# Points of coordinate in a $3$-d space

How can I find the coordinates $(x,y,z)$ in a $3d$ space when,

A) the unknown point is $(x,y,z)$.

B) the known point is $(a,b,c)$.

C) the distance between the two points is $D$.

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I really don't know what you wanna say, try to reformulate your question –  Dominic Michaelis Feb 24 '13 at 10:49

Since we know $(x,y,z)$ is $D$ away from $(a,b,c)$, if we know what distance we are using, it is computable. Suppose we are using Euclidean distance. Let us for a moment assume $(a,b,c)$ to be on the origin, then $(x,y,z)$ would be a sphere with radius $D$. Imagine you take a point in space, and string of length $D$ attached to it, you make the other end of string move arbitrarily in space (make it tight), then wherever you reach it will be a solution, which in the end forms a sphere. $$x^2+y^2+z^2 = D^2$$ But then $(a,b,c)$ might not be on the origin, so we shift our picture, resulting $$(x-a)^2+(y-b)^2+(z-c)^2 = D^2$$ After a second look it is basically saying that $(D>0)$, the Euclidean distance between these two points is $D$. $$\sqrt{(x-a)^2+(y-b)^2+(z-c)^2} = D$$ This equation has infinitely many solutions in $\mathbb{R}^3$ (the 3-D plane in your words), because every point on the sphere would be a solution, and a sphere has infinitely many points on it. So you solution should be $$\{(x,y,z)\;| \;(x-a)^2+(y-b)^2+(z-c)^2 = D^2\}$$ The set of all $(x,y,z)$ that satisfies $(x-a)^2+(y-b)^2+(z-c)^2 = D^2$. For example points in this set $(a,b,c+D),(a,b,c-D),(a,b+D,c)$ and so on.
Impossible. There's not enough information. The point $(x,y,z)$ could lie anywhere on a sphere whose center is $(a,b,c)$ and whose radius is $D$. So, there are an infinite number of possible solutions. One of them is the point $(a+D, b, c)$. Another is the point $(a, b, c+D)$. And so on.