Prove that, for any three vector a, b, c [a+b,b+c,c+a]=2[abc] - Mathematics

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Prove that, for any three vector `veca,vecb,vecc [vec a+vec b,vec b+vec c,vecc+veca]=2[veca vecb vecc]`

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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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2013-2014 (March) Delhi Set 1

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