If $\vec{a}, \vec{b}, \vec{c}$ are three mutually perpendicular unit vectors, then $|\vec{a} + \vec{b} + \vec{c}|$ is equal to

Given three vectors $\vec{a}, \vec{b}, \vec{c}$ where:

• Each is a unit vector: $|\vec{a}| = 1$, $|\vec{b}| = 1$, $|\vec{c}| = 1$
• Each vector is mutually perpendicular to the others

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To find $|\vec{a} + \vec{b} + \vec{c}|$:

$|\vec{a} + \vec{b} + \vec{c}|^2 = (\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c})$

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Expanding the dot product:

$|\vec{a} + \vec{b} + \vec{c}|^2 = \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} + \vec{c} \cdot \vec{b} + \vec{c} \cdot \vec{c}$

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For unit vectors:

$\vec{a} \cdot \vec{a} = |\vec{a}|^2 = 1$

$\vec{b} \cdot \vec{b} = |\vec{b}|^2 = 1$

$\vec{c} \cdot \vec{c} = |\vec{c}|^2 = 1$

For perpendicular vectors, the dot product equals zero:

$\vec{a} \cdot \vec{b} = 0$, $\vec{a} \cdot \vec{c} = 0$, $\vec{b} \cdot \vec{c} = 0$

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Substituting these values:

$|\vec{a} + \vec{b} + \vec{c}|^2 = 1 + 0 + 0 + 0 + 1 + 0 + 0 + 0 + 1$

$|\vec{a} + \vec{b} + \vec{c}|^2 = 3$

$|\vec{a} + \vec{b} + \vec{c}| = \sqrt{3}$

Therefore, $|\vec{a} + \vec{b} + \vec{c}| = \sqrt{3}$