2.5 Equations of Lines and Planes in Space

Equations for a Line in Space

Let’s first explore what it means for two vectors to be parallel. Recall that parallel vectors must have the same or opposite directions. If two nonzero vectors, $\text{u}$ and $\text{v},$ are parallel, we claim there must be a scalar, $k,$ such that $\text{u}=k\text{v}.$ If $\text{u}$ and $\text{v}$ have the same direction, simply choose $k=\frac{\Vert \text{u}\Vert }{\Vert \text{v}\Vert }.$ If $\text{u}$ and $\text{v}$ have opposite directions, choose $k=-\frac{\Vert \text{u}\Vert }{\Vert \text{v}\Vert }.$ Note that the converse holds as well. If $\text{u}=k\text{v}$ for some scalar $\text{k},$ then either $\text{u}$ and $\text{v}$ have the same direction $\left(k>0\right)$ or opposite directions $\left(k<0\right),$ so $\text{u}$ and $\text{v}$ are parallel. Therefore, two nonzero vectors $\text{u}$ and $\text{v}$ are parallel if and only if $\text{u}=k\text{v}$ for some scalar $\text{k}.$ By convention, the zero vector $0$ is considered to be parallel to all vectors.

As in two dimensions, we can describe a line in space using a point on the line and the direction of the line, or a parallel vector, which we call the direction vector (Figure 2.63). Let $L$ be a line in space passing through point $P\left(x_0,y_0,z_0\right).$ Let $\text{v}=\langle a,b,c\rangle$ be a vector parallel to $L.$ Then, for any point $Q\left(x,y,z\right)$ on line $L$, we know that $\overrightarrow{PQ}$ is parallel to $\text{v}.$ Thus, as we just discussed, there is a scalar, $t,$ such that $\overrightarrow{PQ}=t\text{v},$ which gives

Figure 2.63 Vector v is the direction vector for $\overrightarrow{PQ}.$

Using vector operations, we can rewrite Equation 2.11 as

Setting $\text{r}=\langle x,y,z\rangle$ and ${\text{r}}_0=\langle x_0,y_0,z_0\rangle ,$ we now have the vector equation of a line:

Equating components, Equation 2.11 shows that the following equations are simultaneously true: $x-x_0=ta,$ $y-y_0=tb,$ and $z-z_0=tc.$ If we solve each of these equations for the component variables $x,y,\text{and}z,$ we get a set of equations in which each variable is defined in terms of the parameter t and that, together, describe the line. This set of three equations forms a set of parametric equations of a line:

If we solve each of the equations for $t$ assuming $a,b,\text{and}c$ are nonzero, we get a different description of the same line:

Because each expression equals t, they all have the same value. We can set them equal to each other to create symmetric equations of a line:

We summarize the results in the following theorem.

The parametric equations of a line are not unique. Using a different parallel vector or a different point on the line leads to a different, equivalent representation. Each set of parametric equations leads to a related set of symmetric equations, so it follows that a symmetric equation of a line is not unique either.

Sometimes we don’t want the equation of a whole line, just a line segment. In this case, we limit the values of our parameter $t.$ For example, let $P\left(x_0,y_0,z_0\right)$ and $Q\left(x_1,y_1,z_1\right)$ be points on a line, and let $\text{p}=\langle x_0,y_0,z_0\rangle$ and $\text{q}=\langle x_1,y_1,z_1\rangle$ be the associated position vectors. In addition, let $\text{r}=\langle x,y,z\rangle .$ We want to find a vector equation for the line segment between $P$ and $Q.$ Using $P$ as our known point on the line, and $\overrightarrow{PQ}=\langle x_1-x_0,y_1-y_0,z_1-z_0\rangle$ as the direction vector equation, Equation 2.12 gives

Using properties of vectors, then

Thus, the vector equation of the line passing through $P$ and $Q$ is

Remember that we didn’t want the equation of the whole line, just the line segment between $P$ and $Q.$ Notice that when $t=0,$ we have $r=p,$ and when $t=1,$ we have $r=q.$ Therefore, the vector equation of the line segment between $P$ and $Q$ is

Going back to Equation 2.12, we can also find parametric equations for this line segment. We have

Then, the parametric equations are

Distance between a Point and a Line

We already know how to calculate the distance between two points in space. We now expand this definition to describe the distance between a point and a line in space. Several real-world contexts exist when it is important to be able to calculate these distances. When building a home, for example, builders must consider “setback” requirements, when structures or fixtures have to be a certain distance from the property line. Air travel offers another example. Airlines are concerned about the distances between populated areas and proposed flight paths.

Let $L$ be a line in the plane and let $M$ be any point not on the line. Then, we define distance $d$ from $M$ to $L$ as the length of line segment $\stackrel{—}{MP},$ where $P$ is a point on $L$ such that $\stackrel{—}{MP}$ is perpendicular to $L$ (Figure 2.64).

Figure 2.64 The distance from point $M$ to line $L$ is the length of $\stackrel{—}{MP}.$

When we’re looking for the distance between a line and a point in space, Figure 2.64 still applies. We still define the distance as the length of the perpendicular line segment connecting the point to the line. In space, however, there is no clear way to know which point on the line creates such a perpendicular line segment, so we select an arbitrary point on the line and use properties of vectors to calculate the distance. Therefore, let $P$ be an arbitrary point on line $L$ and let $\text{v}$ be a direction vector for $L$ (Figure 2.65).

