Orthonormal basis . Can I have more than one basis for the subspace?

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Required to find an orthonormal basis for the following subspace of $\Bbb R^4$: $$\left\{\left.\begin{pmatrix}x \\ y \\ z \\ w\end{pmatrix}\right\vert x-y-z+w=0 \ \ \text{and} \ \ x+z=0\right\}.$$

I know that to find the othonormal basis, it is required that i find the basis for the subspace, then I use Gram Schmidt process. Afterwards Ill normalize the vectors I get from the GS process and that should give me the orthonormal basis. My question really is, can I get more than one basis for the subspace? I got $\{(3, -2, 1, 0) , ( 0, 1, 0, 1)\}$ whereas my teacher got $\{(-1, -2, 1, 0), (0, 1, 0, 1) \}$. Note: we both used different free variables.

Also, If I used $x$ and $y$ as free variables and get my basis, then i use $x$ and $z$ as free variables to get my basis. They both are legit bases right ?

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Think of $\mathbb R^2$ - the real plane - since your subspace clearly has dimension $2$ and is therefore a plane.

Take an arbitrary unit vector at the origin. Can you find a second unit vector at right-angles to the first?

That is essentially the configuration you are looking for. It doesn't compute the basis vectors for you, but I hope it helps you to visualise the problem.

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Yes: any $k$ independent vectors are the basis of a space of dimension $k$. (They are from this space, of course).