# Shortest Distance Between Two Lines in 3D Space | Class 12 Maths

In 3D space, two lines can either intersect each other at some point, parallel to each other or they can neither be intersecting nor parallel to each other also known as skew lines.

- In the case of intersecting lines the shortest distance between them is 0.
- For parallel lines, the length of the line joining the two parallel lines or the length of the line perpendicular to both the parallel lines has the shortest distance.
- In the case of skew lines, the shortest distance is the line perpendicular to both of the given lines.

Note:The alphabets written in bold represent vector. ‘x’ denotes cross product(vector product).

**Shortest Distance Between **Two** Parallel Lines **

Considering 2 lines in vector form as:

**v1** = **a1** + c * **b**

**v2** = **a2** + d * **b**

Here, c and d are the constants.

**b = **parallel vector to both the vectors **v1** and **v2**

**a1, a2 ** are the position vector of some point on **v1 **and **v2 **respectively

Shortest distance = |b x (a2 – a1)| / |b|

### Examples

**Example** **1: For the following lines in 3D space.**

**v1 = i – 2j + i – j + k**

**v2 = i – 3j + k + i – j + k**

**Find the shortest distance between these lines?**

**Solution:**

v1: i – 2j + i – j + k

v2: i – 3j + k + i – j + k

b = i – j + k

a1 = i -2j

a2 = i – 3j + k

a2 – a1 = -j + k

|b| = √3 = 1.73

|b x(a2–a1)| = √2 = 1.41

Shortest distance = |b x(a2–a1)|/|b| = 1.41/1.73 = 0.815

**Example 2: For the following lines in 3D space.**

**v1 = i – j – k + 2i – 3j + 4k**

**v2 = 2i – 3j + k + 6i – 9j + 12k**

**Find the shortest distance between these lines?**

**Solution: **

The vector can be written in form as:

v1 = i – j – k + 2i – 3j + 4k

v2 = 2i – 3j + k + 3 * (2i – 3j + 4k)

b = 2i – 3j + 4k

|b| = √(2)^{2}+ (-3)^{2}+ (4)^{2}= 5.385

a1=i–j–k

a2= 2i– 3j+k

a2–a1=i– 2j+ 2k

b x(a2–a1) = 2i–k

|bx(a2–a1)| = √(2)^{2}+ (1)^{2 }= 2.236

Now applying the shortest distance formula for parallel lines = |bx(a2–a1)|/|b| = 2.236/5.385 = 0.415

**Example 3: Given two lines in the cartesian format as:**

**V1: (x – 2)/2 = (y – 1)/3 = (z)/4**

**V2: (x – 3)/4 = (y – 2)/6 = (z – 5)/8**

**Find the shortest distance between these lines.**

**Solution:**

The displacement vector ofV1is2i + 3j + 4k, forV2is4i + 6j + 8k

The displacement vectorV2is a multiple ofV1as,

4i + 6j + 8k= 2 * (2i + 3j + 4k)

So the two given lines are parallel to each other.

a1 = 2i + j + 0k

a2 = 3i + 2j + 5k

a2 – a1 = i + j +5k

b = 2i + 3j + 4k

|b| =√(2)^{2}+ (3)^{2}+ (4)^{2 }= 5.385

b x (a2 – a1)=11i – 6j – k

|b x (a2 – a1)|= 12.569

shortest distance = |b x (a2 – a1)|/|b|= 12.569/5.385 = 2.334

**Shortest Distance Between Skew Lines**

Considering 2 lines in vector form as:

**v1** = **a1** + c *** b1**

**v2** = **a2** + d * **b2**

here, c and d are the constants.

The shortest distance 2 skew lines = |(b1 x b2)(a2 – a1)|/|(b1 x b2)|

Note: If two lines are intersecting then’s the shortest distance considering the two lines skew will automatically come out to be zero.

### Examples

**Example 1: Given two lines in vector form as:**

**V1: i – j + 2i + j + k**

**V2: i + j + 3i – j – k**

**Find the shortest distance between these lines.**

**Solution:**

The given lines are skew lines.

b1 = 2i + j + k

b2 = 3i – j – k

a2 = i + j

a1 = i – j

a2 – a1 = 2j

(b1 x b2) = 5j – 5k

Shortest distance = |(b1 x b2)(a2 – a1)|/|(b1 x b2)| = 10/7.07 = 1.41

**Example 2: Given two lines in vector form as:**

**V1: 2i – j + 5 * (3i – j + 2k)**

**v2: i – j + 2k + 2* (i + 3j + 4k)**

**Find the shortest distance between these lines.**

**Solution:**

The given lines are skew lines.

Shortest distance = |(b1xb2)(a2–a1)|/|(b1xb2)|

b1 = 3i – j + 2k

b2 = i + 3j + 4k

a1 = 2i – j

a2 = i – j + 2k

a2 – a1 = -i + 2k

(b1 x b2)=-10i – 10j + 10k

|b1 xb2| = 17.320

|(b1 x b2)(a2 –a1)| = 40

Shortest distance = 40/17.320 = 2.309

**Example 3: Given 2 lines in the cartesian form, find the shortest distance between them.**

**V1: (x – 1)/2 = (y – 1)/3 = (z)/4**

**V2: (x)/1 = (y – 2)/2 = (z – 1)/3 **

**Solution:**

a1 = i + j

a2 = -2j + k

b1 = 2i + 3j + 4k

b2 = i + 2j + 3k

a2 – a1 = -3i – j + k

(b1 x b2) = i – 2j + k

|b1 x b2| = 2.44

Shortest distance =|(i – 2j + k)(-3i – j + k)|/2.44 = 0

Since, shortest distance is zero it means these two lines are intersecting lines.