Let ^a,^b be unit vectors. If →c be a vector such that the angle between ^a and →c is π12, and ^b=→c+2(→c×^a), then ∣∣6→c∣∣2 is equal to
A
6(3+√3)
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B
6(3−√3)
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C
3+√3
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D
6(√3+1)
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Solution
The correct option is A6(3+√3) We have, ^b=→c+2(→c×^a) ⇒^b⋅→c=→c⋅→c+2(→c×^a)⋅→c ⇒^b⋅→c=|→c|2+2[→c^a→c] ⇒^b⋅→c=|→c|2⋯(1)
And ^b−→c=2(→c×^a)
Squaring on both sides, we get |^b|2+|→c|2−2^b⋅→c=4|→c|2|^a|2sin2π12 ⇒1+|→c|2−2|→c|2=4|→c|2(√3−12√2)2 ⇒1−|→c|2=|→c|2(2−√3) ⇒1=|→c|2(3−√3) ⇒36|→c|2=363−√3=6(3+√3)