Section 3.4 Vector Equations of Planes
In \(\mathbb{R}^3\text{,}\) we can also form a plane. However, it is not immediately clear how to form the Cartesian equation of a plane in space. Recall that a line is defined by a point, and a slope or direction. However, planes do not have a clear notion of slope, because the inclination of a plane depends on the direction you are travelling.
First, we will consider the equation of a line in \(\mathbb{R}^2\text{,}\) from the perspective of perpendicularity, and the dot product. Then, this reasoning will generalize to a plane in \(\mathbb{R}^3\text{.}\)
Subsection 3.4.1 Standard Form of a Line, and Normal Vectors
Recall that previously, we specified a line using a point on the line \(\vec{r}_0\text{,}\) and a direction vector \(\vec{v}\text{.}\) Alternatively, we could have instead specified a vector that is perpendicular to the line \(\vec{n}\text{,}\) called a “normal vector”. Then, for any point \(\vec{r}\) on the line, it has the property that \(\vec{r} - \vec{r}_0\) is perpendicular to \(\vec{n}\) (previously, recall \(\vec{r} - \vec{r}_0\) was parallel to \(\vec{v}\)). Using dot products, this means that,
\begin{equation*}
\vec{n} \bullet (\vec{r} - \vec{r}_0) = 0
\end{equation*}
This is another form of the equation of a line. If \(\vec{r} = (x,y), \vec{r} = (x_0, y_0)\text{,}\) and \(\vec{n} = (a,b)\text{,}\) then this is equiavlent to,
\begin{align*}
(a,b) \bullet (x - x_0, y - y_0) \amp = 0\\
a(x - x_0) + b(y - y_0) \amp = 0
\end{align*}
This can be written more simply by expanding, and collecting the constant terms,
\begin{align*}
ax - ax_0 + by - by_0 \amp = 0\\
ax + by \amp = ax_0 + by_0
\end{align*}
You may recognize this as the standard form of a line, typically written as \(ax + by = c\text{.}\) Also, in linear algebra, we prefer this form, because it connects to the form of a linear equation in terms of a linear system.
Conversely, the line \(ax + by = c\) has normal vector \(\vec{n} = (a,b)\text{.}\)
Subsection 3.4.2 Planes in \(\mathbb{R}^3\)
In a similar way, it turns out that a plane can be specified using a point on the plane, and a vector pointing out of the plane, again called a normal vector. Let \(\vec{r}_0 = (x_0, y_0, z_0)\) be a point in space, and let \(\vec{n} = (a,b,c)\) be a normal vector for the plane. Then, for any point \(\vec{r} = (x,y,z)\) on the plane, it has the property that \(\vec{r} - \vec{r}_0\text{,}\) which is a vector that line parallel to the plane, is perpendicular to \(\vec{n}\text{.}\)
\begin{equation*}
\vec{n} \bullet (\vec{r} - \vec{r}_0) = 0
\end{equation*}
This is the equation of the plane in vector form. Notice that this is the same as the equation of a line, except the vectors involved have 3 coordinates instead of 2. In terms of Cartesian coordinates,
\begin{equation*}
a(x - x_0) + b(y - y_0) + c(z - z_0) = 0
\end{equation*}
Again, expanding, and collecting the constant terms, we get,
\begin{align*}
ax - ax_0 + by - by_0 + cz - cz_0 \amp = 0\\
ax + by + cz = ax_0 + by_0 + cz_0
\end{align*}
Then, this can be written in the standard form,
\begin{equation*}
\boxed{ax + by + cz = d}
\end{equation*}
where \(d = ax_0 + by_0 + cz_0\) (or \(d = \vec{n} \bullet \vec{r}_0\)). Notice the similarities to the standard form of a line, \(ax + by = c\text{.}\) Conversely, given the equation of a plane \(ax + by + cz = d\text{,}\) the normal vector can be simply read off as \((a,b,c)\text{.}\) Of course, the normal vector of a plane is not unique, as any multiple of \((a,b,c)\) is also a normal vector of the plane.
Theorem 3.4.1. Equation of a plane.
Let \(\vec{r}_0 = (x_0, y_0, z_0)\) be a point, \(\vec{n} = (a,b,c)\) be a non-zero vector. Then, the equation of the plane through \(\vec{r}_0\) and perpendicular to \(\vec{n}\) has equation,
\begin{equation*}
\vec{n} \bullet (\vec{r} - \vec{r}_0) = 0
\end{equation*}
Or,
\begin{equation*}
ax + by + cz = d
\end{equation*}
where \(d = \vec{n} \bullet \vec{r}_0\text{.}\) Conversely, the graph of the equation \(ax + by + cz = d\) is a plane perpendicular to \(\vec{n} = (a,b,c)\text{,}\) provided that \(a, b, c \neq 0\text{.}\)
If at least of the the scalars \(a,b,c\) are non-zero, then the equation represents a plane. Otherwise, the normal vector is the zero vector, and the equation becomes \(0 = 0\text{,}\) representing all of \(\mathbb{R}^3\text{.}\)
There exists exactly one plane passing through \(\vec{r}_0\) that is perpendicular to \(\vec{n}\text{.}\)
From another perspective, by first starting with all points \((x,y,z)\) in 3-dimensional space \(\mathbb{R}^3\text{,}\) and then by requiring the constraint equation \(ax + by + cz = d\text{,}\) this constraint generally acts to reduce the number of degrees of freedom by 1, i.e. the result is a 2-dimensional object. It turns out that if this equation is linear (like the equation of a plane is), then the result will be a 2-dimensional flat surface, i.e. a plane. Similarly, in \(\mathbb{R}^2\text{,}\) a single linear equation reduces the 2-dimensional plane to a 1-dimensional line.
