9.5: Lines and Planes in Space (2024)

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Motivating Questions

How are lines in \(\mathbb{R}^3\) similar to and different from lines in \(\mathbb{R}^2\text{?}\)
What is the role that vectors play in representing equations of lines, particularly in \(\mathbb{R}^3\text{?}\)
How can we think of a plane as a set of points determined by a point and a vector?
How do we find the equation of a plane through three given non-collinear points?

In single variable calculus, we learn that a differentiable function is locally linear. In other words, if we zoom in on the graph of a differentiable function at a point, the graph looks like the tangent line to the function at that point. Linear functions played important roles in single variable calculus, useful in approximating differentiable functions, in approximating roots of functions (see Newton's Method), and approximating solutions to first order differential equations (see Euler's Method). In multivariable calculus, we will soon study curves in space; differentiable curves turn out to be locally linear as well. In addition, as we study functions of two variables, we will see that such a function is locally linear at a point if the surface defined by the function looks like a plane (the tangent plane) as we zoom in on the graph.

Consequently, it is important for us to understand both lines and planes in space, as these define the linear functions in \(\mathbb{R}^2\) and \(\mathbb{R}^3\text{.}\) (Recall that a function is linear if it is a polynomial function whose terms all have degree less than or equal to 1. For example, \(x\) defines a single variable linear function and \(x+y\) a two variable linear function, but \(xy\) is not linear since it has degree two, the sum of the degress of its factors.) We begin our work by considering some familiar ideas in \(\mathbb{R}^2\) but from a new perspective.

Preview Activity \(\PageIndex{1}\)

We are familiar with equations of lines in the plane in the form \(y = mx+b\text{,}\) where \(m\) is the slope of the line and \((0,b)\) is the \(y\)-intercept. In this activity, we explore a more flexible way of representing lines that we can use not only in the plane, but in higher dimensions as well.

To begin, consider the line through the point \((2,-1)\) with slope \(\frac{2}{3}\) as shown in Figure \(\PageIndex{1}\).

Suppose we increase \(x\) by 1 from the point \((2,-1)\text{.}\) How does the \(y\)-value change? What is the point on the line with \(x\)-coordinate \(3\text{?}\)
Suppose we decrease \(x\) by 3.25 from the point \((2,-1)\text{.}\) How does the \(y\)-value change? What is the point on the line with \(x\)-coordinate \(-1.25\text{?}\)
Now, suppose we increase \(x\) by some arbitrary value \(3t\) from the point \((2,-1)\text{.}\) How does the \(y\)-value change? What is the point on the line with \(x\)-coordinate \(2+3t\text{?}\)
Observe that the slope of the line is related to any vector whose \(y\)-component divided by the \(x\)-component is the slope of the line. For the line in this exercise, we might use the vector \(\langle 3,2 \rangle\text{,}\) which describes the direction of the line. Explain why the terminal points of the vectors \(\mathbf{r}(t)\text{,}\) where
\[ \mathbf{r}(t) = \langle 2,-1 \rangle + \langle 3,2 \rangle t, \nonumber \]

trace out the graph of the line through the point \((2,-1)\) with slope \(\frac{2}{3}\text{.}\)
Now we extend this vector approach to \(\mathbb{R}^3\) and consider a second example. Let \(\mathcal{L}\) be the line in \(\mathbb{R}^3\) through the point \((1,0,2)\) in the direction of the vector \(\langle 2, -1, 4 \rangle\text{.}\) Find the coordinates of three distinct points on line \(\mathcal{L}\text{.}\) Explain your thinking.
Find a vector in the form
\[ \mathbf{r}(t) = \langle x_0, y_0, z_0 \rangle + \langle a,b,c \rangle t \nonumber \]

whose terminal points trace out the line \(\mathcal{L}\) that is described in (e). That is, you should be able to locate any point on the line by determining a corresponding value of \(t\text{.}\)

Lines in Space

In two-dimensional space, a non-vertical line is defined to be the set of points satisfying the equation

\[ y = mx + b, \nonumber \]

for some constants \(m\) and \(b\text{.}\) The value of \(m\) (the slope) tells us how the dependent variable changes for every one unit increase in the independent variable, while the point \((0,b)\) is the \(y\)-intercept and anchors the line to a location on the \(y\)-axis. Alternatively, we can think of the slope as being related to the vector \(\langle 1, m \rangle\text{,}\) which tells us the direction of the line, as shown on the left in Figure \(\PageIndex{2}\). Thus, we can identify a line in space by fixing a point \(P\) and a direction \(\mathbf{v}\text{,}\) as shown on the right. Since we also have vectors in space to provide direction, this same idea of a point and a direction determining a line works in \(\mathbb{R}^n\) for any \(n\text{.}\)

