how to find the basis of a plane or a line?
Find a basis for the plane $x-2y+3z=0$ in $R^3$. Then find a basis for the intersection of that plane with the $xy$ plane.
Is there a proper/algebraic way of finding the basis of a plane?
Just by looking at it a basis could be $(2, 1, 0)$ because any multiple of that will give you $0$ when you substitute, but how do I find this without guessing?
would I use the same process when finding the basis of a line?
Any hints on how to figure out the second part of the question?
$\endgroup$ 31 Answer
$\begingroup$One thing you can identify is that $z = \frac{2y-x}{3}$, then the points that are going to satisfy the equality will be of the form $$ \left(x, y, \frac{2y-x}{3}\right). $$ For you to be able to cover all of such points, you would need to have two different vectors satisfying above such that they are not a multiple of each other.
You can see that indeed we can decompose the above vector as $$ \left(x, y, \frac{2y-x}{3}\right)=x\left(1,0,-\frac{1}{3}\right)+y\left(0,1,\frac{2}{3}\right). $$ which gives you an obvious basis $$ (v_1,v_2)=\left(1,0,-\frac{1}{3}\right)+\left(0,1,\frac{2}{3}\right). $$ There are many different ways of constructing a vector given in the first characterization, which would result in how the basis vector are aligned with respect to each other.
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