JEE Main & Advanced JEE Main Paper (Held On 15 April 2018) Slot-I

  • question_answer
    If \[\overrightarrow{a},\overrightarrow{b}\], and \[\overrightarrow{c}\] are unit vectors such that \[\overrightarrow{a}+2\overrightarrow{b}+2\overrightarrow{c}=\overrightarrow{0}\], the \[|\overrightarrow{a}\times \overrightarrow{c}|\] is equal to                         [JEE Online 15-04-2018]

    A) \[\frac{1}{4}\]                        

    B) \[\frac{\sqrt{15}}{4}\]           

    C) \[\frac{15}{16}\]                                

    D) \[\frac{\sqrt{15}}{16}\]

    Correct Answer: B

    Solution :

    \[\overrightarrow{a}+2\overrightarrow{b}+2\overrightarrow{c}=0\] \[\overrightarrow{a}+2\overrightarrow{c}=-2\overrightarrow{b}\] \[\Rightarrow (\overrightarrow{a}+2\overrightarrow{c}).\overrightarrow{a}+2\overrightarrow{c}=-2\overrightarrow{b}.(-2\overrightarrow{b})\] \[\overrightarrow{a}.\overrightarrow{a}+4\overrightarrow{c}.\overrightarrow{c}=4\overrightarrow{b}.\overrightarrow{b}\] \[1+4+4\overrightarrow{a}.\overrightarrow{c}=4\] \[\overrightarrow{a}.\overrightarrow{c}=\frac{-1}{4}\] \[=\overrightarrow{a}.\overrightarrow{c}\widehat{c}+\overrightarrow{a}\times \overrightarrow{c}=\overrightarrow{a}\] \[|\overrightarrow{a}.\overrightarrow{c}{{|}^{2}}+|\overrightarrow{a}\times \overrightarrow{c}{{|}^{2}}=1\] \[\frac{1}{16}+|\overrightarrow{a}\times \overrightarrow{c}{{|}^{2}}=1\] \[|\overrightarrow{a}\times |\overrightarrow{c}{{|}^{2}}=\frac{15}{16}\] \[|\overrightarrow{a}\times |\overrightarrow{c}|=\frac{\sqrt{15}}{4}\]


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