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Zenmo

  • one year ago

@Small Question, don't need to solve@ Find the general form of the equation of the plane with the given characteristics. Passes through (2, 2, 1) and (-1, 1, -1) and is perpendicular to 2x-3y+z=3.

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  1. Zenmo
    • one year ago
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    For the points (2, 2, 1) and (-1, 1, -1). How do I know, which point is the initial point? Whichever z variable value is greater?

  2. Zenmo
    • one year ago
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    |dw:1435136878321:dw|

  3. dan815
    • one year ago
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    either can be initial

  4. dan815
    • one year ago
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    so firsst find your normal line

  5. dan815
    • one year ago
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    mmaybe if i show u visually it will be easy for u to do it yourself

  6. Zenmo
    • one year ago
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    So if the solution in the book is 7x+y-11z-5=0, but I put the answer as -7x-y+11z+5=0. I would still be correct?

  7. dan815
    • one year ago
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    |dw:1435144221416:dw|

  8. ganeshie8
    • one year ago
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    x = -2 is equivalent to -x = 2

  9. dan815
    • one year ago
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    you want to find your normal line of the plane, move that normal like to a point, then see the other line connecting p1 and p2, then u have 2 lines and u can define your plane now

  10. dan815
    • one year ago
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    u can take the cross of these 2 lines to get your new normal <a,b,c> then the plane equation is ax+by+cz=d and u can find d but subbing in one of your points <x,y,z?

  11. dan815
    • one year ago
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    if that plane quations seems a little magical, u can get it more lgoically by computing the dot product of your normal line with one of your vectors in your plane, <x-x1,y-y1,z-z1> <a,b,c> dot <x-x1,y-y1,z-z1> =0 will also yield the same plane equation

  12. ganeshie8
    • one year ago
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    \[u\times v\] gives the normal vector right

  13. dan815
    • one year ago
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    yup

  14. Zenmo
    • one year ago
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    ok, I got it now. Thanks for clarification. X = -2 is the same as -X = 2. Which, is the same as u x v = - ( v x u).

  15. dan815
    • one year ago
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    pictures r nice :)

  16. ganeshie8
    • one year ago
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    Yes, direction of normal vector is not affected by multiplying the vector by a nonzero scalar

  17. Zenmo
    • one year ago
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    But, just to make sure. I can put the answer as -7x-y+11z+5=0, even though the solution is 7x+y-11z-5=0?

  18. ganeshie8
    • one year ago
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    both are same, but `7x+y-11z-5=0` looks more neat

  19. ganeshie8
    • one year ago
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    14x+2y-2z-10=0 also works

  20. Zenmo
    • one year ago
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    so to get 7x+y-11z-5=0. I would have to start with the point (2, 2, 1) then. Which, comes back to the question of what initial point to use.

  21. ganeshie8
    • one year ago
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    it is an equation, multiplying the same thing both sides doesn't change it in mathematically

  22. Zenmo
    • one year ago
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    Or, whenever I get a negative constant for variable X, I'll just divide it by -1?

  23. ganeshie8
    • one year ago
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    why do you care what point you start wid

  24. ganeshie8
    • one year ago
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    you end up wid same answer either way right

  25. Zenmo
    • one year ago
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    Yea, but the signs would be different, hence: -7x-y+11z+5=0 vs 7x+y-11z-5=0. So, either format works, but the positive one looks neater.

  26. Zenmo
    • one year ago
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    Just really really making sure

  27. ganeshie8
    • one year ago
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    okay i got you, then just multiply -1 in the end to make it look neat :)

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