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Question

# If OABC is a tetrahedron such that OA2+BC2=OB2+CA2=OC2+AB2, then

A
AB is perpendicular to OC
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B
BC is perpendicular to OA
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C
CA is perpendicular to OB
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D
AB is perpendicular to CA
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Solution

## The correct options are A BC is perpendicular to OA B CA is perpendicular to OB C AB is perpendicular to OCGiven eqOA2+BC2=OB2+CA2=OC2+AB2taking OA2+(OC−OB)2=OB2+(OA−OC)2OA2+OC2+OB2−2OC⋅OB=OB2+OA2+OC2−2OA⋅OCOA2+OC2+OB2−(OB2+OA2+OC2)=2OC⋅OB−2OA⋅OCOC⋅(OB−OA)=0OC⋅AB=0i.e. AB and OC is perpendicularGiven eqOA2+BC2=OB2+CA2=OC2+AB2taking OB2+(OA−OC)=OC2+(OB−OA)2OA2+OC2+OB2−2OC⋅OA=OB2+OA2+OC2−2OA⋅OBOA2+OC2+OB2−(OB2+OA2+OC2)=2OC⋅OA−2OA⋅OBOA⋅(OC−OB)=0OA⋅BC=0i.e. BC and OA is perpendicularGiven eqOA2+BC2=OB2+CA2=OC2+AB2taking OA2+(OC−OB)=OC2+(OB−OA)2OA2+OC2+OB2−2OC⋅OB=OB2+OA2+OC2−2OA⋅OBOA2+OC2+OB2−(OB2+OA2+OC2)=2OC⋅OB−2OA⋅OBOB⋅(OC−OA)=0OB⋅CA=0i.e. CA and OB is perpendicular

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