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In 3d space, a right-corner can intersect a 2d plane, like lopping off the corner of a cube to create a tetrahedron. Can higher dimensional 'right-corners' in $N$ dimensions, where all dimensions are 90 degrees perpendicular to each other, be rotated through space to also intersect a 2d plane? Or can they only intersect the $N-1$ space?

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    $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Commented Aug 9, 2022 at 0:57

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If a 2D plane could "cut off" a corner of an N - dimensional hypercube, then the intersection would make up a face of the corner, i.e. parts of its surface. But a surface has dimensions N-1, while the intersection of a plane with the hypercube can have dimensions at most 2

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  • $\begingroup$ Thanks, user33.. I was doing a 'thought experiment' and was trying to see what the fewest dimensions I could reduce any N-d shape to. Your summary got me back on track. $\endgroup$
    – Jim
    Commented Aug 10, 2022 at 14:02
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This is a math question, but consider a square. Consider a line in the (1,1) direction, which points to the opposite corner of the square. A line perpendicular to this line cuts the corner of the square, leaving a right triangle.

Consider a cube. Consider a line in the (1,1, 1) direction, which points to the opposite corner of the cube. A plane perpendicular to this line cuts the corner of the cube, leaving a right tetrahedron.

Consider a hypercube... Can you keep going?

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