What means to complete a pair of vectors $(w, s)$ to an arbitrary basis of $R^d$? I found in an article this : Let $B = (b_1, b_2, . . . , b_d)$ be an orthonormal basis of $R^d$ such that $<b1, b2 >=< w,x >$ (where $< ... >$ denotes linear span). In order to construct $B$, first complete $(w, x)$ to an arbitrary basis of $R^d$ and then apply Gram-Schmidt orthonormalization process.
My question is : what means to complete a pair of vectors $(w, s)$ to an arbitrary basis of $R^d$ ? 
I read this how to complete arbitrary basis knowing 2 orthonormal vectors of Rd (d > 2)$2$-orthonormal-vectors-of-rd-d-2
but I didn't understand how to choose randomly these vectors : "Practically, one could choose $d−2$ random vectors $x_1,…,x_{d−2}$ (e.g. be choosing each coordinate following a Gaussian distribution)."
Please use as many details as possible in your explanations.  

To complete a pair of vectors $(w,s)$ to an arbitrary basis of $\mathbb R^d$ means to find an ordered basis of $\mathbb R^d$ such that $w,s$ are the first two vectors of it. Of course there are a lot of differen possible choices. Not at all completely random however. 
The third vector $v_3$ should be chosen as not to belong to $\langle w,s\rangle$.
The (eventual) fourth vector $v_4$ should be chosen as not to belong to $\langle w,s, v_3\rangle$ and so on...
In this way each vector is not a linear combination of the previous ones and this is enough to show that they are all linearly independent. When you've got $d$ of them you have a basis. 
Nothing you've done so far guarantees you that this basis is orthonormal. At this point you apply Gram-Schmidt.
