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Question

If a=α^i+β^j+3^k,b=β^iα^j^k and
c=^i2^j^k
such that ab=1 and bc=3, then 13((a×b)c) is equal to

A
06
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B
6
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C
6.0
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D
6.00
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Solution

a=α^i+β^j+3^kb=β^iα^j^kc=^i2^j^k

Now,
ab=12αβ3=1αβ=2(1)
Also,
bc=3β+2α+1=32αβ=4
Using equation (1), we get
2α+2α=4α2+2α+1=0α=1β=2
Therefore,
13((a×b)c)=∣ ∣123211121∣ ∣=1(12)2(2+1)+3(41)=36+9=6

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