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Distance between two parallel lines

  • Thread starter Kernul
  • Start date
  • #1
211
7

Homework Statement


Refering to this exercise I started https://www.physicsforums.com/threads/checking-if-the-following-lines-are-coplanar.885948/ I now have to find the plane containing the two intersecting lines and the plane containing the two parallel lines. In the case of the parallel lines, though, I have to find their distance.

Homework Equations




The Attempt at a Solution


The two planes I guess were easy.
For the two intersecting lines(##r## and ##s##), I just have to take their directional vector, do the vector product(that would give me the direction the plane is facing), and then calculate the intersection point of the two lines to then use for the plane equation.
##\vec v_r \times \vec v_s =
\begin{vmatrix}
\hat i & \hat j & \hat k \\
-3 & -1 & -2 \\
3 & 0 & 3
\end{vmatrix} = (-3, 3, 3)##
and
##\begin{cases}
x - 3y + 3 = 0 \\
2y + z - 5 = 0 \\
x + z - 4 = 0 \\
x - 3y + z + 2 = 0
\end{cases}##
We then exclude one of the equations.
##\begin{cases}
x - 3y + 3 = 0 \\
2y + z - 5 = 0 \\
x + z - 4 = 0
\end{cases}
\begin{cases}
x = 3 \\
y = 2 \\
z = 1
\end{cases}##
So we have a point ##P_i (3, 2, 1)##.
The plane equation will be:
##\pi_{r, s} : -3x + 3y + 3z + d = 0##
What we need is ##d##, which can be found out thanks to the point we just found.
##d = -[(-3)*3 + 3*2 + 3*1] = 0##
And so ##\pi_{r, s} : -3x + 3y + 3z = 0##

We will have to do the same thing for the parallel lines, but not with both the directional vectors, since they go the same direction. What I did is using a point from both lines and subtract one from the other, obtaining a vector that is not parallel to them but that we will need to get the plane. We proceed the same way as before, using a point of one of the two lines. I then find myself with ##\pi : 2x - 8y + 5z - 7 = 0##

Now it comes my problem: the distance between the two parallel lines.
I know that you just have to find the distance between one of the lines and a point of the other line.
Let's start from the line ##t##. The directional vector and a point are respectively ##\vec v_t = (-12, -4, 8)## and ##P_t (-1, -1, 1)##
Now I start by finding a plane orthogonal to the ##r##(with ##\vec v_r = (-3, -1, 2)##) line passing for a point belonging to it ##P_r (0, 1, 3)##.
Similar to before, I would have ##\pi : -3x - y + 2z - 5##
Now I have to find the projection of the point ##P_r## on the line ##t##(for example calling it ##P'##) as an intersection between the line and the plane.
##\begin{cases}
-3x - y - 2z - 5 = 0 \\
4x - 2y + 5z - 3 = 0 \\
2y + z + 1 = 0
\end{cases}
\begin{cases}
x = -\frac{5}{3} \\
y = -\frac{11}{9} \\
z = \frac{13}{9}
\end{cases}##
Getting the point ##P' (-\frac{5}{3}, -\frac{11}{9}, \frac{13}{9})##
Now I simply have to find the distance between ##P_r## and ##P'## as:
##d(P_r, P') = \sqrt{(0 - (-\frac{5}{3}))^2 + (1 - (-\frac{11}{9}))^2 + (3 - \frac{13}{9})^2}##
The problem is that I end up with something like this ##\sqrt{\frac{821}{81}}## which is just odd and I don't know if I did it correctly. Could someone tell me if my way of resolving it is correct at least?
 

Answers and Replies

  • #2
Math_QED
Science Advisor
Homework Helper
2019 Award
1,603
643
I did not check your work, but there are methods for finding planes and distances that are a lot easier.
 
  • #3
211
7
I did not check your work, but there are methods for finding planes and distances that are a lot easier.
Then could you please tell me an easier method? This is the only one I know.
 

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