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It takes four noncoplanar points to define a 3-space. Explain whether each of the following is true or not and explain why:

a) Two skew lines define a 3-space.

True because skew lines are Two or more lines which have no intersections but are not parallel. Since two lines in the plane must intersect or be parallel, skew lines can exist only in three or more dimensions.

b) A plane and a point not on the plane define a 3-space.

True. My example is a tetrahedron. If you have four points and they are not on the same plane (not coplanar), then they define a space. So the simplest Platonic solid is a tetrahedron (a four-sided die), which has four corners.

c) Two parallel lines and a point not on either line define a 3-space.

I think it is false but I have no reason.

d) Two intersecting lines and a point not on either line define a 3-space.

I don't know about this one.

Please if you can check my work (add more if you'd like) and help me with c and d that would be helpful.

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  • $\begingroup$ c) - false if point belongs to a plane that contains both lines. same for d) $\endgroup$ – Vasya Sep 11 '18 at 15:31
  • $\begingroup$ Can you draw pictures in a plane for (c) and (d)? $\endgroup$ – Ethan Bolker Sep 11 '18 at 15:31
  • $\begingroup$ Yes I would be able to draw pictures but I am having trouble using GeoGebra. $\endgroup$ – Donna Sep 11 '18 at 15:36
  • $\begingroup$ @Vasya I am still not understanding why they are both false. Can you elaborate? $\endgroup$ – Donna Sep 11 '18 at 15:36
  • $\begingroup$ @Donna: two parallel lines define a plane, the same is true for two intersecting lines. If you have a point that belongs to the same plane but not on both lines, it's still only a plane and not a 3d space. One plane is 2d. $\endgroup$ – Vasya Sep 11 '18 at 16:41

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