Let →a=2^i+^j−2^k,→b=^i+^j. If ^cis a vector such that →a.→c=|→c|,|→a+→c|=2√3 and angle between →a×→b and →c is 30∘, then |(→a×→b).→c|=
A
12
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B
3√32
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C
3
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D
32
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Solution
The correct option is B3√32 Given →a=2^i+^j−2^k,→b=^i+^j |→a|=3,given→a.→c=|→c||→a+→c|=2√3⇒|→a+→c|2=12⇒|→a|2+|→c|2+2→a.→c=12 ⇒9+|→c|2+2|→c|=12⇒|→c|2+2|→c|−3=0⇒|→c|=1 and |(→a×→b).→c|=|→a×→b||→c|cos30∘ =|→a×→b|√32=3√32 [here →a×→b=∣∣
∣
∣∣^i^j^k21−2110∣∣
∣
∣∣ =^i(2)−^j(2)+^k(1)=2^i−2^j+^k