Distance between two skew lines

Distance between two 3D lines calculator

Parametric line equation	L₁	x = + t y = + t z = + t

		x = + s y = + s z = + s

General line equation	L₁	x + = y + = z +

	L₂	x + = y + = z +

Lines defined by 4 points	L₁	x₁ y₁ z₁
		x₂ y₂ z₂

	L₂:	x₃ y₃ z₃
		x₄ y₄ z₄

Distance between the lines:
Connecting line intersections:
Values of t and s:
Angle between the lines:
Connecting line vector (n):
Connecting line unit vector:

Steps to calculate distance between two 3D lines

Presentation of 3D lines

Steps to calculate the distance between two 3D lines

Finding the distance d between two 3D lines L₁ and L₂ that are given by the parametric equations:

L₁:

L₂:

The lines can also be described as:

L₁ = r₁ + t e₁

L₂ = r₂ + s e₂

Step (1)

Calculate the cross product of the direction numbers e₁ = (a₁, b₁, c₁) and e₂ = (a₂, b₂, c₂). n = e₁ ✕ e₂. The result is a vector perpendicular to both lines (i, j, k are the vector notation in the x, y, and z direction).:

Note: if e₁ ✕ e₂ = 0 then the lines are parallel, and this method of calculation is not applicable.

Step (2) Find the norm of the vector (is a scalar value):

Step (3) The unit vector in this direction is:

Step (4) Find a point P on L₁ where t = 0:

Step (5) Find a point Q on L₂ where s = 0:

Step (6) Find vector PQ connecting P to Q by subtracting (Q ⎯ P):

Step (7)

The absolute value of the dot product of vector n with PQ and divide by the norm of |n| will give the required distance d between the lines:

D=|((b_1 c_2-b_2 c_1 )(x_2-x_1 )-(a_1 c_2-a_2 c_1 )(y_2-y_1 )+(a_1 b_2-a_2 b_1 )(z_2-z_1 ))/√((b_1 c_2-b_2 c_1 )^2+(a_1 c_2-a_2 c_1 )^2+(a_1 b_2-a_2 b_1 )^2 )|

Step (8)

The shortest line intersections parameters values, (when the lines are not intersecting each other are):

t=((e_2×n)∙(r_2-r_1 ))/(n∙n)=((e_2×n)∙PQ)/(n∙n) s=((e_1×n)∙(r_2-r_1 ))/(n∙n)=((e_1×n)∙PQ)/(n∙n)

Unique cases

Given two lines with two same axis direction numbers equal to 0, in this case the x and y axis. This indicates that all x and y points of both lines are constant and therefore parallel to the z axis.

L_1=(■(x_1@y_1@z_1 ))+t(■(0@0@c_1 )) L_2=(■(x_2@y_2@z_2 ))+s(■(0@0@c_2 ))

From the drawing we see that both lines (green points directed to us) in the x-y plane. it is obvious that the distance between the lines will be:

The same way we can find the distance between two lines that are parallel to the y and z axis.

Let analyse the case that the none 0 axes are not in the same direction L₁ and L₂.

L_1: (■(x@y@z))=(■(x_1@y_1@z_1 ))+t(■(0@0@c_1 )) L_2: (■(x@y@z))=(■(x_2@y_2@z_2 ))+s(■(0@b_2@0))

From the drawing we can see that line L₁ is located parallel to the z axis and line L₂ is parallel to y axis. Therefore, the angle between the lines is 90 degree and the distance between the lines is simply

D=x_2-x_1

Based on the example above we can derive the distance equations for other similar combinations of direction numbers:

L_1: (x,y,z)=(x_1,y_1,z_1)+t(0,0,c_1 ) L_2: (x,y,z)=(x_2,y_2,z_2)+t(0,b_2,0)

L_1: (x,y,z)=(x_1,y_1,z_1)+t(0,b_1,0) L_2: (x,y,z)=(x_2,y_2,z_2)+s(a_2,0,0)

L_1: (■(x@y@z))=(■(x_1@y_1@z_1 ))+t(■(a_1@b_1@0))

In this case only the z direction number is zero. hence all the points of the line are located on the plane parallel to the x-y plane, where z = z₁. The distance between the lines can be found by the steps described earlier.

