# finding two polynomials that their roots are a given line.

Given a field $F$ and $A = F^3$. we define $L$ to be the line that goes through the points: $(8,1,-1)$, $(5,0,-1)$. My object is to find two polynomials $q(X_1,X_2,X_3)$, $p(X_1,X_2,X_3)$ in $F[X_1,X_2,X_3]$ of degree $\leq 1$ such that $L$ is the set of zeros of $p$ and $q$.

thanks. benny

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You basically are looking for the equation of a line in three-dimensional space (as an intersection of two planes). – Joel Cohen Jan 15 '12 at 12:42
joel, do you mean I am looking for the intersection of the planes: 8X1+X2-X3=0 and 5X1-X3=0? – benjamin Jan 15 '12 at 12:52

## 1 Answer

Since everything turns out to generalize directly from $\mathbb R^3$, it helps to think geometrically. As Joel says, what you're looking for is the equations for two planes such that their intersection is the line. There are many solutions to this; any two different planes that contain your two points will work.

You know how to find an equation for a plane given a normal vector and a point in it, right? So the task is just to find to non-parallel vectors perpendicular to the line.

The direction of the line is $D=P_1-P_2=(3,1,0)$. It is easy enough to find some perpendiculars by guessing, such as $n_1=(1,-3,0)$ and $n_2=(0,0,1)$.

If we don't want to guess (in a general field it might not be immediately clear that our guesses are not parallel) we can be a bit more systematic by finding some vector $e$ that is simply not parallel to $D$ and then using the cross product to get normals $n_1=d\times e$, $n_2=d\times n_1$.

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