Question

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

Answer 1

Hint: 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 $$\overrightarrow{a}\times \overrightarrow{b}=\left|a\right|\left|b\right|\sin \theta \hat{n}$$, where $$\hat{n}$$ 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,
$$\Rightarrow \overrightarrow{a}\times \overrightarrow{b}=\left|a\right|\left|b\right|\sin \theta \hat{n}$$
As,
$$\Rightarrow \overrightarrow{b}\times \overrightarrow{ST}=\left|\overrightarrow{b}\right|\left|\overrightarrow{ST}\right|\sin \theta \cdot \hat{n}$$ ...(1)
Also the distance ST can be written as,
$$\overrightarrow{ST}=\overrightarrow{a_2}-\overrightarrow{a_1}$$ ...(2)

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

Since $$\left|\hat{n}\right|=1$$ 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
$$\Rightarrow \left|\overrightarrow{b}\times \left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)\right|=\left|\overrightarrow{b}\right|\left|PT\right|$$
$$\therefore \left|PT\right|={\displaystyle \frac{\left|\overrightarrow{b}\times \left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)\right|}{\left|\overrightarrow{b}\right|}}$$
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 $${\displaystyle \frac{\left|\overrightarrow{b}\times \left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)\right|}{\left|\overrightarrow{b}\right|}}$$.

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.