Math 344: Calculus III
12.5 Lines and Planes
12.5.1 
Span
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Lines and a Planes are examples of subspaces and affine spaces of a vector space like $\mathbb{R}^n$. They are the graphs of linear or affine functions and our study of the graphs of functions always begins with these types of functions. A line or plane that passes through the origin of the vector space; that is of the zero vector lies in the line or plane, then it is a subspace of the vector space. If not, then we think of the line or plane as a subspace that has been translated away from the origin. This is an affine space of a vector space. We can discuss linear functions and subspaces first.
A linear subspace; that is a line or plane, translated away from the origin is an affine space. A linear equation that has been translated away from the origin is called an affine function. This will make the equations of lines and planes easier to understand. First we will study linear equations and their graphs. Then we will study affine equations.
12.5.2 Lines and Linear Equations¶
Definition¶
A line is the span of one vector. The graph of a linear equation
$$ \def\ihat{\mathbf{\hat{\mmlToken{mi}[mathvariant="bold"]{ı}}}} \def\jhat{\mathbf{\hat{\mmlToken{mi}[mathvariant="bold"]{ȷ}}}} \def\khat{\mathbf{\hat{k}}} \mathbf{r}(t) = t\mathbf{m} $$
is the set of all possible multiples of our slope vector, or direction vector, $\mathbf{m}$. The image of such a linear equation $\mathbf{r}(t) = t\mathbf{m}$ is one dimensional because the domain is one dimensional. Such a function $\mathbf{r}:\mathbb{R}\rightarrow\mathbb{R}^n$ has one input and several outputs. We usually consider the output to be a vector in the codomain $\mathbb{R}^n$ and graph the image in $\mathbb{R}^n$.
This is true in three dimensional space and in any Euclidean vector space $\mathbb{R}^n$.
Example 1¶
Find the equation of the line that passes through the point $P(4,4,1)$ and the origin.
The direction vector is $\mathbf{m} = \overrightarrow{OP} = P - O = \langle 4,4,1 \rangle$. The linear equation is given by
$$
\mathbf{r}(t) = t\mathbf{m} = t\langle 4,4,1\rangle = \langle 4t, 4t, t \rangle = \begin{bmatrix} 4t \\ 4t \\ t \end{bmatrix}
$$
Exercise 1¶
Find the equation of the line that passes through the vector $\mathbf{p} = 5\ihat + \jhat + 4\khat$
The direction vector is $\mathbf{p} = \langle 5, 1, 4 \rangle$. The linear equation is given by
Check Your Work
$$ \mathbf{r}(t) = \mathbf{p}t = \langle 5, 1, 4 \rangle t = \begin{bmatrix} 5t \\ t \\ 4t \end{bmatrix} $$12.5.3 Lines and Affine Equations¶
We translate a linear equation away from the origin using a translation vector $\mathbf{r}_0$ that plays the role that the $y$-intercept does for the equation of a line in slope-intercept form
$$ y = mx + b $$
The vertical shift introduced in the equation shifts the graph up or down $b$. Likewise the translation vector shifts the line $\mathbf{r}_0$.
Example 2¶
Determine the vector equation of the line that passes through the point $\langle 6, 5, 0 \rangle$ and is parallel to the vector $\ihat - 4\jhat + 3\khat$.
$$
\mathbf{r}(t) = \mathbf{m}t + \mathbf{r}_0 = \langle 1, -4, 3 \rangle t + \langle 6, 5, 0 \rangle
$$
Exercise 2¶
Find the equation of the line that passes through the points $A(-2, 2, 1)$ and $B
(0, 5, 6)$.
