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Question
If \[\vec{a,} \vec{b,} \vec{c}\] are three mutually perpendicular unit vectors, then prove that \[\left| \vec{a} + \vec{b} + \vec{c} \right| = \sqrt{3}\]
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Solution
\[\text{ Given that } \vec{a} , \vec{b} \text{ and } \vec{c} \text{ are unit vectors }.\]
\[So,\left| \vec{a} \right|=1,\left| \vec{b} \right|=1 \text{ and } \left| \vec{c} \right|=1\]
\[\text{ Since they are mutually perpendicular },\]
\[ \vec{a} . \vec{b} = \vec{b} . \vec{c} = \vec{c} . \vec{a} = 0\]
\[\text{ Now },\]
\[ \left| \vec{a} + \vec{b} + \vec{c} \right|^2 = \left| \vec{a} \right|^2 + \left| \vec{b} \right|^2 + \left| \vec{c} \right|^2 + 2 \vec{a} . \vec{b} + 2 \vec{b} . \vec{c} + 2 \vec{c} . \vec{a} \]
\[ = 1 + 1 + 1 + 0 + 0 + 0\]
\[ = 3\]
\[ \therefore \left| \vec{a} + \vec{b} + \vec{c} \right| = \sqrt{3}\]
\[\]
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