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Math Help - Intersection of planes

  1. #1
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    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
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  2. #2
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    There is nothing to do.
    The fact the plane contains (0,0,0) gives the answer away.
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  3. #3
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    Hi,

    Sorry for the dumb question...how?

    Kind regards,

    Kepler
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  4. #4
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    It intersects all three at 0.
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  5. #5
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    Quote Originally Posted by kepler View Post
    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= <x_1- 0, y_1- 0, z_1- 0> and B- C= <x_2- 0, y_2- 0, z_2- 0>, and take their cross product.
    Last edited by mr fantastic; January 6th 2011 at 01:05 PM. Reason: Fixed math tags.
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  6. #6
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    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
    Last edited by kepler; January 3rd 2011 at 04:55 AM.
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  7. #7
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    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
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  8. #8
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    Quote Originally Posted by kepler View Post
    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}
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  9. #9
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    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
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  10. #10
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    Quote Originally Posted by kepler View Post
    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.
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