Let →a,→b and →c be three vectors such that |→a|=√3,|→b|=5,→b.→c=10 and the angle between →b and →c is π3. If →a is perpendicular to vector →b×→c, then |→a×(→b×→c)| is equal to
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Solution
|→a×(→b×→c)|=|→a||→b×→c|sinθ where θ is angle between →aand→b×→c θ=π2 (given) ⇒|→a×(→b×→c)|=√3|→b×→c|sinπ3 ⇒|→a×(→b×→c)|=√3×5×|→c|×√32 ⇒|→a×(→b×→c)|=152|→c| Now, |→b||→c|cosθ=10 ⇒5|→c|12=10 ⇒|→c|=4 ⇒|→a×(→b×→c)|=30