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The points are A (1.0, -1) B (2.3.1) C (0.2, -3) Determine point P that’s from the same distance from A, B and C and at a distance √5 from the plane ABC. ![So this is how I solved it , but I have to do something with Bisector planes) (https://i.stack.imgur.com/FqKCt.jpg)

user755662
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  • Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or closed. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Mar 01 '20 at 09:45
  • @user755662 I solved your problem. If you want to see my solution, show please your attempts. – Michael Rozenberg Mar 01 '20 at 10:25
  • Hi thanks I upload it. I did it in other way but I must do it with help of Bisector planes and must get 2 answers. – user755662 Mar 01 '20 at 17:54
  • My boek says CO(151/50,23/10,-98/50) or (-49/50,23/10,1/25) – user755662 Mar 01 '20 at 17:55

1 Answers1

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I would consider the following strategy:

  1. Determine the normal vector n to the plane (Calculate (A-B) x (C-B)). Multiply by $\frac{\sqrt5}{|n|}$.

  2. Find a point $\tilde{P}$ on the plane ABC that has equal distance from A,B and C. You can do so by intersecting the medians to $\overline{AB}$ and $\overline{BC}$.

  3. $P=\tilde{P}+n\frac{\sqrt5}{|n|}$

HexaGone
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  • Isn’t there any way to solve it with Bisector planes https://etc.usf.edu/clipart/42100/42128/dihedralbis_42128_lg.gif – user755662 Mar 01 '20 at 14:31