Your question is a bit vague. If this doesn't answer your question, feel free to explain what you're looking for in the comments.
You mention:
i need to proof that normalization make the different length vectors to be a unite length
There is nothing to prove, really. If you normalize a (non-zero) vector, you divide the vector by its length or norm. This does not change the direction, only the length. The vector you end up with will be, precisely because you divided by its own length, a vector of unit length (length 1).
A few formulas; if a vector $\vec x = \left( x_1,x_2,\ldots,x_n\right)$, then its norm or length is given by:
$$\left\| \vec x \right\| = \sqrt{x_1^2 + x_2^2 + \ldots + x_n^2}$$
Note that $\left\| \vec x \right\|$ is not a vector, but a real number. If $\vec x \ne \vec 0$, then $\left\| \vec x \right\| \ne 0$ and you can divide:
$$\frac{1}{\left\| \vec x \right\|}\vec x$$
The resulting vector has the same direction as $\vec x$, but has length 1 since:
$$\left\| \frac{1}{\left\| \vec x \right\|}\vec x \right\| = \frac{1}{\left\| \vec x \right\|}\left\| \vec x \right\| = 1$$
As for the part about the angle formed by two vectors, the angle is related to these vectors via the dot product. For vectors $\vec x$ and $\vec y$ and the smallest angle $\alpha$ formed by these vectors, you have:
$$\vec x \cdot \vec y = \left\| \vec x \right\|\left\| \vec y \right\|\cos \alpha $$
But if you know the coordinates of these vectors, this dot product can also be found as:
$$\vec x \cdot \vec y = \sum_{i=1}^n x_iy_i = x_1y_1 + x_2y_2 + \ldots + x_ny_n $$
Combining both formulas for $\vec x \cdot \vec y$ allows to solve for $\cos\alpha$ and hence find the angle $\alpha$ between $\vec x$ and $\vec y$. Note that normalizing the vectors doesn't change the direction so it leaves the angle unchanged as well. The norms become easier, obviously!