All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Points and Planes (Posted on 2004-12-17)
Given four non-coplanar points, how many different planes exist that are equidistant from all four points?

 See The Solution Submitted by Bractals Rating: 3.3333 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 I'm pretty sure... | Comment 3 of 10 |

I'm pretty sure there are 7.

Call the points A, B, C, and D

Its pretty clear that taking one of these points it is possible to find a plane parallel to the plane containing the other three and equidistant to the first point.

So this is 4 planes

There are also planes that separate the points by pairs.  To see them take the line determined by the first pair A and B.  There are an infinity of planes equidistant from this line and point C (they all contain the line which is the midpoint of the perpendicular from C to line AB.)   Only one of these is also the same distance from D (and from the line CD.)

There are 3 of these planes:  separating

AB from CD,

AC from BD, and

Grand total = 7 planes.

-Jer

 Posted by Jer on 2004-12-17 17:13:50

 Search: Search body:
Forums (0)