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mags093

can anyone help? Find the distance the point P(1, -3, -5), is to the plane through the three points Q(4, -4, 0), R(8, 1, 3), and S(3, 0, 2).

  • one year ago
  • one year ago

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  1. amistre64
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    find 2 vectors from the given points to cross and find a normal to the plane

    • one year ago
  2. amistre64
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    then you can define a line using the given point and the unit normal vector

    • one year ago
  3. amistre64
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    that line will pierce the plane at the shortest distance ... since it creates a perp line; and the distance between the given point and the piercing point is the solution

    • one year ago
  4. mags093
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    so i find |PR| and |PS|?

    • one year ago
  5. amistre64
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    |dw:1359740566102:dw|

    • one year ago
  6. amistre64
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    im sure theres a formula for this, but as i said prior, I got no idea what it would be :)

    • one year ago
  7. mags093
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    ok thanks :) so i find the plane and then use the normal vector to find the distance of the point?

    • one year ago
  8. amistre64
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    Q(4, -4, 0) R(8, 1, 3) -S( 3, 0, 2) -S(3, 0, 2) ------------------------- <1,-4-2> <5,1,1> yes :)

    • one year ago
  9. mags093
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    howd you get -S?

    • one year ago
  10. mags093
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    i get where ya got it now :) but how would i find is the point is on it?

    • one year ago
  11. amistre64
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    -S is just subtracting the point S from the other 2 points to define the 2 vectors to cross

    • one year ago
  12. amistre64
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    normal I find is: <2,11,-21> we can either make that a unit vector or just use it as is to find the piercing point P(1, -3, -5) x = 1 + 2t y = -3+11t z = -5 - 21t lets use any of the 3 given points in the plane to define the plane equation with; ill use S(3,0,2) i spose 1(x-3)+11(y-0)-21(z-2)=0 and using the xyz values in terms of t 1(1 + 2t-3)+11(-3+11t-0)-21(-5 - 21t-2)=0 and the calculations give me t = -28/121 use that to define the piercing point if you want and distance it from the given P

    • one year ago
  13. amistre64
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    or we can unit the normal vector and just solve for t :) divide normal by sqrt(566) x = 1 + 2t/sqrt(566) y = -3+11t/sqrt(566) z = -5 - 21t/sqrt(566) 1(1 + 2t/sqrt(566)-3)+11(-3+11t/sqrt(566)-0)-21(-5 - 21t/sqrt(566)-2)=0 do it your way and lets see if you get something close to 5 :)

    • one year ago
  14. mags093
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    why sqrt9566)

    • one year ago
  15. amistre64
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    the normal is <2,11,-21> which has a length of sqrt(566) so to make a unit normal we have to divide all the parts by its length

    • one year ago
  16. amistre64
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    then its just a matter of finding out how many t units there are

    • one year ago
  17. mags093
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    i still dont understand how to find the distance of P to the plane?

    • one year ago
  18. amistre64
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    my idea, and granted its not like the textbook since I cant recall the textbook formula :) is to define the normal of the plane you want to measure from. Create a line equation using the P point so that you can determine the point on the plane that this line will pierce thru at. The distance between point P and the piercing point is the distance P is from the plane.

    • one year ago
  19. amistre64
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    |dw:1359741666062:dw|

    • one year ago
  20. mags093
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    the normal of the plane is sqrt(566), how do i create a line for P?

    • one year ago
  21. amistre64
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    thats the length of the vector that is found when crossing the plane vectors.

    • one year ago
  22. amistre64
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    Q(4, -4, 0) R(8, 1, 3) -S( 3, 0, 2) -S(3, 0, 2) ------------------------- <1,-4-2> <5,1,1> these are the 2 plane vectors I came up with they cross to get <2,11,-21> which has a length of sqrt(566)

    • one year ago
  23. mags093
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    ok but how do i get the distance of P then?

    • one year ago
  24. mags093
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    do i minus 92,11,-21) from the point i was given?

