Thread: Finding points on a plane

I'm brand new here, so feel free to move this post if need be.

I have a plane defined by its unit normal (A, B, C) and a distance (D). I need to define this plane by three points instead. I know the equation for a plane is Ax+By+Cz=D, and it's easy enough if two of the coefficients are zero (the third coefficient is one and the point is dist along that axis). Is there a reliable formula for finding an x, y and z that satisfies this equation, particularly when A, B and C are all nonzero?

Subtract 2 pairs, maybe A - B and A - C. A Normal to the plane would ahve to be orthogonal to both these vectors. I think there is a cross product in your future.
If you are not convinced, try a different pair, maybe B - A and C - A, and see if you an manage the same equation.

Originally Posted by TKHunny

Subtract 2 pairs, maybe A - B and A - C. A Normal to the plane would ahve to be orthogonal to both these vectors. I think there is a cross product in your future.
If you are not convinced, try a different pair, maybe B - A and C - A, and see if you an manage the same equation.

I don't quite understand. A, B and C are all scalars for i, j and k along their respective axes, correct? In which case, I don't quite understand how subtracting two scalars results in a vector to perform the cross product on.

Am I understanding this correctly? You have found that the equation of the plane is Ax+ By+ Cz= D and you want to "define this plane by three points instead." In other words, you just want to find three points that lie on that plane. It you take x= y= 0 then z= D/C so (0, 0, D/C) is a point on the plane. If x= z= 0, then By= D so y= D/B. (0, D/B, 0) is a point on that plane. I'll leave the third point to you.

I think TKHunny was misunderstanding the problem, thinking that A, B, and C were three points that you wished to use to find the equation of the plane.

Originally Posted by HallsofIvy

Am I understanding this correctly? You have found that the equation of the plane is Ax+ By+ Cz= D and you want to "define this plane by three points instead." In other words, you just want to find three points that lie on that plane. It you take x= y= 0 then z= D/C so (0, 0, D/C) is a point on the plane. If x= z= 0, then By= D so y= D/B. (0, D/B, 0) is a point on that plane. I'll leave the third point to you.

I think TKHunny was misunderstanding the problem, thinking that A, B, and C were three points that you wished to use to find the equation of the plane.

Thanks for that, I think that leap of logic was exactly what I was missing. Yes, I have the A, B, C and D and I need to find x, y and z to satisfy the equation. Didn't occur to me that if I set all other axes of the point to 0 that the third one would be D divided by the remaining coefficient.

However, this method falls apart if A B or C are zero. So now I know how to find a point if two of them are zero, or if none of them are zero.

If, say, B is 0, y does not enter into the equation so y could be anything. Take x and y to be whatever you want, solve the eqation for z. A different value of y with the same x and z will be another point on the plane. You could not do that with a third value of y and the same x and z because that would give three points on a line which do not define a plane, but you could pick a new value for x and solve for z to get a point not on the line determined by the first two points.

Originally Posted by HallsofIvy

I think TKHunny was misunderstanding the problem, ...

Why should today be different? :-)