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Question

If a=i2j3k,b=2i+jk,c=i+3j2k then (a×b)×c is

A
(3i+16j+19k)
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B
4(i+3j+4k)
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C
5(i3j4k)
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D
4(i+3j+4k)
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Solution

The correct option is D 5(i3j4k)
Given:a=^i2^j3^k,b=2^i+^j^k and c=^i+3^j2^k
a×b=∣ ∣ ∣^i^j^k123211∣ ∣ ∣
=(2+3)^i(1+6)^j+(1+4)^k
=5^i5^j+5^k
(a×b)×c=∣ ∣ ∣^i^j^k555132∣ ∣ ∣
=(1015)^i(105)^j+(15+5)^k
=5^i+15^j+20^k
=5(^i3^j4^k)

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