# plane determined by 2 vectors

i have 2 perpendicular vectors in space . How can i determine the plane determined by the 2 vectors?

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The plane determined by two noncollinear vectors $\mathbf{v}_1$ and $\mathbf{v}_2$ is the collection of all vectors of the form $\alpha\mathbf{v}_1 + \beta\mathbf{v}_2$, with $\alpha$ and $\beta$ scalars.
If by "space" you happen to mean $\mathbb{R}^3$, $\mathbf{v}_1=(a,b,c)$ and $\mathbf{v}_2=(r,s,t)$, then $(a,b,c)\times(r,s,t)$ (the cross product) is perpendicular to both $(a,b,c)$ and $(r,s,t)$, hence perpendicular to the plane they determine, so it will be the normal to the plane. One you have the normal, presumably you know how to get the usual equation of the plane.
@Badescu: The plane perpendicular to $(a,b,c)$ is all vectors $(x,y,z)$ that are perpendicular to $(a,b,c)$. That means, you must have $(a,b,c)\cdot (x,y,z)=0$. That gives the equation. –  Arturo Magidin Oct 26 '10 at 14:41