﻿ Intersection of three planes
 Intersection of 3 planes ▲
 Plane 1: x + y + z + = 0 Plane 2: x + y + z + = 0 Plane 3: x + y + z + = 0
 Plane defined by 3 points: xa ya za xb yb zb xc yc zc
Intersection type:
Intersection coordinate: det + rank:
NOTES:
Intersection line 1 equation:
Intersection line 2 equation:
Intersection line 3 equation:
n1 x n2         |             Angle
n1 x n3         |             Angle
n2 x n3         |             Angle
 Solved problems: Intersection of 3 planes Three parallel planes
 Intersection of 3 planes summary ▲
Three planes intersection.
 Plane 1:     A1 x + B1 y + C1 z + D1 = 0 Plane 2:     A2 x + B2 y + C2 z + D2 = 0 Plane 3:     A3 x + B3 y + C3 z + D3 = 0
Normal vectors to planes are:
 n1 = iA1 + jB1 + kC1 n2 = iA2 + jB2 + kC2 n3 = iA3 + jB3 + kC3
For intersection line equation between two planes see two planes intersection.
In order to find which type of intersection lines formed by three planes, it is required to analyse the ranks   Rc   of the coefficients matrix and the augmented matrix   Rd.
 Rank: Rank: If vectors:    n1 × n2 = 0    then the planes are parallel (cross product).
The angle between two planes from the vectors dot product is:  Plane - P1 Plane - P2 Plane - P3
Condition for a single point of intersection is: rank Rc = 3rank Rd = 3
Intersection point is given by Cramer's rule: In vector analysis:     n1 · (n2 × n3) ≠ 0 Condition for one line intersection (two coincide planes) are:
rank   Rc = 2         and         Rd = 2
In vector analysis:     n2 × n3 = 0         n1 × n3 = n1 × n2 ≠ 0 Condition for one line intersection is:
rank   Rc = 2         and         Rd = 2
In vector analysis:     n1 × n2 = n1 × n3 = n2 × n3 ≠ 0
Note:   n1 is the same as   −n1   but pointing to the opposite direction.
The reason for this is the fact that:     n1 × n2 = −n2 × n1 Condition for two lines intersection (two parallel planes) is:
rank   Rc = 2         and         Rd = 3
In vector analysis:     n2 × n3 = 0         n1 × n3 = n1 × n2 ≠ 0 Condition for three lines intersection is:
rank   Rc = 2         and         Rd = 3
All values of the cross product of the normal vectors to the planes are not 0 and are pointing to the same direction.
n1 × n2 = n1 × n3 = n2 × n3 ≠ 0 Condition for no intersection (three parallel planes) are:
rank   Rc = 1         and         Rd = 2
Check number of similar planes Rs:         Rs = 0        (See example)
In vector analysis:     n1 × n2 = n1 × n3 = n2 × n3 = 0 Condition for no intersection (three parallel planes, 2 coincides) are:
rank   Rc = 1         and         Rd = 2
Check number of similar planes Rs:         Rs = 2         (See example)
In vector analysis:     n1 × n2 = n1 × n3 = n2 × n3 = 0 Condition for no intersection (three coincide planes) are:
rank   Rc = 1         and         Rd = 1
In vector analysis:     n1 × n2 = n1 × n3 = n2 × n3 = 0
The cross product is equal to:   Example - Intersection of 3 planes ▲
 Investigate the intersection values between the three planes: Plane 1:     2x − y + z + 3 = 0 Plane 2:     2x + y + z + 2 = 0 Plane 3:     − 4x + 2y − 2z − 6 = 0

First find the rank of the coefficient matrix by upper triangularization. Rc = 2 Rd = 2
The ranks were found by two steps first subtructing the second row from the first row and the second step to subtruct the fourth row from the first row multiplyed by two.
If we look on the augmented matrix we see that we have two equations with three unknown so we can choos one value for example z = t and because B1/D1 ≠ 0 we have the solution for y the value y = 0.5 now the value of x will be from the first raw:
 2x - y + z = -3 x = (y - z -3)/2 x = (0.5 -t -3)/2 = -1.25 - 0.5t
Thus is a line presented pay parameter t:
L{x,y,z}={− 1.25 − 0.5t,    0.5,    t}    same as    {− 1.25 − t,    0.5,    2t}     After multiplying both t by 2.

We could find the intersection line by vectors calculation, the planes are described by the vectors in i, j and k direction
 n1 = 2i − j + k n2 = 2i + j + k n3 = − 4i + 2j − 2k
Now we can perform the cross product on each pair of the vectors.   L2 is equal to 0 that is beacaus the two plans 1 and 3 are parallel.
We also see that lines L1 and L2 are pointing to the same direction    − 4i + 8k = − 2i + 4k    (vectors can be factored without changing their direction).
Now that we have the intersection line direction we need a point on the line in order to set the line equation, beacause   Rd = 2   we must have the value of   y   from the Rd matrix:    y = 1/2 = 0.5
now we can choos an arbitrary value to z let say   z = 0   than   x = − 1.25t   or parametric line equation:
L{x,y,x} = {− 1.25 − 2t,    0.5,    4t}
 Because   Rc = 2   and   Rd = 2   then two planes are parallel and beacause one cross product of the vectors are   0   the intersection looks like: The angle between plane 1 and plane 2 is found from the dot product:     A · B = |A||B| cos θ Intersection of 3 parallel planes ▲
Given three planes by the equations:     x + 2y + z − 1 = 0       2x + 4y + 2z − 6 = 0       4x + 8y + 4z − n = 0
Determine the locations of the planes to each other in the case that   n = 4   and second time   n = 8.

The matrix of the   x y and z   coefficients after multiplying first row by 2 and subtracting from second row and then multiplying first row by 4 and subtracting from third row we get the matrix: Rank Rc = 1    we see that all the planes are parallel (1)
In order to find the number of coincides planes we have to analyze the augmented matrix with   n = 4 after dividing second row by 2 and third row by 4 and repeating the process done on the coefficients matrix we get the matrix: Rank Rd = 2 (2)
In the case that n = 8 we have the augmented matrix: Rank Rd = 2 (3)
We can clearly see that if we get Rd = 2 we have to check for similar planes in order to find the number of coincides planes, from the center matrix (2) we see that the first and third planes are coincides so  Rs = 2,  from the center matrix (3) we see that all the planes are distinct   Rs = 3
Notice: If all the planes are similar then  Rd = 1  and we don't have to check for planes similarity.