Let →a=2→i+→j−2→k and →b=→i+→j. If →c is a vector such that →a.→c=|→c|, |→c−→a|=2√2 and the angle between →a×→b and →c is 30∘, then ∣∣(→a×→b)×→c∣∣ is equal to
A
23
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B
32
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C
2
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D
3
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Solution
The correct option is A32 ∣∣(→a×→b)×→c∣∣=∣∣→a×→b∣∣|→c|sin30∘ Now,