DeGuaTheorem::usage="It is the three-dimensional version of Pythagorean theorem.If a tetrahedron is orthocentric (three ortogonal edges), the square of the opposite faceto the corner of the right angle, that is, the inclined face, is equal to the sum of the squares of the the other three faces.Reference: Personal.";
Development :
Vertices:
In[]:=
P1={1,2,3};P2={4,2,3};P3={1,6,3};P4={1,2,15};​​
Calculation of square areas:
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P
2
P
3
P
4
2
area
==
2
1
2
Norm[(P3-P2)(P4-P2)]
​​
P
1
P
2
P
3
2
area
==
2
1
2
Norm[(P3-P1)(P2-P1)]
​​
P
1
P
3
P
4
2
area
==
2
1
2
Norm[(P4-P1)(P3-P1)]
​​
P
1
P
2
P
4
2
area
==
2
1
2
Norm[(P4-P1)(P2-P1)]
2
area
P
2
P
3
P
4
936;
2
area
P
1
P
2
P
3
36;
2
area
P
1
P
3
P
4
576;
2
area
P
1
P
2
P
4
324;
In[]:=
324+576+36
936;
These calculations verify the theorem
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gr=Show[{​​Graphics3D[{Black,​​Ball[P1,0.04],​​Text[Style[Subscript[P,1],14,Bold],P1,{0,-1.2}],​​Ball[P2,0.04],​​Text[Style[Subscript[P,2],14,Bold],P2,{0,-1.2}],​​Ball[P3,0.04],​​Text[Style[Subscript[P,3],14,Bold],P3,{0,-1.2}],​​Ball[P4,0.04],​​Text[Style[Subscript[P,4],14,Bold],P4,{0,-1.2}],​​Green,Opacity[0.4],EdgeForm[Thick],​​Tetrahedron[{P1,P2,P3,P4}]}]},AxesLabel->{x,y,z},AxesOrigin{0,0,0},AxesTrue,BoxedFalse,BoxRatiosAutomatic,PlotRangeAll,ViewPoint{1.3,-2.4,2.},ImageSize700]
Out[]=
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ResourceFunction["SpinShow"][gr,30]
Out[]=