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If I have a vertical plane, and if there are few points bit away from the plane (nearly 5 to 10 cm), then how to know the + or - distance that point is making with the given plane.

Any explanation please.

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What do you mean by "inside" and "outside" of a plane? –  Eckhard Apr 5 '13 at 14:31
    
If your plane is $ax+by+cz+d=$, simply enter the coordinates of your point $P(p,q,r)$ in the equation and look if the equation $ap+bq+cr+d=0$ is verified... For the distance with the plane, it's just $$\frac{|ap+bq+cr+d|}{\sqrt{a^2+b^2+c^2}}$$ –  mwoua Apr 5 '13 at 14:33
    
@mwoua: thank you for the response. for clarification i want to ask; for any vertical plane (not parallel either X or Y axes), does the + value always give by the point which is always far than its plane projection? or is there any relation..? –  gnp Apr 5 '13 at 15:22
    
I still do not understand. The + or - value is arbitrary. You could for example take the plane $ax+by+cz+d=0$ or the plane $-ax-by-cz-d=0$ which is the same and then your "distance" (without the absolute value, in order to introduce a sort of direction) to the plane for $P(p,q,r)$ is, for the first plane $$\frac{ap+bq+cr+d}{\sqrt{a^2+b^2+c^2}}$$ while for the second, the "distance" is the opposite $$-\frac{ap+bq+cr+d}{\sqrt{a^2+b^2+c^2}}$$ As you see, + or - doesn't mean anything but a sort of orientation (that you choose). –  mwoua Apr 5 '13 at 22:16
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