# Thread: Intersection of 3 planes

1. ## 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?

2. 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

3. Hello, skeske1234!

Find the intersection of $\displaystyle P_1, P_2, P_3$ and describe geometrically.

. . $\displaystyle \begin{array}{cccc}P_1 & x+y-z&=&3 \\ P_2 & 2x-y+z&=&5 \\ P_3 & x-2y+2z&=&6 \end{array}$

We have: .$\displaystyle \left|\begin{array}{ccc|c} 1 & 1 & \text{-}1 & 3 \\ 2 & \text{-}1 & 1 & 5 \\ 1 & \text{-}2 & 2 & 6 \end{array}\right|$

$\displaystyle \begin{array}{c}\\ R_2 - R_1 \\ \\ \end{array} \left|\begin{array}{ccc|c} 1 & 1 & \text{-}1 & 3 \\ 1 & \text{-}2 & 2 & 2\\ 1 & \text{-}2 & 2 & 6 \end{array}\right|$

$\displaystyle \begin{array}{c} \\ \\ R_3-R_2 \end{array} \left|\begin{array}{ccc|c}1 & 1 & \text{-}1 & 3\\ 1 & \text{-}2 & 2 & 2 \\ {\color{red}0} & {\color{red}0} & {\color{red}0} & {\color{red}4}\end{array} \right|$

The bottom row says: .$\displaystyle 0 \:=\:4$

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:

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