A) \[\frac{\mathbf{c}}{p}-\frac{(\mathbf{b}\cdot \mathbf{c})\mathbf{a}}{{{p}^{2}}}\]
B) \[\frac{\mathbf{a}}{p}-\frac{(\mathbf{c}\cdot \mathbf{a})\mathbf{b}}{{{p}^{2}}}\]
C) \[\frac{\mathbf{b}}{p}-\frac{(\mathbf{a}\cdot \mathbf{b})\mathbf{c}}{{{p}^{2}}}\]
D) \[\frac{\mathbf{c}}{{{p}^{2}}}-\frac{(\mathbf{b}\cdot \mathbf{c})\mathbf{a}}{p}\]
Correct Answer: A
Solution :
We have, \[p\mathbf{r}+(\mathbf{r}\cdot \mathbf{b})\mathbf{a}=\mathbf{c}\] ... (i) \[\Rightarrow \] \[p(\mathbf{r}\cdot \mathbf{b})+(\mathbf{r}\cdot \mathbf{b})(\mathbf{a}\cdot \mathbf{b})=\mathbf{c}\cdot \mathbf{b}\] \[\Rightarrow \] \[p(\mathbf{r}\cdot \mathbf{b})=\mathbf{c}\cdot \mathbf{b}[\because \mathbf{a}\bot \mathbf{b}\therefore \mathbf{a}\cdot \mathbf{b}=0]\] \[\Rightarrow \] \[\mathbf{r}\cdot \mathbf{b}=\frac{\mathbf{c}\cdot \mathbf{b}}{p}\] On substituting the value of \[\mathbf{r}\cdot \mathbf{b}\] in Eq. (i), we get \[p\mathbf{r}+\left( \frac{\mathbf{c}\cdot \mathbf{b}}{p} \right)\mathbf{a}=\mathbf{c}\] \[\Rightarrow \] \[\mathbf{r}=\frac{\mathbf{c}}{p}-\frac{\mathbf{c}\cdot \mathbf{b}}{{{p}^{2}}}\mathbf{a}\]
You need to login to perform this action.
You will be redirected in
3 sec
