Let →a=2^i+^j−2^k and →b=^i+^j. Let →c be a vector such that |→c−→a|=3,|(→a×→b)×→c|=3 and the angle between →c and →a×→b be 30∘. Then →a⋅→c is equal to:
A
258
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B
2
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C
5
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D
18
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Solution
The correct option is B2 |→a|2=4+1+4=9 and→a×→b=2^i−2^j+^k ⇒|→a×→b|=9 Also |→c−→a|=3 ⇒|→c|2+|→a|2−2→c⋅→a=9 ⇒|→c|2−2→c⋅→a=0⋯(1) Also,|(→a×→b)×→c|=3 ⇒|→a×→b||→c|sin30∘=3 ⇒3|→c|×12=3 ⇒|→c|=2 From(1),→c⋅→a=→a⋅→c=2