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# Reply for the maths problem that introduced by Calvin Lin

Calvin Lin: You have introduced a problem that describes about a minimum number of sets of lines.

If every line is perpendicular at most one line, and since the lines have placed arbitrary: then within the the possible space volume, the angle theta should obey 0<theta< pi/2 and the azimuthal angle should obey 0<azimuthal angle< pi/2 as usual notations in spherical polar coordinates. Otherwise there is a possibility to be at least 2 lines perpendicular to one line(for 3-D case) Then there is a contradiction to the the statement: every line is perpendicular at most one line.(since lines have placed arbitrary). But within above constrains for angles, no any two lines perpendicular to each other. If we divide the set of lines into N groups according to the criteria: every line is perpendicular at most one line in its group, we have to divide the set of lines into 2014 groups. We can't think about 2-D case. Because in one plane the maximum number of lines is 3.

Note by Geerasee Wijesuriya
3 years, 4 months ago

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The statement is true for any set of 2014 lines (through the origin). It is possible for 2 of those lines to be perpendicular, which would give $$\theta = \pi / 2$$. It is also possible for 1 line to be perpendicular to 2 other lines, like in the case of the x, y and z axis.

It is possible for 2 lines in the same group to be perpendicular to each other. For example, consider the set of 3 lines which are the x, y and z axis, where each of these lines is perpendicular to the other. 2 groups are sufficient - in group 1, we have the x and y axis, and in group 2 we have the z axis. You can check that this satisfies the condition. Staff · 3 years, 4 months ago