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 around a point, the graph will look like the tangent line to the function at that point. In Figure 9.5.1 you can look at the graph of a curve and zoom in around a point on that curve to see that eventually the graph of the function will look linear. You can use the sliders at the top of change how zoomed in the plot is displayed or change the location on the curve that you are examining. You should visually verify at a several points that this function is differentiable because at any point shown, the graph is locally linear.
Recall that linear functions have the form \(f(x)=ax+b\text{.}\) These functions play important roles in single variable calculus and are useful in approximating differentiable functions, in approximating roots of functions (Newton’s Method), and approximating solutions to first order differential equations (Euler’s Method). In multivariable calculus, we will study curves in space and we will see that connection between local linearity and differentiability exists for curves in three (or more) dimensions. As we extend our study of functions to include those of two variables in Chapter 11, we will see that 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.
Because of their important role in understanding curves and surfaces in \(\R^3\text{,}\) it is important for us to understand both lines and planes in space, as can both be thought of as generalizations of lines in two dimensions to three-dimensional space. In your study of one-variable functions, you probably worked with slope, direction, and measuring change using lines several times before you generalized those ideas for a curve in general. Similarly, we will want to understand how we talk about direction and measuring change along flat objects in \(\R^3\) (lines and planes) before we start generalizing to curved objects. In this section, we discuss how we can use vectors to describe lines in space. The next section will be dedicated to understanding planes.
We will start our work on lines by considering some familiar ideas in \(\R^2\) but from a new perspective. You are probably familiar with equations of lines in the \(xy\)-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 is useful in the \(xy\)-plane and higher dimensions. To begin, consider the line through the point \((2,-1)\) with slope \(\frac{2}{3}\) as shown in Figure 9.5.2.
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 \(2+3t\text{?}\)
Remember that the horizontal component of a vector describes the “run” and the vertical component measures the “rise” of the vector. So 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 activity, we might use the vector \(\langle 3,2 \rangle\) to describe the direction of the line. We will look at the following vector-valued function of one variable:
Write a few sentences that compare the endpoints of your vectors from the previous task to the plot of the line through the point \((2, -1)\) with a slope of \(\frac{2}{3}\) (Figure 9.5.2). In particular you should address what aspects of the line are related to \(\langle 2, -1 \rangle\) or \(\langle 3,2 \rangle\text{.}\)
The way most people draw a line is more connected to the ideas in Preview Activity 9.5.1 than slope or algebraic forms. Take a minute and draw a plot of the line described by \(x+y-1=0\text{.}\) You should not continue reading until you have made a plot (by hand) of \(x+y-1=0\) that you would be proud to share with classmates.
You may have done some algebra or other calculation to get to slope-intercept form or another familiar algebraic structure, but when you went to draw the line you likely did the following steps:
Found a point to start your plot (maybe the \(y\)-intercept).
In other words, you made a starting point, extended indefinitely far in a particular direction, then added the opposite direction. This is the essence of the general description for a line we will give below.
In two-dimensional space, a non-vertical line is defined to be the set of points satisfying the equation
\begin{equation*}
y = mx + b,
\end{equation*}
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 9.5.5. Thus, we can identify a line in space by fixing a point \(P\) and a direction \(\vv\text{,}\) as shown on the right. Since we also have vectors in \(\R^n\) to provide direction, this same idea of a point and a direction determining a line works in \(\R^n\) for any \(n\text{.}\)
Two-dimensional axes showing a line passing through the point \((0,1)\text{.}\) The vector \(\vv=\langle 1,m\rangle\) is depicted on top of the line with its initial point at the point \((0,1)\) and its terminal point on the line.
The vector \(\vv\) in Definition 9.5.6 is called a direction vector for the line. As we saw in Preview Activity 9.5.1, any vector parallel to \(\vv\) will have the form \(t \vv\) for some scalar \(t\text{.}\) As a result, any vector emanating from the point \(P\) in a direction parallel to the vector \(\vv\) will be of the form
Figure 9.5.7 shows the plot of a line in two-space. The vector \(\overrightarrow{OP}\) is illustrated in blue and points from the origin to our starting point \(P\text{.}\) The direction vector \(t \vv\) as in Equation (9.5.1) is shown in red with its initial point at point \(P\text{.}\) The line is traced out by the terminal points of the green vector \(\overrightarrow{OP}+t\vv\text{.}\) In other words, the tips (terminal points) of the green vectors trace out the line as \(t\) changes. You can use the slider to change the value of \(t\) for this example line. You should pay attention to how \(\overrightarrow{OP} + \vv t\) changes as you vary \(t\text{.}\) Be sure to see what happens with \(t=0\) as well as for some negative values of \(t\text{.}\)
To summarize, the terminal points of the vectors of the form in (9.5.1) define a linear function \(\vr\) in space of the following form, which is valid in any dimension and gives some geometric insight into lines in higher dimensions.
