Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $D$.

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

2 Answers 2

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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.