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JEE Main & Advanced JEE Main Paper (Held On 10-Jan-2019 Morning)

  • question_answer
     Let \[\overrightarrow{a}=2\widehat{i}+{{\lambda }_{1}}\widehat{j}+3\widehat{k},\text{ }\overrightarrow{b}=4\widehat{i}+(3-{{\lambda }_{2}})\widehat{j}+6\widehat{k}\] and \[\overrightarrow{c}=3\widehat{i}+6\widehat{j}+({{\lambda }_{3}}-1)\widehat{k}\] be three vectors such that \[\overrightarrow{b}=2\overrightarrow{a}\] and \[\overrightarrow{a}\] is perpendicular to\[\overrightarrow{c}\]. Then a possible value of \[({{\lambda }_{1}},\,\,{{\lambda }_{2}},\,\,{{\lambda }_{3}})\] is - [JEE Main Online Paper (Held On 10-Jan-2019 Morning]

    A) (1, 5, 1)                                    

    B) (1, 3, 1)

    C) \[\left( -\frac{1}{2},\,\,4,\,\,0 \right)\]     

    D)                  \[\left( \frac{1}{2},\,\,4,\,\,-2 \right)\]

    Correct Answer: C

    Solution :

    \[\overrightarrow{a}.\overrightarrow{c}=0\,\,\Rightarrow \,\,6+6\lambda +3({{\lambda }_{3}}-1)=0\] \[\Rightarrow \,\,{{\lambda }_{3}}\,=\,-2{{\lambda }_{1}}-1\] \[\overrightarrow{b}=2\overrightarrow{a}\,\,\Rightarrow \,\,3-{{\lambda }_{2}}=2{{\lambda }_{1}}\] \[\Rightarrow \,\,{{\lambda }_{2}}=3-2{{\lambda }_{1}}\] \[Let\,\,\,{{\lambda }_{1}}=1;\,\,{{\lambda }_{2}}=1;\text{ }{{\lambda }_{3}}=-3\] \[{{\lambda }_{1}}=-\frac{1}{2};\,\,{{\lambda }_{2}}=4;\text{ }{{\lambda }_{3}}=0\] \[{{\lambda }_{1}}=\frac{1}{2};\,\,{{\lambda }_{2}}=2;\text{ }{{\lambda }_{3}}=-2\]


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