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Math Help - Intersection of 3 planes

  1. #1
    Senior Member
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    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?
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  2. #2
    MHF Contributor
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    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
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  3. #3
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    Hello, skeske1234!

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

    . . \begin{array}{cccc}P_1 & x+y-z&=&3 \\<br />
P_2 & 2x-y+z&=&5 \\ P_3 & x-2y+2z&=&6 \end{array}

    We have: . \left|\begin{array}{ccc|c}<br />
1 & 1 & \text{-}1 & 3 \\ 2 & \text{-}1 & 1 & 5 \\ 1 & \text{-}2 & 2 & 6 \end{array}\right|


    \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|


    \begin{array}{c} \\  \\  R_3-R_2 \end{array}<br />
\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: . 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:

    Code:
              \   /
               \ /
                *
               ? \
              ?   \
             /     \
        - - * - - - * - -
           /         \
          /           \
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