Question

# For non-coplanar vectors A,B and C, |[ABC]|=|A||B||C| holds if and only if

A
AB=BC=CA=0
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B
AB=0=BC
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C
AB=0=CA
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D
BC=0=CA
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Solution

## The correct options are A A⋅B=B⋅C=C⋅A=0 B A⋅B=0=B⋅C C A⋅B=0=C⋅A D B⋅C=0=C⋅A Let θ be the angle between A and B and ϕ the angle between C and A×B i.e. the angle between C and the perpendicular to the plane containing A and B. Then|(A×B)⋅C|=|A×B||C|cosϕ=|A||B||C|sinθcosϕso if the given relation holds, we have sinθcosϕ=1 since|sinθ||cosϕ|≤1 so we must have sinθ=1,cosϕ=1 i.e.θ=π/2,ϕ=0. The former implies that A and B areperpendicular so that A⋅B=0 On the hand, if ϕ is zero, C must be perpendicular to both A and Bso that B⋅C=0=A⋅C.

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