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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2004

  • question_answer
        Let \[(1,b,{{b}^{2}})\]and a unit vector \[(1,c,{{c}^{2}})\] be coplanar, if \[\text{abc}\] is perpendicular to \[\Delta ABC\] then \[2\overrightarrow{AC}=3\overrightarrow{CB}\]is equal to:

    A)  \[2\overrightarrow{OA}+3\overrightarrow{OB}\]                           

    B)  \[4\overrightarrow{OC}\]

    C)  \[-\overrightarrow{OC}\]                           

    D)  \[\overrightarrow{OC}\]

    Correct Answer: A

    Solution :

                    \[\overrightarrow{c}\]coplanar with\[\overrightarrow{a},\overrightarrow{b}\] \[\therefore \]  \[\overrightarrow{c}=x\overrightarrow{a}+y\overrightarrow{b}\] \[\Rightarrow \]               \[\overrightarrow{c}=x(2\hat{i}+\hat{j}+\hat{k})+y(\hat{i}+2\hat{j}-\hat{k})\]                 \[\overrightarrow{c}=(2x+y)\hat{i}+(x+2y)\hat{j}+(x-y)\hat{k}\] \[\because \]     \[\overrightarrow{a}.\overrightarrow{b}=0\] \[\Rightarrow \]               \[2(2x+y)+x+2y+x-y=0\] \[\Rightarrow \]               \[y=-2x\] \[\therefore \]  \[\overrightarrow{c}=-3x\hat{j}+3x\hat{k}=3x(-\hat{j}+\hat{k})\] \[\because \]     \[|c|=1\] \[\Rightarrow \]               \[9{{x}^{2}}+9{{x}^{2}}=1\] \[\Rightarrow \]               \[x=\pm \frac{1}{3\sqrt{2}}\] \[\Rightarrow \]               \[\overrightarrow{c}=\frac{1}{\sqrt{2}}(-\hat{j}+\hat{k})\]


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