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I'm working on an exercise in linear algebra and I'm stumped by part of it. Say I have a line in 3 dimensions that can be represented as $\operatorname{Span}\lbrace[1,-2,-2]\rbrace$. The book says this line can also be represented as the solution set of a pair of homogeneous equations. I'm assuming this would be the intersection of two planes.

The book gives the two equations as $\lbrace [x,y,z] \in \mathbb{R}^3 : [4,-1,1] \cdot [x,y,z] = 0, [0,1,1] \cdot [x,y,z]=0\rbrace$

I found the first plane on my own. I picked $[1,1,1]$ as an arbitrary point and then calculated the vector normal to both $[1,1,1]$ and $[1,-2,-2]$. That vector is $[0,1,1]$. But I couldn't figure out how to find the plane represented by $[4,-1,1]$. It seems to me that infinite planes could intersect along $[1,-2,-2]$, so what is special about $[4,-1,1]$ and how can I find it?

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1 Answer 1

There's infinitely many planes as you say, although the ones you find happen to be incorrect because $[0,1,1]\cdot[1,-2,-2] = -4\not = 0$ (just go through the calculations again, you should spot it). There's nothing special about $[4,-1,1]$, the two normal vectors of the planes you're looking for can be any $u,v\in\mathbb R^3$ such that $\mathrm {rank}(u,v,[1,-2,-2]) = 3$, which is equivalent to $\det(u,v,[1,-2,-2])\not= 0$. Maybe there's an extra condition you're missing which could limit the possibilities?

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