Matrix with orthonormal base I have the two following given vectors:
$\vec{v_{1} }=\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$
 $\vec{v_{2} }=\begin{pmatrix} 3 \\ 0 \\ -3 \end{pmatrix}  $
I have to calculate matrix $B$ so that these vectors in $\mathbb{R}^{3}$ construct an orthonormal basis.
The solution is: 
$$B=\begin{pmatrix} 0 & -\frac{\sqrt{2} }{2}  & \frac{\sqrt{2} }{2}  \\ 1 & 0 & 0 \\ 0 & -\frac{\sqrt{2} }{2}  & -\frac{\sqrt{2} }{2}   \end{pmatrix}$$
I really don't have any idea how to get this matrix. I'm also confused because I only have 2 vectors.

Maybe these calculations would help you.
We need to find vertor $\vec{v}_3$ such that $\vec{v}_3\perp\vec{v}_1$ and $\vec{v}_3\perp \vec{v}_2$, i.e. 
$$
\begin{cases}
(\vec{v}_1, \vec{v}_3) = 0, \\
(\vec{v}_2, \vec{v}_3) = 0.
\end{cases}
$$
Here $(\vec{x},\vec{y})$ is a scalar product of vectors $\vec{x}$ and $\vec{y}$.
If we denote $\vec{v}_3$ as $(x_1,x_2,x_3)^T$ we get the system
$$
\begin{cases}
0\cdot x_1 + 1\cdot x_2 + 0\cdot x_3 = 0, \\
3\cdot x_1 +0\cdot x_2 - 3\cdot x _3 = 0
\end{cases} \iff 
\begin{cases}
x_2 = 0, \\
3x_1 - 3x_3 = 0
\end{cases}\iff 
\begin{cases}
x_2 = 0, \\
x_1 = x_3.
\end{cases}
$$
So vector $\vec{v}_3$ is depends on one parameter $x$ and has form $(x,0,x)^T$. 
Then we need to normalize this system, i.e. calculate vectors $\vec{u}_i = \dfrac{\vec{v}_i}{||\vec{v}_i||}$. We get
$$
\vec{u}_1 = \frac{1}{\sqrt{1^2}}
\begin{pmatrix}
0 \\ 1 \\ 0
\end{pmatrix} = 
\begin{pmatrix}
0 \\ 1 \\ 0
\end{pmatrix};
$$
$$
\vec{u}_2 = \frac{1}{\sqrt{3^2 + (-3)^2}}
\begin{pmatrix}
3 \\ 0 \\ -3
\end{pmatrix} = 
\begin{pmatrix}
\frac{\sqrt{2}}{2} \\ 0 \\ -\frac{\sqrt{2}}{2}
\end{pmatrix};
$$
$$
\vec{u}_3 = \frac{1}{\sqrt{x^2 + x^2}}
\begin{pmatrix}
x \\ 0 \\ x
\end{pmatrix} = 
\begin{pmatrix}
\frac{\sqrt{2}}{2} \\ 0 \\ \frac{\sqrt{2}}{2}
\end{pmatrix}.
$$
One may see that system of vectors $(\vec{u}_1,\vec{u}_2,\vec{u}_3)$ is orthonormal.
