In this article, we will discuss how to find the distance between two parallel and skew lines.

The shortest distance between two lines refer to how far away two lines are located from each other. In other words, we can say that the shortest distance between two lines in a plane is the minimum distance between any two points that are present on both the lines. The shortest distance is the measure of the length of a perpendicular line between two lines.

In geometry, we come across different lines such as parallel lines and skew lines. We have formulas to calculate the distance between two parallel and skew lines. In this article, we will focus on using formulas to find the minimum distance between these types of lines.

Distance Between Parallel Lines

What are parallel lines?

Two lines that are parallel never meet each other and they have the same slopes.

How to find the distance between two parallel lines?

Before proceeding to discuss the method to find the distance between two parallel lines, let us first see what is meant by the distance.

It is the length of the perpendicular line that stretches from any point on one line to the other one.
To find the distance between two parallel lines, you should follow the following method:

  • See whether the equations are given in slope intercept form or not
  • Slope value should be common for both the lines if they are given in the slope intercept form. Determine the values of c_1 and c_2
  • After finding the values, substitute them in the slope-intercept equation to calculate y.
  • In the last step, substitute all the values in the distance formula discussed below to find the distance between two parallel lines

 

The formula for distance between two parallel lines is given below if the lines are in the slope intercept form. Remember that the slope intercept form of the line is y = mx + c.

d = \frac {|c_2 - c_1|} {\sqrt{1 + m^2}}

Here, c_1 is the constant of line 1 and c_2 is the constant of line 2. m represents the slope of

If the equations of the parallel lines are given in the ax + by + d, then the formula is slightly different than the first one:

d = \frac {|d_2 - d_1|} {\sqrt{a^2 + b^2}}

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Example 1

Find the distance between the parallel lines having the following equations:

Line 1 :y = 5x + 5

Line 2 : y = 5x - 7

Solution

These lines have the same slopes, so we will use the following formula to compute the distance between them:

 d = \frac {|c_2 - c_1|} {\sqrt{1 + m^2}}

Here, c_1 = 5, c_2 = -7, and m = 5. Substitute these values in the above formula:

 d = \frac {|-7 - 5|} {\sqrt{1 + 5^2}}

d = \frac {|-12|} {\sqrt{26}}

d = \frac {|-12|} {5.09} = 2.35 units

Hence, the distance between two lines is 2.35 units.

 

Example 2

Find the distance between the parallel lines having the following equations:

Line 1 : y = 3x + 2

Line 2 : y = 3x - 1

Solution

These lines have the same slopes, so we will use the following formula to compute the distance between them:

 d = \frac {|c_2 - c_1|} {\sqrt{1 + m^2}}

Here, c_1 = 2, c_2 = -1, and m = 3. Substitute these values in the above formula:

 d = \frac {|-1 - 2|} {\sqrt{1 + 3^2}}

d = \frac {|-3|} {\sqrt{10}}

d = \frac {3} {3.16} = 0.94 units

Hence, the distance between two lines is 0.94 units.

 

Example 3

Find the distance between the following two parallel lines:

Line 1 : 4x + 2y + 5

Line 2 : 4x + 2y + 9

Solution

Since the equations are given in the form ax + by + d, hence we will use the following formula to compute the distance:

d = \frac {|d_2 - d_1|} {\sqrt{a^2 + b^2}}

Here, d_1 = 5, d_2 = 9, a = 4 and b = 2. Substitute these values in the above formula to get the distance:

d = \frac {|9 - 5|} {\sqrt{4^2 + 2^2}}

d = \frac {|4|} {\sqrt{20}}

d = \frac {|4|} {4.47}

d = 0.89 units

Hence, the distance between these lines is 0.89 units.

 

Example 4

Find the distance between the following two parallel lines:

Line 1 : 5x + 2y + 1

Line 2 : 5x + 2y + 7

Solution

Since the equations are given in the form ax + by + d, hence we will use the following formula to compute the distance:

d = \frac {|d_2 - d_1|} {\sqrt{a^2 + b^2}}

Here, d_1 = 1, d_2 = 7, a = 5 and b = 2. Substitute these values in the above formula to get the distance:

d = \frac {|7 - 1|} {\sqrt{4^2 + 2^2}}

d = \frac {|6|} {\sqrt{20}}

d = \frac {|6|} {4.47}

d = 1.34 units

Hence, the distance between these lines is 1.34 units.

 

Example 5

Find the distance between the following two parallel lines:

Line 1 : 3x + 2y + 8

Line 2 : 3x + 2y + 9

Solution

Since the equations are given in the form ax + by + d, hence we will use the following formula to compute the distance:

d = \frac {|d_2 - d_1|} {\sqrt{a^2 + b^2}}

Here, d_1 = 8, d_2 = 9, a = 3 and b = 2. Substitute these values in the above formula to get the distance:

d = \frac {|9 - 8|} {\sqrt{3^2 + 2^2}}

d = \frac {|1|} {\sqrt{13}}

d = \frac {|1|} {3.60}

d = 0.27 units

Hence, the distance between these lines is 0.27 units.

 

Example 6

Find the distance between the following two parallel lines:

Line 1 : 7x + y + 8

Line 2 : 7x + y + 9

Solution

Since the equations are given in the form ax + by + d, hence we will use the following formula to compute the distance:

d = \frac {|d_2 - d_1|} {\sqrt{a^2 + b^2}}

Here, d_1 = 8, d_2 = 9, a = 7 and b = 1. Substitute these values in the above formula to get the distance:

d = \frac {|9 - 8|} {\sqrt{7^2 + 1^2}}

d = \frac {|1|} {\sqrt{50}}

d = \frac {|1|} {7.07}

d = 0.14 units

Hence, the distance between these lines is 0.14 units.

Distance Between Skew Lines

These lines do not intersect each other at any point, but they are also not parallel to each other. These lines are also called agonic lines and most of the time these lines exist in three or more dimensions.  In the figure below, the lines r and s are skew lines.

Distance between skew lines

The general formula to calculate the distance between two skew lines of the form l_1 =a_1 +tb_1 and l_2 = a_2 + tb_2 is given below:

d = \frac{ (\overrightarrow {a_2} - \overrightarrow {a_1}) (\overrightarrow{b_2} - \overrightarrow {b_1})} {(\overrightarrow {b_1} \times \overrightarrow {b_2})}

 

Example

Find the distance between the following lines:

r = 2i -  j + k + \lambda (2 + j + 2k) and s = 3i + 2j + 3k + \mu (2i + j + 2k)

Solution

Here, a_1 = 2i -j + k, a_2 = 3i +2 j + 3k and values of b_1 and b_2 are same, i.e., b_1 = b_2. We will use the following formula to compute the distance:

d = |\frac{\overrightarrow{b} ( \overrightarrow{a_2 - a_1})} {|\overrightarrow{b}|}|

\overrightarrow{a_2 - a_1} = i + 3j + 2k = (1, 3, 2)

\overrightarrow{b} \times \overrightarrow{a_2 - a_1} =\begin {bmatrix} i & j & k \\ 2 & 1 & 2\\ 1 & 3 & 2 \\ \end {bmatrix} = \sqrt{45}

|\overrightarrow{b}| = 3

d = \frac{\sqrt{45}}{3} = 2.23

Hence, the distance between the lines is 2.23 units.

 

 

 

 

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Rafia Shabbir