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Question

If a=2^i+3^j+^k,b=^i2^j+^k and c=3^i+^j+2^k,
find [abc].

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Solution

Given,
a=2^i+3^j+^k
b=^i2^j+^k
c=3^i+^j+2^k
The value of [abc] is given by (a×b)c

a×b=^i^j^k231121

=(3×11(2))^i(2×11×1)^j+(2(2)3×1)^k
=5^i^j7^k

Now, (a×b)c
=(5^i^j7^k)(3^i+^j+2^k)
=15114
=30


Alternate:
[abc] is equal to the determinant of the matrix that has the three vectors either as its rows or its columns.
[abc]=∣ ∣231121312∣ ∣

=2(41)3(2+3)+1(16)=30




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