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mags093

  • one year ago

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).

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

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

  3. amistre64
    • one year ago
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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

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

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

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

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

  8. amistre64
    • one year ago
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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 :)

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

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

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

  12. amistre64
    • one year ago
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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

  13. amistre64
    • one year ago
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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 :)

  14. mags093
    • one year ago
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    why sqrt9566)

  15. amistre64
    • one year ago
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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

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

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

  18. amistre64
    • one year ago
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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.

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

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

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

  22. amistre64
    • one year ago
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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)

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

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

  25. amistre64
    • one year ago
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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?

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

  27. amistre64
    • one year ago
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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?

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

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

  30. amistre64
    • one year ago
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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?

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

  32. mags093
    • one year ago
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    ya i do

  33. amistre64
    • one year ago
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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?

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

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

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

  37. amistre64
    • one year ago
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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:\[D=\frac{2(1)+11(-3)-21(-5)+36}{\sqrt{2^2+11^2+21^2}}=5\sqrt{\frac{2}{283}}=4.6237~ or~ so\]

  38. amistre64
    • one year ago
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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

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

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

  41. mags093
    • one year ago
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    wolfram alpha?

  42. amistre64
    • one year ago
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    yep

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

  44. mags093
    • one year ago
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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).

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

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

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

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

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

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

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

  52. amistre64
    • one year ago
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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>

  53. amistre64
    • one year ago
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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

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

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

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

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

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

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

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

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

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

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

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

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

  66. amistre64
    • one year ago
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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 :)

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

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

  69. amistre64
    • one year ago
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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

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

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

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

  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

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

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

  76. amistre64
    • one year ago
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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

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

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

  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

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

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

  82. amistre64
    • one year ago
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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

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

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

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

  86. amistre64
    • one year ago
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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) ??

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

  88. mags093
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    no thats wrong :/

  89. amistre64
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    ....buggers :/

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

  91. amistre64
    • one year ago
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    yes

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

  93. mags093
    • one year ago
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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

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

  95. mags093
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    still wrong :(

  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|

  97. amistre64
    • one year ago
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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)

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

  99. amistre64
    • one year ago
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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

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

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

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

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