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 If a2+b2+c2=1 where a,b,cR, then the maximum  value of (4a3b)2+(5b4c)2+(3c5a)2 is 

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a
25
b
50
c
144
d
None of these

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detailed solution

Correct option is B

Sol.  (b) Let r→1=ai^+bj^+ck^,r→2=3i^+4j^+5k^ r→1×r→22≤r→12r→22--------(1) Now r→1×r→2=i^j^k^abc345=i^(5b−4c)+j^(3c−5a)+k^(4a−3b) So, from (1) (5b−4c)2+(3c−5a)2+(4a−3b)2≤50


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