Lines in the \(xy\)-plane are commonly described using the slope-intercept form \(y=mx + b\text{.}\) The vector form of the line, as described above, is an alternative way to represent lines that has the two advantages. First, in two dimensions, we are able to represent vertical lines, whose slope \(m\) is not defined, using a vertical direction vector. For example, the vertical line described by \(x=2\) could be described in vector form with a direction vector such as \(\vv=\langle 0, 1\rangle\text{,}\) giving \(\vr(t)= \langle 2,0\rangle +t \langle 0,1\rangle\text{.}\) The second advantage of the vector form of a line is that this description of lines works in any dimension whereas the concept of slope of a line does not generalize to three (or more) dimensions.
We will now find a vector form of the line that is the \(y\)-axis in \(\R^3\text{.}\) We need to pick a point on our line and a direction vector for our line. A convenient point would be the origin. We can use \(\vj=\langle 0,1,0 \rangle\) as our direction vector. This gives a vector form of \(\vr_1(t)=\langle 0,0,0 \rangle + t\langle 0,1,0 \rangle\) which can be combined to a single vector form of \(\vr_1(t)=\langle 0,t,0 \rangle\text{.}\) Since the \(y\)-axis consists of points with \(x\)- and \(z\)-coordinates of zero while the \(y\)-coordinate can be any real number, this form should be consistent with your intuition.
We had many other choices we could have made for the initial point and the direction vector used to create the vector form of the \(y\)-axis. We could have selected the point \((0,\pi,0)\) and the direction vector given by \(\langle 0,-\frac{11}{7},0 \rangle\text{.}\) While these are likely not choices you would have made, this point and direction vector still describe the necessary information to give the vector form of the \(y\)-axis. The associated vector form of the \(y\)-axis would be \(\vr_2(t)=\langle 0,\pi,0\rangle + t \langle 0,-\frac{11}{7},0 \rangle=\langle 0,\pi-\frac{11}{7}t,0 \rangle \text{.}\) It is unlikely that anyone else has ever described the \(y\)-axis with this equation but the terminal points of the output vectors for \(\vr_2(t)\) do trace out the entire \(y\)-axis. In the next chapter, we will explore how the context in which we wish to arrive at the vector form of a line may influence the choice of initial point and direction vector.
To look at a line that is less familiar, we now consider the line \(\mathcal{L}_1\) in \(\R^3\) passing through \((1,2,1)\) and in the direction \(\langle 1,1,-\frac{1}{2} \rangle \text{.}\) The corresponding vector form is \(\vr(t)= \langle 1,2,1 \rangle + \langle 1,1,-\frac{1}{2} \rangle t\text{.}\)Figure 9.5.9 shows a plot of this line with a slider that you can use to change the value for \(t\) and examine how the different aspects of this vector form look. Notice that when \(t=1\text{,}\) we have \(\vr(1) = \langle 2,3,1/2\rangle\text{,}\) so the point \(P=(2,3,1/2)\) lies on \(\mathcal{L}_1\text{.}\)
You can verify visually using Figure 9.5.9 that the line \(\mathcal{L}_1\) from the previous part does not intersect the \(y\)-axis. In two dimensions, two lines were parallel if they did not intersect. However, this idea does translate to higher dimensions because it does not seem appropriate to say that \(\mathcal{L}_1\) is parallel to the \(y\)-axis. In general, two lines in \(\R^n\) are parallel if they can be described using parallel direction vectors. (Recall that two vectors are parallel if they are non-zero scalar multiples of each other.) This definition is based in the perspective that parallel lines should move in the same direction, but that direction can be described by any parallel vector.
We formalize the discussion in the last part of the previous activity by giving names to the two nonintersecting configurations that lines in \(\R^n\) for \(n\geq 3\) can have.
Two lines in \(\R^n\) are said to be parallel provided that their direction vectors are parallel. Two lines that do not intersect and whose direction vectors are not parallel are said to be skew.
Let \(P_1 = (1,2,-1)\) and \(P_2 = (-2,1,-2)\) and let \(\mathcal{L}\) be the line in \(\R^3\) through \(P_1\) and \(P_2\text{.}\) Note that Figure 9.5.9 shows a similar example of a line in 3D defined by two points.
Consider the vector equation \(\vs(t) = \langle -5, 0, -3 \rangle + t \langle 6, 2, 2 \rangle.\) What is the direction of the line given by \(\vs(t)\text{?}\) Is this new line parallel to line \(\mathcal{L}\text{?}\)
Do \(\vr(t)\) and \(\vs(t)\) represent the same line, \(\mathcal{L}\text{?}\) Write a couple of sentences to justify why you think \(\vr(t)\) and \(\vs(t)\) do or do not describe the same set of points.