Subsection 3.4.3 Planes Using Two Vectors: Cross Product
Another way of specifying a plane is to specify a point \(\vec{r}_0\) on the plane, and two non-parallel (non-zero) vectors \(\vec{v}_1, \vec{v}_2\text{.}\) Then, a normal vector to this plane is the cross product \(\vec{v}_1 \times \vec{v}_2\text{.}\)
Subsection 3.4.4 Parametric Equation of a Plane
Similarly to lines, we can specify a plane in parametric form. Consider a point on the plane \(\vec{r}_0\)
Subsection 3.4.5 Planes Using 3 Points
Yet another method to specify a plane is to specify 3 distinct points on the plane. If the points are \((a_{11}, a_{12}, a_{13}), (a_{21}, a_{22}, a_{23}), (a_{31}, a_{32}, a_{33})\text{,}\) then a plane containing these three points, of the form \(ax + by + cz = d\text{,}\) must satisfy the system,
\begin{equation*}
\begin{array}{rrrrr}
aa_{11} \amp + ba_{12} \amp + ca_{13} \amp = \amp d \\
aa_{21} \amp + ba_{22} \amp + ca_{23} \amp = \amp d \\
aa_{31} \amp + ba_{32} \amp + ca_{33} \amp = \amp d
\end{array}
\end{equation*}
This is a linear system in the 4 variables \(a, b, c, d\text{.}\) Here, there are 4 unknowns, and only 3 equations, so we expect that there are more than one solution. Performing row reduction, we can solve the system, and determine the coefficients of the plane's equation.
This works, as long as the points are not colinear.
Another method, which uses vectors, is to directly determine the normal vector. Recall that if we a point \(\vec{r}_0\) on the plane, and two vectors such that \(\vec{r}_0 + \vec{v}_1\) and \(\vec{r}_0 + \vec{v}_2\) are on the plane, then the cross product \(\vec{v}_1 \times \vec{v}_2\) is a normal vector of the plane. Let \(\vec{r}_0, \vec{r}_1, \vec{r}_2\) be the 3 points. Then, we can use one point (say, \(\vec{r}_0\)) as the starting point. Then, the two points \(\vec{r}_1 - \vec{r}_0\) and \(\vec{r}_2 - \vec{r}_0\) are vectors which are parallel to the plane. Then, their cross product is a normal vector for the plane,
\begin{equation*}
\vec{n} = \brac{\vec{r}_1 - \vec{r}_0} \times \brac{\vec{r}_2 - \vec{r}_0}
\end{equation*}
Subsection 3.4.6 Summary
In summary, a plane can be defined by,
A point and a vector perpendicular to the plane.
Two vectors, that are not parallel.
Three points, which are not collinear.
Subsection 3.4.7 Planes Parallel to the Coordinate Planes
In \(\mathbb{R}^3\text{,}\) the coordinate planes are the planes spanned by the axes, the \(xy\)-plane, \(xz\)-plane, and the \(yz\)-plane.
For example, the \(xy\)-plane is the set of all points with \(z\)-coordinate equal to 0, and so it's Cartesian equation is \(z = 0\text{.}\) Similarly, the \(yz\)-plane and \(xz\)-plane have equations \(x = 0\) and \(y = 0\text{,}\) respectively.
Slightly more generally, the equations \(x = a\text{,}\) \(y = a\text{,}\) and \(z = a\text{,}\) for some \(a\text{,}\) represent a plane parallel to the coordinate planes, but \(a\) units above or below.
Subsection 3.4.8 Intersecting Planes
In almost all cases, two planes will intersect, and this intersection will be a line. Let \(a_1 x + b_1 y + c_1 z = d_1\) and \(a_2 x + b_2 y + c_2 z = d_2\) be the equation of two planes, with normal vectors \(\vec{n}_1 = \veciii{a_1}{b_1}{c_1}, \vec{n}_2 = \veciii{a_2}{b_2}{c_2}\text{.}\) Then, the line of intersection will be perpendicular to both normal vectors \(\vec{n}_1\) and \(\vec{n}_2\text{.}\) Thus, it will be parallel to their cross product. In other words, \(\vec{v} = \vec{n}_1 \times \vec{n}_2\) is a direction vector for the line of intersection.
Then, it is sufficient to find a single point on the line. To do this, we can solve the system of equations,
\begin{align*}
a_1 x + b_1 y + c_1 z \amp = d_1\\
a_2 x + b_2 y + c_2 z \amp = d_2
\end{align*}
This is a system with 3 variables and only 2 equations, so we expect there to be infinitely many solutions. In this case, we only need a single solution, so it is easiest to choose a convenient value (typically letting \(x = 0\text{,}\) \(y = 0\text{,}\) or \(z = 0\) in the two equations, and solving for the other two variables, resulting a solution, say \(\vec{r}_0\text{.}\) Then, the vector parametric equation of the line is,
\begin{equation*}
\vec{r} = \vec{r}_0 + t \vec{v}
\end{equation*}
Subsection 3.4.9 Parallel Planes
Two planes are parallel if their normal vectors are parallel (or anti-parallel). If two planes are parallel, then they don't intersect.
