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For two planes:

$(\vec{p} - \vec{p_0})^T \vec{n_p} = 0$

$(\vec{q} - \vec{q_0})^T \vec{n_q} = 0$

Prove if the normals are orthogonal, i.e. $\vec{n_p}^T \vec{n_0} = 0$, then all vectors (except potentially the vectors along the intersection of two planes) $\vec{p}-\vec{p_0}$ and $\vec{q} - \vec{q_0}$ are orthogonal.

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    $\begingroup$ There seems to be a condition missing, since the statement is not true as explained in dmh's answer. Where did you get this problem from? $\endgroup$
    – Carmeister
    Sep 22, 2023 at 4:35

1 Answer 1

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Isn't this not true? Consider the hyperplanes in $\mathbb R^3$ with normals $(1, 0, 0)$ and $(0, 1, 0)$, respectively. These normals are orthogonal, but both planes contain the vector $(0, 0, 1)$.

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  • $\begingroup$ Let's exclude the line at the intersection of the planes. $\endgroup$
    – user3180
    Sep 21, 2023 at 19:29
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    $\begingroup$ @user3180 the planes still contain the vectors $(0,1,1)$ and $(1,0,1)$. $\endgroup$
    – Toffomat
    Sep 21, 2023 at 19:37

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