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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. – Qiaochu Yuan Jun 2 '12 at 15:27
$f$ is a map $\mathbb R^3\to \mathbb R$ – palio Jun 2 '12 at 15:35
Then take $f(x, y, z) = (ax + by + cz - d)^2 + (ex + fy + gz - h)^2$. – Qiaochu Yuan Jun 2 '12 at 15:40
this is a polynomial!! what about a line being an algebraic curve so it is given by two equations $f_i(x,y,z)=0$? – palio Jun 2 '12 at 17:07
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 $\mathbb{R}$.) – Qiaochu Yuan Jun 2 '12 at 17:14

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 $f_i(x,y,z)=0$ – palio Jun 2 '12 at 17:11

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).

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