# 2.5 Equations of lines and planes in space  (Page 2/19)

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## Parametric and symmetric equations of a line

A line $L$ parallel to vector $\text{v}=⟨a,b,c⟩$ and passing through point $P\left({x}_{0},{y}_{0},{z}_{0}\right)$ can be described by the following parametric equations:

$x={x}_{0}+ta,y={y}_{0}+tb,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}z={z}_{0}+tc.$

If the constants $a,b,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}c$ are all nonzero, then $L$ can be described by the symmetric equation of the line:

$\frac{x-{x}_{0}}{a}=\frac{y-{y}_{0}}{b}=\frac{z-{z}_{0}}{c}.$

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.

## Equations of a line in space

Find parametric and symmetric equations of the line passing through points $\left(1,4,-2\right)$ and $\left(-3,5,0\right).$

First, identify a vector parallel to the line:

$\text{v}=⟨-3-1,5-4,0-\left(-2\right)⟩=⟨-4,1,2⟩.$

Use either of the given points on the line to complete the parametric equations:

$x=1-4t,y=4+t,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}z=-2+2t.$

Solve each equation for $t$ to create the symmetric equation of the line:

$\frac{x-1}{-4}=y-4=\frac{z+2}{2}.$

Find parametric and symmetric equations of the line passing through points $\left(1,-3,2\right)$ and $\left(5,-2,8\right).$

Possible set of parametric equations: $x=1+4t,y=-3+t,z=2+6t;$

related set of symmetric equations: $\frac{x-1}{4}=y+3=\frac{z-2}{6}$

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 $\text{p}=⟨{x}_{0},{y}_{0},{z}_{0}⟩$ and $\text{q}=⟨{x}_{1},{y}_{1},{z}_{1}⟩$ be the associated position vectors. In addition, let $\text{r}=⟨x,y,z⟩.$ 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 $\stackrel{\to }{PQ}=⟨{x}_{1}-{x}_{0},{y}_{1}-{y}_{0},{z}_{1}-{z}_{0}⟩$ as the direction vector equation, [link] gives

$\text{r}=\text{p}+t\left(\stackrel{\to }{PQ}\right).$

Using properties of vectors, then

$\begin{array}{cc}\hfill \text{r}& =\text{p}+t\left(\stackrel{\to }{PQ}\right)\hfill \\ & =⟨{x}_{0},{y}_{0},{z}_{0}⟩+t⟨{x}_{1}-{x}_{0},{y}_{1}-{y}_{0},{z}_{1}-{z}_{0}⟩\hfill \\ & =⟨{x}_{0},{y}_{0},{z}_{0}⟩+t\left(⟨{x}_{1},{y}_{1},{z}_{1}⟩-⟨{x}_{0},{y}_{0},{z}_{0}⟩\right)\hfill \\ & =⟨{x}_{0},{y}_{0},{z}_{0}⟩+t⟨{x}_{1},{y}_{1},{z}_{1}⟩-t⟨{x}_{0},{y}_{0},{z}_{0}⟩\hfill \\ & =\left(1-t\right)⟨{x}_{0},{y}_{0},{z}_{0}⟩+t⟨{x}_{1},{y}_{1},{z}_{1}⟩\hfill \\ & =\left(1-t\right)\text{p}+t\text{q}.\hfill \end{array}$

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

$\text{r}=\left(1-t\right)\text{p}+t\text{q}.$

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

$\text{r}=\left(1-t\right)\text{p}+t\text{q},0\le t\le 1.$

Going back to [link] , we can also find parametric equations for this line segment. We have

$\begin{array}{ccc}\hfill \text{r}& =\hfill & \text{p}+t\left(\stackrel{\to }{PQ}\right)\hfill \\ \hfill ⟨x,y,z⟩& =\hfill & ⟨{x}_{0},{y}_{0},{z}_{0}⟩+t⟨{x}_{1}-{x}_{0},{y}_{1}-{y}_{0},{z}_{1}-{z}_{0}⟩\hfill \\ & =\hfill & ⟨{x}_{0}+t\left({x}_{1}-{x}_{0}\right),{y}_{0}+t\left({y}_{1}-{y}_{0}\right),{z}_{0}+t\left({z}_{1}-{z}_{0}\right)⟩.\hfill \end{array}$

Then, the parametric equations are

$x={x}_{0}+t\left({x}_{1}-{x}_{0}\right),y={y}_{0}+t\left({y}_{1}-{y}_{0}\right),z={z}_{0}+t\left({z}_{1}-{z}_{0}\right),0\le t\le 1.$

## Parametric equations of a line segment

Find parametric equations of the line segment between the points $P\left(2,1,4\right)$ and $Q\left(3,-1,3\right).$

By [link] , we have

$x={x}_{0}+t\left({x}_{1}-{x}_{0}\right),y={y}_{0}+t\left({y}_{1}-{y}_{0}\right),z={z}_{0}+t\left({z}_{1}-{z}_{0}\right),0\le t\le 1.$

Working with each component separately, we get

$\begin{array}{cc}\hfill x& ={x}_{0}+t\left({x}_{1}-{x}_{0}\right)\hfill \\ & =2+t\left(3-2\right)\hfill \\ & =2+t,\hfill \end{array}$
$\begin{array}{cc}\hfill y& ={y}_{0}+t\left({y}_{1}-{y}_{0}\right)\hfill \\ & =1+t\left(-1-1\right)\hfill \\ & =1-2t,\hfill \end{array}$

and

$\begin{array}{cc}\hfill z& ={z}_{0}+t\left({z}_{1}-{z}_{0}\right)\hfill \\ & =4+t\left(3-4\right)\hfill \\ & =4-t.\hfill \end{array}$

Therefore, the parametric equations for the line segment are

$x=2+t,y=1-2t,z=4-t,0\le t\le 1.$

Find parametric equations of the line segment between points $P\left(-1,3,6\right)$ and $Q\left(-8,2,4\right).$

$x=-1-7t,y=3-t,z=6-2t,0\le t\le 1$

## 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.

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