    • one year ago
  25. amistre64
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    if we apply the normal as a direction vector to the stated point P we create a line equation L = (1,-3,-5) + t<2,11,-21> we can use this parametrically as: x = 1 +2t y = -3 +11t z = -5 -21t do you agree?

    • one year ago
  26. mags093
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    ya but i still dont understand where i get the distance of P?

    • one year ago
  27. amistre64
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    do you agree that this line equation is perpendicular to the plane, and that it will pierce thru the plane at some point?

    • one year ago
  28. mags093
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    does it not pierce through at the point (1,-3,-5) ?

    • one year ago
  29. amistre64
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    no, that is the given point that is floating someplace above the plane

    • one year ago
  30. amistre64
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    lets take the normal vector and define the plane equation with it 2(x-xo) + 11(y-yo) - 21(z-zo) = 0 I will use the one of the points that are already on the plane, I chose point S (3,0,2) 2(x-3) + 11(y-0) - 21(z-2) = 0 do you agree that this equation will define all xyz point on the plane?

    • one year ago
  31. mags093
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    so do i sub in that point into the line to get the distance to the plane?

    • one year ago
  32. mags093
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    ya i do

    • one year ago
  33. amistre64
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    now for any given point on the line we have the xyz equation for them in terms of "t" so lets put the line equation point values into the plane equation and solve for t x = 1 +2t y = -3 +11t z = -5 -21t 2(x-3) + 11(y-0) - 21(z-2) = 0 2((1 +2t)-3) + 11((-3 +11t)-0) - 21((-5 -21t)-2) = 0 this is a longer way to do it, if I could recall the formula it would be simpler prolly, but this works fine as well do you agree that the value of t will define us a point x,y,z that is common to both the line and the plane?

    • one year ago
  34. amistre64
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    the formula seems to be: http://mathworld.wolfram.com/Point-PlaneDistance.html \[D=\frac{ax_o+by_o+cz_o+d}{\sqrt{a^2+b^2+c^2}}\]

    • one year ago
  35. mags093
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    i worked that out and got this? 2(-3-6t) +11-21(10+42t)=0 is this right?

    • one year ago
  36. mags093
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    i dont understand how to use that formula?

    • one year ago
  37. amistre64
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    a,b,c is the normal vector components: <2,11,-21> xo,yo,zo is the stated point: (1,-3,-5) d is the constant on the plane equation, so i found d=36 from the plane equation giving me:\[D=\frac{2(1)+11(-3)-21(-5)+36}{\sqrt{2^2+11^2+21^2}}=5\sqrt{\frac{2}{283}}=4.6237~ or~ so\]

    • one year ago
  38. amistre64
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    if we work out my idea : 2((1 +2t)-3) + 11((-3 +11t)-0) - 21((-5 -21t)-2) = 0 t = -55/283 which would define the point x,y,z that is common to both line and plane. or, if we unitify the normal to begin with by dividing it by its length sqrt(2^2+11^2+21^2) we get: 2((1 +2t/sqrt(566))-3) + 11((-3 +11t/sqrt(566))-0) - 21((-5 -21t/sqrt(566))-2) = 0 and t equals the distance required of 55 sqrt(2/283) just like the formula

    • one year ago
  39. mags093
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    ya your ans right :) i tried working it out myself and got it wrong though :/

    • one year ago
  40. amistre64
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    i got help from the wolf so i didnt have to kill an eraser ;)

    • one year ago
  41. mags093
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    wolfram alpha?

    • one year ago
  42. amistre64
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    yep

    • one year ago
  43. amistre64
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    maybe, my smarts tend to fade exponentially as the day progresses ;)

    • one year ago
  44. mags093
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    ok well this is it anyway :) Find the distance the point P(1,5,5) is to the line through the two points Q(1,2,5), and R(4,4,2).

    • one year ago
  45. mags093
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    do i just get the normal vector and use the same formula then?

    • one year ago
  46. amistre64
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    you do something similar; but we dont have a plane to play with now.

    • one year ago
  47. mags093
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    its the one i tried earlier but couldnt get it!

