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Question

If a,b,c and A,B,CR(0} such that aA+bB+cC+(a2+b2+c2)(A2+B2+C2)=0, then value of aBbA+bCcB+cAaC is

A
3
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B
4
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C
5
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D
6
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Solution

The correct option is A 3
aA+bB+cC+(a2+b2+c2)(A2+B2+C2)=0

aA+bB+cC=(a2+b2+c2)(A2+B2+C2)

squaring both sides, we get
(aA+bB+cC)2=((a2+b2+c2)(A2+B2+C2))2

(aA+bB+cC)2=(a2+b2+c2)(A2+B2+C2)

a2A2+b2B2+c2C2+2AbB+2bBcC+2cCaA=a2(A2+B2+C2)+b2(A2+B2+C2)+c2
(A2+B2+C2)

a2A2+b2B2+c2C2+2AbB+2bBcC+2cCaA=a2A2+a2B2+a2C2+b2A2+b2B2+b2C2+
c2A2+c2B2+c2C2

2AabB+2bBcC+2cCaA=a2B2+a2C2+b2A2+b2C2+c2A2+c2B2

2AabB+2bBcC+2cCaA=a2B2+b2A2+a2C2+c2A2+b2C2+c2B2

0=a2B22AabB+b2A2+a2C22cCaA+c2A2+b2C22bBcC+c2B2

0=(aBbA)2+(aCcA)2+(bCBc)2

(aBbA)2=0,(aCcA)2=0,(bCBc)2=0

aBbA=0,aCcA=0,bCBc=0

aB=bA,aC=cA,bC=Bc

Substituting the values of aB=bA,aC=cA,bC=Bc in aBbA+bCcB+cAaC we get

The value of aBbA+bCcB+cAaC=bAbA+cBcB+cAcA=1+1+1=3

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