Parametric parallel lines

Question

What is the distance between the parallel lines given by $\begin{pmatrix} 1 \\ 4 \end{pmatrix} + t \begin{pmatrix} 4 \\ 3 \end{pmatrix}$and $\begin{pmatrix} -5 \\ 6 \end{pmatrix} + s \begin{pmatrix} 4 \\ 3 \end{pmatrix}$?

I understand that, for each of these lines, I can create systems of equations with the variables $t$ and $s$, respectively, but I am stuck on how to evaluate the distance between these two lines.

Help with this endeavor would be much appreciated!

Answer 1

Here's one way. The second line has point $\pmatrix{-13\\0}$, when $s=2$.

A line perpendicular to the parallel lines has slope $-\dfrac34$.

Take the point we mentioned and the slope we mentioned to get the equation of a line

perpendicular to the parallel lines, and find where it intersects the first line.

Then find the distance from the point we mentioned to the point you found.

Answer 2

The first line, say $L_{1}$ can be written as $y=\frac{3x}{4}+\frac{13}{4}$ and the second line $L_{2}$ is given by $y=\frac{3x}{4}+\frac{39}{4}$. If point $(x_{1},y_{1})$ lies on $L_{1}$ and point $(x_{2},y_{2})$ lies on $L_{2}$ the distance between these two points is $\sqrt{(x_{1}-x_{2})^2+(y_{1}-y_{2})^2}$.

To find the shortest distance between the two lines you have to find a line perpendicular to $L_{1}$ (or $L_{2}$) and find where this line meets $L_{2}$ (or $L_{1}$) and then apply the formula above.

Answer 3

The classical way is projecting $\begin{pmatrix} -5 \\ 6 \end{pmatrix}- \begin{pmatrix} 1 \\ 4 \end{pmatrix} = \begin{pmatrix} -6 \\ 2 \end{pmatrix}$ onto a normalized vector orthogonal to the given lines:

$$\frac{1}{5}\begin{pmatrix} 3 \\ -4 \end{pmatrix} \Rightarrow d= \frac{1}{5}\left\lvert \begin{pmatrix} -6 \\ 2 \end{pmatrix}\cdot \begin{pmatrix} 3 \\ -4 \end{pmatrix}\right\rvert =\frac{26}{5}=5.2$$

Answer 4

Pick a point $P_0$ on one of the lines and two points $P_1$ and $P_2$ on the other one. The distance between the lines is the altitude of the triangle $\triangle{P_0P_1P_2}$ from $P_0$. Using the formula $A=\frac12bh$ for the area of a triangle, you can find $h$: for this triangle we have $b=P_1P_2$, so you just need some way to compute the area of the triangle given the three vertices. One way to do that is to use a determinant: $$A=\frac12\begin{vmatrix}P_1-P_0\\P_2-P_0\end{vmatrix}.$$ The two factors of $1/2$ cancel, giving $$h = {1\over\lVert P_2-P_1\rVert}\begin{vmatrix}P_1-P_0\\P_2-P_0\end{vmatrix}.$$ (Well, really the absolute value of this expression.)

Note that since the line that we’re taking two points from is given in the parametric form $\mathbf Q+t\mathbf v$, we can take advantage of linearity of the determinant: setting $P_1=Q$ and $P_2=Q+\mathbf v$, we have $P_2-P_1=\mathbf v$ and the above determinant is $$\begin{vmatrix}Q-P_0\\Q+\mathbf v-P_0\end{vmatrix} = \begin{vmatrix}Q-P_0\\\mathbf v\end{vmatrix},$$ which requires even less computation