} \right] \bullet \left[ \begin{array}{c} x - x_0 \\ y - y_0 \\ z - z_0 \end{array} \right] &= 0 \\ a(x - x_0) + b (y - y_0) + c (z-z_0) &= 0 \end{aligned}\]
We can also write this equation as \[ax + by + cz = ax_0 + by_0 + cz_0\nonumber \]
Notice that since \(P_0\) is given, \(ax_0+by_0+cz_0\) is a known scalar, which we can call \(d\). This equation becomes \[ax + by + cz = d\nonumber \]
Let \(\vec{n} = \left[ \begin{array}{c} a \\ b \\ c \end{array} \right]\) be the normal vector for a plane which contains the point \(P_0 = (x_0, y_0, z_0)\).Then if \(P=(x,y,z)\) is an arbitrary point on the plane, the scalar equation of the plane is given by \[ax + by + cz = d\nonumber \] where \(a,b,c,d \in \mathbb{R}\) and \(d = ax_0 + by_0 + cz_0\).
Consider the following equation.
Find an equation of the plane containing \(P_0 = (3, -2, 5)\) and orthogonal to \(\vec{n} = \left[ \begin{array}{r} -2 \\ 4 \\ 1 \end{array} \right]\).
Solution
The above vector \(\vec{n}\) is the normal vector for this plane. Using Definition \(\PageIndex{2}\) , we can determine the vector equation for this plane. \[\begin{aligned} \vec{n} \bullet (\overrightarrow{0P} - \overrightarrow{0P_0}) &= 0 \\ \left[ \begin{array}{r} -2 \\ 4 \\ 1 \end{array} \right] \bullet \left(\left[ \begin{array}{c} x \\ y \\ z \end{array} \right] - \left[ \begin{array}{r} 3 \\ -2 \\ 5 \end{array} \right] \right) &= 0 \\ \left[ \begin{array}{r} -2 \\ 4 \\ 1 \end{array} \right] \bullet \left[ \begin{array}{c} x - 3 \\ y + 2 \\ z - 5 \end{array} \right] &= 0 \end{aligned}\]
Using Definition \(\PageIndex{3}\) , we can determine the scalar equation of the plane. \[-2x + 4y + 1z = -2(3) + 4(-2) + 1(5) = -9\nonumber \]
Hence, the vector equation of the plane is \[\left[ \begin{array}{r} -2 \\ 4 \\ 1 \end{array} \right] \bullet \left[ \begin{array}{c} x - 3 \\ y + 2 \\ z - 5 \end{array} \right] = 0\nonumber \] and the scalar equation is \[-2x + 4y + 1z = -9\nonumber \]
Below is a video on finding the equation of a plane given a point and a line.
Below is a video on finding the equation of a plane given three points on the plane.
Below is a video on finding the equation of the plane that is orthogonal to a given line.
Suppose a point \(P\) is not contained in a given plane. We are then interested in the shortest distance from that point \(P\) to the given plane. Consider the following example.
Find the shortest distance from the point \(P = (3,2,3)\) to the plane given by
\(2x + y + 2z = 2\), and find the point \(Q\) on the plane that is closest to \(P\).
Solution
Pick an arbitrary point \(P_0\) on the plane. Then, it follows that \[\overrightarrow{QP} = proj_{\vec{n}}\overrightarrow{P_0P}\nonumber \] and \(\| \overrightarrow{QP} \|\) is the shortest distance from \(P\) to the plane. Further, the vector \(\overrightarrow{0Q} = \overrightarrow{0P} - \overrightarrow{QP}\) gives the necessary point \(Q\).
From the above scalar equation, we have that \(\vec{n} = \left[ \begin{array}{c} 2 \\ 1 \\ 2 \end{array} \right]\). Now, choose \(P_0 = (1, 0, 0)\) so that \(\vec{n} \bullet \overrightarrow{0P} = 2 = d\). Then, \(\overrightarrow{P_0P} = \left[ \begin{array}{c} 3 \\ 2 \\ 3 \end{array} \right] - \left[ \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right] = \left[ \begin{array}{c} 2 \\ 2 \\ 3 \end{array} \right]\).
Next, compute \(\overrightarrow{QP} = proj_{\vec{n}}\overrightarrow{P_0P}\). \[\begin{aligned} \overrightarrow{QP} &= proj_{\vec{n}}\overrightarrow{P_0P} \\ &= \left( \frac{ \overrightarrow{P_0P} \bullet \vec{n}}{\| \vec{n} \| ^2}\right)\vec{n} \\ &= \frac{12}{9} \left[ \begin{array}{r} 2 \\ 1 \\ 2 \end{array} \right] \\ &= \frac{4}{3} \left[ \begin{array}{r} 2 \\ 1 \\ 2 \end{array} \right] \end{aligned}\]
Then, \(\| \overrightarrow{QP} \| = 4\) so the shortest distance from \(P\) to the plane is \(4\).
Next, to find the point \(Q\) on the plane which is closest to \(P\) we have \[\begin{aligned} \overrightarrow{0Q} &= \overrightarrow{0P} - \overrightarrow{QP} \\ &= \left[ \begin{array}{r} 3 \\ 2 \\ 3 \end{array} \right] - \frac{4}{3} \left[ \begin{array}{r} 2 \\ 1 \\ 2 \end{array} \right] \\ &= \frac{1}{3} \left[ \begin{array}{r} 1 \\ 2 \\ 1 \end{array} \right]\end{aligned}\]
Therefore, \(Q = (\frac{1}{3}, \frac{2}{3}, \frac{1}{3} )\).
Below is a video on determining the distance between a point and a plane.