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Shortest Distance Between Two Lines in 3D Space | Class 12 Maths
  • Last Updated : 11 Feb, 2021
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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 (a2a1)| = √2 = 1.41

Shortest distance = |b x (a2a1)|/|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 = ijk

a2 = 2i – 3j + k

a2 a1 = i – 2j + 2k

b x (a2a1) = 2ik

|b x (a2a1)| = √(2)2 + (1)2 = 2.236

Now applying the shortest distance formula for parallel lines = |b x (a2a1)|/|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 of V1 is 2i + 3j + 4k, for V2 is 4i + 6j + 8k

The displacement vector V2 is a multiple of V1 as, 

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 = |(b1 x b2)(a2a1)|/|(b1 x b2)|

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 x b2| = 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.

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