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Question

Let a,b,c are vectors such that |a|=2,|b|=3,|c|=3 and (3a2b) is perpendicular to c,(3b2c) is perpendicular to a and (3c2a) is perpendicular to b. Then the value of |a+b+c|=

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Solution

Given :
(3a2b)c=03ac2bc=0(i),
(3b2c)a=03ab2ac=0(ii)
and
(3c2a)b=03bc2ab=0(iii)
Adding (i),(ii) and (iii):
we get, (ab+bc+ac)=0(A)
Also,
|a+b+c|2=|a|2+|b|2+|c|2+2(ab+bc+ac)=4+9+3 [using (A)][|a|=2,|b|=3,|c|=3]=16|a+b+c|=4

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