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

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

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

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

The 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 .

O' is any point in the interior of a triangle ABC. OD bot BC , OE bot AC and OF bot AB , Show that OA^2 + OB^2 + OC^2 overset(~)n OD^2 overset(~)n OF^2 = AF^2 + BD^2 + CE^2

OABC is a tetrahedron D and E are the mid points of the edges vec OA and vec BC. Then the vector vec DE in terms of vec OA,vec OB and vec OC