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JEE Main & Advanced Sample Paper JEE Main Sample Paper-32

  • question_answer
    \[=2\hat{i}-\hat{j}+\hat{k},\overrightarrow{q}=\hat{i}+2\hat{j}-\hat{k}\]and \[=\hat{i}+\hat{j}-2\hat{k}.\]. If \[=\overrightarrow{p}+\lambda \overrightarrow{q}\] and projection of \[\]on \[\overrightarrow{r}\] is \[\frac{4}{\sqrt{6}},\] then \[\lambda \] equals

    A) \[1\]      

    B)                                    \[\frac{1}{5}\]

    C) \[\frac{3}{5}\]                   

    D)    \[\frac{2}{5}\]

    Correct Answer: A

    Solution :

    \[\vec{V}=(2,-1,1)+\lambda (1,2,-1)\]                 Given \[\frac{\vec{V}.\vec{r}}{|\,\vec{r}|}=\frac{4}{\sqrt{6}}\Rightarrow \frac{(\vec{p}.\vec{r})+\lambda (\vec{q}.\vec{r})}{|\,\,\vec{r}|}=\frac{4}{\sqrt{6}}\] \[\therefore \,\,\frac{-1+\lambda (5)}{\sqrt{6}}=\frac{4}{\sqrt{6}}\,\,\,\,\,\,\,\,\,\,\therefore \,\,\lambda =1\]


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