Check Your Work
$$ \begin{align*} \mathbf{m} &= \overrightarrow{AB} = \langle 0, 5, 6 \rangle - \langle -2, 2, 1 \rangle = \langle 2, 3, 5 \rangle \\ \\ \mathbf{r}_0 &= \langle -2, 2, 1 \rangle \\ \\ \mathbf{r}(t) &= \mathbf{m}t + \mathbf{r}_0 = \langle 2, 3, 5 \rangle t + \langle -2, 2, 1\rangle \end{align*} $$12.5.4 Parametric Form¶
Given an affine equation
$$
\mathbf{r}(t) = \mathbf{m}t + \mathbf{r}_0
$$
we can represent the dependent variable (vector) by the vector $\langle x, y, z \rangle$, and the independent variable with our variable $t$. Our equation becomes
$$
\langle x, y, z \rangle = \mathbf{r}(t) = \mathbf{m}t + \mathbf{r_0}
$$
Similarly we can represent the translation vector $\mathbf{r}_0 = \langle x_0, y_0, z_0 \rangle$. We usually represent the direction vector $\mathbf{m} = \langle a, b, c \rangle$. Hence our equation becomes
$$
\begin{align*}
\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \langle x, y, z \rangle &= \mathbf{r}(t) = \mathbf{m}t + \mathbf{r_0} \\
\\
&= \langle a, b, c \rangle t + \langle x_0, y_0, z_0 \rangle \\
\\
&= \langle at + x_0, bt + y_0, ct + z_0 \rangle \\
\\
&= \begin{bmatrix} at + x_0 \\ bt + y_0 \\ ct + z_0 \end{bmatrix} = \begin{bmatrix} x(t) \\ y(t) \\ z(t) \end{bmatrix}
\end{align*}
$$
For these two vectors to be equal, each of their coordinates must be equal. In this case we call the independent variable a parameter.
Definition¶
The parametric equations for a line passing through the point $(x_0,y_0,z_0)$ and parallel to the direction vector $\langle a, b, c \rangle$ are given by the system of equations
$$ \begin{align*} x &= x(t) = at + x_0 \\ y &= y(t) = bt + y_0 \\ x &= z(t) = ct + z_0 \\ \end{align*} $$
The real-valued functions $x(t)$, $y(t)$, and $z(t)$ are called the coordinate functions for function $\mathbf{r}$. The slope of each coordinate function, the scalars $a$, $b$, and $c$; that is, the coordinates of the direction vector $\langle a, b, c\rangle$ are called the direction numbers.
Example 3¶
Find parametric equations for the line that passes through the point $(0,5,6)$ and is parallel to the vector $-2\ihat + 2\jhat + \khat$.
$$\begin{align*}
\mathbf{m} &= \langle -2, 2, 1 \rangle \\
\mathbf{r}_0 &= \langle 0, 5, 6 \rangle \\
\mathbf{r}(t) &= \mathbf{m}t + \mathbf{r}_0 = \langle -2, 2, 1 \rangle t + \langle 0, 5, 6 \rangle \\
\\
x(t) &= -2t \\
y(t) &= 2t + 5 \\
z(t) &= t + 6
\end{align*}$$
Exercise 3¶
Find the parametric equations of the line that passes through the points $A(2,5,7)$ and $B(-2,5,-2)$.
Check Your Work
$$ \begin{align*} \mathbf{m} &= \overrightarrow{AB} = \langle -2, 5, -2 \rangle - \langle 2, 5, 7 \rangle = \langle -2 - 2, 5 - 5, -2 - 7 \rangle = \langle -4, 0, -9 \rangle \\ \mathbf{r}_0 &= \langle 2, 5, 7 \rangle \\ \mathbf{r}(t) &= \langle -4, 0, -9 \rangle t + \langle 2, 5, 7 \rangle \\ \\ x(t) &= -4t + 2 \\ y(t) &= 5 \\ z(t) &= -9t + 7 \end{align*} $$12.5.5 Symmetric Equations¶
If the direction numbers of a line with parametric equations
$$ \begin{align*}
x(t) &= at + x_0 \\
y(t) &= bt + y_0 \\
z(t) &= ct + z_0
\end{align*}$$
are all nonzero, then we can solve each equation for the independent variable $t$
$$ t = \dfrac{x - x_0}{a} \qquad\qquad t = \dfrac{y - y_0}{b} \qquad\qquad t = \dfrac{z - z_0}{c} $$
Definition¶
The symmetric equations of a line with nonzero direction numbers is given by
$$ \dfrac{x - x_0}{a} = \dfrac{y - y_0}{b} = \dfrac{z - z_0}{c} $$
If one or more of the direction numbers is zero, then we have a constant function for the coordinate function. For example
$$ \begin{align*} x &= 8t - 1 \\ y &= 6 \\ z &= -t - 2 \\ \end{align*} $$
This yields the symmetric equations
$$ y = 6, \qquad \dfrac{x+1}{8} = \dfrac{z + 2}{-1} $$
Example 4¶
Find the parametric equations for the line that passes through the point $P(-1,4,2)$ and is perpendicular to the line $\langle 3, 7 -1 \rangle t + \langle 0, 6, 3 \rangle$.
