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I Know it is impossible to do so since the parametric equation for a plane is the intersection of $2$ planes.For example:

$x$ $=$ $\frac{-5}{4t}+\frac{1}{4}$;

$y=\frac{3}{4t}+\frac{5}{4}$;

$z=t$

But when I combine any of the 2 equations above, I can only get the relationship between $x $ and $y$; $x$ and $z$ or $y$ and $z$, that is, $3$ lines in two dimensions. So but the intersection of $2$ planes yields for only $1$ line. So, don't they contradict with one another?

And I also have 2 associated questions:

  1. How to use the parametric equation of a line in three dimensions to get the equations of intersecting planes?

2.How to use the parametric equation of a line in three dimensions to get the equations of the line derive from the intersection of 2 planes?

TIWARI
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JimfyWinsy
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    The relationship you find between $x$ and $y$ (for example) can be considered a plane in $3$ dimensions—it's just that the equation of that plane doesn't happen to have $z$ in it. So you can find three planes that the line must lie in; any two of them intersect in the line. – Greg Martin Jun 22 '15 at 04:20
  • A line in three dimensions requires two equations. A single equation will "always" give a surface (if it's a first degree equation, that surface is a plane). I say "always", because you can transform several equations into one (or make equations with no solutions) by noting that a square is never negative. $x=y,x=z$ can, via $x-y=0,x-z=0$, be transformed into $(x-y)^2+(x-z)^2=0$. – Arthur Jun 22 '15 at 04:43
  • Sorry, but here there's a wrong edition.The parameter t should be in the numerator,and multiply by the scalars. – JimfyWinsy Jun 22 '15 at 07:50

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