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Question

Prove that for any three vectors a, b, c [b + c c + a a + b] = 2 [abc].

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Solution

For any vector a,b and c , prove that
[a+bb+cc+a]=2[abc]
sowing LHS :-
[a+bb+cc+a]=(a+b).[(b+c)×(c+a)]
[abc=a.(b×c]
=(a+b).[(b×c)+(b×a)+(c×c)+(c×a)]

[c×c=|c||c|sinO^n=o
=(a+b).[(b×c)+(b×a)+o+(c×a)]
=a.(b×c)+b.(b×c)+a.(b×a)+b.(b×a)+a.(c×a)+b.(c×a)
=[a,b,c]+[bbc]+[aba]+[aca]+[bca]

[bbc]=[cbb]
=c.(b×b)
As (b×b)=o
=c.o
=o

Using property
[bbc]=0
[aba]=0
[bba]=0
[aca]=0

=[abc]+o+o+o+o+[bca]
=[abc]+[bca]
=[abc]+[abc]
=2[abc]=R.H.S

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