Definition \(\PageIndex{1}\)

A line in space is the set of terminal points of vectors emanating from a given point \(P\) that are parallel to a fixed vector \(\mathbf{v}\text{.}\)

The fixed vector \(\mathbf{v}\) in the definition is called a direction vector for the line. As we saw in Preview Activity \(\PageIndex{1}\), to find an equation for a line through point \(P\) in the direction of vector \(\mathbf{v}\text{,}\) observe that any vector parallel to \(\mathbf{v}\) will have the form \(t \mathbf{v}\) for some scalar \(t\text{.}\) So, any vector emanating from the point \(P\) in a direction parallel to the vector \(\mathbf{v}\) will be of the form

\[ \overrightarrow{OP} + \mathbf{v} t\label{eq_9_5_lines_1}\tag{\(\PageIndex{1}\)} \]

for some scalar \(t\) (where \(O\) is the origin).

The vector form of a line

The vector form of a line through the point \(P\) in the direction of the vector \(\mathbf{v}\) is

\[ \mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v},\label{eq_9_5_line_vect}\tag{\(\PageIndex{2}\)} \]

where \(\mathbf{r}_0\) is the position vector \(\overrightarrow{OP}\) from the origin to the point \(P\text{.}\)

Of course, it is common to represent lines in the plane using the slope-intercept equation \(y=mx + b\text{.}\) The vector form of the line, described above, is an alternative way to represent lines that has the following two advantages. First, in two dimensions, we are able to represent vertical lines, whose slope \(m\) is not defined, using a vertical direction vector, such as \(\mathbf{v}=\langle 0, 1\rangle\text{.}\) Second, this description of lines works in any dimension even though there is no concept of the slope of a line in more than two dimensions.

Activity \(\PageIndex{2}\)

Let \(P_1 = (1,2,-1)\) and \(P_2 = (-2,1,-2)\text{.}\) Let \(\mathcal{L}\) be the line in \(\mathbb{R}^3\) through \(P_1\) and \(P_2\text{,}\) and note that three snapshots of this line are shown in Figure \(\PageIndex{5}\).

Find a direction vector for the line \(\mathcal{L}\text{.}\)
Find a vector equation of \(\mathcal{L}\) in the form \(\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}\text{.}\)
Consider the vector equation \(\mathbf{s}(t) = \langle -5, 0, -3 \rangle + t \langle 6, 2, 2 \rangle.\) What is the direction of the line given by \(\mathbf{s}(t)\text{?}\) Is this new line parallel to line \(\mathcal{L}\text{?}\)
Do \(\mathbf{r}(t)\) and \(\mathbf{s}(t)\) represent the same line, \(\mathcal{L}\text{?}\) Explain.

The Parametric Equations of a Line

The vector form of a line, \(\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}\) in Equation (\(\PageIndex{2}\)), describes a line as the set of terminal points of the vectors \(\mathbf{r}(t)\text{.}\) If we write this in terms of components letting

\[ \mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle, \ \ \ \mathbf{r}_0 = \langle x_0, y_0, z_0 \rangle, \ \ \ \text{ and } \ \ \ \mathbf{v} = \langle a, b, c \rangle, \nonumber \]

The parametric equations of a line

The parametric equations for a line through the point \(P = (x_0, y_0, z_0)\) in the direction of the vector \(\mathbf{v} = \langle a,b,c \rangle\) are

\[ x(t) = x_0 + at, \ \ \ \ \ y(t) = y_0 + bt, \ \ \ \ \ z(t) = z_0 + ct. \nonumber \]

Notice that there are many different parametric equations for the same line. For example, choosing another point \(P\) on the line or another direction vector \(\mathbf{v}\) produces another set of parametric equations. It is sometimes useful to think of \(t\) as a time parameter and the parametric equations as telling us where we are on the line at each time. In this way, the parametric equations describe a particular walk taken along the line; there are, of course, many possible ways to walk along a line.

Activity \(\PageIndex{3}\)

Let \(P_1 = (1,2,-1)\) and \(P_2 = (-2,1,-2)\text{,}\) and let \(\mathcal{L}\) be the line in \(\mathbb{R}^3\) through \(P_1\) and \(P_2\text{,}\) which is the same line as in Activity \(\PageIndex{2}\).

