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.
 R_{c} = 2 
 R_{d} = 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 B _{1}/D _{1} ≠ 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
n_{1} = 2i − j + k 
n_{2} = 2i + j + k 
n_{3} = − 4i + 2j − 2k 
Now we can perform the cross product on each pair of the vectors.
L_{2} is equal to 0 that is beacaus the two plans 1 and 3 are parallel.
We also see that lines L_{1} and L_{2} 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 R_{d} = 2 we must have the value of y from the R_{d} 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 R_{c} = 2 and R_{d} = 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 = AB cos θ