The direction vector for the specified line is $\mathbf{m}_1 = \langle 3, 7, -1 \rangle$. We need a direction vector perpendicular to $\mathbf{m}_1$. Thus we need a direction vector $\mathbf{m}_2 = \langle a, b, c \rangle$ such that
$$ \mathbf{m}_1\cdot\mathbf{m}_2 = 0 $$
That is
$$ \langle 3, 7 -1 \rangle\cdot\langle a, b, c \rangle = 3a + 7b - c = 0 $$
This is only one equation and three unknowns. There are infinitely many solutions for $a$, $b$, and $c$. We only want one so to make arithmetic easy let us set $b=0$ because multiplying by seven is annoying. We have
$$ 3a - c = 0 $$
If we set $a = 1$, then $c = 3$! So a direction vector perpendicular to our given line is the vector $\langle 1, 0, 3 \rangle$. Now we can solve for the symmetric equations.
$$ \langle x, y, z \rangle = \mathbf{r}(t) = \mathbf{m}t + \mathbf{r}_0 = \langle 1, 0, 3 \rangle t + \langle -1, 4, 2\rangle $$
$$ y = 4,\qquad x + 1 = \dfrac{z - 2}{3} $$
Exercise 4¶
Determine the symmetric equations for the line parallel to the line $\langle 2, 6, 8 \rangle t + \langle 3, -4, -4 \rangle$ passing through the point $A(5,5,0)$.
Check Your Work
$$ \begin{align*} \mathbf{m} &= \langle 2, 6, 8 \rangle \\ \mathbf{r}_0 &= \langle 5, 5, 0 \rangle \\ \mathbf{r}(t) &= \langle 2, 6, 8 \rangle t + \langle 5, 5, 0 \rangle \\ \\ \dfrac{x - 5}{2} &= \dfrac{y - 5}{6} = \dfrac{z}{8} \end{align*} $$12.5.6 Line Segments¶
The equation of a line segment from vector $\mathbf{r}_0\in\mathbb{R}^n$ to vector $\mathbf{r}_1\in\mathbb{R}^n$, has direction vector
$$ \mathbf{m} = \mathbf{r}_1 - \mathbf{r}_0 $$
and passes through the point $\mathbf{r}_0$. However we do not want the entire line, we want only the line segment from $\mathbf{r}_0$ to $\mathbf{r}_1$. Thus we define our function
$$ \mathbf{r}\ :\ [0,1]\rightarrow\mathbb{R}^n $$
by
$$\mathbf{r}(t) = \mathbf{m}t + \mathbf{r}_0 = \left(\mathbf{r}_1 - \mathbf{r}_0\right)t + \mathbf{r}_0 = \mathbf{r}_1t + \mathbf{r}_0(1-t) $$
This equation is a parameterization of the line segment because there a infinitely many ways to create an equation whose graph is our line segment from $\mathbf{r}_0$ to $\mathbf{r}_1$. This is just one of the simplest.
