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Question

Let a=^j^k and c=^i^j^k, then the vector b satisfying a×b+c=0 and a.b=3 is

A
^i+^j2^k
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B
2^i^j+2^k
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C
^i^j2^k
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D
^i+^j2^k
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Solution

The correct option is B ^i+^j2^k
a×b=c
a×(a×b)=a×c
(a.b)a(a.a)b=c×a
b=1(a.a)[3a+a×c]
a.a=(^j^k)(^j^k)=1+1=2
a×c=∣ ∣ ∣^i^j^k011111∣ ∣ ∣
=^i(11)^j(0+1)+^k(01)
=2^i^j^k
b=12[3(^j^k)+(2^i^j^k)]
=12[3^j3^k2^i^j^k]
=12[2^i+2^j4^k]
=^i+^j2^k

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