What is the most accurate method to get intersection point in 3D? I have been given 3D point data, belonging to different planar segments. Points are not exactly laid on the planes so that I have fitted best planes using least square solutions.
Now, I want to find the intersection point where more than $2$ planes are intersected. I found there are two methods for that:


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*By computing intersection lines using $2$ plane pairs, and then finding the final point using some adjustment process;

*By using 3D planes itself and finding the intersection point.
So, now I would like to know which method will give the most accurate result. I would guess that the two methods will give two different results.
Please explain it to me.
 A: I think neither methods will lead to the most accurate solution: If one first estimates the planes and then find the intersection point, one does not enforce directly the constraint that all planes intersect exactly in a single point.
If I understand correctly, this is what we know:


*

*There is a set of $M$ planes (three or more) which intersect in a single point $\mathbf a$.
(This is our parametric model.)

*The planes are represented by a set of $K$ points $\mathbf x^{(k)}$ affected by zero-mean Gaussian noise (our data).

*We assume that we know which point belong to which plane. Thus we have a mapping $I: \{1,...,K\} \rightarrow \mathbb N$ which maps a point index $k$  onto a plane index $I(k)$. (If we don't know the mapping $I$, we can find it using RANSAC or a similar method.)


Maximizing the likelihood of the model parameters is equivalent to minimizing the geometric error between the data and the model.
Our model might look as follows:
All planes intersect in a single point $\mathbf a = (a_1,a_2,a_3)^\top$.
Each plane has a different orientation specified by a normal vector $\mathbf n^{(m)}$. Now our $M$ planes are represented by $(3+3M)$ parameters: $$\mathbf p = (a_1,a_2,a_3,n_1^{(1)},n_2^{(1)},n_3^{(1)} ,...,n_1^{(M)},n_2^{(M)},n_3^{(M)})^\top$$
(However, the problem has only $3+2M$ degree of freedom since the length of the normals are insignificant. Thus, we have a gauge freedom of $M$ dimension. If necessary, this free gauge can be removed by using a minimal/two-dimensional parametrisation of the normals. A good possibility is to restrict the normals to lie on a sphere as described in Hartley, Zisserman: "Multiple View Geometry", Second edition, A 6.9.3.)
Now the geometric error we wish to minimize is:
$$ S = \sum_{1=k}^K  [d(\mathbf x^{(k)}, \mathbf a,\mathbf n^{(I(k))})]^2 $$
Here, $d(\mathbf x, \mathbf a,\mathbf n)$ is the distance between a point $\mathbf x$ and a plane $(\mathbf a,\mathbf n)$.
We can find the optimal plane parameters $\mathbf p$ by jointly minimizing $S$ wrt. to $\mathbf p$.
