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s3a

  • 4 years ago

Question about equation of plane, parametric equations and stuff like that: http://f.imgtmp.com/HIhr6.jpg Help would be GREATLY appreciated!

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  1. amistre64
    • 4 years ago
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    personally id create 2 vectors and cross them for the normal vector

  2. amistre64
    • 4 years ago
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    can you type up the points for me, its hard to keep track of them

  3. s3a
    • 4 years ago
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    that's to get the first part?

  4. amistre64
    • 4 years ago
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    yeah, plane equation

  5. s3a
    • 4 years ago
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    P(-4,-6,2) and Q(-8,-7,0), R(7,3,5)

  6. amistre64
    • 4 years ago
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    good, now what im gonna do is define 2 vectors by subtracting one point from the other 2

  7. s3a
    • 4 years ago
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    wait, first let me finish part 1

  8. amistre64
    • 4 years ago
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    Q(-8,-7,0) R( 7, 3, 5) - P(-4,-6,2) - P(-4,-6,2) ------------------------- -<4, 1, 2> <11,9,3> these are 2 vectors that reside on your plane between your points

  9. amistre64
    • 4 years ago
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    lets cross them for a normal vector <4, 1, 2> <11,9,3> ---------- x = (3-18) = -15 y = -(12-22) = 10 z = (36-11) = 25 <-15,10,25> is the normal vector which we can scale down to work with -------- -5 <3,-2,-5> looks good to me

  10. amistre64
    • 4 years ago
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    now, given any point on the plane; they gave you 3 to play with; we can construct the equation as such: 3(x-Px)-2(y-Py)-5(z-Pz)=0

  11. s3a
    • 4 years ago
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    2 secs, i was searching for something. now i can read properly.

  12. amistre64
    • 4 years ago
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    id use a simple point like (-8,-7,0) 3(x--8)-2(y--7)-5(z-0)=0 3x+24-2y-14-5z=0 3x-2y-5z+10=0 if i did it right that should be a match to your 3 points

  13. s3a
    • 4 years ago
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    ok so <3,-2,-5> is the normal vector and 3(x-Px)-2(y-Py)-5(z-Pz)=0 is from dot producting it?

  14. amistre64
    • 4 years ago
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    in effect yes.

  15. s3a
    • 4 years ago
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    ya i get that part now thank you. :)

  16. s3a
    • 4 years ago
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    what about the line?

  17. amistre64
    • 4 years ago
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    gotta dbl chk my plane equation first, im prone to error at my age :)

  18. amistre64
    • 4 years ago
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    wolfram likes it :)

  19. amistre64
    • 4 years ago
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    so line PQ ... let me see

  20. amistre64
    • 4 years ago
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    its a vector addition of P and Q right?

  21. amistre64
    • 4 years ago
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    |dw:1327363860012:dw|

  22. amistre64
    • 4 years ago
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    vector subtraction that is and the points themselvs act as defining the vectors

  23. s3a
    • 4 years ago
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    i just pictured like a toddler helping me on the other side :P

  24. s3a
    • 4 years ago
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    and sorry technical issues make it hard to respond fast here

  25. amistre64
    • 4 years ago
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    so, P-Q, i did above = <4,1,2> and this needs to be scaled to any length to create the line t<4,1,2> now all we have to do is attach it to either P or Q

  26. s3a
    • 4 years ago
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    i'm guessing your older rather than younger

  27. amistre64
    • 4 years ago
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    the best thing about today is that im one step closer to death :)

  28. s3a
    • 4 years ago
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    there is a comma between P and Q if that matters.

  29. s3a
    • 4 years ago
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    lol ... well eat healthy, etc

  30. amistre64
    • 4 years ago
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    using -8,-7,0 to anchor our line vector to we get: x = -8 + 4t y = -7 + 1t z = 0 + 2t should be the parametric stuff

  31. amistre64
    • 4 years ago
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    if you plug in P for the x y z parts you should come out to t being the same for all parametrics

  32. amistre64
    • 4 years ago
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    P=-4,-6,2 -4 = -8 +4t 4 = 4t t = 1 -7+1 = -6 0+2 = 2... that works for me :)

  33. s3a
    • 4 years ago
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    ok so you used Q to anchor and P works too. just confirmed.