The vector form of a line, \(\vr(t) = \vr_0 + t\vv\) in Equation (9.5.2), describes a line as the set of terminal points of the vectors \(\vr(t)\text{.}\) If \(\vr_0 = \langle x_0, y_0, z_0 \rangle\) and \(\vv = \langle a, b, c \rangle\text{,}\) then we can view this vector form in terms of components and get the following:
\begin{align*}
\vr(t) \amp= \langle x(t), y(t), z(t) \rangle\\
\amp= \langle x_0, y_0, z_0 \rangle + t \langle a, b, c \rangle\\
\amp= \langle x_0+at, y_0+bt, z_0+ct \rangle
\end{align*}
Splitting this vector form into each component equation gives
These equations describe the coordinates of the points on the line separately where each is coordinate is a function of \(t\text{.}\) The variable \(t\) represents an arbitrary scalar and is called a parameter. In particular, we use the following language.
Notice that there are many different parametric equations for the same line. For example, choosing another point \(P\) on the line or another (parallel) direction vector \(\vv\) produces another set of parametric equations. In many physical applications, it is 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 way to walk along the line. There are, of course, many possible ways to walk along the same line, including in opposite directions on at different speeds. In mathematical terms, we say that the parameterization of a line is not unique.
Let \(P_1 = (1,2,-1)\) and \(P_2 = (-2,1,-2)\text{,}\) and let \(\mathcal{L}\) be the line in \(\R^3\) through \(P_1\) and \(P_2\text{.}\) (This is the same line as in Activity 9.5.2.)
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. To find a point of intersection, you can set the coordinate equations of each line equal to each other try to solve for \(t\) and \(s\text{.}\)
Remember that the two lines need to go through the same \((x,y,z)\) point but do not need to have the same parameter value at that point (which is why we used different variable names for the parameters \(t\) and \(s\)).
Before we move on to our discussion of planes in the next section, we will talk about some terminology that may helpful in describing features and measurements later. A line is a one-dimensional, flat graph in three (or \(n\)) dimensions. We say the line is a one-dimensional graph because there is only one direction (forward/backward) to go while staying on the graph. We say that a line is flat because the direction you are allowed to go doesn’t change as you move along the graph. Note that flat does not mean horizontal. A circle would be another example of a one-dimensional graph that can be drawn in two or more dimensional space (because there is only one direction to move along the circle). However, a circle is not flat since the direction you go to stay on the figure changes as you move along the graph. As we will see in the next section, a plane in three dimensional space is an example of a flat, two-dimensional graph. If you were an ant on a plane, you would have two directions (forward/backward or left/right) to move in while staying on the graph. However, the directions that the graph extends do not change when you change locations on the plane.
While lines in \(\R^3\) do not have a slope, like lines in \(\R^2\) 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 \(\vr(t) = \vr_0 + t\vv\) define a linear function \(\vr\) in space through the terminal point of the vector \(\vr_0\) in the direction of the vector \(\vv\text{,}\) tracing out a line in space.
Find the vector and parametric equations for the line through the point P(4, 3, -4) and parallel to the vector \(5\mathbf i - 2\mathbf j + 3\mathbf k\text{.}\)
Let \(P_1 = (1,2,-1)\) and \(P_2 = (-2,1,-2)\text{,}\) and let \(\mathcal{L}\) be the line in \(\R^3\) through \(P_1\) and \(P_2\) as in Activity 9.5.2.
What value of the parameter \(t\) makes \((x(t), y(t), z(t)) = P_1\text{?}\) What value of \(t\) makes \((x(t), y(t), z(t)) = P_2\text{?}\)
Now, consider a segment that lies on a different line: parameterize the segment that connects point \(R=(4,-2,7)\) to \(Q=(-11,4,27)\) in such a way that \(t = 0\) corresponds to point \(Q\text{,}\) while \(t = 2\) corresponds to \(R\text{.}\)
This exercise explores key relationships between a pair of lines. Consider the following two lines: one with parametric equations \(x(s) = 4-2s\text{,}\)\(y(s) = -2 + s\text{,}\)\(z(s) = 1 + 3s\text{,}\) and the other being the line through \((-4, 2, 17)\) in the direction \(\vv = \langle -2, 1, 5 \rangle\text{.}\)
Find a direction vector for the first line, which is given in parametric form.
Show that the two lines intersect at a single point by finding the values of \(s\) and \(t\) that result in the same point. Then find the point of intersection.
Find the acute angle formed where the two lines intersect, noting that this angle will be given by the acute angle between their respective direction vectors.