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vishal_kothari Group Title

A,B,C,D are four distinct points in three space. Suppose each of the angles ABC, BCD, CDA, and DAB are right angles. Show that all four points lie in the same plane.

  • 2 years ago
  • 2 years ago

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  1. badreferences Group Title
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    Is heuristic reasoning okay, or should I make a rigorous proof?

    • 2 years ago
  2. vishal_kothari Group Title
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    easy proof...

    • 2 years ago
  3. badreferences Group Title
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    Not for me. :P

    • 2 years ago
  4. vishal_kothari Group Title
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    yes for u only..

    • 2 years ago
  5. AravindG Group Title
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    this is easy

    • 2 years ago
  6. AravindG Group Title
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    tell me which text u refer

    • 2 years ago
  7. vishal_kothari Group Title
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    yeah...

    • 2 years ago
  8. AravindG Group Title
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    do u have balagopal book?

    • 2 years ago
  9. vishal_kothari Group Title
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    yeah..

    • 2 years ago
  10. AravindG Group Title
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    this is there in it!!

    • 2 years ago
  11. AravindG Group Title
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    refer page 116

    • 2 years ago
  12. vishal_kothari Group Title
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    ya but the proof is very easy....

    • 2 years ago
  13. badreferences Group Title
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    Proof by probabilistic assumption. :P There are no sets of regular non-planes that adhere to the rules you specify.

    • 2 years ago
  14. vishal_kothari Group Title
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    kkk

    • 2 years ago
  15. badreferences Group Title
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    Lol, is that really an answer? Laaame.

    • 2 years ago
  16. vishal_kothari Group Title
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    no...it needs 3 d geometry....

    • 2 years ago
  17. vishal_kothari Group Title
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    what are u typing?

    • 2 years ago
  18. badreferences Group Title
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    A rigorous mathematical proof by extending vectors. XD

    • 2 years ago
  19. badreferences Group Title
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    Not including non-Euclidian, of course, because I don't know that stuff.

    • 2 years ago
  20. Ishaan94 Group Title
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    I am not sure about this, It's been time since I have done 3d. If you can take A,B,C,D to be following vectors \(\vec{a},\vec{b},\vec{c} \) and \(\vec{d}\), and then apply the dot product rule.

    • 2 years ago
  21. AravindG Group Title
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    use balagopal book

    • 2 years ago
  22. Ishaan94 Group Title
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    \[AB = \vec{b} - \vec{a}\]\[BC= \vec{c} - \vec{b}\] \[\angle ABC=90 \implies (\vec{b} - \vec{a})(\vec{c} - \vec{b})=0 \] Something like this, and then applying the same procedure for every angle. In the end you might get something.

    • 2 years ago
  23. badreferences Group Title
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    Gah, I wish I had a tablet.

    • 2 years ago
  24. vishal_kothari Group Title
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    take it then...

    • 2 years ago
  25. vishal_kothari Group Title
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    bye..

    • 2 years ago
  26. badreferences Group Title
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    lol Ishaan has the right idea. We take what we know... the vector sum being zero, and each individual vector extending in another component from the previous, alternating and opposite. We can determine then, through mathematics that I tried HTML typing, that the only possible unit vector components are ones that cancel out.

    • 2 years ago
  27. badreferences Group Title
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    We can probably determine the last part through annoying set elimination by contradiction or through matrix algebra.

    • 2 years ago
  28. vishal_kothari Group Title
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    thanks....

    • 2 years ago
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