# Getting 3rd dimension, Concept( edited)

I have following problem. It's just a concept, so I cann't provide any code.

I have points A,B,C,D,E whose third dimensions in space are to be determined.

known:

1. Distances (r) and vertical angles (vt) of all points(A, etc) from Point P1. Note: P1 cann't be directly measured.

2. 2 points P2,P3.

unknown:

1. Position (x1,y1,z1) of P1.
2. 3rd dimension of points/ horizonal angles of points A,B,C,D,E w.r.t. P1.

To calculate these unknowns I can take this conditions. Assume P1, P2,P3 are not collinear and P1P2 & P1P3 intersect each other. They can be taken as three point forming an arbitrary plane. So again P1 can be fixed.

To make concept clear. North
| |
B(r2,vt2) angle from vertical axis (vt \| / A(r1,vt1)----------------------r-------P1(x1,y1,z1) /\ / \ / \ P2(x2,y2,z2)---P3(x3,y3,z3)

How can I have A(x,y,z)/A(r,horizonatl anle,vt)?? Note: Angles are from P1. Do I need some other information or other condition? regards,

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Your question is very unclear. What do the points $P_2$ and $P_3$ have to do with anything? Are they related in some way to $P_1$, and/or to $A$? – Gerry Myerson May 12 '12 at 6:24
@ Gerry Myerson P2 & P3 are reference for P1. They make a plane. – user31177 May 12 '12 at 6:42
Is the "vertical axis" normal to the plane $P_1,P_2,P_3$? – Neal May 12 '12 at 6:44
Think of $P_1$ as the pointy end of an ice cream cone. All the points on the rim of the cone have the same distance from $P_1$, and they all make the same angle from the vertical axis. So you need some more information to single one point out. – Gerry Myerson May 12 '12 at 6:44
@ Gerry Myerson please see edited. – user31177 May 12 '12 at 7:20