Two lines will be parallel in space when all three direction numbers are the same:

a₁ = a₂ and b₁ = b₂ and c₁ = c₂

3D lines - Example 1

Question:

Find the distance between the lines: L1: x = 3 + 4t, y = − 2, z = 5 − 2t and the line L2: x = 3 + s, y = 2 + 2s, z = − 1 + 7s.

Solution:

Step (1) Cross product of the direction numbers is:

Step (2) The norm of the vector is:

Step (3) The unit vector in the line direction is:

Step (4) A point P on L₁ where t = 0 is at: (3, ⎯ 2, 5)

Step (5) A point Q on L₂ where s = 0 is at: (3, 2, ⎯ 1)

Step (6) (Q ⎯ P) = (0, 4 ⎯ 6)

Step (7) Finally the distance between the lines is:

3D lines - Example 2

Question:

Find the intersection point of the lines:

(x+3)/2=y/4=(z+2)/3 (x-4)/1=(y-1)/(-11)=(z-3)/(-4)

Solution:

We will solve the problem by two ways first we shall write the parameterized equations of the line:

L_1: (■(x@y@z))=(■(-3@0@-2))+t(■(2@4@3)) L_2: (■(x@y@z))=(■(4@1@3))+s(■(1@-11@-4))

Because the x and y coordinates lines L₁ and L₂ are the same, we can find the values of s and t.

x: −3 + 2t = 4 + s

y: 0 + 4t = 1 −11s

x: 2t − s = 7

y: 4t + 11s = 1

Now solving by Cramer's rule or by substitution we get:

t=|■(7&-1@1&11)|/|■(2&-1@4&11)| =78/26=3 s=|■(2&7@4&1)|/|■(2&-1@4&11)| =(-26)/26=-1

Once we fount the values of t and s the only thing remained is to check if the values of z in both lines are equal if yes then the lines intersect.

L₂ z: −2 + 3 * 3 = 7 and L₂ z: 3 + 4 = 7 so the lines intersect.

To find the intersection point, we can substitute the value of t = 3 into the parametric equation to get: P_i (3, 12, 7).

Another way to solve the problem is to calculate the distance between the lines, if the distance is 0 then the lines intersect.

The vectors of the points on the line and the direction numbers are:

r₁ = (−3, 0, −2) r₂ = (4, 1, 3) e₁ = (2, 4, 3) e₂ = (1, −11, −4)

The vector in the direction of the shortest line (perpendicular to both lines) is:

The unit vector in this direction is:

The connecting vector between the points of both lines is:

PQ = r₂ − r₁ = (4 + 3, 1 − 0, 3 + 2) = 7i + j + 5k

Now the only step remains is to find the distance between the lines is to project PQ to the shortest line direction:

d=(7i+j+5k)∙(0.52i+0.33j-0.79k)=0

As expected, we get 0 distance and the lines are intersecting each other.

3D lines - Example 3

Question:

Find the distance between the lines:

L_1: (■(x@y@z))=(■(6@8@4))+t(■(6@7@0)) L_2: (■(x@y@z))=(■(6@8@2))+t(■(6@8@4))

Solution:

The vectors of the points on the line and the direction numbers are:

r₁ = (6, 8, 4) r₂ = (6, 8, 2) e₁ = (6, 7, 0) e₂ = (6, 8, 4)

The vector in the direction of the shortest line (perpendicular to both lines) is:

n=e_1×e_2=|■(i&j&k@6&7&0@6&8&4)|=28i-24j+6k

The unit vector in this direction is:

n ̂=n/|n| =(28i-24j+6k)/√(28^2+24^2+6^2 )=0.749i-0.642j+0.161k

The connecting vector between the points of both lines is:

PQ = r₂ − r₁ = (6 − 6, 8 − 8, 2 − 4) = 0i + 0j + −2k

Now the only step remains is to find the distance between the lines is to project PQ to the shortest line direction:

d=(0i+0j+2k)∙(0.749i-0.642j+0.161k)=0.32