Projection of a point from one plane onto another

[cptolemy]

Hi,

In a 3D plane, I have another plane P1 equal to Ax+By+Cz=0 (D=0 since one of its points is (0,0,0) )

If I have the coordinates (x1,y1,z1) in the first plane, what are the cordinates of this point in the P1 plane?

I know the equation of the intersection line. But my calculations are going wrong, I'm afraid... And I thing it's because of the inclination rotations routines...

And I also think that a starting x-axis vector must be defined - from the intersection line.

Can someone help?

Kind regards,

CPtolemy

 

[Aceix]

Hi,

In a 3D plane, I have another plane P1 equal to Ax+By+Cz=0 (D=0 since one of its points is (0,0,0) )

If I have the coordinates (x1,y1,z1) in the first plane, what are the cordinates of this point in the P1 plane?

I know the equation of the intersection line. But my calculations are going wrong, I'm afraid... And I thing it's because of the inclination rotations routines...

And I also think that a starting x-axis vector must be defined - from the intersection line.

Can someone help?

Kind regards,

CPtolemy

The equation of the first plane?

 

[cptolemy]

Hi,

The first plane is the 3D origin normal axis. So it can be, for instance, Ax + By = k (or any other constant), I'm picturing a 3D set - the normal x-y-z axis. Then, in that coordinate system I've created a new plane with an equation referred to that one.

Regards,

CPtolemy

 

[Mark44]

In a 3D plane

Do you mean "in a 3D space"? A plane is a subset of three-dimensional space.

Hi,

The first plane is the 3D origin normal axis.

This makes no sense.

So it can be, for instance, Ax + By = k (or any other constant),
I'm picturing a 3D set - the normal x-y-z axis.

?
What does "normal x-y-z axis" mean. There is no x-y-z axis.

Then, in that coordinate system I've created a new plane with an equation referred to that one.

Regards,

CPtolemy

 

[cptolemy]

Hi,

Very technical - still no answer for helping out. I do believe you understood the question. Pointing out terminology is not helpful and just fills these posts with rubbish...

But thanks anyway.

Regards,

CPtolemy

 

[Mark44]

Very technical - still no answer for helping out. I do believe you understood the question. Pointing out terminology is not helpful and just fills these posts with rubbish...

If you use the terminology incorrectly, it makes it more difficult to understand what you're asking, and to give an accurate answer.

When you start a thread off with "In a 3D plane, I have another plane P1 equal to Ax+By+Cz=0" without telling us about the first plane, you haven't formulated your question very well.

Your question can very likely be answered with a bit of trig and some basic vectors. Based on your misuse of terminology, it's difficult to tell what your level of mathematics knowledge is.

 

[cptolemy]

Ok,

Let's start from scratch. If I have a plane P1 defined by an equation, isn't that referred to a origin reference frame? That's the frame I'm talking about. P1 equation depends of that initial frame. My question is if I have a 3 dimensional point in that original frame - or reference system if you like - what are its coordinates in P1.
I assume I must define first an x-axis vector - the intersection line.
I've already solved the problem with a set of rotation equations: first a rotation over the z-axis (with the angle of the intersection line / the x-axis origin of P1 in the frame), and then over the x-axis (with the angle of the inclination of P1). But the formulas become complex, even if they work. I'm trying to find an easy way of doing this.

Regards,

CPtolemy

 

[Mark44]

Ok,

Let's start from scratch. If I have a plane P1 defined by an equation, isn't that referred to a origin reference frame?

If I understand your question, then no. The equation of the plane is relative to the standard coordinate axis system. Are you trying to come up with a different coordinate system, one where two of the axes lie in the plane?

The general form of the equation of a plane is Ax + By + Cz + D = 0. The values of x, y, and z are measured along the usual x-, y-, and z-axes. If you are talking about a different set of axes, they could be called the x'-, y'-, and z'-axes, to distinguish them from the usual axes.

That's the frame I'm talking about. P1 equation depends of that initial frame. My question is if I have a 3 dimensional point in that original frame - or reference system if you like - what are its coordinates in P1.

A point in space has three coordinates. Do you know the coordinates of the point? As for your question -- "what are its coordinates in P1?" -- I don't understand what you're asking. Are you asking what the coordinates are in the new coordinate system?

I assume I must define first an x-axis vector - the intersection line.

What intersection line? The intersection of the plane with some other plane?

I've already solved the problem with a set of rotation equations: first a rotation over the z-axis (with the angle of the intersection line / the x-axis origin of P1 in the frame), and then over the x-axis (with the angle of the inclination of P1). But the formulas become complex, even if they work. I'm trying to find an easy way of doing this.

If what you're calling P1 doesn't go through the origin, I don't think that rotations will do what you're asking. This wikipedia article might be helpful: https://en.wikipedia.org/wiki/Affine_transformation.