Relation between a point and a plane

Question

I have a plane i reprensent by its normal $n$ and a point on it $p$. I calculate $d = -n^Tp$, and i consider "point $x$ is above the plane" if $n$ is pointing at $x$.

How to determine if a point $x$ is above or below the plane?

I thought that if $n^Tx-d>=0$ then $x$ is above the plane but that seem to not hold to any plane...

Vedaad Shakib · Answer 1 · 2014-11-23T21:14:13.353

1

Let's pretend $\vec{xp}$ is a vector representing the point on the plane and $\vec{x0}$ is a vector representing the point you want to determine the location in respect to the plane.

You could simply check if $(\vec{x0}-\vec{xp})\cdot \vec{n} \geq 0$ (where $\cdot$ is dot product of vector).

If it is greater than 0, it is in the direction of the normal. If it is less than 0, the point is in the opposite direction of the vector.

edited Nov 23 '14 at 21:14

answered Nov 23 '14 at 21:07

Vedaad Shakib

111

is x0 any point on the plane or is it the projection of xp on the plane? – Ofek Ron Nov 23 '14 at 21:08
$\vec{x0}$ is any point on the plane. – Vedaad Shakib Nov 23 '14 at 21:09
do you know of a way to compute that using d and n only? – Ofek Ron Nov 23 '14 at 21:11
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Using $x0$ for the OP's $p$, and $xp$ for the OP's $x$, is guaranteed to cause confusion. – TonyK Nov 23 '14 at 21:12
Yes. Substitute $\vec{x0}$ in the equation with $\frac{d}{-n^T}$ – Vedaad Shakib Nov 23 '14 at 21:12
@TonyK Just edited. – Vedaad Shakib Nov 23 '14 at 21:14

score 1 · Accepted Answer · answered Nov 23 '14 at 21:14

1

You want $d = n^Tp$, not $d = -n^Tp$. (Or you can keep your definition of $d$, but then your check should be $n^Tx+d \ge 0$, not $n^Tx-d \ge 0$.)

answered Nov 23 '14 at 21:14

TonyK

64,559

this works thanks! – Ofek Ron Nov 23 '14 at 21:20

Relation between a point and a plane

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