superprismatic Posted August 2, 2009 Report Share Posted August 2, 2009 The diagonals of a square are mutually perpendicular. Is the same true for the spacial diagonals of a cube? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 3, 2009 Report Share Posted August 3, 2009 On 8/2/2009 at 8:44 PM, superprismatic said: The diagonals of a square are mutually perpendicular. Is the same true for the spacial diagonals of a cube? Reveal hidden contents Nope. Consider a unit cube with its vertices at (0,0,0), (0,0,1), (0,1,0), (0,1,1), (1,0,0), (1,0,1), (1,1,0), and (1,1,1). Two of the diagonals in question are between the vertices of (0,0,0) and (1,1,1) and (0,0,1) and (1,1,0). Those four vertices define a rectangle with a height of 1, and a width of sqrt(2), so those diagonals are the diagonals of this rectangle as well. Since this rectangle is not a square, its diagonals are not perpendicular. Quote Link to comment Share on other sites More sharing options...
0 superprismatic Posted August 3, 2009 Author Report Share Posted August 3, 2009 On 8/3/2009 at 12:46 AM, Chuck said: Reveal hidden contents Nope. Consider a unit cube with its vertices at (0,0,0), (0,0,1), (0,1,0), (0,1,1), (1,0,0), (1,0,1), (1,1,0), and (1,1,1). Two of the diagonals in question are between the vertices of (0,0,0) and (1,1,1) and (0,0,1) and (1,1,0). Those four vertices define a rectangle with a height of 1, and a width of sqrt(2), so those diagonals are the diagonals of this rectangle as well. Since this rectangle is not a square, its diagonals are not perpendicular. You are correct, but there's a much, much simpler explanation! Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 3, 2009 Report Share Posted August 3, 2009 Reveal hidden contents We know that right angles in 3d are the 3 axes, but a cube has 4 diagnols, so they can't be mutually perpendicular. The shape that does is 8 sided pair of tetrahedrons made from connecting the points +1 and -1 on each axis Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 3, 2009 Report Share Posted August 3, 2009 easier answer Reveal hidden contents take the two diagonals of a square ...perpendicular make that square the top of a cube. and drop one side (the same side of each) of the two diagonals on the square and point them to the corners of the cube. They dont reach. you have to stretch them to reach, thus changing the angle. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 3, 2009 Report Share Posted August 3, 2009 On 8/3/2009 at 2:37 AM, Doctor Moshe said: Reveal hidden contents We know that right angles in 3d are the 3 axes, but a cube has 4 diagnols, so they can't be mutually perpendicular. The shape that does is 8 sided pair of tetrahedrons made from connecting the points +1 and -1 on each axis Pardon my use of tetrahedron previously, I meant square pyramids Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 3, 2009 Report Share Posted August 3, 2009 On 8/2/2009 at 8:44 PM, superprismatic said: The diagonals of a square are mutually perpendicular. Is the same true for the spacial diagonals of a cube? Reveal hidden contents Consider the plane that is defined by any two diagonals of a cube. The diagonals of the cube are that of a rectangle that consists of two parallel edges of the cube and two parallel diagonals of a face. Thus, the diagonals of this non-square rectangle are not perpendicular. Quote Link to comment Share on other sites More sharing options...
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The diagonals of a square are mutually perpendicular. Is the same true for the spacial diagonals of a cube?
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