Equation of a plane through three points

[srn]

edit: should probably be in precalc sorry about that

Homework Statement

I have three points, two known and one partially known, only the x coordinate (3D)

Point one (A): (-0,2; -0,2; -0,2)
Point two (B): (-0.13; -0.15; 0)
Point three (C): (-0,2; unknown; unknown)

I also know that the distance from A to C is 0,2 and that the distance from B to C is 0,15.

Question: what are the other two components of point C.

The Attempt at a Solution

As I see it, the lines going from A and from B with the above magnitudes and crossing each other, describe a circle that is perpendicular to AB. So far, there are infinitely many solutions. However, with the x coordinate known, I think that it brings the amount of solutions down to exactly two.

I've composed the vector AC and AB and their cross product. This should then be perpendicular to the plane and I've used that with the equation

a(x-x0) + b(y-y0) + c(z-z0) = 0

Where a, b and c are the components of AC X AB and x0, y0, z0 a random point which I took as A.

I then replaced the x with x = -0,2.

Then I took |AC| = 0,2 and |BC| = 0,15. I worked this out and set it equal to each other, then I got something in y and z, set one of them equal to the other and then filled it in in the above equation of the plane.

This didn't yield anything at all. Can someone please explain where I have gone wrong?

 

[rock.freak667]

I think you should just have C= (-0.2,y₁,z₁) and then find AC and BC in terms of y₁ and z₁.

Then use the fact that |AC|=0.2 and |BC|=0.15. You will get two equations in y₁ and z₁ which you can solve.

 

[srn]

Ah, not so hard after all :(

j := (y+.2)^2+(z+.2)^2 = 0.04
k := 0.07^2+(y+.15)^2+z^2 = .02255;

> solve({j, k}, {z, y});

{y = -.2815230378, z = -0.01736924054}, {y = -0.4212402098e-1, z = -0.07721899475}

That should be it then? Does anyone know how to draw a plane through 3 points in maple? Can't find anything on google.