question_answer

Find a unit vector perpendicular to each of the vectors \[\vec{a}+\vec{b}\] and \[\vec{a}-\vec{b},\] where \[\vec{a}=3\hat{i}+2\hat{j}+2\hat{k}\] and \[\vec{b}=\hat{i}+2\hat{j}-2\hat{k}.\]

Answer:

Given, \[\overrightarrow{a}=3\hat{i}+2\hat{j}+2\hat{k}\] and \[\overrightarrow{b}=\hat{i}+2\hat{j}-2\hat{k}\] \[\therefore \]\[\overrightarrow{a}+\overrightarrow{b}=(3\,\hat{i}+2\hat{j}+2\hat{k})+(\hat{i}+2\hat{j}-2\hat{k})=4\,\hat{i}+4\hat{j}\] \[\overrightarrow{a}-\overrightarrow{b}=(3\hat{i}+2\hat{j}+2\hat{k})-(\hat{i}+2\hat{j}-2\hat{k})=2\hat{i}+4j\hat{k}\] Let the vector \[\overrightarrow{c}\]be perpendicular to \[(\overrightarrow{a}+\overrightarrow{b})\]and \[(\overrightarrow{a}-\overrightarrow{b})\]. \[\therefore \]      \[\overrightarrow{c}=(\overrightarrow{a}+\overrightarrow{b})\times (\overrightarrow{a}-\overrightarrow{b})\] \[=16\hat{i}-16\hat{j}-8\hat{k}\] \[\therefore \]Unit vector perpendicular to \[(\overrightarrow{a}+\overrightarrow{b})\]and \[(\overrightarrow{a}-\overrightarrow{b})\] \[=\frac{\overrightarrow{c}}{|\overrightarrow{c}|}=\frac{16\hat{i}-16\hat{j}-8\hat{k}}{\sqrt{{{(16)}^{2}}+{{(-\,16)}^{2}}+{{(-\,8)}^{2}}}}\] \[=\frac{16\hat{i}-16\hat{j}-8\hat{k}}{\sqrt{256+256+64}}=\frac{16\hat{i}-16\hat{j}-8\hat{k}}{\sqrt{576}}\] \[=\frac{8\,\,(2\hat{i}-2\hat{j}-\hat{k}}{24}=\frac{2\hat{i}-2\hat{j}-\hat{k}}{3}\] \[=\frac{2}{3}\hat{i}-\frac{2}{3}\hat{j}-\frac{1}{3}\hat{k}\]