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anonymous

  • 5 years ago

calculate plane equation (1,0,0), (0,1,0),(0,0,1)

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  1. amistre64
    • 5 years ago
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    yay.... given three point we can create 2 vectors and cross product for the noral right?

  2. anonymous
    • 5 years ago
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    yes. but please do it. i'm being too lazy =P

  3. amistre64
    • 5 years ago
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    LOL.... happy to, i need the practice ;)

  4. amistre64
    • 5 years ago
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    lets choose a starting point; (1,0,0) looks fine; and create a vector from that point another and then the other

  5. anonymous
    • 5 years ago
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    cheers!

  6. amistre64
    • 5 years ago
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    (1,0,0) (0,1,0) ------- <1,1,0> = v1 (1,0,0) (0,0,1) ------ <1,0,1> = v2 right?

  7. amistre64
    • 5 years ago
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    we cross these 2 vectors to get a normal vector to the plane: < i, j, k> <1,1,0> <1,0,1> ---------- +(1-0) | - (1-0) | = <1,1,-1> for the normal +(0-1) |

  8. anonymous
    • 5 years ago
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    just the pane eq...

  9. amistre64
    • 5 years ago
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    now attach the normal to a point... once you fix my typo lol <1,-1,-1> = n ; and P(1,0,0) 1(x-1) -1(y+0) -1(z+0) = 0

  10. amistre64
    • 5 years ago
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    x-1 -y -z = 0 x-y-z-1 = 0 should satisfy it

  11. amistre64
    • 5 years ago
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    i wonder if i did my point addition right...

  12. anonymous
    • 5 years ago
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    thanks alot1 =)

  13. amistre64
    • 5 years ago
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    was it correct?

  14. anonymous
    • 5 years ago
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    i dunno =P its part of a long question.

  15. amistre64
    • 5 years ago
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    well, let me dbl check me thoughts and verify my results :)

  16. anonymous
    • 5 years ago
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    yes plz =)

  17. anonymous
    • 5 years ago
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    it correct! thumbs up!

  18. amistre64
    • 5 years ago
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    yay!! ..... i was like 84% sure :)

  19. anonymous
    • 5 years ago
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    x+y+z-1=0 the normal vector shd be 1,1,1. Use intuition. If you want to use the cross product to find the normal vector, it shd be: (1,-1,0)×(0,1,-1)=(1,1,1). Bear in mind that you should subtract one vector from another, NOT add them together. The next step should be verifying your answer, in this case, try sub. (1,0,0), (0,1,0), (0,0,1) into your equation to see if it makes sense.

  20. amistre64
    • 5 years ago
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    thats what my doubts were :) I saw that in the back of my head ..

  21. amistre64
    • 5 years ago
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    but i couldnt seem to get the crossproduct of the vectors; even with subtraction; to conform...

  22. amistre64
    • 5 years ago
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    (1,0,0) (1,0,0) -(0,1,0) -(0,0,1) ------- ------ <1,-1,0> <1,0,-1> <1,-1,0> <1,0,-1> <1-0,-(-1-0),0-1> = <1,1,-1>

  23. amistre64
    • 5 years ago
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    1(x-1) +1(y+0)-1(z+0) = 0 x + y -z -1 = 0 ....

  24. amistre64
    • 5 years ago
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    (0,0,1) = -2 then

  25. anonymous
    • 5 years ago
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    So the true answer is x+y+z=1 for your k component in the post 4 min ago, it should be (0)(1)-(-1)(1)=1

  26. amistre64
    • 5 years ago
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    .... i forgot how to multiply lol....

  27. anonymous
    • 5 years ago
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    ^.^ Are you from USA?

  28. amistre64
    • 5 years ago
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    I am

  29. amistre64
    • 5 years ago
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    i kept breezing over that k part and needed a 2nd set of eyes to overcome my inherent stupidity :) thanx..

  30. anonymous
    • 5 years ago
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    "add oil" Guess from where I come from.

  31. amistre64
    • 5 years ago
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    venezula?

  32. amistre64
    • 5 years ago
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    cape fear?

  33. amistre64
    • 5 years ago
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    madagascar... gotta be madagascar :)

  34. anonymous
    • 5 years ago
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    no =] it should not be hard to guess if you google "add oil"

  35. amistre64
    • 5 years ago
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    its cantonese; soo... you from canton?

  36. anonymous
    • 5 years ago
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    Hong kong actually

  37. anonymous
    • 5 years ago
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    so add oil

  38. amistre64
    • 5 years ago
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    thnx :) you too

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