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I have a plane represented as a 3D dataset and a normal to that plane I'd like to represent that plane in hesse normal form in order to find the d (distance) in:

$\vec{r}.\vec{n}$ = d.

where $\vec{n}$ is the normal vector, and $\vec{r}$ is the position vector of point $\vec{P}$ that lies on the plane (I took the centroid point). Assuming that the origin is $(0,0,0)$, then the equation would be $\vec{P}\cdot\vec{n}$ = d.

is that correct?

If the answer is yes, then I'm even more confused, since if I used another point that lies on the plane and dot product it with the normal then I'll get a different d result. Then how come that the hesse normal form is fullfilled by each point lying on the plane?

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Do you really mean $\vec r\cdot\vec P$? What is the difference between $\vec r$ and $\vec P$ in your interpretation? – Rahul Jan 25 '13 at 14:36
You already have the equation $\vec r\cdot\vec n=d$. Just use that. Or do you not know the value of $\vec n$, is that the problem? – Rahul Jan 25 '13 at 14:37
well r⃗ is a location vector for point P⃗ , i.e. r⃗ =OP→ where O is the origin vector so that r⃗ =P⃗ −O⃗ . but since the origin is (0,0,0) then r⃗ =P⃗ .btw $\vec{n}$ is also available – Moataz Elmasry Jan 25 '13 at 15:01
up vote 1 down vote accepted

If $\vec r$ is a point on the plane, and $\vec n$ is the normal to the plane, then $\vec r\cdot\vec n=d$ is constant for all points on the plane. It is not possible to get a different value of $\vec r\cdot\vec n$ for other points on the plane. If you take another point $\vec s$ and get $\vec s\cdot\vec n\ne\vec r\cdot\vec n$, then $\vec s$ does not lie on the plane through $\vec r$ normal to $\vec n$, because it means that $(\vec s-\vec r)\cdot\vec n\ne0$, so the vector from $\vec r$ to $\vec s$ is not normal to $\vec n$.

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sorry r⃗ ⋅P⃗ is a typo mistake, I corrected it – Moataz Elmasry Jan 25 '13 at 15:02

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