Let →a=−^i+4^k,→b=5^i+2^k and →c=−3^i+^k. If →a=x→b+y→c, then the value of (x,y) is
A
(1,1)
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B
(−2,2)
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C
(1,2)
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D
(2,−1)
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Solution
The correct option is C(1,2) Given : →a=x→b+y→c ⇒−^i+4^k=x(5^i+2^k)+y(−3^i+^k)
Equating the cofficient of like vectors, we get ⇒5x−3y=−1⋯(i)2x+y=4⋯(ii)
Solving (i) and (ii), we get x=1,y=2