I have a plane, defined by a normal vector $n$ and a point $p$. I also have a point $a = (x, y, z)$. Based on this information, how do I know if the point $a$ exists somewhere past the plane in the direction of the normal?
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Take the dot product $n \cdot (a-p)$: if the sign is positive then $a$ exists somewhere past the plane in the direction of the normal $n$. If such dot product is zero then $a$ is in the plane and in the case the dot product is negative is in the opposite side of the plane w.r.t the normal $n$.
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You can check if the point is on the plane by
$$ d = \vec{n} \cdot (\vec{a}-\vec{p}) = 0 $$
where $\cdot$ is the vector dot product, and $d$ is the distance of the point to the plane in the direction of the normal.
answered Jun 18 '15 at 14:14
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1$\begingroup$ I don't want to know if the point is on the plane, but rather past the plane in the direction of the normal. $\endgroup$ – Ren Jun 18 '15 at 14:19
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1$\begingroup$ Oops, i forgot to add the part where you check if $d$ is positive or negative. My bad. $\endgroup$ – John Alexiou Jun 18 '15 at 14:43
