The OABC is a tetrahedron such that `OA^2+BC^2=OB^2+CA^2=OC^2+AB^2`,then

If OABC is a tetrahedron such that OA^2 + BC^2 = OB^2 + CA^2 = OC^2 + AB^2 then

If OABC is a tetrahedron such that OA^(2)+BC^(2)=OB^(2)+CA^(2)=OC^(2)+AB^(2) then

If OABC is a tetrahedron such that OA^(2)+BC^(2)=OB^(2)+CA^(2)=OC^(2)+AB^(2)= then

If OABC is a tetrahedron such that OA^(2)+BC^(2)=OB^(2)+CA^(2)=OC^(2)+AB^(2) then

If OABC be a regular tetrahedron such that OA^(2)+BC^(2)=OB^(2)+CA^(2)=OC^(2)+AB^(2) then

OABC is a tetrahedron such that OA=OB=OC=k and each of the edges OA,OB and OC is inclined at an angle theta with the other two the range of theta is

O is any point inside a rectangle. Prove that OA^2+OC^2=OB^2+OD^2 .

In a tetrahedron OABC, the edges are of lengths, |OA|=|BC|=a,|OB|=|AC|=b,|OC|=|AB|=c. Let G_1 and G_2 be the centroids of the triangle ABC and AOC such that OG_1 _|_ BG_2, then the value of (a^2+c^2)/b^2 is

In a tetrahedron OABC, the edges are of lengths, |OA|=|BC|=a,|OB|=|AC|=b,|OC|=|AB|=c. Let G_1 and G_2 be the centroids of the triangle ABC and AOC such that OG_1 _|_ BG_2, then the value of (a^2+c^2)/b^2 is