Find parametric equations of the line \(\mathcal{L}\text{.}\)
Does the point \((1, 2, 1)\) lie on \(\mathcal{L}\text{?}\) If so, what value of \(t\) results in this point?
Consider another line, \(\mathcal{K}\text{,}\) whose parametric equations are
\[ x(s) = 11 + 4s, \ \ y(s) = 1-3s, \ \ z(s) = 3 + 2s. \nonumber \]

What is the direction of the line \(\mathcal{K}\text{?}\)
Do the lines \(\mathcal{L}\) and \(\mathcal{K}\) intersect? If so, provide the point of intersection and the \(t\) and \(s\) values, respectively, that result in the point. If not, explain why.

Planes in Space

Now that we have a way of describing lines, we would like to develop a means of describing planes in three dimensions. We studied the coordinate planes and planes parallel to them in Section 9.1. Each of those planes had one of the variables \(x\text{,}\) \(y\text{,}\) or \(z\) equal to a constant. We can note that any vector in a plane with \(x\) constant is orthogonal to the vector \(\langle 1,0,0 \rangle\text{,}\) any vector in a plane with \(y\) constant is orthogonal to the vector \(\langle 0,1,0 \rangle\text{,}\) and any vector in a plane with \(z\) constant is orthogonal to the vector \(\langle 0,0,1 \rangle\text{.}\) This idea works in general to define a plane.

Definition \(\PageIndex{6}\)

A plane \(p\) in space is the set of all terminal points of vectors emanating from a given point \(P_0\) perpendicular to a fixed vector \(\mathbf{n}\text{,}\) as shown in Figure \(\PageIndex{5}\).

The definition allows us to find the equation of a plane. Assume that \(\mathbf{n}=\langle a,b,c\rangle\text{,}\) \(P_0 = (x_0, y_0, z_0)\text{,}\) and that \(P=(x,y,z)\) is an arbitrary point on the plane. Since the vector \(\overrightarrow{P_0P}\) lies in the plane, it must be perpendicular to \(\mathbf{n}\text{.}\) This means that

\begin{align*} 0 =\mathstrut & \mathbf{n}\cdot\overrightarrow{P_0P}\\[4pt] =\mathstrut & \mathbf{n}\cdot \big(\langle x,y,z \rangle - \langle x_0, y_0, z_0\rangle\big)\\[4pt] =\mathstrut & \mathbf{n} \cdot \langle x-x_0, y-y_0, z-z_0 \rangle\\[4pt] =\mathstrut & a(x-x_0) + b(y-y_0) + c(z-z_0). \end{align*}

The fixed vector \(\mathbf{n}\) perpendicular to the plane is frequently called a normal vector to the plane. We may now summarize as follows.

Equations of a plane

The scalar equation of the plane with normal vector \(\mathbf{n} =\langle a,b,c \rangle\) containing the point \(P_0 = (x_0, y_0,z_0)\) is
\[ a(x-x_0) + b(y-y_0) + c(z-z_0) = 0.\label{eq_9_5_plane_norm}\tag{\(\PageIndex{3}\)} \]
The vector equation of the plane with normal vector \(\mathbf{n} =\langle a,b,c \rangle\) containing the points \(P_0 = (x_0, y_0,z_0)\) and \(P = (x,y,z)\) is
\[ \mathbf{n} \cdot \overrightarrow{P_0P} = 0.\label{eq_9_5_plane_norm_vector}\tag{\(\PageIndex{4}\)} \]

We may take the scalar equation of a plane a little further and note that since

\[ a(x-x_0) + b(y-y_0) + c(z-z_0) = 0, \nonumber \]

it equivalently follows that

\[ ax + by + cz = ax_0+by_0+cz_0. \nonumber \]

That is, we may write an equation of a plane as \(ax+by+cz = d\) where \(d = \mathbf{n}\cdot\langle x_0,y_0,z_0\rangle\text{.}\)

For instance, if we would like to describe the plane passing through the point \(P_0=(4, -2,1)\) and perpendicular to the vector \(\mathbf{n} = \langle 1, 2, 1 \rangle\text{,}\) we have

\[ \langle 1,2,1 \rangle\cdot \langle x,y,z\rangle = \langle 1,2,1 \rangle\cdot \langle 4,-2,1\rangle \nonumber \]

\[ x + 2y + z = 1. \nonumber \]

Notice that the coefficients of \(x\text{,}\) \(y\text{,}\) and \(z\) in this description give a vector perpendicular to the plane. For instance, if we are presented with the plane

\[ -2x + y - 3z = 4, \nonumber \]

we know that \(\mathbf{n} = \langle -2, 1, -3\rangle\) is a vector perpendicular to the plane.

