Question

How do you find the distance between two parallel lines in 3-dimensional space?

Answer 1

In the above question, we are given two parallel lines in a 3-dimensional space. We have to find the distance between those two given lines. Recall the formula of cross product of two vectors. The cross product of two vectors is itself a vector and is given by the formula $\overrightarrow a \times \overrightarrow b = \left| a \right|\left| b \right|\sin \theta \widehat n$, where $\widehat n$ is the unit vector in the perpendicular direction of both vectors. This formula will be useful in finding the required distance, let see how.

Complete step by step answer:
Given that, two parallel lines that lie in a 3-dimensional space. Let the two parallel lines be ${l_1}$ and ${l_2}$. Let the equations of the two parallel lines be,
${l_1} \Rightarrow \overrightarrow r = \overrightarrow {{a_1}} + \lambda \overrightarrow b$
And
${l_2} \Rightarrow \overrightarrow r = \overrightarrow {{a_2}} + \mu \overrightarrow b$
Where $\overrightarrow {{a_1}}$ and $\overrightarrow {{a_2}}$ are points on ${l_1}$ and ${l_2}$ and $\overrightarrow b$ is the line parallel to both ${l_1}$ and ${l_2}$ .

A diagram of both the lines is shown above where the distance between ${l_1}$ and ${l_2}$ is PT. Consider the vectors $\overrightarrow {ST}$ and $\overrightarrow b$ , their cross product can be written using the formula,
$\Rightarrow \overrightarrow a \times \overrightarrow b = \left| a \right|\left| b \right|\sin \theta \,\widehat n$
As,
$\Rightarrow \overrightarrow b \times \overrightarrow {ST} = \left| {\overrightarrow b } \right|\left| {\overrightarrow {ST} } \right|\sin \theta \cdot \widehat n$ ...(1)
Also the distance ST can be written as,
$\overrightarrow {ST} = \overrightarrow {{a_2}} - \overrightarrow {{a_1}}$ ...(2)

Now from the diagram, we have
$\Rightarrow \sin \theta = \left| {\dfrac{{PT}}{{ST}}} \right|$
That gives,
$\Rightarrow \left| {ST} \right|\sin \theta = \left| {PT} \right|$
Multiplying both sides by $\left| {\overrightarrow b } \right| \cdot \widehat n$ , we get
$\Rightarrow \left| {\overrightarrow b } \right|\left| {ST} \right|\sin \theta \cdot \widehat n = \left| {\overrightarrow b } \right| \cdot \left| {PT} \right| \cdot \widehat n$
Now, using the equation ...(1) we can write the above equation as
$\Rightarrow \overrightarrow b \times \overrightarrow {ST} = \left| {\overrightarrow b } \right|\left| {PT} \right| \cdot \widehat n$
Taking modulus of both sides,
$\Rightarrow \left| {\overrightarrow b \times \overrightarrow {ST} } \right| = \left| {\overrightarrow b } \right|\left| {PT} \right| \cdot \left| {\widehat n} \right|$

Since $\left| {\widehat n} \right| = 1$ that gives,
$\Rightarrow \left| {\overrightarrow b \times \overrightarrow {ST} } \right| = \left| {\overrightarrow b } \right|\left| {PT} \right|$
Again, putting $\overrightarrow {ST} = \overrightarrow {{a_2}} - \overrightarrow {{a_1}}$ we get
$\Rightarrow \left| {\overrightarrow b \times \left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right)} \right| = \left| {\overrightarrow b } \right|\left| {PT} \right|$
$\therefore \left| {PT} \right| = \dfrac{{\left| {\overrightarrow b \times \left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right)} \right|}}{{\left| {\overrightarrow b } \right|}}$
That is the required distance between the two parallel lines ${l_1}$ and ${l_2}$.

Therefore, the distance between two parallel lines in a 3-dimensional space is given by $\dfrac{{\left| {\overrightarrow b \times \left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right)} \right|}}{{\left| {\overrightarrow b } \right|}}$.

Note: In three-dimensional geometry, skew lines are two lines that do not intersect and also are not parallel. As a result they do not lie in the same plane. A simple example of a pair of skew lines is the pair of lines through opposite edges of a regular tetrahedron. While intersecting lines and parallel lines lie in the same plane i.e. are coplanar.