Let a be a unit vector, b=2→i+→j−→k and c=→i+3→k. Then, maximum value of [→a→b→c] is
A
−1
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B
√10+√6
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C
√10−√6
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D
√59
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Solution
The correct option is D√59 Given, b=2→i+→j−→k and c=→i+3→k Now, b×c=∣∣
∣
∣∣→i→j→k21−1103∣∣
∣
∣∣ =→i(3−0)−→j(6+1)+→k(0−1) =3→i−7→j−→k Now, abc=a⋅b×c =a(b×c)cosθ =1√32+72+12cosθ =√59cosθ ⇒abcmax=√59⋅1 (∵ maximum value of cosθ is 1) Hence, maximum value is √59.