Shortest Distance between Two Non-Intersecting Lines

Before we proceed towards the shortest distance between two lines, we first try to find out the distance formula for two points.

Formula of Distance

If there are two points say A(x₁, y₁) and B(x₂, y₂), then the distance between these two points is given by √[(x₁-x₂)² + (y₁-y₂)²].   

View the following video for more on distance formula:

Straight Line

A straight line in space is characterized by the intersection of two planes which are not parallel and hence the equation of straight line is in fact the solution of the system consisting of the equation of the planes:

a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0

Such form of equation is also termed as the unsymmetrical form. 

We now try to find the equation of the straight line in symmetrical form:

A(x₁, y₁, z₁) be a given point on the straight line and l, m and n are the dc’s, then its equation is given by 

(x-x₁)/l = (y-y₁)/m = (z-z₁)/n = k (say)

Clearly, any general point on this line at a distance ‘k’ from the point A(x₁, y₁, z₁) is given by P(x₁ + lk, y₁ + mk, z₁ + nk).
 

Shortest Distance Between Two Non-Intersecting Lines

Now we discuss the condition for non-intersecting lines. Two lines are called non intersecting if they do not lie in the same plane. The straight line which is perpendicular to each of non-intersecting lines is called the line of shortest distance. And length of shortest distance line intercepted between two lines is called length of shortest distance. Let us discuss the method of finding this line of shortest distance.

Method: Let the equation of two non-intersecting lines be

(x–x₁) / l₁ = (y–y₁) /m₁ = (z–z₁) /n₁ = r₁ (say)                               ……(1)

And (x–x₂)/ l₂ = (y–y₂) /m₂ = (z–z₂) /n₂ = r₂ (say)                          ……(2)

Any point on line (1) is of the form P (x₁ + l₁r₁, y₁ + m₁r₁, z₁ + n₁r₁)

and on line (2) is Q (x₂ + l₂r₂, y₂ + m₂r₂, z₂ + n₂r₂).

Let PQ be the line of shortest distance. Its direction ratios will be

[(l₁r₁ + x₁ – x₂ – l₂r₂), (m₁r₁ + y₁ – y₂ – m₂r₂), (n₁r₁ + z₁– z₂ – n₂r₂)]

This line is perpendicular to both the given lines. By using the condition of perpendicularity we obtain 2 equations in r₁ and r₂.

And hence by solving these, values of r₁ and r₂ can be found. Subsequently he points P and Q can be found. The distance PQ is shortest distance.

The shortest distance can be found by PQ = 

Remark: If any straight line is given in general form then it can be transformed into symmetrical form and we can further proceed.

Illustration:

Find the shortest distance between the lines, 

(x–3)/3 = (y –8)/–1 = (z–3)/1,

(x + 3)/–3 = (y+7)/2 = (z–6)/4.

Also find the equation of the line of shortest distance. 

Solution:

Given lines are 

(x – 3)/3 = (y – 8)/–1 = (z– 3)/1 = r₁ (say)          ……(1)

(x + 3)/–3 = (y +7)/2 = (z – 6)/4 = r₂ (say)          ……(2)

Any point on line (1) is of the form P (3r₁ + 3, 8 – r₁, r₁ + 3)

and on line (2) is of the form Q (–3 – 3r₂, 2r₂ – 7, 4r₂ + 6).

If PQ is line of shortest distance, then direction ratios of PQ

= (3r₁ + 3) – (–3 – 3r₂), (8 – r₁) – (2r₂ – 7), (r₁+ 3) – (4r₂ + 6)

i.e. 3r₁ + 3r₂ + 6, –r₁ – 2r₂ + 15, r₁ – 4r₂ – 3

As PQ is perpendicular to lines (1) and (2)

∴ 3(3r₁ + 3r₂ + 6) – 1(–r₁ – 2r₂ + 15) + 1(r₁ – 4r₂ - 3) = 0

⇒11r₁ + 7r₂ = 0                                              ……(3)

and –3(3r₁ + 3r₂ + 6) + 2(–r₁ – 2r₂ + 15) + 4(r₁ – 4r₂ - 3) = 0

i.e. 7r₁ + 11r₂ = 0                                              ……(4)

On solving equations (3) and (4), we get r₁ = r₂= 0.

So, point P = (3, 8, 3) and Q = (–3, –7, 6).

∴ Length of shortest distance PQ = √{(–3–3)² + (–7–8)² + (6–3)²} = 3√30

Direction ratios of shortest distance line are 2, 5, –1.

∴ Equation of shortest distance line is

x–3/2 = y–8/5 = z–3/–1.

Illustration:

Consider the lines

L₁ = (x+1)/3 = (y+2)/1 = (z+1)/2

L₂ = (x-2)/1 = (y+2)/2 = (z-3)/3

Find the unit vector perpendicular to both L₁ and L₂. Also find the shortest distance between the two. (2008)

Solution:

We may represent the given lines in vector form as

L₁ = (-i-2j-k) + λ (3i+j+2k)

L₂ = (2i-2j+3k) + μ (i+2j+3k)

Since, the vector is perpendicular to both L₁ and L₂ and so by solving with the help of determinants we obtain it as -i-7j+5k.

Hence, the required unit vector is (-i-7j+5k)/√[(-1)² + (-7)² + (5)²]

                                  = 1/5√3. (-i-7j+5k)

The shortest distance between L₁ and L₂ is

|[(2-(-1))i + (2-2)j + (3-(-1))k] . (-i-7j+5k)/ 5√3|

= | (3i + 4k). (-i-7j+5k)/ 5√3|

 = 17/ 5√3
 

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