question_answer

Let \[\vec{a}=2\hat{i}+\hat{k},\] \[\vec{b}=\hat{i}+\hat{j}+\hat{k}\] and \[\vec{c}=4\hat{i}-3\hat{j}+7\hat{k}\] be three vectors. Fine a vector \[\vec{r}\] which satisfies \[\vec{r}\times \vec{b}=\vec{c}\times \vec{b}\] and \[x+y=v\]

Answer:

Given, \[\vec{a}=2\hat{i}+\hat{k},\] \[\vec{b}=\hat{i}+\hat{j}+\hat{k},\] \[\vec{c}=4\hat{i}-3\hat{j}+7\hat{k}\] and for vector \[\vec{r},\vec{r}\times \vec{b}=\vec{c}\times \vec{b}\]and \[\vec{r}.\,\vec{a}=0\] Now, consider \[\vec{r}\times \vec{b}=\vec{c}\times \vec{b}\] \[\Rightarrow \] \[\vec{r}\times \vec{b}-\vec{c}\times \vec{b}=0\] \[\Rightarrow \] \[(\vec{r}-\vec{c})\times \vec{b}=0\] \[\Rightarrow \] \[\vec{r}-\vec{c}\] is parallel to \[\vec{b}.\] Let \[\vec{r}-\vec{c}=\lambda \,\vec{b}\] for some scalar \[\lambda .\]             \[\Rightarrow \]   \[\vec{r}=\vec{c}+\lambda \,\vec{b}\]                              ?(i) Also, it is given that,             \[\vec{r}\cdot \vec{a}=0\] \[\therefore \]      \[(\vec{c}+\lambda \,\vec{b})\cdot \vec{a}=0\]                  [using Eq.(i)] \[\Rightarrow \]   \[\vec{c}\cdot \vec{a}+\lambda (\vec{b}\cdot \vec{a})=0\] \[\Rightarrow \]   \[\lambda =\frac{\vec{c}\cdot \vec{a}}{\vec{b}\cdot \vec{a}}=\frac{-[(4\hat{i}-3\hat{j}+7\hat{k})\cdot (2\hat{i}+\hat{k})]}{[(\hat{i}+\hat{j}+\hat{k})\cdot (2\hat{i}+\hat{k})]}\]             \[=\frac{-\,(8+7)}{2+1}=\frac{-\,15}{3}=-\,5\]               Now, putting \[\lambda =-\,5\] in Eq. (i), we get \[\vec{r}=\vec{c}-5\vec{b}=(4\hat{i}-3\hat{j}+7\hat{k})-5(\hat{i}+\hat{j}+\hat{k})\] \[=-\hat{i}-8\hat{j}+2\hat{k}\]