1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

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

A
3
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
5
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
6
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is A 3aA+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=a2B2−2AabB+b2A2+a2C2−2cCaA+c2A2+b2C22bBcC+c2B2⇒0=(aB−bA)2+(aC−cA)2+(bC−Bc)2⇒(aB−bA)2=0,(aC−cA)2=0,(bC−Bc)2=0⇒aB−bA=0,aC−cA=0,bC−Bc=0⇒aB=bA,aC=cA,bC=BcSubstituting the values of aB=bA,aC=cA,bC=Bc in aBbA+bCcB+cAaC we getThe value of aBbA+bCcB+cAaC=bAbA+cBcB+cAcA=1+1+1=3

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
(a ± b ± c)²
MATHEMATICS
Watch in App
Explore more
Join BYJU'S Learning Program