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If anyone got time. Me and my friends appreciate the help We got a problem with the following task:

Determine the minimum distance between lines

Line 1:
$$\begin{eqnarray*} x &=&1+t \\ y &=&2+2t. \end{eqnarray*}$$
Line 2:
$$\begin{eqnarray*} x &=&2-s \\ y &=&1-2s. \end{eqnarray*}$$

Both of these lines intersect the line $x + y = 0$. Determine the respective intersection point and cutting angle. Thanks for taking your time to read this and we hope you could help us out.

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1) Distance between the two lines.

The lines are parallel and distinct, so this makes sense. Let us pick the point $P_2=(2,1)$ on line $2$. What you want is the orthogonal projection of $P_2$ on line $1$. There is a formula for that (see other answers), but let's do it as if we didn't know that.

First, you need to determine an equation of the line orthogonal to line $1$ pasing by $P_2$. Since the direction of line $1$ is given by the vector $(1,2)$, the direction of this orthogonal line can be given by $(2,-1)$ (that's a natural way to get an orthogonal vector to $(1,2)$). Then recall this orthogonal line passes through $P_2$ and write the corresponding parameterization.

Now find the intersection of line $1$ and the new orthogonal line you have just determined. Call it $P_1$. The distance between line $1$ and line $2$ is the distance bete ween the points $P_1$ and $P_2$. I assume you know how to compute this given their coordinates.

2)Intersections with $x+y=0$.

Just plug the parameterized coordinates of line $1$ and line $2$ in this equation. Then solve for the parameter.

3) Angle.

A vector perpendicular to the line $x+y=0$ is $w=(1,1)$. A vector perpendicular to line $1$ is $u_1=(2,-1)$. The angle $\theta_1$ between line $1$ and the former is the angle between $w$ and $u_1$. You can use $$ \cos\theta_1=\frac{(u_1,w)}{\|u_1\|\|w\|}. $$

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You can make a function $d(t)$ which measures the distance from line1 to line2 at a given time. By taking the derivative and setting it equal to 0 you can find a $t$ where the distance is minimal.

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