Figure 2.65 Vectors $\overrightarrow{PM}$ and v form two sides of a parallelogram with base $\Vert \text{v}\Vert$ and height $d,$ which is the distance between a line and a point in space.

By Area of a Parallelogram, vectors $\overrightarrow{PM}$ and $\text{v}$ form two sides of a parallelogram with area $\Vert \overrightarrow{PM}\times \text{v}\Vert .$ Using a formula from geometry, the area of this parallelogram can also be calculated as the product of its base and height:

We can use this formula to find a general formula for the distance between a line in space and any point not on the line.

Example 2.47

Calculating the Distance from a Point to a Line

Find the distance between the point $M=\left(1,1,3\right)$ and line $\frac{x-3}{4}=\frac{y+1}{2}=z-3.$

Solution

From the symmetric equations of the line, we know that vector $\text{v}=\langle 4,2,1\rangle$ is a direction vector for the line. Setting the symmetric equations of the line equal to zero, we see that point $P\left(3,\mathrm{-1},3\right)$ lies on the line. Then,

$$\overrightarrow{PM}=\langle 1-3,1-(\mathrm{-1}),3-3\rangle =\langle \mathrm{-2},2,0\rangle .$$

To calculate the distance, we need to find $\overrightarrow{PM}\times \text{v}\text{:}$

$$\begin{array}{cc}\hfill \overrightarrow{PM}\times \text{v}& =\left|\begin{array}{ccc}\hfill \text{i}& \hfill \text{j}\hfill & \hfill \text{k}\hfill \\ \hfill \mathrm{-2}& \hfill 2\hfill & \hfill 0\hfill \\ \hfill 4& \hfill 2\hfill & \hfill 1\hfill \end{array}\right|\hfill \\ & =\left(2-0\right)\text{i}-\left(\mathrm{-2}-0\right)\text{j}+\left(\mathrm{-4}-8\right)\text{k}\hfill \\ & =2\text{i}+2\text{j}-12\text{k}.\hfill \end{array}$$

Therefore, the distance between the point and the line is (Figure 2.66)

$$\begin{array}{cc}\hfill d& =\frac{\Vert \overrightarrow{PM}\times \text{v}\Vert }{\Vert \text{v}\Vert }\hfill \\ & =\frac{\sqrt{2^2+2^2+{12}^2}}{\sqrt{4^2+2^2+1^2}}\hfill \\ & =\frac{2\sqrt{38}}{\sqrt{21}}.\hfill \end{array}$$

Figure 2.66 Point $\left(1,1,3\right)$ is approximately $2.7$ units from the line with symmetric equations $\frac{x-3}{4}=\frac{y+1}{2}=z-3.$

Relationships between Lines

Given two lines in the two-dimensional plane, the lines are equal, they are parallel but not equal, or they intersect in a single point. In three dimensions, a fourth case is possible. If two lines in space are not parallel, but do not intersect, then the lines are said to be skew lines (Figure 2.67).

Figure 2.67 In three dimensions, it is possible that two lines do not cross, even when they have different directions.

To classify lines as parallel but not equal, equal, intersecting, or skew, we need to know two things: whether the direction vectors are parallel and whether the lines share a point (Figure 2.68).

Figure 2.68 Determine the relationship between two lines based on whether their direction vectors are parallel and whether they share a point.

Equations for a Plane

We know that a line is determined by two points. In other words, for any two distinct points, there is exactly one line that passes through those points, whether in two dimensions or three. Similarly, given any three points that do not all lie on the same line, there is a unique plane that passes through these points. Just as a line is determined by two points, a plane is determined by three.

This may be the simplest way to characterize a plane, but we can use other descriptions as well. For example, given two distinct, intersecting lines, there is exactly one plane containing both lines. A plane is also determined by a line and any point that does not lie on the line. These characterizations arise naturally from the idea that a plane is determined by three points. Perhaps the most surprising characterization of a plane is actually the most useful.

Imagine a pair of orthogonal vectors that share an initial point. Visualize grabbing one of the vectors and twisting it. As you twist, the other vector spins around and sweeps out a plane. Here, we describe that concept mathematically. Let $\text{n}=\langle a,b,c\rangle$ be a vector and $P=\left(x_0,y_0,z_0\right)$ be a point. Then the set of all points $Q=\left(x,y,z\right)$ such that $\overrightarrow{PQ}$ is orthogonal to $\text{n}$ forms a plane (Figure 2.69). We say that $\text{n}$ is a normal vector, or perpendicular to the plane. Remember, the dot product of orthogonal vectors is zero. This fact generates the vector equation of a plane: $\text{n}·\overrightarrow{PQ}=0.$ Rewriting this equation provides additional ways to describe the plane:

Figure 2.69 Given a point $P$ and vector $\text{n},$ the set of all points $Q$ with $\overrightarrow{PQ}$ orthogonal to $\text{n}$ forms a plane.

As described earlier in this section, any three points that do not all lie on the same line determine a plane. Given three such points, we can find an equation for the plane containing these points.