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Addition of Vectors

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Question

If $\vec{a,} \vec{b,} \vec{c}$ are three mutually perpendicular unit vectors, then prove that $|\vec{a} + \vec{b} + \vec{c}| = \sqrt{3}$ .

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Solution

$Given that \vec{a}, \vec{b} and \vec{c} are unit vectors. So, |\vec{a}| =1, |\vec{b}| =1 and |\vec{c}| =1 Since they are mutually perpendicular, \vec{a} . \vec{b} = \vec{b} . \vec{c} = \vec{c} . \vec{a} = 0 Now, {|\vec{a} + \vec{b} + \vec{c}|}^{2} = {|\vec{a}|}^{2} + {|\vec{b}|}^{2} + {|\vec{c}|}^{2} + 2 \vec{a} . \vec{b} + 2 \vec{b} . \vec{c} + 2 \vec{c} . \vec{a} = 1 + 1 + 1 + 0 + 0 + 0 = 3 ∴ |\vec{a} + \vec{b} + \vec{c}| = \sqrt{3}$

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