How can we define scalar product so that those three vectors will form orthonormal basis? How can we define scalar product on $\mathbb{R}^3$ so that vectors $\overrightarrow{x} = \begin{pmatrix}1\\0\\-1\end{pmatrix}, \overrightarrow{y} = \begin{pmatrix}1\\0\\1\end{pmatrix}, \overrightarrow{z} = \begin{pmatrix}1\\-1\\0\end{pmatrix}$ will form orthonormal basis of $\mathbb{R}^3$?
I tried to do it via orthonormality condition, i.e. $\langle x_i | x_j \rangle = \delta_{i,j}$, but I'm having a hard time coming with some reasonable scalar product.
Edit:
Changed title, so that question is how can we find the following scalar product.
P.S. I've seen this post, but I didn't get it, unfortunately.
 A: Let $A$ be the $3 \times 3$ matrix whose columns are your vectors. If indeed your vectors are linearly independent, the matrix $A \cdot A^T$ is invertible (because $A$ is) and then you can define an inner product $g(\vec{u}, \vec{v})$ on $\mathbb{R}^3$ by the formula
$$ g(\vec{u}, \vec{v}) := \vec{u}^T (A \cdot A^T)^{-1} \vec{v}. $$
For this inner product, the columns $A\vec{e_i}$ of $A$ will form an orthnormal basis because
$$ g(A\vec{e_i}, A\vec{e_j}) = \vec{e_i}^T A^T (A \cdot A^T)^{-1} A \vec{e_j} = \vec{e_i}^T A^T (A^T)^{-1} A^{-1} A \vec{e_j} = \vec{e_i} \vec{e_j}^T = \delta_{ij}. $$
This is indeed an inner product because the matrix $(A \cdot A^T)^{-1}$ is positive definite (being the inverse of a positive definite matrix).
A: This is always possible if the three vectors are linearly independent and span the space.
So the only question you need to ask yourself is: are they linearly independent? 
(Because three linearly independent vectors in $\mathbb{R}^3$ automatically span, since it is 3-dimensional.)
A proof of this fact can be found on p.201 of Winitzki's Linear Algebra via the Exterior Product.
The book is available online, for free, here: https://sites.google.com/site/winitzki/linalg
A: Yes. In fact, for every basis $v_1,\ldots,v_n$ of $\mathbb{R}^n$ one can define an inner product by declaring this basis to be orthonormal.
