If →a=2^i+3^j−4^k,→b=^i+^j+^k and →c=4^i+2^j+3^k, then ∣∣∣→a×(→b×→c)∣∣∣=
A
√10
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B
1
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C
2
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D
√5
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Solution
The correct option is D√5 We know that∣∣∣→a×(→b×→c)∣∣∣=∣∣∣(→a⋅→c)→b−(→a⋅→b)→c∣∣∣ ⇒∣∣∣(→a⋅→c)→b−(→a⋅→b)→c∣∣∣=∣∣2(^i+^j+^k)−(1)(4^i+2^j+3^k)∣∣
Where, (→a⋅→c)=2,(→a⋅→b)=−1 =|−2^i−^k|=√4+1=√5