    • one year ago
  48. amistre64
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    can you define the vector from Q to R for me?

    • one year ago
  49. mags093
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    how do i define a vector? is that when i multiply them out?

    • one year ago
  50. amistre64
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    a vector is defined by subtracting one point from the other.

    • one year ago
  51. mags093
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    so its (3,2,-3) i think :)

    • one year ago
  52. amistre64
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    if we move the points so that one of them is at the origin (0,0,0) the components of the vector are defined by the nonorigined point Q(1,2,5) R(4,4,2) -Q(1,2,5) -Q(1,2,5) --------------------- (0,0,0) (3,2,-3) so yes the vector between them is <3,2,-3>

    • one year ago
  53. amistre64
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    lets move the point P as well to define its vector as QP Q(1,2,5) R(4,4,2) P(1,5,5) -Q(1,2,5) -Q(1,2,5) -Q(1,2,5) -------------------------------- (0,0,0) (3,2,-3) (0,3,0) all we have done is move things on the graph, we havent change the nature of the problem itself

    • one year ago
  54. amistre64
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    now how do you find the angle between 2 vectors?

    • one year ago
  55. mags093
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    is it xi,yj,zk ?

    • one year ago
  56. amistre64
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    |dw:1359744085632:dw|

    • one year ago
  57. mags093
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    is it the length of the vectors * coz theta?

    • one year ago
  58. amistre64
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    \[cos\alpha=\frac{\vec u \cdot \vec v}{|\vec u|~| \vec v|}\] \[\alpha=cos^{-1}\frac{\vec u \cdot \vec v}{|\vec u|~| \vec v|}\]

    • one year ago
  59. amistre64
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    u = <0,3,0>; |u| = 3 v = <3,2,-3>; |v| = sqrt(20) --------- u.v= 0+6+0 = 6

    • one year ago
  60. mags093
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    so its cos^-1(6) and then we find the angle ya?

    • one year ago
  61. amistre64
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    this gives us: \[\alpha=cos^{-1}\frac{6}{3\sqrt(20)}\] the distance is equal to |u| sin a \[|u|~sin \alpha=|u|~sin\left(cos^{-1}\frac{6}{3\sqrt(20)}\right)\] \[3~sin \alpha=3~sin\left(cos^{-1}\frac{2}{\sqrt(20)}\right)\]

    • one year ago
  62. amistre64
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    http://www.wolframalpha.com/input/?i=3sin%28arccos%282%2Fsqrt%2820%29%29%29

    • one year ago
  63. mags093
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    so the angle should be 6/sqrt(5) ? it says thats wrong?

    • one year ago
  64. amistre64
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    you are not asked to find an angle are you, just the distance between a point and a line

    • one year ago
  65. mags093
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    oh ya just the distance not the angle

    • one year ago
  66. amistre64
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    http://mathworld.wolfram.com/Point-LineDistance3-Dimensional.html this is the distance formula for this then if they need an exact answer :)

    • one year ago
  67. amistre64
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    \[d=\frac{|uxv|}{|u|}\]

    • one year ago
  68. amistre64
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    or i think that shows |v| for my setup

    • one year ago
  69. amistre64
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    | (3,2,-3) x (0,3,0)| = |(9,0,9)| = sqrt(162) and |v| = sqrt(20) sqrt(162/20) = 9/sqrt(10) if I seen it right

    • one year ago
  70. mags093
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    ok so x1= Q and x2=R what do i need the t for?

    • one year ago
  71. amistre64
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    the t is just demonstrating the process of how they derived the formula.

    • one year ago
  72. amistre64
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    the formula is the key, and knowing which vector parts to put in which place is important as well :)

    • one year ago
  73. amistre64
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    x0 = p x1 = q x2 = r so you need the vectors from: x0-x1; x0-x2; and x2-x1

    • one year ago
  74. mags093
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    |dw:1359745138247:dw|

    • one year ago
  75. mags093
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    the line = (1,2,5) + t(3,2,-3) right?

