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I'm in trouble... I have to find the cartesian equation of the plane $\sigma$ through three points and orthogonal to another plane $\pi$... How can I solve this?

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  • $\begingroup$ I think it is impossible if the 3 points are arbitrary. $\endgroup$ – KittyL Feb 15 '15 at 10:12
  • $\begingroup$ Hi! Can you add more context to your question? For example, how is the plane $\pi$ assigned?... $\endgroup$ – MattAllegro Feb 15 '15 at 10:13
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    $\begingroup$ @MattAllegro let be $\pi$ in vectorial form... I haven't got any data $\endgroup$ – Groucho Marx Feb 15 '15 at 10:18
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    $\begingroup$ @MattAllegro for example, if I have to find $\sigma$ through $A=(2,4,4)$ and $B=(2,-2,-4)$ and orthogonal to $\pi:4y-3z-4=0$, the plane is $x-2=0$... but, if I have three points instead of two? $\endgroup$ – Groucho Marx Feb 15 '15 at 10:22
  • $\begingroup$ I'm supposing that if for the case before mentioned I have to calculate the vector $AB=(0,-6,-8)$, then for three points (let have another point $C=(1,3,0)$) I have to calculate the vector $ABC=(1,9,8)$? Is this right? $\endgroup$ – Groucho Marx Feb 15 '15 at 10:35
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Three points already define a plane. If that plane is not perpendicular to the other plane, there is nothing you can do.

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