how do I calculate the normal from a point Ive read this post Normal Vector and I know it explains how to calculate it at the top, but I don't understand it, can someone explain it a bit better for me with examples please?
 A: For simplicity I will derive it for $y=f(x)$. 
The first thing is to find the tangent vector of the curve. In vector form the curve can be written as $$\vec r=x \vec e_i + f(x) \vec e_j$$
Since I want to find the tangent I apply an increment to the vector such as
$$\vec r+\Delta\vec r=(x+\Delta x)\vec e_i+f(x+\Delta x)\vec e_j$$
By using Taylor approximation in first order
$$\vec r+\Delta\vec r=(x+\Delta x)\vec e_i+\bigg(f(x)+\frac{df}{dx}\Delta x\bigg)\vec e_j$$
The tangent vector can be written as
$$\Delta\vec r=\Delta x\vec e_i+\bigg(\frac{df}{dx}\Delta x\bigg)\vec e_j$$
Now I define a unit vector $\vec n=a\vec e_i+b\vec e_j$ perpendicular to $\Delta \vec r$ which must satisfy the following conditons (I could have used other conditions but these are simple)
Perpendicular $\Rightarrow a\Delta x+b\frac{df}{dx}\Delta x=0$
Unit vector $\Rightarrow a^2+b^2=1$
Solving these equations reveals
$a=\frac{\frac{df}{dx}}{\sqrt{\frac{df}{dx}^2+1}}\quad b=-\frac{1}{\sqrt{\frac{df}{dx}^2+1}}$   or   $a=-\frac{\frac{df}{dx}}{\sqrt{\frac{df}{dx}^2+1}}\quad b=\frac{1}{\sqrt{\frac{df}{dx}^2+1}}$
In common practice people take the first one as unit normal vector
$$\vec n=\frac{\frac{df}{dx}}{\sqrt{\frac{df}{dx}^2+1}}\vec e_i-\frac{1}{\sqrt{\frac{df}{dx}^2+1}}\vec e_j=\frac{1}{\sqrt{\frac{df}{dx}^2+1}}\bigg(\frac{df}{dx}\vec e_i-\vec e_j\bigg)$$
PS: If you want to generalize it with three coordinates don't forget to use the orthogonality to tangent plane condition. If not your system will be underdetermined.