    • one year ago
  76. amistre64
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    almost had that lol, ... lets do this :) define the direction vector from Q to R as the normal vector of a plane containing P QR = <3,2,-3> and P = (1,5,5) plane: 3(x-1)+2(y-5)+3(z-5) = 0 the line containing QR is then defined as: x=1+3t y=2+2t z=5-3t <--- -3 not -5 :) 3((1+3t)-1)+2((2+2t)-5)+3((5-3t)-5) = 0 t = 3/2 x=2+3(3/2) = 11/2 y=4+2(3/2) = 10/2 z=10-5(3/2) = -5/2

    • one year ago
  77. mags093
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    is z not equal to 5 -3(t)?

    • one year ago
  78. mags093
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    oh ya you changed that :)

    • one year ago
  79. amistre64
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    it is, just typoed it ;) sqrt((1-11/2)^2+(2-10/2)^2+(5+5/2)^2) http://www.wolframalpha.com/input/?i=sqrt%28%281-11%2F2%29%5E2%2B%282-10%2F2%29%5E2%2B%285%2B5%2F2%29%5E2%29

    • one year ago
  80. mags093
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    where did you getbthe new x,y,z from? :)

    • one year ago
  81. amistre64
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    the new xyz is the point where the line QR pierces thru the plane

    • one year ago
  82. amistre64
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    by putting the line parts for xyz into the plane equation, you define a point that is both on the line and in the plane

    • one year ago
  83. mags093
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    oh ok :) it says its wrong though? :/

    • one year ago
  84. amistre64
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    yeah, i prolly tripped up on the math along the way :/

    • one year ago
  85. mags093
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    oh ok :) ill try it now :) i did it the d formula and got -8+d/sqrt(4) ?

    • one year ago
  86. amistre64
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    cos a = u.v/|u||v| u.v = 6; |u||v| = 3 sqrt20 cos a = 2/sqrt(20) is the correct answer by chance, 6/sqrt(5) ??

    • one year ago
  87. mags093
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    is it suppose to be 10-3(3/2) for z?

    • one year ago
  88. mags093
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    no thats wrong :/

    • one year ago
  89. amistre64
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    ....buggers :/

    • one year ago
  90. mags093
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    is z= 10 -3(3/2) rather than 10-5(3/2) ??

    • one year ago
  91. amistre64
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    yes

    • one year ago
  92. amistre64
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    and with any luck, we get something like 9/sqrt(11) ... crosses fingers

    • one year ago
  93. mags093
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    http://www.wolframalpha.com/input/?i=sqrt%28%281-13%2F2%29%5E2%2B%282-7%29%5E2%2B%285-11%2F2%29%5E2%29

    • one year ago
  94. mags093
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    could that be right?

    • one year ago
  95. mags093
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    still wrong :(

    • one year ago
  96. amistre64
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    distance Q from plane containg P is:\[\frac{3(1)+2(2)-3(5)+2}{\sqrt{3^2+2^2+3^2}}\] and distance Q is from P is 3 giving me: |dw:1359746813395:dw|

    • one year ago
  97. amistre64
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    Q(1,2,5) R(4,4,2) P(1,5,5) -P(1,5,5) -P(1,5,5) -P(1,5,5) -------------------------------- (0,-3,0) (3,-1,-3) (0,0,0) (0,-3,0) x (3,-1,-3) = (9,0,9) ; 9 sqrt(2) 9sqrt(2)/|QR| |QR| = sqrt(3^2+2^2+3^2) = sqrt(22) 9sqrt(2/22) = 9/sqrt(11)

    • one year ago
  98. mags093
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    thats right :) howd you figure it out?

    • one year ago
  99. amistre64
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    i had to see which points the formula was refering to and i worked it out from there. I worked it out earlier using the distance of Q to the plane and the distance from Q to P; and pythaging it

    • one year ago
  100. mags093
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    oh ok thanks :)

    • one year ago
  101. amistre64
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    im going to bed after that one :) good luck

    • one year ago
  102. mags093
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    ok thanks :) bye

    • one year ago
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