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Math Help - Intersection of three planes

  1. #1
    Junior Member Hasan1's Avatar
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    Arrow Intersection of three planes

    Solve each of the following systems of equations. Give a geometrical interpretation of each system and its solution.

    1. x - y + 3z = 4
    2. x + y + 2z = 2
    3. 3x + y + 7z = 9

    OK so the first thing I did was check to see if any of the equations were scalar multiples, to see if any of the planes were coincident, which they weren't. Then I checked their normals to see if they were scalar multiples to see if the planes were parallel, they weren't.
    Then I took the triple scalar product to see if the normals were coplanar. They were. From this I said the planes either have one line of intersection, or they have no solution, because they have three separate intersections as a triangular prism.

    I do not know how to determine which possibility it is, the answer says it has no solution, it forms a triangular prism.

    Anybody know?
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  2. #2
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    One way to see this is to manually calculate the solution of these three equations. I can use the coefficient matrix to determine if any solutions exist.

    \left( \begin{array}{ccc}<br />
1 & -1 & 3 \\<br />
1 & 1 & 2 \\<br />
3 & 1 & 7 \end{array} \right)<br />
\left( \begin{array}{c}<br />
x  \\<br />
y \\<br />
z  \end{array} \right) = <br />
\left( \begin{array}{c}<br />
4  \\<br />
2 \\<br />
9  \end{array} \right)

    For this equation to have a solution it be true that the first matrix (call it A) has an inverse A^{-1}. Since then:

    A^{-1}A\left( \begin{array}{c}<br />
x  \\<br />
y \\<br />
z  \end{array} \right) = <br />
\left( \begin{array}{ccc}<br />
1 & 0 & 0 \\<br />
0 & 1 & 0 \\<br />
0 & 0 & 1 \end{array} \right)<br />
\left( \begin{array}{c}<br />
x  \\<br />
y \\<br />
z  \end{array} \right) =<br />
\left( \begin{array}{ccc}<br />
x & 0 & 0 \\<br />
0 & y & 0 \\<br />
0 & 0 & z \end{array} \right)=<br />
A^{-1}\left( \begin{array}{c}<br />
4  \\<br />
2 \\<br />
9  \end{array} \right)

    But if A^{-1} does not exist then there is no solution. So when doesn't A^{-1} exist? A having a determinant equal to zero would do it. And if this case the determinant does equal zero.
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  3. #3
    Senior Member
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    Berkeley, Illinois
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    You can also use Cramers' Rule for 3 unknowns as well.


    3 Unknown Calculator Cramer's Rule

    The math work involved is all there if you wish to see how a no solution is developed. If you already know Cramers' Rule, then disregard this message.
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  4. #4
    Junior Member Hasan1's Avatar
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    Thanks, but we never learned about determinants or cramer's rule in class, the most we learned about matrices was gaussian and gauss-jhordan elimination, so i don't really understand those solutions. Is there a way to determine if it has no solution manually?
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  5. #5
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    Add the first and second equations. Then add the first and third equations. You will have 2 equations in x and z:

    2x+5z = 6 and
    4x+10z = 13

    This is an impossibility. To see this more clearly observe that equation 2 can be written

    2x+5z = 6.5

    We know there are no values of x and y that make 2x+5z = 6.5 AND 6. If there were solutions you could use substitution or elimination to solve for x,y, and z.

    For example:
    x+y-2z = 2
    x-y+z = 6
    2x-2y-3z=2

    we eliminate y from 2 of the equations like we did before (adding 2 equations)

    2x-z = 8
    4x-7z=6

    now multiply the first equation by 2 and subtract the second from it.
    This gives:

    0+5z = 10 which implies that z=2.

    Plug this into either of the above 2 equations:
    2x-z = 2x - 2 = 8 which gives x = 5.

    Plug z=2 and x=5 into one of the original equations and solve for y:
    x+y-2z = 5 + y - 4 = 2, y=1.

    So the solution is a single point (5,1,2).
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