12.5.7 Skew Lines¶
In two dimensional Euclidean space $\mathbb{R}^2$, lines must either be parallel or intersect at exactly one point. Parallel lines usually never intersect, but in some cases it turns out that both lines are really the same line so the intersect at all their points. In larger dimensional Euclidean spaces, non-parallel lines may still pass by each other without ever intersecting. These lines are called skew.
Example 5¶
Show that the lines $L_1$ and $L_2$ with parametric equations
$$ \begin{align*} L_1&: &\ x &= -3s + 4 &\qquad y &= 3s &\qquad z &= 2s + 2 \\ \\ L_2&: &\ x &= 6t - 2 &\qquad y &= 3 &\qquad z &= 2t - 1 \end{align*} $$
are skew. The direction vectors are $\mathbf{m}_1 = \langle -3, 3, 2 \rangle$ and $\mathbf{m}_2 = \langle 6, 0, 3 \rangle$. The cosine of the angle $\theta$ between them is given by
$$ \cos(\theta) = \dfrac{\mathbf{m}_1\cdot\mathbf{m}_2}{|\mathbf{m}_1|\,|\mathbf{m}_2|} = \dfrac{\langle -3, 3, 2 \rangle\cdot\langle 6, 0, 3 \rangle}{|\langle -3, 3, 2 \rangle|\,|\langle 6, 0, 3 \rangle|} = \dfrac{(-3)(6) + (3)(0) + (2)(3)}{\sqrt{9 + 9 + 4}\sqrt{36 + 0 + 9}} = \dfrac{-12}{\sqrt{22}\sqrt{45}} \neq \pm 1 $$
This means that the angle $\theta$ is not a multiple of $\pi$ so these direction vectors are not colinear (parallel). We need to show that the lines do not intersect. To accomplish this we try solve for values of $s$ and $t$ that result in the same vector $\langle x, y, z\rangle$ in both sets of equations. This gives us the system of equations
$$ \begin{align*} -3s + 4 &= 6t - 2 \\ 3s &= 3 \\ 2s + 2 &= 2t - 1 \end{align*} $$
Inspecting this system of equations one obtains $s = 1$. Substituting $1$ for $s$ results in the equivalent system of equations
$$ \begin{align*} 1 &= 6t - 2 \\ 4 &= 2t - 1 \\ \end{align*} $$
Solving each of these two equations results in
$$ \begin{align*} t &= \frac{1}{2} \\ \\ t &= \frac{5}{2} \end{align*} $$
These values for $t$ are different so there is no single set of values $(s,t)$ that solves all three equations in our system. Thus the lines do not intersect.
Since the lines $L_1$ and $L_2$ are not parallel and do not intersect, they are skew.
12.5.8 The Vector Equation of a Plane¶
A plane is the span of two nonparallel vectors.
$$ \left\{ \mathbf{r} = a\mathbf{u} + b\mathbf{v}\ :\ a,b\in\mathbb{R}\,\right\} $$
If vector $\mathbf{n}$ is perpendicular to both vectors $\mathbf{u}$ and $\mathbf{v}$, then $\mathbf{n}\cdot\mathbf{u} = \mathbf{n}\cdot\mathbf{v} = 0$. This means that for every vector on the plane we have
$$ \mathbf{n}\cdot\left(a\mathbf{u} + b\mathbf{v}\right) = a\left(\mathbf{n}\cdot\mathbf{u}\right) + b\left(\mathbf{n}\cdot\mathbf{v}\right) = a\left(0\right) + b\left(0\right) = 0 $$
We know such a vector exists because $\mathbf{n} = \mathbf{u}\times\mathbf{v}$ is a vector orthogonal to both vectors $\mathbf{u}$ and $\mathbf{v}$. Furthermore $\mathbf{n}\neq\mathbf{0}$ when they are not parallel. We call such a vector $\mathbf{n}$ a normal vector.
