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xEnOnn

  • 3 years ago

Is it possible to convert from an explicit equation back to its implicit equation? Suppose I have the following explicit equation: \[\left \{ (\frac{1}{2}-\frac{3}{2}t, -\frac{1}{2}+\frac{1}{2}t,t)|t \in \mathbb{R}\right \}\] Possible to change this back to its implicit form?

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  1. JamesJ
    • 3 years ago
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    Do you, mean for example, as z = t, then x = 1/2 ( 1 - 3z) y = 1/2 ( z - 1) ?

  2. xEnOnn
    • 3 years ago
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    oh..I meant the above explicit form of the equation was derived from the system of equations (implicit form): \[x+y+z=0\]\[x-y+2z=1\] So after getting the explicit form, can I change the explicit form back to this implicit form?

  3. xEnOnn
    • 3 years ago
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    But how do I find the equations of the planes? Is there a systematic way to do this? Probably a linear algebra way? I tried to solve for its nullspace and all of that but none of the ways lead me back to the implicit form of the 2 equations.

  4. JamesJ
    • 3 years ago
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    Oh wait. No. Think about this for a minute. The number of planes that contain one line in 3-D space is infinite.

  5. JamesJ
    • 3 years ago
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    Therefore there are an infinite number of planes that can intersect to give that plane.

  6. JamesJ
    • 3 years ago
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    Hence once given such a line, there is no unique solution to the question "what are the planes that intersect at that line."

  7. JamesJ
    • 3 years ago
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    So coming back to your question: yes, you can find any number of "implicit forms". But you may not find the original pair of planes/pair of equations.

  8. xEnOnn
    • 3 years ago
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    So it's kind of like a one way thing where I could change from an implicit form to an explicit form. After which, from the explicit form, I cannot change back to the original implicit form any more?

  9. JamesJ
    • 3 years ago
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    You can change back to A implicit form. But it many not be THE implicit form with which you started. So if you like, the operation from implicit forms to explicit forms is not a one-to-one operation.

  10. xEnOnn
    • 3 years ago
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    ahh.. Thanks you so much! :)

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