# Inside/outside test of any point on 2D plane?

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.

• PLease edit per the comments to explain what you mean by "inside/outside" . Dec 7, 2021 at 15:51
• Inside mean behind and outside mean front of observer. Dec 7, 2021 at 15:55
• – D.W.
Dec 30, 2021 at 20:07