question_answer

Express the vector \[\vec{a}=(5\hat{i}\,-2\hat{j}\,+5\hat{k})\] as sum of two vectors such that one is parallel to the vector \[\vec{b}=(3\hat{i}\,+\hat{k})\] and the other is perpendicular to \[\vec{b}.\]       

Answer:

It is given that,\[{{\overrightarrow{a}}_{1}}\parallel \overrightarrow{b}\]and \[{{\overrightarrow{a}}_{1}}\bot \overrightarrow{b},\]where \[\overrightarrow{a}={{\overrightarrow{a}}_{1}}+{{\overrightarrow{a}}_{2}}\] \[\therefore \]      \[{{\overrightarrow{a}}_{1}}=\lambda \overrightarrow{b}\] \[\therefore \]      \[{{\overrightarrow{a}}_{2}}=\overrightarrow{a}-{{\overrightarrow{a}}_{1}}=\overrightarrow{a}-\lambda \overrightarrow{b}\] \[{{\overrightarrow{a}}_{2}}\bot \overrightarrow{b}\] \[\therefore \]      \[{{\overrightarrow{a}}_{2}}\cdot \overrightarrow{b}=0\Rightarrow (\overrightarrow{a}-\lambda \overrightarrow{b})\cdot \overrightarrow{b}=0\] \[\Rightarrow \]   \[\overrightarrow{a}\cdot \overrightarrow{b}-\lambda \overrightarrow{b}\cdot \overrightarrow{b}=0\] \[\Rightarrow \]               \[\lambda =\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{\overrightarrow{b}\cdot \overrightarrow{b}}\] Now,     \[\overrightarrow{a}=5\,\hat{i}-2\hat{j}+5\hat{k}\]and \[\overrightarrow{b}=3\,i+k\] \[\therefore \]      \[\overrightarrow{a}\cdot \overrightarrow{b}=15+0+5=20\]and \[\overrightarrow{b}\cdot \overrightarrow{b}=9+0+1=10\] \[\therefore \]      \[\lambda =\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{\overrightarrow{b}\cdot \overrightarrow{b}}=\frac{20}{10}=2\] \[\therefore \]      \[{{\overrightarrow{a}}_{1}}=\lambda \overrightarrow{b}=2\,(3\,\hat{i}+\hat{k})=6\,\hat{i}+2\hat{k}\] And \[{{\overrightarrow{a}}_{2}}=\overrightarrow{a}-{{\overrightarrow{a}}_{1}}=(5\,\hat{i}-2\hat{j}+5\hat{k})-\,(6\,\hat{i}+2\hat{k})\] \[=-\,\hat{i}-2\hat{j}+3\hat{k}\]