Activity \(\PageIndex{4}\)

Write a scalar equation of the plane \(p_1\) passing through the point \((0, 2, 4)\) and perpendicular to the vector \(\mathbf{n}=\langle 2, -1, 1\rangle\text{.}\)
Is the point \((2, 0, 2)\) on the plane \(p_1\text{?}\)
Write a scalar equation of the plane \(p_2\) that is parallel to \(p_1\) and passing through the point \((3, 0, 4)\text{.}\) (Hint: Compare normal vectors of the planes.)
Write a parametric description of the line \(l\) passing through the point \((2,0,2)\) and perpendicular to the plane \(p_3\) described by the equation \(x+2y-2z = 7\text{.}\)
Find the point at which \(l\) intersects the plane \(p_3\text{.}\)

Just as two distinct points in space determine a line, three non-collinear points in space determine a plane. Consider three points \(P_0\text{,}\) \(P_1\text{,}\) and \(P_2\) in space, not all lying on the same line as shown in Figure \(\PageIndex{6}\).

Observe that the vectors \(\overrightarrow{P_0P_1}\) and \(\overrightarrow{P_0P_2}\) both lie in the plane \(p\text{.}\) If we form their cross-product

\[ \mathbf{n} = \overrightarrow{P_0P_1} \times \overrightarrow{P_0P_2}, \nonumber \]

we obtain a normal vector to the plane \(p\text{.}\) Therefore, if \(P\) is any other point on \(p\text{,}\) it then follows that \(\overrightarrow{P_0P}\) will be perpendicular to \(\mathbf{n}\text{,}\) and we have, as before, the equation

\[ \mathbf{n} \cdot \overrightarrow{P_0P} = 0.\label{eq_9_5_plane_vect}\tag{\(\PageIndex{5}\)} \]

Activity \(\PageIndex{5}\)

Let \(P_0 = (1,2,-1)\text{,}\) \(P_1 = (1, 0 ,-1)\text{,}\) and \(P_2 = (0,1,3)\) and let \(p\) be the plane containing \(P_0\text{,}\) \(P_1\text{,}\) and \(P_2\text{.}\)

Determine the components of the vectors \(\overrightarrow{P_0P_1}\) and \(\overrightarrow{P_0P_2}\text{.}\)
Find a normal vector \(\mathbf{n}\) to the plane \(p\text{.}\)
Find a scalar equation of the plane \(p\text{.}\)
Consider a second plane, \(q\text{,}\) with scalar equation \(-3(x-1) + 4(y+3) + 2(z-5)=0\text{.}\) Find two different points on plane \(q\text{,}\) as well as a vector \(\mathbf{m}\) that is normal to \(q\text{.}\)
The angle between two planes is the acute angle between their respective normal vectors. What is the angle between planes \(p\) and \(q\text{?}\)

Summary

While lines in \(\mathbb{R}^3\) do not have a slope, like lines in \(\mathbb{R}^2\) they can be characterized by a point and a direction vector. Indeed, we define a line in space to be the set of terminal points of vectors emanating from a given point that are parallel to a fixed vector.
Vectors play a critical role in representing the equation of a line. In particular, the terminal points of the vector \(\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}\) define a linear function \(\mathbf{r}\) in space through the terminal point of the vector \(\mathbf{r}_0\) in the direction of the vector \(\mathbf{v}\text{,}\) tracing out a line in space.
A plane in space is the set of all terminal points of vectors emanating from a given point perpendicular to a fixed vector.
If \(P_1\text{,}\) \(P_2\text{,}\) and \(P_3\) are non-collinear points in space, the vectors \(\overrightarrow{P_1P_2}\) and and \(\overrightarrow{P_1P_3}\) are vectors in the plane and the vector \(\mathbf{n} = \overrightarrow{P_1P_2} \times \overrightarrow{P_1P_3}\) is a normal vector to the plane. So any point \(P\) in the plane satisfies the equation \(\overrightarrow{PP_1} \cdot \mathbf{n} = 0\text{.}\) If we let \(P = (x,y,z)\text{,}\) \(\mathbf{n} = \langle a,b,c \rangle\) be the normal vector, and \(P_1 = (x_0,y_0,z_0)\text{,}\) we can also represent the plane with the equation
\[ a(x-x_0) + b(y-y_0) + c(z-z_0) = 0. \nonumber \]