# 2.5 Equations of lines and planes in space  (Page 3/19)

 Page 3 / 19

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 $L$ ( [link] ).

When we’re looking for the distance between a line and a point in space, [link] 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 $L$ ( [link] ).

By [link] , vectors $\stackrel{\to }{PM}$ and $\text{v}$ form two sides of a parallelogram with area $‖\stackrel{\to }{PM}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}‖.$ Using a formula from geometry, the area of this parallelogram can also be calculated as the product of its base and height:

$‖\stackrel{\to }{PM}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}‖=‖\text{v}‖d.$

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.

## Distance from a point to a line

Let $L$ be a line in space passing through point $P$ with direction vector $\text{v}.$ If $M$ is any point not on $L,$ then the distance from $M$ to $L$ is

$d=\frac{‖\stackrel{\to }{PM}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}‖}{‖\text{v}‖}.$

## Calculating the distance from a point to a line

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

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

$\stackrel{\to }{PM}=⟨1-3,1-\left(-1\right),3-3⟩=⟨-2,2,0⟩.$

To calculate the distance, we need to find $\stackrel{\to }{PM}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}\text{:}$

$\begin{array}{cc}\hfill \stackrel{\to }{PM}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}& =|\begin{array}{ccc}\hfill \text{i}& \hfill \text{j}\hfill & \hfill \text{k}\hfill \\ \hfill -2& \hfill 2\hfill & \hfill 0\hfill \\ \hfill 4& \hfill 2\hfill & \hfill 1\hfill \end{array}|\hfill \\ & =\left(2-0\right)\text{i}-\left(-2-0\right)\text{j}+\left(-4-8\right)\text{k}\hfill \\ & =2\text{i}+2\text{j}-12\text{k}.\hfill \end{array}$

Therefore, the distance between the point and the line is ( [link] )

$\begin{array}{cc}\hfill d& =\frac{‖\stackrel{\to }{PM}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}‖}{‖\text{v}‖}\hfill \\ & =\frac{\sqrt{{2}^{2}+{2}^{2}+{12}^{2}}}{\sqrt{{4}^{2}+{2}^{2}+{1}^{2}}}\hfill \\ & =\frac{2\sqrt{38}}{\sqrt{21}}.\hfill \end{array}$

Find the distance between point $\left(0,3,6\right)$ and the line with parametric equations $x=1-t,y=1+2t,z=5+3t.$

$\sqrt{\frac{10}{7}}$

## 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    ( [link] ).

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 ( [link] ).

## Classifying lines in space

For each pair of lines, determine whether the lines are equal, parallel but not equal, skew, or intersecting.

1. ${L}_{1}:x=2s-1,y=s-1,z=s-4$
${L}_{2}:x=t-3,y=3t+8,z=5-2t$
2. ${L}_{1}\text{:}$ $x=\text{−}y=z$
${L}_{2}:\frac{x-3}{2}=y=z-2$
3. ${L}_{1}:x=6s-1,y=-2s,z=3s+1$
${L}_{2}:\frac{x-4}{6}=\frac{y+3}{-2}=\frac{z-1}{3}$
1. Line ${L}_{1}$ has direction vector ${\text{v}}_{1}=⟨2,1,1⟩;$ line ${L}_{2}$ has direction vector ${\text{v}}_{2}=⟨1,3,-2⟩.$ Because the direction vectors are not parallel vectors, the lines are either intersecting or skew. To determine whether the lines intersect, we see if there is a point, $\left(x,y,z\right),$ that lies on both lines. To find this point, we use the parametric equations to create a system of equalities:
$2s-1=t-3;\phantom{\rule{1em}{0ex}}s-1=3t+8;\phantom{\rule{1em}{0ex}}s-4=5-2t.$

By the first equation, $t=2s+2.$ Substituting into the second equation yields
$\begin{array}{ccc}\hfill s-1& =\hfill & 3\left(2s+2\right)+8\hfill \\ \hfill s-1& =\hfill & 6s+6+8\hfill \\ \hfill 5s& =\hfill & -15\hfill \\ \hfill s& =\hfill & -3.\hfill \end{array}$

Substitution into the third equation, however, yields a contradiction:
$\begin{array}{ccc}\hfill s-4& =\hfill & 5-2\left(2s+2\right)\hfill \\ \hfill s-4& =\hfill & 5-4s-4\hfill \\ \hfill 5s& =\hfill & 5\hfill \\ \hfill s& =\hfill & 1.\hfill \end{array}$

There is no single point that satisfies the parametric equations for ${L}_{1}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{L}_{2}$ simultaneously. These lines do not intersect, so they are skew (see the following figure).
2. Line L 1 has direction vector ${\text{v}}_{1}=⟨1,-1,1⟩$ and passes through the origin, $\left(0,0,0\right).$ Line ${L}_{2}$ has a different direction vector, ${\text{v}}_{2}=⟨2,1,1⟩,$ so these lines are not parallel or equal. Let $r$ represent the parameter for line ${L}_{1}$ and let $s$ represent the parameter for ${L}_{2}\text{:}$
$\begin{array}{cccc}\begin{array}{cc}x\hfill & =r\hfill \\ y\hfill & =\text{−}r\hfill \\ z\hfill & =r\hfill \end{array}\hfill & & & \begin{array}{cc}x\hfill & =2s+3\hfill \\ y\hfill & =s\hfill \\ z\hfill & =s+2.\hfill \end{array}\hfill \end{array}$

Solve the system of equations to find $r=1$ and $s=-1.$ If we need to find the point of intersection, we can substitute these parameters into the original equations to get $\left(1,-1,1\right)$ (see the following figure).
3. Lines ${L}_{1}$ and ${L}_{2}$ have equivalent direction vectors: $\text{v}=⟨6,-2,3⟩.$ These two lines are parallel (see the following figure).

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