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Question
Prove that, for any three vector `veca,vecb,vecc [vec a+vec b,vec b+vec c,vecc+veca]=2[veca vecb vecc]`
Solution
We have :
`[vec a+vec b,vec b+vec c,vecc+veca]`
`=(veca+vecb).[(vecb+vecc)xx(vecc+veca)]`
`=(veca+vecb).[vecbxxvecc+vecbxxveca+vecc xx vecc+vec c xx vec a] (By distributive law)`
`=(veca+vecb).[vecbxxvecc+vecbxxveca+vec c xx vec a] (∵vecc xx vecc=0)`
`=veca.(vecbxxvecc)+vecb(vecb xx vecc)+vec a(vecb xx veca+vecb.(vecb xx veca))+veca.(vecc xxveca)+vecb.(vecc xx veca)`
`=[veca,vecb,vecc]+[vecb,vecb,vecc]+[veca,vecb,veca]+[vecb,vecb,veca]+[veca,vecc,veca]+[vecb,vecc,veca]`
`=[veca,vecb,vecc]+[vecb,vecc,veca] " (∵scalar triple product with two equal vectors is 0) "`
`=[veca,vecb,vecc]+[veca,vecb,vecc] (because [vecb,vecc,veca]=[veca,vecb,vecc])`
`=2[veca,vecb,vecc]`
Hence,
`[vec a+vec b,vec b+vec c,vecc+veca]=2[veca vecb vecc]`
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