This gives us an algebraic equation of a plane:
Equation of a Plane¶
For nonzero vector $\mathbf{n}$ the graph of the equation
$$ \mathbf{n}\cdot\mathbf{r} = 0 $$
is a plane passing through the origin. This is a linear equation and the graph is a subspace of $\mathbb{R}^n$. Vector $\mathbf{n}$ is perpendicular to every vector on the plane and is called a normal vector for the plane.
We can translate every vector on the plane by adding vector $\mathbf{r}_0$ just as we did for a line.
Vector Equation of a Plane¶
For nonzero vector $\mathbf{n}\in\mathbb{R}^n$ and $\mathbf{r}_0\in\mathbb{R}^n$ the graph of the equation
$$ \mathbf{n}\cdot\left(\mathbf{r}-\mathbf{r}_0\right) = 0 $$
is a plane in $\mathbb{R}^n$ perpendicular to its normal vector $\mathbf{n}$ and passing through the vector $\mathbf{r}_0$.
The equation of a plane given here is not a parameterization. Every vector that satisfies the vector equation $\mathbf{n}\cdot\left(\mathbf{r} - \mathbf{r}_0\right) = 0$ is a vector on the plane. The first equation
$$ \mathbf{r}(s,t) = s\,\mathbf{u} + t\,\mathbf{v} $$
is a parameterization of the plane. The graph is two dimensional because this function $\mathbf{r}\,:\,\mathbb{R}^2\rightarrow\mathbb{R}^n$ has two inputs.
12.5.9 The Scalar Equation of a Plane¶
As with the line, if we define $\mathbf{n} = \langle a, b, c \rangle$, $\mathbf{r} = \langle x, y, z \rangle$, and $\mathbf{r}_0 = \langle x_0, y_0, z_0 \rangle$, then we obtain
$$ \begin{align*} \mathbf{n}\cdot\left(\mathbf{r} - \mathbf{r}_0\right) &= 0 \\ \\ \langle a, b, c \rangle\cdot\left(\langle x, y, z \rangle - \langle x_0, y_0, z_0 \rangle\right) &= 0 \\ \\ \langle a, b, c \rangle\cdot\langle x-x_0, y-y_0, z-z_0\rangle &= 0 \\ \\ a(x-x_0) + b(y-y_0) + c(z-z_0) &= 0 \\ \\ ax + by + cz &= d = ax_0 + by_0 + cz_0 \end{align*} $$
Scalar Equation of a Plane¶
For a normal vector $\,\mathbf{n} = \langle a,b,c \rangle$, the equation
$$ ax + by + cz = d $$
is the scalar equation of a plane. If $d=0$, then $\mathbf{r}_0=\mathbf{0}$, the plane passes through the origin, and the equation is linear. If $d\neq 0$, then the plane does not include the origin, and the equation of the plane is affine.
12.5.10 Distances¶
Consider a plane with equations $ax + by + cz = d$ and a point $A(x_0, y_0, z_0)$ not on that plane. What is the distance between the point and the plane? By distance we mean perpendicular distance. That is the shortest distance between the point and the plane; the length of the line segment from point $A$ to the a point on the plane $P$ so that vector $\overrightarrow{PA}$ is perpendicular to the plane.
One may think of $\mathbf{P}$ as a vector on the plane or a point on the plane. In any case the vector $\overrightarrow{PA}$ must be perpendicular to the plane in order that its length be our distance between the point and the plane. Thus $\overrightarrow{PA}$ and normal vector $\mathbf{n}$ are parallel. We need not ever compute the coordinates of $\mathbf{P}$. Constructing the vector from $\mathbf{r}_0$ on the plane to point $A$, the projection of vector $\overrightarrow{\mathbf{r}_0A}$ onto the normal vector $\mathbf{n}$ is vector $\overrightarrow{PA}$. Consider the right rectangle in the illustration. Hence the length of this vector is the distance from point $A$ to the plane. The length of a projection vector is the component. Thus the distance from the point to the plane is given by
$$ d = \left|\overrightarrow{PA}\right| = \text{comp}_{\mathbf{n}}\overrightarrow{\mathbf{r}_0A} = \dfrac{\overrightarrow{\mathbf{r}_0A}\cdot\mathbf{n}}{|\mathbf{n}|} $$
Some textbooks reduce this using vector algebra to a compact formula, however we will use this picture repeatedly to solve problems throughout this course. This illustration is more useful than memorizing yet another formula.
