Let →a=2^i+λ1^j+3^k,→b=4^i+(3−λ2)^j+6^k and →c=3^i+6^j+(λ3−1)^k be three vectors such that →b=2→a and →a is perpendicular to →c. Then the possible value of (λ1,λ2,λ3) is :
A
(−12,4,0)
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B
(1,5,1)
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C
(1,3,1)
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D
(12,4,−2)
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Solution
The correct option is A(−12,4,0) →b=2→a 3−λ2=2λ1 2λ1+λ2=3⋯(1) →a⊥→c
So, 6+6λ1+3λ3−3=0 ⇒6λ1+3λ3+3=0 ⇒2λ1+λ3+1=0⋯(2)
So, (λ1,3−2λ1,−1−2λ1)≡(λ1,λ2,λ3)
Clearly (−12,4,0) is the correct option