  34. s3a
    • 4 years ago
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    sorry for being delayed.

  35. amistre64
    • 4 years ago
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    'sok, this computers getting slow as well

  36. amistre64
    • 4 years ago
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    now the next one is perp to L and hits R

  37. s3a
    • 4 years ago
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    wait before that.

  38. s3a
    • 4 years ago
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    i didn't get what you did when you found a value for t and why you did it.

  39. s3a
    • 4 years ago
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    also, i don't recommend you use CRT. rather use LCD screens because they don't harm your eyes or are negligible in comparison. plasmas make radiation that dissipates before harming you so it's acceptable but i'd go with lcd anyways.

  40. s3a
    • 4 years ago
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    just saying.

  41. amistre64
    • 4 years ago
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    since im anchored to Q and it has to go thru P then some value of "t" has to remain the same for each parametric in order to come up with the P values. I solved for Px in place of x to see if we can get a t that worked

  42. amistre64
    • 4 years ago
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    when t=1 we get Px so i just subbed in t=1 for the others to see if they matched Py and Pz

  43. amistre64
    • 4 years ago
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    if we anchored to P, then we would have used Q to test it out with :)

  44. s3a
    • 4 years ago
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    i see it now :D

  45. s3a
    • 4 years ago
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    so what about the last part?

  46. amistre64
    • 4 years ago
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    the last part I believe uses this line vector as a normal vector of this other plane; and this other plane has to hit R so we should just place Rx Ry Rz in the plane equation

  47. amistre64
    • 4 years ago
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    |dw:1327364753446:dw|

  48. s3a
    • 4 years ago
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    passing through means parallel but more precise?

  49. s3a
    • 4 years ago
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    i mean covering it wholly

  50. amistre64
    • 4 years ago
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    sooo, n<4,1,2>, calibrate with R(x,y,z) 4(x-Rx)+1(y-Ry)+2(z-Rz)=0 expanded of course

  51. amistre64
    • 4 years ago
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    we want a plane perp to L, not parallel

  52. amistre64
    • 4 years ago
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    which means L is our norm for this perp plane we need a point on that plane to anchor to, they say use R

  53. s3a
    • 4 years ago
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    o R is a point.

  54. s3a
    • 4 years ago
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    makes sense since a point R is given :P

  55. amistre64
    • 4 years ago
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    yeah, it was what (7,3,5) right?

  56. amistre64
    • 4 years ago
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    4(x-7)+1(y-3)+2(z-5)=0 4x-28+y-3+2z-10=0 4x+y+2z-10-28-3=0 4x+y+2z-41=0 maybe?

  57. s3a
    • 4 years ago
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    R(7,3,5) yes

  58. s3a
    • 4 years ago
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    o too late lol

  59. amistre64
    • 4 years ago
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    im quick for an old man lol

  60. s3a
    • 4 years ago
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    lol eye of the tiger

  61. amistre64
    • 4 years ago
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    im sure well be going over these in calc3 this term.

  62. amistre64
    • 4 years ago
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    trying to get me a masters in math before i die :)

  63. s3a
    • 4 years ago
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    go for it lol and eat healthy!

  64. amistre64
    • 4 years ago
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    :) ill try. good luck

  65. amistre64
    • 4 years ago
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    we got the part 3 right?

  66. s3a
    • 4 years ago
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    i think that;'s wrong. brb

  67. amistre64
    • 4 years ago
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    pretty confident its good. the line vector from part 2 is our normal, and we anchored it with pointR; unless i missed the multiplication and adding parts :) that happens

  68. s3a
    • 4 years ago
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    I'm at part three if you're going to help.

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