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JEE Main & Advanced Sample Paper JEE Main Sample Paper-3

  • question_answer
    Given that \[\vec{a}\] is \[\bot \]to \[\vec{b}\]and p is a non zero scalar if \[p\vec{r}+(\vec{r}.\vec{b})\vec{a}=\vec{c}\]then \[\vec{r}\]equals

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

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

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

    D)  None

    Correct Answer: A

    Solution :

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


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