If →a,→b, and →c are unit vectors such that →a+2→b+2→c=→0, then |→a×→c| is equal to :
A
√154
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B
14
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C
1516
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D
√1516
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Solution
The correct option is A√154 →a+2→b+2→c=→0⇒→a+2→c=−2→b⇒(→a+2→c)⋅(→a+2→c)=(−2→b)⋅(−2→b)⇒1+4→a⋅→c+4=4⇒→a⋅→c=−14
Let the angle between →a and →c be θ ⇒cosθ=−14⇒sinθ=±√154
Now, |→a×→c|=|→a||→c|sinθ=√154