1. ## Intersection of planes

Hi,

I have a problem that I really don't know how to solve. I have a regular 3 dimensions axis (x,y,z) with the center C(0,0,0) and 2 points, A(x1,y1,z1) and B(x2,y2,z2).

How can I calculate the intersection of the plane defined by A, B and C with the x,y,z axis in this last one?

Please give me a mathematical formula - if possible - without matrices; I'm not very good at that...

Kind regards,

Kepler

2. There is nothing to do.
The fact the plane contains $(0,0,0)$ gives the answer away.

3. Hi,

Sorry for the dumb question...how?

Kind regards,

Kepler

4. It intersects all three at 0.

5. Originally Posted by kepler
Hi,

I have a problem that I really don't know how to solve. I have a regular 3 dimensions axis (x,y,z) with the center C(0,0,0) and 2 points, A(x1,y1,z1) and B(x2,y2,z2).

How can I calculate the intersection of the plane defined by A, B and C with the x,y,z axis in this last one?

Please give me a mathematical formula - if possible - without matrices; I'm not very good at that...

Kind regards,

Kepler
A plane, in three dimensions, can be written as $\alpha(x- x_0)+ \beta(y- y_0)+ \gamma(z- z_0)= 0$ where $(x_0, y_0, z_0)$ is a point on the plane and $<\alpha, \beta, \gamma>$ is a vector perpendicular to the plane.

You already know three points in the plane: $(0, 0, 0)$, $(x_1, y_1, z_1)$, and $(x_2, y_2, z_2)$. It would be simplest of course to let $(x_0, y_0, z_0)= (0, 0, 0)$.

To find a vector perpendicular to the plane, find two vectors in the plane, say $A- C= $ and $B- C= $, and take their cross product.

6. Hi again Plato,

That's obvious. But what I need to know is this: imagine a circle or an elipse or a rectangle, with center at (0,0,0) that passes through A and B. Where's does intercepts - the line - in x,y,z (alias in x,y)?

Regards,

Kepler

7. ## Continuing...

Hi again,

I've solved the problem of the plane equation and the xy line equation for the interception Now I have a bigger problem...

The plane has an inclination - how do I calculate that? And even worst, how do I calculate the line that goes through that maximum inclination?

Can someone help me out?

Kind regards,

Kepler

8. Originally Posted by kepler
Hi again,

I've solved the problem of the plane equation and the xy line equation for the interception Now I have a bigger problem...

The plane has an inclination - how do I calculate that? And even worst, how do I calculate the line that goes through that maximum inclination?

Can someone help me out?

Kind regards,

Kepler
Dear Kepler,

If you know normal vectors to the two planes, say $\underline{n_1}~and~\underline{n_2}$ then you can easily find the angle $(\theta)$ between them using the definition of the dot product. This will be the angle between the two planes.

$\underline{n_1}.\underline{n_2}=\mid \underline{n_1}\mid\mid\underline{n_2}\mid\cos\the ta\Rightarrow \cos\theta=\displaystyle\frac{\underline{n_1}.\und erline{n_2}}{\mid \underline{n_1}\mid\mid\underline{n_2}\mid}$

9. Dear Sudharaka,

Thank you very much for your reply - it was helpful. Just one question. What's the equation of the xy plane in the 3D format?

Kind regards,

Kepler

10. Originally Posted by kepler
Dear Sudharaka,

Thank you very much for your reply - it was helpful. Just one question. What's the equation of the xy plane in the 3D format?

Kind regards,

Kepler
z = 0.