# Distance between two parallel lines

• Kernul
In summary, the conversation discusses finding the plane containing two intersecting lines and two parallel lines. The method used involves finding the directional vector and using the vector product to determine the direction of the plane. The conversation also addresses finding the distance between two parallel lines using projections and the distance formula. However, the speaker mentions that there are easier methods for finding planes and distances.
Kernul

## 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.

## 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?

I did not check your work, but there are methods for finding planes and distances that are a lot easier.

Math_QED said:
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.

## 1. What is the formula for finding the distance between two parallel lines?

The formula for finding the distance between two parallel lines is the absolute value of the difference between the y-intercepts of the lines, divided by the square root of the sum of the squares of their slopes.

## 2. How do you determine if two lines are parallel?

To determine if two lines are parallel, you can compare their slopes. If the slopes are equal, then the lines are parallel. Alternatively, you can also check if the lines have the same y-intercept.

## 3. Can the distance between two parallel lines ever be negative?

No, the distance between two parallel lines cannot be negative. The distance is always a positive value, as it represents the shortest distance between the two lines.

## 4. What is the relationship between the distance between two parallel lines and their slopes?

The distance between two parallel lines is inversely proportional to their slopes. This means that as the slopes of the lines increase, the distance between them decreases, and vice versa.

## 5. Can you use the Pythagorean theorem to find the distance between two parallel lines?

No, the Pythagorean theorem cannot be directly applied to find the distance between two parallel lines. However, it can be used in a different way by considering the distance between the lines as the hypotenuse of a right triangle formed by connecting two points on the lines with a perpendicular line segment.

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