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Math Help - Need help solving basic linear system

  1. #1
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    Need help solving basic linear system

    Find the shortest distance between the solutions of the system A and the solutions of the system B

    A
    7x-2y+3z=-2
    -3x+y+2z+5=0

    I got this one, it's x+7z+12

    B

    -6x+2y+4z+5=0
    3x-y-2z-2.5=0
    -3x+y+2z+2.5=0

    The answer is supposedly
    z=t
    y=-41-23t
    x=-12-7t

    When I do this however:

    -6 2 4 5
    3 -1 -2 -2.5
    -3 1 2 2.5
    >
    1 -2/6 -4/6 -5/6
    0 0 0 0
    0 0 0 0

    Now, the solution from here says

    v=(-7, -23,1) u=(3,-1,-2) and computes u.v

    Where do these values come from?

    I also need help with the rest of the steps. I know after you get x, y, you can just use the distance formula and plug it in for 3x-y-2z-2.5, but how do I calculate for x and y?
    Last edited by shibble; August 6th 2010 at 08:25 PM.
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  2. #2
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    Bump, really need to get this down. Just need help on how the values of x,y,z were determined for B, and where the values of v=(-7,-32,1) came from.

    Thanks a lot
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  3. #3
    MHF Contributor

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    I responded to this on a different forum but-
    Quote Originally Posted by shibble View Post
    Find the shortest distance between the solutions of the system A and the solutions of the system B

    A
    7x-2y+3z=-2
    -3x+y+2z+5=0

    I got this one, it's x+7z+12
    This makes no sense. What is "x+ 7z+ 12"? Not the answer certainly- the answer must be values for x, y, and z.

    If you multiply the second equation by 2 and add to the first you eliminate y: x+ 7z+ 10= -2 or x+ 7z+12= 0. Is that what you meant? You weren't finished. From that , x= -7z- 12. Putting that back into the second equation -3(-7z- 12)+ y+ 2z+ 5= 0 so y= 3(-7z- 12)- 2z- 5= -23z- 41. Since there are only two equations in three values, you cannot go any further. You could write the solution in terms of parametric equations, x= -7t- 12, y= -23t- 41, z= t.

    Geometrically, the solution is any point on that line.

    B

    -6x+2y+4z+5=0
    3x-y-2z-2.5=0
    -3x+y+2z+2.5=0

    The answer is supposedly
    z=t
    y=-41-23t
    x=-12-7t
    Are you sure you don't have this confused with the solution to A?

    When I do this however:

    -6 2 4 5
    3 -1 -2 -2.5
    -3 1 2 2.5
    >
    1 -2/6 -4/6 -5/6
    0 0 0 0
    0 0 0 0
    Obviously, the second equation is just the first equation multiplied by -1/2 and the third equation is just the first equation multiplied by 1/2. You really only have one equation and so can only solve for one value in terms of the other two.

    Since -3x+y+2z+2.5=0, y= 3x- 2z- 2.5. You can write that as parametric equations in terms of two parameters. x= u, y= 3u- 2v- 2.5, z= v. Geometrically, that is a plane.

    Now, the solution from here says

    v=(-7, -23,1) u=(3,-1,-2) and computes u.v

    Where do these values come from?

    I also need help with the rest of the steps. I know after you get x, y, you can just use the distance formula and plug it in for 3x-y-2z-2.5, but how do I calculate for x and y?
    (- 7, -23, 1) is the "direction vector" of the straight line satisfying the first equations. That is, it is a vector pointing along the line. Any time you have a straight line given by parametric equations x= At+ x_0, y= Bt+ y_0, z= Ct+ y_0, the vector A\vec{i}+ B\vec{j}+ C\vec{k}, which can be written as (A, B, C), points in the same direction as the line.

    (3, -1, -2) is the "normal vector" to the plane satisfying the second set of equations. If a plane is written in the form Ax+ By+ Cz= d, then A\vec{i}+ B\vec{j}+ C\vec{k}, which can be written as (A, B, C), is a vector perpendicular to the plane.

    Once we have the line x= -7t- 12, y= -23t- 41, z= t and the plane -3x+y+2z+2.5=0, the first thing I would do is see if any point satisfies both of those. Look at -3(-7t- 12)+ (-23t- 41)+ 2(t)+ 2.5= 0. That becomes 21t+ 36- 23t- 41+ 2t+ 2.5= 0 or (21- 23+ 2)t+ 36- 41+ 2.5= 0t- 2.5= 0. That is never true so the line and plane are parallel. There is a specific distance between the two.

    To find the distance between the line and plane, first choose a point on the line. The simplest thing to do is to let t= 0 in the equations of the line: x= -12, y= -41, z= 0 so (-12, -41, 0) is on the line. The shortest distance from that point to the plane is along the perpendicular and we already know that any perpendicular to the plane is parallel to the vector (3, -1, -2). A line through the point (-12, -41, 0) in the direction of the vector (3, -1, -2) is, as I said before, given by x= 3t- 12, y= -t- 41, z= -2t. To find where the that line intersects the plane, put the x, y, z values into the equation ofthe plane:
    -3(-3t- 12)+ (-t- 41)+ 2(-2t)+ 2.5= (9- 1- 4)t+ 36- 41+ 2.5= 4t- 2.5= 0. t= 2.5/4= .625.

    Put that value of t into x= 3t- 12, y= -t- 41, z= -2t to find the point where the perpendicular line from (-12, -41, 0) passes through the plane. The distance between that point and the point (-12, -41, 0) is the distance from the line to the plane, the distance from the solutions set for the first set of equations and the solution set for the second set of equations.
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