Paul's Online Notes
Home / Calculus III / 3-Dimensional Space / Equations of Lines
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Section 12.2 : Equations of Lines

5. Determine the intersection point of the line through the points $$\left( {1, - 2,13} \right)$$ and $$\left( {2,0, - 5} \right)$$ and the line given by $$\vec r\left( t \right) = \left\langle {2 + 4t, - 1 - t,3} \right\rangle$$ or show that they do not intersect.

Show All Steps Hide All Steps

Start Solution

Because we don’t have the equation for the first line that will be the first thing we’ll need to do. The vector between the two points (and hence parallel to the line) is,

$\vec v = \left\langle {1,2, - 18} \right\rangle$

Using the first point listed the equation of the first line is then,

$\vec r\left( t \right) = \left\langle {1, - 2,13} \right\rangle + t\left\langle {1,2, - 18} \right\rangle = \left\langle {1 + t, - 2 + 2t,13 - 18t} \right\rangle$ Show Step 2

If the two lines do intersect then they must share a point in common. In other words there must be some value, say $$t = {t_1}$$, and some (probably) different value, say $$t = {t_2}$$, so that if we plug $${t_1}$$ into the equation of the first line and if we plug $${t_2}$$ into the equation of the second line we will get the same $$x$$, $$y$$ and $$z$$ coordinates.

Show Step 3

This means that we can set up the following system of equations.

\begin{align*}1 + {t_1} & = 2 + 4{t_2}\\ - 2 + 2{t_1} & = - 1 - {t_2}\\ 13 - 18{t_1} & = 3\end{align*}

If this system of equations has a solution then the lines will intersect and if it doesn’t have a solution then the lines will not intersect.

Show Step 4

Solving a system of equations with more equations than unknowns is probably not something that you’ve run into all that often to this point. The basic process is pretty much the same however with a couple of minor (but very important) differences.

Start off by picking any two of the equations (so we now have two equations and two unknowns) and solve that system. For this problem let’s just take the first and third equation. We’ll worry about the second equation eventually.

Note that for this system the third equation should definitely be used here since we can quickly just solve that for $${t_1}$$.

Solving a system of two equations and two unknowns is something everyone should be familiar with at this point so we’ll not put in any real explanation to the solution work below.

${t_1} = \frac{5}{9}\hspace{0.5in} \to \hspace{0.5in}1 + \frac{5}{9} = 2 + 4{t_2} \hspace{0.5in} \to \hspace{0.5in} {t_2} = - \frac{1}{9}$ Show Step 5

Now, recall that to get this solution we used the first and third equations. What this means is that if we use this value of $${t_1}$$ and $${t_2}$$ in the equations of the first and second lines respectively then the $$x$$ and $$z$$ coordinates will be the same (remember we used the $$x$$ and $$z$$ equations to find this solution….).

At this point we need to recall that we did have another equation that also needs to be satisfied at these values of $$t$$. In other words, we need to plug $${t_1} = \frac{5}{9}$$ and $${t_2} = - \frac{1}{9}$$ into the second equation and see what we get. Doing this gives,

$- 2 + 2\left( {\frac{5}{9}} \right) = - \frac{8}{9} = - 1 - \left( { - \frac{1}{9}} \right)$

Okay, the two sides are the same. Just what does this mean? In terms of systems of equations it means that $${t_1} = \frac{5}{9}$$ and $${t_2} = - \frac{1}{9}$$ is a solution to the system of equations in Step 3.

In terms of whether or not the lines intersect we now know that at $${t_1} = \frac{5}{9}$$ and $${t_2} = - \frac{1}{9}$$ the two lines will have the same $$x$$, $$y$$ and $$z$$ coordinates (since they satisfy all three equations).

In other words, these two lines do intersect.

Before leaving this problem let’s note that it doesn’t matter which two equations we solve in Step 4. Different sets of equations will lead to the same values of $${t_1}$$ and $${t_2}$$ leading to the two lines intersecting.