J & K CET Engineering J and K - CET Engineering Solved Paper-2009

  • question_answer If \[2\text{ }\vec{a}+3\text{ }\vec{b}+\vec{c}=0,\] then \[\vec{a}\times \vec{b}+\vec{b}\times \vec{c}+\vec{c}\times \vec{a}\]is equal to

    A)  \[6\,(\vec{b}\times \vec{c})\]       

    B)  \[3\,(\vec{b}\times \vec{c})\]

    C)  \[2\,(\vec{b}\times \vec{c})\]        

    D)  \[\vec{0}\]

    Correct Answer: B

    Solution :

    Given, \[2\vec{a}+3\vec{b}+\vec{c}=\vec{0}\] \[\Rightarrow \] \[2\vec{a}+3\vec{b}=-\vec{c}\] Taking cross product with \[\vec{a}\] and \[\vec{b}\] respectively, we get \[2(\vec{a}\times \vec{a})+3(\vec{a}\times \vec{b})=-\vec{a}\times \vec{c}\] \[\Rightarrow \] \[3(\vec{a}\times \vec{b})=\vec{c}\times \vec{a}\] ?..(i) and \[2(\vec{b}\times \vec{a})+3(\vec{b}\times \vec{b})=-\vec{b}\times \vec{c}\] \[\Rightarrow \] \[2(\vec{a}\times \vec{b})=\vec{b}\times \vec{c}\] ?(ii) Now, \[\vec{a}\times \vec{b}+\vec{b}\times \vec{c}+\vec{c}\times \vec{a}\] \[=\vec{a}\times \vec{b}+\vec{b}\times \vec{c}+3(\vec{a}\times \vec{b})\]   [using Eq. (i)] \[=4(\vec{a}\times \vec{b})+\vec{b}\times \vec{c}\] \[=2(\vec{b}\times \vec{c})+\vec{b}\times \vec{c}\] [using Eq. (ii)] \[=3(\vec{b}\times \vec{c})\]

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