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JEE Main & Advanced Sample Paper JEE Main Sample Paper-36

  • question_answer
    Given that \[\overrightarrow{a}\] is \[1\] to \[\overrightarrow{b}\] and \[p\] is a nonzero scalar if \[p\overrightarrow{r}+(\overrightarrow{r}+\overrightarrow{b})\overrightarrow{a}=\overrightarrow{c}\] then \[\overrightarrow{r}\] equals

    A) \[\overrightarrow{c}/p-\left[ (\overrightarrow{b}.\overrightarrow{c})\overrightarrow{a} \right]/{{p}^{2}}\]

    B) \[\overrightarrow{a}/p-\left[ (\overrightarrow{c}.\overrightarrow{a})\overrightarrow{b} \right]/{{p}^{2}}\]

    C) \[\overrightarrow{b}/p-\left[ (\overrightarrow{a}.\overrightarrow{b})\overrightarrow{c} \right]/{{p}^{2}}\]

    D)  None

    Correct Answer: A

    Solution :

                 \[p\overrightarrow{r}+(\overrightarrow{r}.\overrightarrow{b})\overrightarrow{a}=\overrightarrow{c}\]                           ... (I) \[p(\overrightarrow{r}.\overrightarrow{b})+(\overrightarrow{r}.\overrightarrow{b})(\overrightarrow{a}.\overrightarrow{b})=\overrightarrow{c}.\overrightarrow{b}\] \[\Rightarrow \overrightarrow{r}.\overrightarrow{b}=\frac{\overrightarrow{c}.\overrightarrow{b}}{\overrightarrow{p}}\],\[\overrightarrow{a}.\overrightarrow{b}=0\] putting in (I), \[\Rightarrow \]\[\overrightarrow{r}=\frac{\overrightarrow{c}}{p}-\frac{(\overrightarrow{c}.\overrightarrow{b})}{{{p}^{2}}}\,\,\,\overrightarrow{a}=\frac{\overrightarrow{c}}{p}-\frac{\left( \overrightarrow{b}.\overrightarrow{c} \right)\overrightarrow{a}}{{{p}^{2}}}\]


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