I have a number of planes (in $\mathbb{R}^3$), each represented by a point $\vec{P_i}$ which lies within each plane and the normal vector $\vec{n_i}$.

If I project a point $\vec{Q}$ (which does not lie within any of the planes) onto each plane, the resulting projected points are $\vec{Q'_i}$.

Is the following valid?

If the planes $k$ and $j$ (given by $\left(\vec{P_k},\vec{n_k}\right)$, and $\left(\vec{P_j},\vec{n_j}\right)$) represent the same plane:

$\Rightarrow \vec{Q'_k} = \vec{Q'_j}$

Otherwise (planes $k$ and $j$ are distinct planes that are either parallel or intersect in a line):

$\Rightarrow \vec{Q'_k} \neq \vec{Q'_j}$


By projecting $\vec Q$ onto eaqch plane, I assume you mean orthogonal projection. So the question is, can one point projected onto two different planes result in the same image point. For that to be possible, the planes in question have to intersect, otherwise they don't have any common points at all. But if they intersect and are not equal, then their normal vecors differ in direction. And if the normal vectors differ, then the lines of projection differ. And for two different lines to intersect the two different planes in the same point, that point must be the only point of intersection of the lines. We already know that the lines intersect in $\vec Q$, though, so this can only happen if $\vec Q$ lies in both planes. Since you stated that it lies in none of the planes, that can happen either, therefore you can't have two equal image points, and your assumption is valid.

  • $\begingroup$ yes I meant orthogonal projection $\endgroup$ – pistermink Sep 1 '15 at 6:28

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