CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2005

  • question_answer
    If \[\vec{a}\] is perpendicular to\[\overrightarrow{b}\]and\[\overrightarrow{c},|\overrightarrow{a}|=2,\]\[|\overrightarrow{b}|=3|\overrightarrow{c}|=4\]and the angle between\[\overrightarrow{b}\]and\[\overrightarrow{c}\]is\[\frac{2\pi }{3},\]then\[[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}]\]is equal to:

    A)  \[4\sqrt{3}\]                                     

    B)  \[6\sqrt{3}\]

    C)  \[12\sqrt{3}\]                  

    D)         \[18\sqrt{3}\]

    E)  \[8\sqrt{3}\]

    Correct Answer: C

    Solution :

    \[\because \]\[[\overrightarrow{a}\,\,\overrightarrow{b}\,\,\overrightarrow{c}]=\overrightarrow{a}.(\overrightarrow{b}\times \overrightarrow{c})\] \[=\overrightarrow{a}.\left( |\overrightarrow{b}|\overrightarrow{c}|\sin \frac{2\pi }{3}\hat{n} \right)\] \[=|\overrightarrow{a}|.|\overrightarrow{b}||\overrightarrow{c}|\left( \sin \frac{2\pi }{3} \right)=2\times 3\times 4\times \frac{\sqrt{3}}{2}\] \[=12\sqrt{3}\]


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