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Given three unit vectors $u_1,u_2,u_3$ in $\mathbb{R}^3$, can we find some body $K \subset \mathbb{R}^3$ (probably convex) such that the following three things hold

(1) $|P_{u_1^\perp}K|=|P_{u_2^\perp}K|=|P_{u_3^\perp}K|=1$

(2) $|P_{span(u_1,u_2)^\perp}K|=|P_{span(u_1,u_3)^\perp}K|=|P_{span(u_2,u_3)^\perp}K|=1$

(3) $|K|=1$

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    $\begingroup$ By considering $v_1,v_2,v_3$ biorthogonal to $v_1,v_2,v_3$ you can get rid of the orthogonal complements. Then $P_{v_1}=P_{v_1}P_{span (v_1,v_2)}$, and so $1=|P_{v_1}P_{span (v_1,v_2)}K|$. Analogously, $1=|P_{v_2}P_{span (v_1,v_2)}K|$. Since $|P_{span (v_1,v_2)}K|=1$ this means that the projections of a set of area $1$ on two lines have length $1$, and so these lines have to be orthogonal. Thus, $v_1,v_2,v_3$ is an orthonormal basis, and so the same is true for $u_1,u_2,u_3$. Or am I missing something? $\endgroup$
    – erz
    Commented Feb 22, 2019 at 3:00
  • $\begingroup$ Thanks for the reply, but I think what you claimed in the end is not true. You can find $K'=P_{span(v_1,v_2)}K$ with $|K'|=1$ and both projections on to $v_1$ and $v_2$ having length equal to 1. Just draw two strips of width one. Their intersection is a parallelogram and has volume bigger than 1. So inside you can find a set with volume 1 and having both projections equal to 1. $\endgroup$
    – STrick
    Commented Feb 27, 2019 at 19:21
  • $\begingroup$ but these are orthogonal projection, not parallel to $v_1$ and $v_2$. Orthogonal projections of your parallelogram will be greater than $1$. $\endgroup$
    – erz
    Commented Feb 28, 2019 at 3:35
  • $\begingroup$ Any inscribed quadrilateral (inscribed in the parallelogram) has orthogonal projections equal to 1. We can easily choose one of them to have area 1. $\endgroup$
    – STrick
    Commented Feb 28, 2019 at 23:44

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