Gram-Schmidt Calculation Does someone knows how to calculate to make these 3 vectors below to a Ortonormal basis:

I know how to use Gram-Schmidt but i lose myself in these calulations?
Because these Calculations contains fractions i get lost?
First i choose this vector below:

But after i should do this according to the Gram-Schmidt formula:

I get lost when i should calculate the Projection formula and take it minus the vector 1 it becomes weird should i use the magnitude in the projection formula because i get in the projection formula and here in
the projection formula i get something divided by something it feels weird??
Should i use the magnitude in the projection formula or?
And after that how should make it to an unit vector??


 A: If you look up Gram-Schimdt process, for example on wikipedia, you might notice that one starts from a set of vectors $\vec v_i$, calculates some orthogonal vectors $\vec u_i$, which then are normalized to get $\vec e_i$. The important part is how one calculates $\vec u_i$:
$$\vec u_i=\vec v_i-\sum_{j=1}^{i-1}\mathrm{proj}_\vec{u_j}(\vec v_i)$$
The projection operation is $$\mathrm{proj}_\vec{u_j}(\vec v_i)=\frac{\vec{u_j}\cdot \vec{v_i}}{\vec{u_j}\cdot\vec{u_j}}\vec{u_j}
$$
This is followed by normalization of the vectors:
$$\vec e_i=\frac{\vec u_i}{\sqrt{\vec u_i\cdot \vec u_i}}$$
Notice that we can write the projection operator in terms of $\vec e_i$, which would reduce the number of calculations:
$$\mathrm{proj}_\vec{u_j}(\vec v_i)=\frac{\vec{u_j}\cdot \vec{v_i}}{\sqrt{\vec{u_j}\cdot\vec{u_j}}}\frac{\vec{u_j}}{\sqrt{\vec{u_j}\cdot\vec{u_j}}}=(\vec e_j\cdot\vec v_i)\vec e_j=\mathrm{proj}_\vec{e_j}(\vec v_i)
$$
So You calculate first $\vec e_1$:
$$\vec e_1=\frac{\vec v_1}{\sqrt{\vec v_1\cdot\vec v_1}}=\frac{(-2,-4,3,0,0)}{\sqrt{29}}$$
Now calculate $\vec e_1\cdot \vec v_2$, and you will get $$\vec u_2=\frac13(1,-7,0,3,0)-\frac{26}{3\cdot 29}(-2,-4,3,0,0)$$
From here is just arithmetic, and repeat a similar procedure for $\vec e_3$.
