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), \[\Rightarrow \]\[\overrightarrow{r}=\frac{\overrightarrow{c}}{p}-\frac{(\overrightarrow{c}.\overrightarrow{b})}{{{p}^{2}}}\,\,\,\overrightarrow{a}=\frac{\overrightarrow{c}}{p}-\frac{\left( \overrightarrow{b}.\overrightarrow{c} \right)\overrightarrow{a}}{{{p}^{2}}}\]You need to login to perform this action.
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