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Suppose I have one plane $Ax+By+Cz+D=0.$ I want to test $(x, y, z)$ is inside of this plane or outside.

My attempt:1 Then draw normal N on that plane which point away from the observer Image1.

If $Ax+By+Cz+D>0$ or $(-A)x+(-B)y+(-C)z+(-D)<0$ then $(x, y, z)$ is outside of the plane.

And if $Ax+By+Cz+D<0$ or $(-A)x+(-B)y+(-C)z+(-D)>0$ then $(x, y, z)$ is inside of the plane.

My attempt:2 Now draw the normal N on that plane which point towards the observer image2.

If $Ax+By+Cz+D<0$ or $(-A)x+(-B)y+(-C)z+(-D)>0$ then $(x, y, z)$ is outside of the plane.

And if $Ax+By+Cz+D>0$ or $(-A)x+(-B)y+(-C)z+(-D)<0$ then $(x, y, z)$ is inside of the plane.

My question is my both attempts are right? Can anybody help me to understand.

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1 Answer 1

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if a point is in a plane it means if you plug in x.y,z in the plains EQ. you get zero. And if you get nonzero the point is not in the plane.

Maybe I am not understanding your question. If you can rephrase your question I may get it.

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  • $\begingroup$ If I say Ax+By+Cz+D>0 then (x, y, z) is the outside of the plane and then (−A)x+(−B)y+(−C)z+(−D)<0 indicates (x, y, z) inside of the plane or outside of the plane or both depending upon implementation? $\endgroup$
    – Alok Maity
    Dec 7, 2021 at 9:54
  • $\begingroup$ both mean not in the plane. only zero answer is in the plane. $\endgroup$
    – kamran
    Dec 7, 2021 at 9:58
  • $\begingroup$ inside mean behind the plane and outside mean front of the plane $\endgroup$
    – Alok Maity
    Dec 7, 2021 at 10:00
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    $\begingroup$ behind and front imply that there is a viewer involved. It is not clear on which side of the plane the viewer is placed. $\endgroup$
    – AJN
    Dec 7, 2021 at 12:42

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