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Determining the distance between the lines is necessary to determine how far apart they are located. Finding the distance is an easy process; we just need to derive a formula that works and helps us calculate. The shortest distance between two non-intersecting lines in the same plane is the shortest of all distances between two points on both lines. We will look in-depth at the shortest distance between two lines here. In this article, you will learn about the shortest distance between the lines, and also learn how the perpendicular is the shortest line, and find the perpendicular distance.
Solved Questions on the Shortest Distance Between the Lines
- Question: The following lines are mentioned in a 3-dimensional space:
x1 = i – j + λ(2i – 3j + 4k)
x2= 2i – 3j + 6k + λ(i – 2j + 7k)
What is the shortest distance between these lines?
Solution: x1 : i – j + λ(2i – 3j + 4k)
x2: 2i – 3j + 6k + λ(i – 2j + 3k)
b = i – j + k
a1 = i -j
a2 = 2i – 3j + 6k
a2 – a1 =i -2j +6 k
|b| = √3 = 1.73
|b x (a2 – a1)| = √42 = 6.48
The shortest distance that can be measured between two lines = |b x (a2 – a1)|/|b|
= 6.48/1.73 = 3.74.
- Question: Mentioned below is the cartesian form of two lines.
V1: (x – 2)/3 = (y – 1)/4 = (z)/2
V2: (x – 3)/6 = (y – 2)/8 = (z – 5)/4
Determine the shortest distance between these lines.
Solution: First, we calculate the displacement vector of V1 and V2
V1 = 3i + 4j + 2k,
V2 = 6i + 8j + 4k
We can easily see that V2 is the multiple of V1,
6i + 8j + 4k = 2 * (3i + 4j + 2k)
Thus, the two 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
Thus, the shortest distance between the lines = |b x (a2 – a1)|/|b|
= 12.569/5.385 = 2.334
- Question: Mentioned below are two lines in the vector form.
V1: i – j + λ(2i + j + k)
V2: i + j + λ(3i – j – k)
What will be the shortest distance between these two lines?
Solution: The following lines are skew lines, meaning they do not intersect.
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
- Question: Mentioned below are two lines in the vector form.
V1: 2i – j + 5 * (3i – j + 2k)
V2: i – j + 2k + 2* (i + 3j + 4k)
What is the shortest distance between these lines?
Solution: The mentioned lines are skewed.
Shortest distance between these lines is = |(b1 x b2)(a2 – a1)|/|(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
Thus, the shortest distance between the lines = 40/17.320 = 2.309
- Question: Look at the two lines in the cartesian form given below. 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 the shortest distance is zero, these two lines are intersecting lines.
- Question: Determine the distance between two parallel lines y = x + 6 and y = x – 2.
Solution: The given equations are of the form, y = mx + c
Here, m = 1, c1 = 6, c2 = -2
d = |c1 – c2|/√(1 + m2)
Therefore, d = 8/√2 or 5.65 units.
Conclusion
We have looked at some solved examples to determine the minimum distance between two straight lines in a plane. With the help of unit vectors, we can determine direction and magnitude in 3-dimensional spaces. We also learnt about the shortest distance between the lines and perpendicular distance.
