Thread: Intersection of 3 planes

Find the intersection of pi1, pi2, pi3, and describe geometrically in each case.
pi1: x+y-z=3
pi2: 2x-y+z=5
pi3: x-2y+2z=6

ok, so this is my work (how does it look?)

1 1 -1 3
2 -1 1 5
1 -2 2 6

1 1 -1 3
2 -1 1 5
2 -4 4 12 << R3 x 2

1 1 -1 3
0 3 -3 -7 << R2 - R3
2 -4 4 12

1 1 -1 3 << R1 -R3
0 3 -3 -7
1 -2 2 6 << x (1/2)

0 3 -3 -3 << R1 - R2
0 3 -3 -7
1 -2 2 6

0 0 0 4
0 3 -3 -7
1 -2 2 6

final matrix
1 -2 2 6
0 3 -3 -7
0 0 0 4

so from above, would it be correct to say that there are no intersections? since 0z=4
but i'm confused because i don't think that it makes sense, since all three planes are not parallel.. meaning that there should be a point of intersection.. what did i do wrong here?

Hi

More simple
R1+R2 => 3x=8
2R1+R3 => 3x=12

There is effectively no intersection
3 planes do not always have an intersection point

Think about 2 planes : or they are parallel or they are secant along a line
In this case the intersection with a 3rd plane is the intersection between the line and this 3rd plane : you can see that it can be empty if the line is parallel to the plane

The intersection between pi1 and pi2 is a line that can be defined by
x=0t+8/3
y=t
z=t-1/3
Its direction is given by u(0,1,1)

A normal vector to pi3 is given by v(1,-2,2)

The dot product between u and v is 0, which means that the line is parallel to pi3

Hello, skeske1234!

Find the intersection of and describe geometrically.

. .

We have: .








The bottom row says: .

There is no point of intersection.



Interpretation:

Taking the planes in pairs, the three lines of intersection are parallel.
The three planes enclose a triangular prism.

The end view looks like this:

Code:

          \   /
           \ /
            *
           ? \
          ?   \
         /     \
    - - * - - - * - -
       /         \
      /           \