Example 6¶
Determine the distance from the plane with equation $x - 3y + 8z = 3$ and the point $A(6,5,3)$
Unfortunately the scalar equation of a plane does not give us a point $\mathbf{r}_0$ on the plane. To obtain a point on the plane, set $z=0$ so that we don't have to divide by eight anywhere. Now we have $x - 3y = 3$, or $x = 3y + 3$. Letting $y=1$ gives us $x=6$. So the point $R_0(6,1,0)$ is a point on this plane. We can also think of this as the translation vector $\mathbf{r}_0 = \langle 6, 1, 0 \rangle$.
We are now in a position to use the picture above. The vector $\overrightarrow{R_0A} = \langle 6, 5, 3 \rangle - \langle 6, 1, 0 \rangle = \langle 0, 4, 3 \rangle$. The component of this vector onto the normal vector $\mathbf{n} = \langle 1, -3, 8 \rangle$ yields
$$ d = \text{comp}_{\mathbf{n}}\langle 0, 4, 3 \rangle = \dfrac{(0)(1) + (4)(-3) + (3)(8)}{\sqrt{1 + 9 + 64}} = \dfrac{12}{\sqrt{74}} $$
12.5.11 The Distance Between Parallel Planes¶
Any two parallel planes must have a the same normal vector. Suppose we have two planes
$$ \begin{align*} \mathbf{n}\cdot\left(\mathbf{r} - \mathbf{r}_0\right) = 0 \\ \\ \mathbf{n}\cdot\left(\mathbf{r} - \mathbf{r}_1\right) = 0 \\ \end{align*} $$
We can compute the distance from point $\mathbf{r}_0$ to plane $\mathbf{n}\cdot\left(\mathbf{r} - \mathbf{r}_1\right) = 0$, or we can compute the distance from point $\mathbf{r}_1$ to plane $\mathbf{n}\cdot\left(\mathbf{r} - \mathbf{r}_0\right) = 0$. In either case we compute the component of vector $\overrightarrow{\mathbf{r}_0\mathbf{r}_1}$ onto $\mathbf{n}$.
12.5.12 The Distance Between Two Skew Lines¶
Consider two lines with vector equations
$$ \begin{align*} \mathbf{r} &= \mathbf{m}_0t + \mathbf{r}_0 \\ \\ \mathbf{r} &= \mathbf{m}_1t + \mathbf{r}_1 \end{align*} $$
Since they are skew we can imagine them lying in two parallel planes. The distance between the lines is also the distance between the two planes. We need a normal vector; that is a vector perpendicular to both planes. Such a vector is $\mathbf{n} = \mathbf{m}_0\times\mathbf{m}_1$. Now we have two planes
$$ \begin{align*} \left(\mathbf{m}_0\times\mathbf{m}_1\right)\cdot\left(\mathbf{r} - \mathbf{r}_0\right) = 0 \\ \\ \left(\mathbf{m}_0\times\mathbf{m}_1\right)\cdot\left(\mathbf{r} - \mathbf{r}_1\right) = 0 \\ \end{align*} $$
Just as in 12.5.11 we require the distance from point $\mathbf{r}_0$ and plane $\left(\mathbf{m}_0\times\mathbf{m}_1\right)\cdot\left(\mathbf{r} - \mathbf{r}_1\right) = 0$. That is we want the component of $\overrightarrow{\mathbf{r}_0\mathbf{r}_1}$ onto $\mathbf{n} = \mathbf{m}_0\times\mathbf{m}_1$. One should draw pictures and compute these distances until the picture above and the applications to all three distance problems becomes familiar. In this way we can avoid several unique formulas in this chapter, and several more later in the course.
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