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Question
If \[\overrightarrow{a} = \hat{i} + \hat{j} , \overrightarrow{b} = \hat{j} + \hat{k} , \overrightarrow{c} = \hat{k} + \hat{i}\], find the unit vector in the direction of \[\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}\].
Solution
Let \[\overrightarrow{a} = \hat{i} + \hat{j} , \overrightarrow{b} = \hat{j} + \hat{k} , \overrightarrow{c} = \hat{k} + \hat{i}\]
Then,
\[\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} = \hat{i} + \hat{j} + \hat{j} + \hat{k} + \hat{k} + \hat{i} = 2\left( \hat{i} + \hat{j} + \hat{k} \right)\]
∴ \[\left| \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} \right| = \sqrt{2^2 + 2^2 + 2^2} = \sqrt{12} = 2\sqrt{3}\] "
Therefore, unit vector in the direction of \[\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} = \frac{2\left( \hat{i} + \hat{j} + \hat{k} \right)}{2\sqrt{3}} = \frac{1}{\sqrt{3}}\left( \hat{i} + \hat{j} + \hat{k} \right)\]
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