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Given $2$ points $p_1(x_1, y_1, z_1)$ and $p_2(x_2, y_2, z_2)$, they are the diagonally opposite vertices of a cuboid. How can we find the other six vertices?

doing this in a $2$d case ( for a rectangle ) is easy, we can find the perpendicular bisector of the diagonal, and it will give the other 2 points. But I can't find a way to start this for 3d.

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  • $\begingroup$ Please type it properly and explain what have you tried? $\endgroup$
    – MANI
    Commented Dec 6, 2019 at 14:21
  • $\begingroup$ I could do this for a square, but don't know where to start for 3d case. @MANI $\endgroup$
    – brownser
    Commented Dec 6, 2019 at 14:27
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    $\begingroup$ Use the same idea. Note that there are many cuboids that satisfy the conditions. $\endgroup$
    – Calvin Lin
    Commented Dec 6, 2019 at 14:32
  • $\begingroup$ @CalvinLin: is it? 2 diagonally opposite points are enough to uniquely determine a cuboid right? can 2 different cuboids have the same diagonal ? $\endgroup$
    – brownser
    Commented Dec 6, 2019 at 14:34
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    $\begingroup$ You can spin a cuboid about the diagonal axis, so that gives you multiple answers. $\endgroup$
    – Calvin Lin
    Commented Dec 6, 2019 at 14:42

1 Answer 1

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If you have such a cuboid, you can obviously rotate it around the line $P_1P_2$ by any amount, so the cuboid is not unique. However, there is a unique cuboid which has its edges parallel to the coordinate axes. We can argue as follows:

The other three vertices that share an edge with $P_1$ will have two coordinates the same as $P_1$ and the other coordinate the same as $P_2$. Similarly, the three vertices that share an edge with $P_2$ will have two coordinates the same as $P_2$ and the third the same as $P_1$, so we get $(x_1,y_1,z_2),(x_1,y_2,z_1),(x_2,y_1,z_1)$, $(x_1,y_2,z_2),(x_2,y_1,z_2),(x_2,y_2,z_1)$.

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  • $\begingroup$ @ChristianBlatter wlog one point is at the origin. So one face $S$ of the cuboid lies in the $xy$ plane, and one edge lies along the $z$-axis. The opposite face to $S$ is parallel to the $xy$-plane and passes through the opposite vertex $P$. The intersection of that plane with the $z$-axis must contain a vertex of the cuboid. So that vertex is uniquely defined. Similarly for the other vertex on the $y$-axis and the other vertex on the $z$-axis. Now switch the origin to $P$ to see that the other three vertices are uniquely defined. $\endgroup$
    – almagest
    Commented Dec 6, 2019 at 16:17
  • $\begingroup$ I don't think the claim "However, there is a unique cuboid which has its edges parallel to the coordinate axes" is true. Assume $P_1=(0,0,0)$ and $P_2=(1,2,3)$, or similar. Then you cannot rotate the cube around the axis $P_1\vee P_2$ such that the three edges ending at $P_1$ are lying in the coordinate axes. $\endgroup$
    – Christian Blatter
    Commented Dec 6, 2019 at 16:38
  • $\begingroup$ So you are challenging existence rather than uniqueness? In the case given, the other vertices are $(1,0,0),(0,2,0),(0,0,3)$ and $(1,2,0),(1,0,3),(0,2,3)$. $\endgroup$
    – almagest
    Commented Dec 6, 2019 at 16:45

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