# Intersection of Normally distributed planes

I have a set of planes in $R^3$. Each plane is defined via the normal of a unit vector that intersects the origin, $(0,0,0)$, and a distance away from from the origin, $d$. However, the parameter $d$ is normally distributed, and I am after the distribution of possible locations after the intersection of all the planes has been performed.

To constrain the problem (and hopefully simply it), it can be assumed that all of the planes approximately intersect a point within $\pm 3\sigma$ of their respective standard deviations. Also, the result should be a Normal distribution in $R^3$ centred at the approximate intersection point.

I have no idea how to go about solving this. If it were just the intersection of planes, or just the convolution between two normal distributions, it would make sense. I'm not keen on having a combinatronics problem to try to approximate the mean, and don't have a clue for the standard deviations. Any hints or guidance in this would be greatly appreciated.

EDIT:

To better define what the planes actually are, I will apologise in advance for butchering any mathematical terms or concepts.

A Gaussian distribution is constructed using a mean and a covariance matrix (for $R^3$). The mean can be considered a $0D$ point in space where the maximum probability lies. For the purposes of this question, I am considering a mean that is a $2D$ plane of maximal likelihood (ignoring the issues of a CDF that can be larger than 1 along certain axes). The intersection of two planes then becomes somewhat equivalent to the convolution of two Gaussian's, except that one of the axes of the mean can become defined (aka, it goes from a $2D$ object to a $1D$ object).

• "I am after the distribution of possible locations after the intersection of all the planes has been performed" Could you clarify what that means? – Paul Apr 23 '18 at 11:19
• I mean this in the kindest way possible: I think you have some work to do making this mathematically precise before you have hope of having your question answered by yourself or anyone else. I can't make sense of what your question is even asking as it's written now. – Aaron Montgomery Apr 23 '18 at 14:33