Question

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

• A. $AB$ is perpendicular to $OC$
• B. $BC$ is perpendicular to $OA$
• C. $CA$ is perpendicular to $OB$
• D. $AB$ is perpendicular to $CA$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $BC$ is perpendicular to $OA$
B $CA$ is perpendicular to $OB$
C $AB$ is perpendicular to $OC$

Given eq

$O{A}^2+B{C}^2=O{B}^2+C{A}^2=O{C}^2+A{B}^2$

taking

$O{A}^2+(OC-OB{)}^2=O{B}^2+(OA-OC{)}^2$

$O{A}^2+O{C}^2+O{B}^2-2OC\cdot OB=O{B}^2+O{A}^2+O{C}^2-2OA\cdot OC$

$O{A}^2+O{C}^2+O{B}^2-(O{B}^2+O{A}^2+O{C}^2)=2OC\cdot OB-2OA\cdot OC$

$OC\cdot (OB-OA)=0$

$OC\cdot AB=0$

i.e. AB and OC is perpendicular

Given eq

$O{A}^2+B{C}^2=O{B}^2+C{A}^2=O{C}^2+A{B}^2$

taking

$O{B}^2+(OA-OC{)}^{=}O{C}^2+(OB-OA{)}^2$

$O{A}^2+O{C}^2+O{B}^2-2OC\cdot OA=O{B}^2+O{A}^2+O{C}^2-2OA\cdot OB$

$O{A}^2+O{C}^2+O{B}^2-(O{B}^2+O{A}^2+O{C}^2)=2OC\cdot OA-2OA\cdot OB$

$OA\cdot (OC-OB)=0$

$OA\cdot BC=0$

i.e. BC and OA is perpendicular

Given eq

$O{A}^2+B{C}^2=O{B}^2+C{A}^2=O{C}^2+A{B}^2$

taking

$O{A}^2+(OC-OB{)}^{=}O{C}^2+(OB-OA{)}^2$

$O{A}^2+O{C}^2+O{B}^2-2OC\cdot OB=O{B}^2+O{A}^2+O{C}^2-2OA\cdot OB$

$O{A}^2+O{C}^2+O{B}^2-(O{B}^2+O{A}^2+O{C}^2)=2OC\cdot OB-2OA\cdot OB$

$OB\cdot (OC-OA)=0$

$OB\cdot CA=0$

i.e. CA and OB is perpendicular