Coordinates of a point outside a plane

[Vivio]

Hello,

If you can get me a hint for solving this matter it would be much appreciated.

I have the 3D coordinates of three points on a plane A, B, C.

There's another point G and we know AG, BG, CG.

My problem is to find the coordinates of point G

Thanks in advance!

 

[mathman]

G=(A+AG,B+BG,C+CG) unless I don't understand what you are saying.

 

[sin(1/x)]

If you are given:

A=(A_1,A_2,A_3)
B=(B_1,B_2,B_3)
C=(C_1,C_2,C_3)

And you want to find G=(G_1,G_2,G_3), then knowing:

1) AG=|A-G|=((A_1-G_1)^2+(A_2-G_2)^2+(A_3-G_3)^2)^(1/2)
2) BG=|B-G|
3) CG=|C-G|

is not enough information.

Let h be the perpendicular distance from G to the plane. Then if A, B, and C and there is another point G' on the other side of the plane with perpendicular distance h from the plane also satisfynig 1, 2 and 3 so the solution is not unique. If A, B, and C are collinear you can find an entire circle of points satisfying 1, 2 and 3.

If you know that the A, B, and C are not collinear and which side of the plane G is on, then you can solve for the components of G using the distance formula or trigonometry.

 

[Vivio]

Thanks mathman and sin(1/x) for your input :shy:.

I've made a drawing with the problem. I hope all the data are there.

Sorry for my omissions.

 

[mathman]

I get it now. G=A + |AG|a = B + |BG|b = C + |CG|c, where a,b,c are (unknown) unit vectors. By eliminating G, you will have 6 linear equations for the coordinates of a,b, and c. Using the fact they are unit vectors gives 3 quadratic equations. Solving for them will give you the vectors a,b,c. There will be 2 true solutions. When you solve the quadratics there will be extra solutions, so you need to check to see if you get the same value for G from a given set a,b,c.