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Finding the length of an intersection Of 2 planes

  1. Jul 11, 2012 #1
    1. The problem statement, all variables and given/known data
    Line of intersection between P1: x+y+z=7 and P2:2x-3y-z=-8 crosses the XZ plane at point A and crosses the YZ plane at point B

    Find the length Of AB


    Okay so first of all i`m having trouble with understanding crossing the `XZ`plane or `YZ`plane.

    does this mean that A x=z=0 A(0,something,0) and B(something,0,0)


    What i`m thinking to do with this problem, is finding the points A and B, an then finding vector AB and finding the magnitude, and that should be the length.

    So the first thing i did was get the equation of the line..

    1 x+y+z=7
    2 2x-3y-z=-8

    subtract eq1-eq2

    -x+4y+0 = 15

    x= 4y-15 (3)

    sub (3) in 1

    4y-15 +y + z =7
    z= 22-4y
    y=t

    x=-15 +4t
    y=t
    z=22-4t

    r=(-15,0,22) + t(4,1,-5) (equation of line of intersection)

    so that means that the direction vector is paralell to the line segment AB

    but now i`m clueless as to how to solve for A or for B... can some one give me a hintÉ
     
  2. jcsd
  3. Jul 11, 2012 #2

    HallsofIvy

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    No, the "xz plane" has points with all values for x and z but y= 0. And the "yz plane has x= 0


    Yep, good plan!

    No, z- (-z)= 2z not 0. To eliminate z, add the two equations.

    Again, once you have found the correct equation for the line of intersection, to find A, set y= 0 to solve for the value of the parameter and the find x and y. To find B, set x= 0, solve for the parameter and use that to find y and z.
     
    Last edited: Jul 12, 2012
  4. Jul 11, 2012 #3

    AH THATS GENIUS!!!! Thank you kindly sir all of your help is much appreciated, and it really means a lot to me that there are people like you in the world who are willing to spread the knowledge of math. It`s quite beautiful actually.
     
  5. Jul 11, 2012 #4

    Okay so

    EQ1 + EQ2
    3x-2y + 0 =-1

    (3) x=(2y-1)/3

    plug 3 in eq1

    (2Y-1)/3 + Y + Z =7
    z=(22-5y)/3

    let y=t

    x=(2t-1)/3
    y=t
    z=(22-5t)/3

    r= (-1/3,0,22/3) + t(2/3,1,-5/3)

    so therefore for point A

    y=0

    therefore t=0

    x=-1/3
    z=22/3

    A(-1/3,0,22/3)

    for point B

    0=(2t-1)/3

    t=1/2

    x=0
    y=1/2
    z=13/2


    So B(0,1/2,13/2)

    so AB=(1/3,1/2,-5/6)

    mAGNITUDEofAB= √((1/3)sqrd +(1/2)sqrd + (-5/6)sqrd) = √(38)/6


    How does this look, personally the only thing i'm concerned with is the fact that i had an ugly parametric equation with the fractions.. but it would seem everything else is correct to me, from how you described to do this.


    * i made an error earlier, but fixed it.
     
    Last edited: Jul 11, 2012
  6. Jul 12, 2012 #5

    HallsofIvy

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    That's the answer I get.
     
  7. Jul 12, 2012 #6
    sweeeeeeeeeeet, Thank you very much for all your help, once again i appreciate it very much and i'm sure everyone else you help feels the same way. :D
     
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