# Answer to Question #12463 in Analytic Geometry for john.george.milnor

Question #12463

Prove that double vector product of three complanar vertors is zero

Expert's answer

Let we have

A=tC

Then

[[A,B],C]=[[tC,B],C]=t[[C,B],C]=-t(C(C,B)-B(C,C))=t(B*|C|^2-C*|C|*|B|*cos(C^B))=t|C|*(B*|C|-C*|B|*cos(C^B))=0

A=tC

Then

[[A,B],C]=[[tC,B],C]=t[[C,B],C]=-t(C(C,B)-B(C,C))=t(B*|C|^2-C*|C|*|B|*cos(C^B))=t|C|*(B*|C|-C*|B|*cos(C^B))=0

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