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In $\mathbb R^3$ we write the equation of a plane as $ax+by+cz=d$. Do we have a similar equation of a line in $\mathbb R^3$? In my knowledge we don't have such equation. All we know is that given a point on a line $L$ and a vector parallel to $L$ we can derive a vector equation and a parametric equation but we don't have an equation in the form $f(x,y,z)=0$ is this correct?

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    Depends on what you allow $f$ to be.2012-06-02
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    $f$ is a map $\mathbb R^3\to \mathbb R$2012-06-02
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    Then take $f(x, y, z) = (ax + by + cz - d)^2 + (ex + fy + gz - h)^2$.2012-06-02
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    this is a polynomial!! what about a line being an algebraic curve so it is given by two equations $f_i(x,y,z)=0$?2012-06-02
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    I don't understand your question. You said "map." $f$ is a map. Yes, lines in $\mathbb{R}^3$ are given by two (linear) equations. They can _also_ be given by one quadratic equation. (This is specific to $\mathbb{R}$.)2012-06-02
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    i'm confused. you are writing a line as the zero set of a polynomial $f(x,y,z)$ isn't this a surface, i mean an algebraic variety of dimension $2$ while a line is an algebraic variety of dimension $1$?2012-06-02

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I believe the nearest thing would be $$\frac{x-a}{d} = \frac{y-b}{e} = \frac{z-c}{f},$$ which is not quite of the form $f(x,y,z)=0$, but it is standard and about as close as you will get to what you want.

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    and we can rewrite the above in the form of two linearly independant equations $f_i(x,y,z)=0$2012-06-02
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No, that's not (quite) correct. You need two (linear) equations to write down a line in the form $f(x, y, z) = 0 $ The kernel of a linear map $\mathbb{R}^3\rightarrow\mathbb{R}^2$ of rank $2$ is one dimensional, just translate it (I'm too lazy right now to recall how to LaTeX a matrix).