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Math Help - Intersection of three planes

  1. #1
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    Intersection of three planes

    Solve the following system of equations and explain what the solution means geometrically:

    1) x-y+z=10
    2) 2x+3y-2z=-21
    3) 0.5x+0.4y+0.25z=-0.5
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  2. #2
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    Quote Originally Posted by skeske1234 View Post
    Solve the following system of equations and explain what the solution means geometrically:
    1) x-y+z=10
    2) 2x+3y-2z=-21
    3) 0.5x+0.4y+0.25z=-0.5
    Here is a suggestion, write #3 as 10x+8y+5z=-10.
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  3. #3
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    ok im stuck here:

    0 0 43 | 382
    0 7 -15 |-95
    -30 -31 0 |125
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  4. #4
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    Quote Originally Posted by skeske1234 View Post
    ok im stuck here:

    0 0 43 | 382
    0 7 -15 |-95
    -30 -31 0 |125
    Using a CAS, I get (1,-5,4).
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  5. #5
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    how did you get that? can you show me please?
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  6. #6
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    Im pretty sure you need to use substituion.

    solve for z in terms of x and y then
    put then in another equation and solve for x in terms of y then
    put it in the last equation and solve for x

    from there you can find y and z. the solution (x,y,z) should be the point at which the planes intersect...i think.

    hope this helps

    i tried and failed otherwise i would show you how
    Last edited by snaes; September 25th 2009 at 12:58 PM. Reason: i tried and failed :(
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  7. #7
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    i tried snaes method and it made it really complicated, so im trying to use the matrices to work with it. i'm still working on it, but to this point, i can't find the answer that plato has still.
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  8. #8
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    sorry about that im working on it now. I tried substition and combinations and here is the result (im off somewhere but i believe this should work)...

    <br />
x-y+z=10
    2x+3y-2z=-21
    10x+8y+5z=-10<br />

    multiply by -1 for first equation and 2 for the second:
    -1 1 -1 = -10
    4 3 -4 = -42
    10 8 5 = -10
    +_________________
    13x + 12y = -41

    same thing only different
    -2 2 -2 = -20
    -8 -12 8 = 84
    10 8 5 = -10
    _____________________
    -2y + 11z = 74

    from those 2 equation we just found find what x and z equal both in terms of y and you get:
    x=\frac{-12y-62}{13}
    z=\frac{54+2y}{11}

    dang i thought it would work...sorry...this comes out to be y=-5.6 something. Good luck.
    Last edited by snaes; September 25th 2009 at 01:29 PM. Reason: lots of stuff
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  9. #9
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    Quote Originally Posted by skeske1234 View Post
    how did you get that? can you show me please?
    Using a CAS.
    Attached Thumbnails Attached Thumbnails Intersection of three planes-show.gif  
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  10. #10
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    Much easier. You win!
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  11. #11
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    yeah, not sure what a "cas" is.. i'm guessing it's technology, which we aren't supposed to use.

    I still can't get this question with matrices. Did you?
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  12. #12
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    CAS means Computer Algebra System, Computer algebra system - Wikipedia, the free encyclopedia

    So not pencil and paper.
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  13. #13
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    a computer is technology,
    the point of the question is to use matrices using pencil and paper, and this is the issue i am having right now.
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  14. #14
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    Without CAS, you can use row-reduction. First do it with just a 2x2 matrix then do this one. I did the first row reductions below quick so need to check them:

    \left[\begin{array}{cccc} 1 & -1 & 1 & 10 \\<br />
                                      2 & 3 & -2 & 21 \\<br />
                                      10 & 8 & 5 & 10<br />
         \end{array}<br />
\right]<br />

    Now: first row times -2, add to second and first row times -10 and add to third gives:

    \left[\begin{array}{cccc} 1 & -1 & 1 & 10 \\<br />
                                      0 & 5 & -4 & 1 \\<br />
                                      0 & 18 & -5 & -110<br />
         \end{array}<br />
\right]<br />

    Second row times 1/5, add to first and second times 18/5 add to third gives:

    \left[\begin{array}{cccc} 1 & 0 & 1/5 & 51/5 \\<br />
                                      0 & 5 & -4 & 1 \\<br />
                                      0 & 0 & 27/5 & -568/5<br />
         \end{array}<br />
\right]<br />

    Keep doing it until you get ones on the diagonal in the first three columns (divide the second row by 5 right).
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  15. #15
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    I did it by reducing it down. Steps are shown below:

    \left[\begin{array}{cccc} 1 & -1 & 1 & 10 \\<br />
                                      2 & 3 & -2 & -21 \\<br />
                                      10 & 8 & 5 & -10<br />
         \end{array}<br />
\right]<br />

    \left[\begin{array}{cccc} 1 & -1 & 1 & 10 \\<br />
                                      0 & 5 & -4 & -41 \\<br />
                                      10 & 8 & 5 & -10<br />
         \end{array}<br />
\right]<br />

    \left[\begin{array}{cccc} 1 & -1 & 1 & 10 \\<br />
                                       0 & 1 & -4/5 & -41/5 \\<br />
                                       10 & 8 & 5 & -10<br />
          \end{array}<br />
 \right]<br />

    \left[\begin{array}{cccc} 1 & 0 & 1/5 & 9/5 \\<br />
                                        0 & 1 & -4/5 & -41/5 \\<br />
                                        10 & 8 & 5 & -10<br />
           \end{array}<br />
  \right]<br />

    \left[\begin{array}{cccc} 1 & 0 & 1/5 & 9/5 \\<br />
                                         0 & 1 & -4/5 & -41/5 \\<br />
                                         0 & 8 & 3 & -28<br />
            \end{array}<br />
   \right]<br />

    \left[\begin{array}{cccc} 1 & 0 & 1/5 & 9/5 \\<br />
                                          0 & 1 & -4/5 & -41/5 \\<br />
                                          0 & 0 & 47/5 & 188/5<br />
             \end{array}<br />
    \right]<br />

    So, 47z = 188 sets z = 4.

    Then:

    1) x + 1/5z = 9/5 equates to x = 1
    2) y - 4/5z = -41/5 equates to y = -5
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