Planes in \(\RR ^3\)

You are probably familiar with the expression “two points determine a line." This means that given two distinct points, there is exactly one line that passes through both of them. There are other ways to describe a line. For example, as we saw in Parametric Equations of Lines, a single point together with a direction vector also determine a line.

You are probably also familiar with the fact that three non-collinear points determine a plane. (This is why photographers use tripods for stability, while four-legged chairs often wobble!) Is there another way to determine a plane?

The diagram below shows several planes perpendicular to vector \(\vec {n}\). In fact, there are infinitely many such planes. Vector \(\vec {n}\) does not determine a plane, but if we know of a point \(P\) contained in the plane, then together, \(P\) and \(\vec {n}\) describe a unique plane.

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Given a point \(P_{0} = P_{0}(x_{0}, y_{0}, z_{0})\) and a nonzero vector \(\vec {n}\), there is a unique plane through \(P_{0}\) with normal \(\vec {n}\).

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This fact can be used to give a very simple description of a plane. Observe that a point \(P = P(x, y, z)\) lies on this plane if and only if the vector \(\overrightarrow {P_{0}P}\) is orthogonal to \(\vec {n}\) (i.e. \(P\) lies in the plane if and only if \(\vec {n} \dotp \overrightarrow {P_{0}P} = 0\)).

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Let \(\vec {n}=\begin{bmatrix}a\\b\\c\end{bmatrix}\). By “head-tail" formula (Formula form:headminustailr3), we have: \(\overrightarrow {P_{0}P} = \begin{bmatrix} x - x_{0}\\ y - y_{0}\\ z - z_{0} \end{bmatrix}\). So, \(P(x, y, z)\) lies in the plane if and only if

\[\begin{bmatrix}a\\b\\c\end{bmatrix}\dotp \begin{bmatrix} x - x_{0}\\ y - y_{0}\\ z - z_{0} \end{bmatrix}=a(x-x_0)+b(y-y_0)+c(z-z_0)=0\]

We summarize this result as a theorem.

As demonstrated in Example ex:planewithnormalvector, we can distribute coefficients \(a\), \(b\) and \(c\) of equation (eq:plane) as follows:

\[ax-ax_0+by-by_0+cz-cz_0=0\]

Setting \(d = ax_{0} + by_{0} + cz_{0}\), shows that every plane with a normal vector \(\vec {n} = \begin{bmatrix} a\\ b\\ c \end{bmatrix}\) has a linear equation of the form

\begin{equation} \label {eq:eqofline} ax + by + cz = d \end{equation}

for some constant \(d\). Conversely, the graph of this equation is a plane with \(\vec {n} = \begin{bmatrix} a\\ b\\ c \end{bmatrix}\) as a normal vector (assuming that \(a\), \(b\), and \(c\) are not all zero).

Linear Equations and their Graphs: From Lines to Hyperplanes

An equation of the form \(ax+by=d\) is a linear equation whose graph is a line in \(\RR ^2\). If we solve for \(y\) in terms of \(x\), we obtain a more familiar form of this equation, \(y=-\frac {a}{b}x+\frac {d}{b}\). The slope of the corresponding line is \(m=-\frac {a}{b}\). Observe that a line perpendicular to this line has the slope \(\frac {b}{a}\). If we interpret \(a\) as a horizontal “run" and \(b\) as a vertical “rise", we see that the vector \(\begin{bmatrix}a\\b\end{bmatrix}\) is perpendicular to the line \(ax+by=d\). You can use the following GeoGebra interactive to solidify your understanding of this.

The idea of coefficients in front of variables forming components of a normal vector should be very familiar to you. Recall that the graph of \(ax+by+cz=d\) is a plane in \(\RR ^3\) with a normal vector \(\vec {n}=\begin{bmatrix}a\\b\\c\end{bmatrix}\).

Lines and planes may seem very different, but they are all graphs of linear equations, just in different dimensions. In general, an equation of the form

\[a_1x_1+a_2x_2+\dots +a_nx_n=d\]

is called a linear equation in \(n\) variables. The graph of such an equation, for \(n>3\), is called a hyperplane. The vector in \(\RR ^n\) whose components are the coefficients \(a_1, a_2, \dots ,a_n\) is orthogonal to every vector in the hyperplane. Unfortunately, hyperplanes are impossible to see, but we can often use insights we gain from working with lines and planes and generalize them to the invisible world of higher dimensions.

Practice Problems

Find an equation of the form \(ax+by+cz=d\) for the plane described below.

The plane has a normal vector \(\vec {n}=\begin{bmatrix}2\\-3\\1\end{bmatrix}\) and the plane passes through the point \((4, 4, -3)\).

Answer:

\[\answer {2}x+\answer {-3}y+\answer {1}z=\answer {-7}\]
Find an equation of the form \(ax+by+cz=d\) for the plane described below.

The plane contains the point \((-1, 3, 0)\) and is parallel to the plane described by \(2x-5y+4z=7\).

Answer:

\[\answer {2}x+\answer {-5}y+\answer {4}z=\answer {-17}\]
Find an equation of the form \(ax+by+cz=d\) for the plane described below.

The plane contains the point \((2, 0, 5)\) and is parallel to the \(xz\)-plane.

Answer:

\[\answer {0}x+\answer {1}y+\answer {0}z=\answer {0}\]
How many planes satisfying the given conditions are there?

Containing points \((1, 2, 3)\) and \((4, 5, 6)\), and having a normal vector \(\vec {n}=\begin{bmatrix}-9\\-9\\-9\end{bmatrix}\).

0 1 Infinitely Many
How many planes satisfying the given conditions are there?

Containing \((0,2, 4)\), \((-1, 5, 3)\), \((4, 9, 1)\).

0 1 Infinitely Many
How many planes satisfying the given conditions are there?

Planes containing \((-1, 3, 5)\), \((1, 3, 6)\), \((3, 3, 7)\).

0 1 Infinitely Many

Text Source

The text in this section is an adaptation of Section 4.2 of Keith Nicholson’s Linear Algebra with Applications. (CC-BY-NC-SA)

W. Keith Nicholson, Linear Algebra with Applications, Lyryx 2018, Open Edition, p 233-234.