Find an orthogonal basis of $R^3$ which contains a vector Find an orthogonal basis of ${\rm I\!R}^3$ which contains the vector $v=\begin{bmatrix}1\\1\\1\end{bmatrix}$.
I want to solve this without the use of the cross-product or G-S process. Please look at my solution and let me know if I did it right.
Let $u$ be an arbitrary vector $u=\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}$ that is orthogonal to $v$.
Thus,
$v\  \bullet\  u = x_1 + x_2 + x_3 = 0$
$0= x_1 + x_2 + x_3$
$x_1= -x_2 -x_3$
Any vector of the form $\begin{bmatrix}-x_2 -x_3\\x_2\\x_3\end{bmatrix}$ will be orthogonal to $v$.
Let $x_2 = x_3 = 1$
So, $u=\begin{bmatrix}-2\\1\\1\end{bmatrix}$ is orthogonal to $v$.
We now have two orthogonal vectors $u$ and $v$.
Put $u$ and $v$ as rows of a matrix, called $A$. Find a basis for $A^\bot = null(A)^T$:  

Digression: I have memorized that when looking for a basis of $A^\bot$, we put the orthogonal vectors as the rows of a matrix, but I do not
  know why we put them as the rows and not the columns. Anyone care to explain the intuition?

$A=\begin{bmatrix}1&1&1\\-2&1&1\end{bmatrix} \sim \begin{bmatrix}1&0&0\\0&1&1\end{bmatrix}$
$x_3 = x_3$
$x_2 = -x_3$
$x_1 = 0$ 
A basis for $null(A)$ or $A^\bot$ with $x_3$ = 1 is: $(0,-1,1)$. Call this $w$.
Therefore, $w$ is orthogonal to both $u$ and $v$ and is a basis which spans ${\rm I\!R}^3$.
By definition of orthogonal vectors, the set $[u,v,w]$ are all linearly independent.
Is this correct? If so, what is a more efficient way to do this? If not, how do you do this keeping in mind I can't use the cross product G-S process?

If you use the same reasoning to get $w=(x_1,x_2,x_3)$ (that you did to get $v$), then $0=v\cdot w=-2x_1+x_2+x_3$.  So in general,  $(\frac{x_2+x_3}2,x_2,x_3)$ will be orthogonal to $v$.  
Now we get $-x_2-x_3=\frac{x_2+x_3}2$ (since $w$ needs to be orthogonal to both $u$ and $v$). 
So, $-2x_2-2x_3=x_2+x_3$.  Then $x_2=-x_3$.
So, say $x_2=1,x_3=-1$.  Then we get $w=(0,1,-1)$.
