JEE Main & Advanced JEE Main Paper (Held on 08-4-2019 Afternoon)

  • question_answer
    Let \[\vec{a}=3\hat{i}+2\hat{j}+x\hat{k}\]and\[\vec{b}=\hat{i}-\hat{j}+\hat{k},\]for some real x. Then \[\left| \vec{a}\times \vec{b} \right|=r\]is possible if : [JEE Main 8-4-2019 Afternoon]

    A) \[3\sqrt{\frac{3}{2}}<r<5\sqrt{\frac{3}{2}}\]       

    B) \[0<r\le \sqrt{\frac{3}{2}}\]

    C) \[\sqrt{\frac{3}{2}}<r\le 3\sqrt{\frac{3}{2}}\]

    D)   \[r\ge 5\sqrt{\frac{3}{2}}\]

    Correct Answer: D

    Solution :

    \[\vec{a}\times \vec{b}=\left| \begin{matrix}    {\hat{i}} & {\hat{j}} & {\hat{k}}  \\    3 & 2 & x  \\    1 & -1 & 1  \\ \end{matrix} \right|\] \[=(2+x)\hat{i}+(x-3)\hat{j}-5k\] \[\left| \vec{a}\times \vec{b} \right|=\sqrt{4+{{x}^{2}}+4x+{{x}^{2}}+9-6x+25}\] \[=\sqrt{2{{x}^{2}}-2x+38}\] \[\Rightarrow \]\[\left| \vec{a}\times \vec{b} \right|\ge \sqrt{\frac{75}{2}}\]\[\Rightarrow \]\[\left| \vec{a}\times \vec{b} \right|\ge 5\sqrt{\frac{3}{2}}\]


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