What are parallel lines in 3D geometry and how is the distance between such lines calculated? This lesson explains how two parallel lines are coplanar and that the distance between them is nothing but the length of the perpendicular between them. We have also provided a solved example for you to understand the calculations.

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**Parallel Lines in 3D Geometry**

In three-dimensional geometry, one of the most crucial elements is a straight line. Any two straight lines can be differently related to each other in the Cartesian plane in the sense that they may be intersecting each other, skewed lines or parallel lines. Note that the distance between two intersecting lines is zero.

In the following section, we shall move on to explore how the distance between parallel lines can be measured. You must make note that the shortest distance between parallel lines is actually the length of the perpendicular between them or joining the two lines.

**Browse more Topics Under Three Dimensional Geometry**

- Angle Between a Line and a Plane
- Angle Between Two Lines
- Coplanarity of Two Lines
- Angle Between Two Planes
- Direction Cosines and Direction Ratios of a Line
- Distance Between Parallel Lines
- The Distance Between Two Skew Lines
- Distance of a Point from a Plane
- Equation of a Plane in Normal Form
- Equation of a Plane Perpendicular to a Given Vector and Passing Through a Given Point
- The Equation of Line for Space
- Equation of Plane Passing Through Three Non Collinear Points
- Intercept Form of the Equation of a Plane
- Plane Passing Through the Intersection of Two Given Planes

## How to measure the distance between two parallel lines?

Let us consider two lines that are parallel to each other, whose equations are given by â€“

$$ \vec{r}_1Â = \vec{a}_1Â + Â \vec{b} $$

$$ \vec{r}_2Â = \vec{a}_2Â + Â \vec{b} $$

You must understand that \( \vec{a}_1 \; and \; \vec{a}_2 \)_{Â }are the position vectors of the points S and T on lines \( l_1_{Â }and l_2 \)_{Â }respectively. Note that these two lines are lying on the same plane i.e. they are coplanar.

Let us drop a perpendicular from point T which lies on the line \( l_2 \)Â onto the other line \( l_1 \) such that the foot of the perpendicular is at point P. So it can be easily understood that the distance between the two lines must actually equal the length of the perpendicular i.e. \( | \vec{PT} | \)

Let the angle between S and T be Éµ, so â€“

$$ \vec{b} \times | \vec{ST}Â | =Â | \vec{b}| \; | \vec{ST}Â | sin Î˜Â . \vec{n} $$

Where \( \vec{n} \) is nothing but the unit vector perpendicular to the common plane containing l_{1} and l_{2}. Now, $$ | \vec{ST}Â | = | \vec{a}_2Â Â – \vec{a}_1 | and | \vec{ST}Â | sin Î˜ = | \vec{PT} | $$

Thus,Â $$ \vec{b} \times (\vec{a}_2Â Â – \vec{a}_1 )Â =Â | \vec{b}| \; | \vec{PT} |. \vec{n}Â $$

In other words, $$ | \vec{b} \times (\vec{a}_2Â Â – \vec{a}_1) | =Â | \vec{b}| \; | \vec{PT} | $$

This means that we have,Â $$ | \vec{PT} |. = | { \vec{b} \timesÂ (\vec{a}_2Â Â – \vec{a}_1Â ) } | / | \vec{b}|$$

Thus, the distance between two parallel lines is given by â€“

$$ d = | \vec{PT} |. = | { \vec{b} \times (\vec{a}_2Â Â – \vec{a}_1Â ) } | / | \vec{b}| $$

Explore the following section for a simple example that will make it clearer how to use this formula.

**Solved Examples for You**

**Question 1: Find the distance between the two lines whose equations are given by:**

**$$ \vec{r}_1Â = \hat{i} â€“ 2 \hat{j} + 4 \hat{k} + Â ( 2 \hat{i} + 3 \hat{j} + 6 \hat{k} ) $$**

**$$ \vec{r}_2Â = 3 \hat{i} + 3 \hat{j} â€“ 5 \hat{k} + Â ( 2 \hat{i} + 3 \hat{j} + 6 \hat{k} ) $$**

**Answer :** It is evident that the lines are parallel. We shall use our formula to arrive at the distance between these lines â€“

$$ \vec{a}_2 – \vec{a}_1Â Â = Â 3 \hat{i} + 3 \hat{j} â€“ 5 \hat{k} â€“ { \hat{i} â€“ 2 \hat{j} + 4 \hat{k} } $$

Or, $$\vec{a}_2 – \vec{a}_1Â = 2 \hat{i} + 1 \hat{j} â€“ 1 \hat{k} $$

Also,Â $$ \vec{b} =Â 2 \hat{i} + 3 \hat{j} + 6 \hat{k} $$

Implies, $$ | \vec{b}|Â = \sqrt ( 2^2 + 3^2 + 6^2)Â = 7 $$

Also, $$ \vec{b} \times (\vec{a}_2 – \vec{a}_1Â ) = \begin{vmatrix} i & j & k \\ 2 & 3 & 6 \\ 2 & 1 & -1 \end{vmatrix}$$

= i (-3 – 6) â€“ j (12 â€“ (-2)) + k (2 – 6)

= -9i + 14j -4k

Now, $$| \vec{b}Â x (\vec{a}_2Â Â – \vec{a}_1) | = \sqrt ( (-9)^2Â + 14^2 + (-4)^2)Â = \sqrt (293)$$

Hence, the distance between the given lines is = \( \sqrt{293} / 7 \) units.

**Question 2: Can we say that parallel lines are equal in measure?**

**Answer:** Parallel lines are certainly equal in measure. If the cutting of two parallel lines takes place by a transversal, then the alternate exterior angles certainly happen to be congruent.

**Question 3: What is the appearance of parallel lines?**

**Answer:** Lines that never intersect are Parallel lines are those that never intersect each other. Parallel lines do not result in the formation of any angles. Parallel lines have similarity to railroad tracks. This means that these lines are always the same distance apart.

**Question 4: Explain how can one find a parallel line?**

**Answer:** Two lines are parallel in case they both have the same slope. For example, ascertain the slope of the line that is parallel to the line 4x â€“ 5y = 12. In order to find the slope of this line, one must get the line into slope-intercept form (y = mx + b). So, one must solve for y. Therefore, the slope of the line 4x â€“ 5y = 12 is m, which is 4/5.

**Question 5: Can we say that parallel lines are coplanar?**

**Answer:** Yes, parallel lines are certainly coplanar and they do not intersect.

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