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Math Help - Intersection of 3 planes

  1. #1
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    Intersection of 3 planes

    Find the intersection of pi1, pi2, pi3, and describe geometrically in each case.

    a) pi1: x-y+2z=7
    pi2: 2x+y-z=3
    pi3: x+y+z=9

    b) pi1: x+y-z=3
    pi2: 2x-y+z=5
    pi3: x-2y+2z=6

    so this is my work:
    a)
    n1: (1,-1,2)
    n2: (2,1,-1)
    n3: (1,1,1)

    b)
    n1: (1,1,-1)
    n2: (2,-1,1)
    n3: (1,-2,2)

    ok, now I can't find any way that both a) and b) have intersections because n1, n2, n3 are never parallel or have n1=kn2=kn3 .. i think i'm doing something wrong here. please help me out and demonstrate the correct method to this question. thanks in advance!
    ----------
    however there is a part c) and this one worked out (i think?):
    pi1: 4x-6y+2z=10
    pi2: 2x-3y+z=0
    pi3: 2x-18y-8z=0
    n1=(4,-6,2)=(2,-3,1)
    n2=(2,-3,1)
    n3=(2,-18,-8)=(1,-9,-4)
    so this is my answer: not sure if it correct
    pi1//pi2 and no common intersections
    Last edited by skeske1234; September 13th 2009 at 09:48 AM.
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  2. #2
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    Quote Originally Posted by skeske1234 View Post
    Find the intersection of pi1, pi2, pi3, and describe geometrically in each case.

    a) pi1: x-y+2z=7
    pi2: 2x+y-z=3
    pi3: x+y+z=9

    b) pi1: x+y-z=3
    pi2: 2x-y+z=5
    pi3: x-2y+2z=6

    so this is my work:
    a)
    n1: (1,-1,2)
    n2: (2,1,-1)
    n3: (1,1,1)

    b)
    n1: (1,1,-1)
    n2: (2,-1,1)
    n3: (1,-2,2)

    ok, now I can't find any way that both a) and b) have intersections because n1, n2, n3 are never parallel or have n1=kn2=kn3 .. i think i'm doing something wrong here. please help me out and demonstrate the correct method to this question. thanks in advance!
    What??? It would be precisely in the situation where they are parallel that they would not have intersections!
    Your equations for (a) are x-y+2z=7, 2x+y-z=3, and x+y+z=9. I think a first obvious thing to do is to add the second and third equations, eliminating z: (2x+y-z)+(x+y+z)= 3x+ 2y= 3+ 9= 12.
    Multiplying the second equation by 2 gives 4x+2y-2z= 6 and adding that to the first equation also eliminates z: (x-y+2z)+ (4x+2y-2z)= 5x+y= 6+7= 13.
    Now multiply that by 2 to get 10x+ 2y= 26 and subtract 3x+2y= 12.
    (10x+2y)-(3x+2y)= 7x= 26- 12= 14. Since 7x= 14, x= 2. Then 3x+2y= 3(2)+ 2y= 6+ 2y= 12 so 2y= 6 and y= 3. Finally, x+y+z= 2+3+z=9 so z= 4. The point of intersection of the three planes is (2, 3, 4).

    ----------
    however there is a part c) and this one worked out (i think?):
    pi1: 4x-6y+2z=10
    pi2: 2x-3y+z=0
    pi3: 2x-18y-8z=0
    n1=(4,-6,2)=(2,-3,1)
    n2=(2,-3,1)
    n3=(2,-18,-8)=(1,-9,-4)
    so this is my answer: not sure if it correct
    pi1//pi2 and no common intersections
    Well, you could try to solve for the intersection and see what happens!
    If you subtract the third equation from the second you eliminate x: (2x-3y+z)- (2x-18y-8z)= 15y+ 9z= 0. If you multiply the second equation by 2, getting 4x-6y+z= 0 and subtract the first, eliminating x again, (4x-6y+2z)-(4x-6y+2z)= 0x+0y+0z= 0= 10- 0. Because that equation reduces to "10= 0" which is false, no matter what x, y, and z are, there is no point that satisifies. Yes, you are correct. pi1 and pi2 are parallel and have no common point.
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  3. #3
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    Hello, skeske1234!

    The intersection of three planes is usually a point.


    Find the intersection of P_1,\:P_2,\:P_3, and describe geometrically in each case.

    (a)\;\;\begin{array}{cccc}P_1\!: &x-y+2z&=& 7 \\ <br />
P_2\!: & 2x+y-z&=& 3 \\<br />
P_3\!: & x+y+z &=& 9 \end{array}
    Solve the system of equations . . .


    We have: . \left|\begin{array}{ccc|c} 1 & \text{-}1 & 2 & 7 \\ 2 & 1 & \text{-}1 & 3 \\ 1 & 1 & 1 & 9 \end{array}\right|


    \begin{array}{c} \\ R_2 - 2R_1 \\ R_3 - R_1 \end{array}\left|\begin{array}{ccc|c}<br />
1 & \text{-}1 & 2 & 7 \\ 0 & 3 & \text{-}5 & \text{-}11 \\ 0 & 2 & \text{-}1 & 2\end{array}\right|


    \begin{array}{c}\\ R_2 - R_3 \\ \\ \end{array}<br />
\left|\begin{array}{ccc|c}<br />
1 & \text{-}1 & 2 & 7 \\ 0 & 1 & \text{-}4 & \text{-}13 \\ 0 & 2 & \text{-}1 & 2 \end{array}\right|


    \begin{array}{c}R_1+R_2 \\ \\ R_3 - 2R_2\end{array}<br />
\left|\begin{array}{ccc|c}<br />
1 & 0 & \text{-}2 & \text{-}6 \\<br />
0 & 1 & \text{-}4 & \text{-}13 \\<br />
0 & 0 & 7 & 28 \end{array}\right|


    . . \begin{array}{c} \\ \\ \frac{1}{7}R_3\end{array}\;\;<br />
\left|\begin{array}{ccc|c}<br />
1 & 0 & \text{-}2 & \text{-}6 \\<br />
0 & 1 & \text{-}4 & \text{-}13 \\<br />
0 & 0 & 1 & 4 \end{array}\right|


    \begin{array}{c}R_1+2R_3 \\ R_2 + 4R_3 \\ \\ \end{array}<br />
\left|\begin{array}{ccc|c}<br />
1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & 4 \end{array}\right|


    The intersection is a point: (2,3,4)

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  4. #4
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    thanks
    Last edited by skeske1234; September 13th 2009 at 09:46 AM.
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