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Question

Let a=2^i+λ1^j+3^k,b=4^i+(3λ2)^j+6^k and c=3^i+6^j+(λ31)^k be three vectors such that b=2a 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=2a
3λ2=2λ1
2λ1+λ2=3 (1)
ac
So, 6+6λ1+3λ33=0
6λ1+3λ3+3=0
2λ1+λ3+1=0 (2)
So, (λ1,32λ1,12λ1)(λ1,λ2,λ3)
Clearly (12,4,0) is the correct option

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