To find the magnitude of the vector sum \(|\vec{a} + \vec{b} + \vec{c}|\) where \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are mutually perpendicular vectors of equal magnitude \(a\), we proceed as follows:
- Firstly, since the vectors are mutually perpendicular, we have: \(\vec{a} \cdot \vec{b} = 0\), \(\vec{b} \cdot \vec{c} = 0\), and \(\vec{c} \cdot \vec{a} = 0\).
- The magnitude of each vector is given as \(|\vec{a}| = |\vec{b}| = |\vec{c}| = a\).
- The magnitude of the sum of the vectors \(\vec{a} + \vec{b} + \vec{c}\) can be found using the formula for the magnitude of a vector sum: \(|\vec{a} + \vec{b} + \vec{c}| = \sqrt{(\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c})}\).
- Expanding the dot product, we have: \((\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} + \vec{c} \cdot \vec{c} + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})\).
- Since \(\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = 0\), this simplifies to: \((\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} + \vec{c} \cdot \vec{c}\).
- Since \(|\vec{a}| = a\), \(|\vec{b}| = a\), and \(|\vec{c}| = a\), we have: \(\vec{a} \cdot \vec{a} = a^2\), \(\vec{b} \cdot \vec{b} = a^2\), and \(\vec{c} \cdot \vec{c} = a^2\).
- Substituting these values, we get: \((\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = a^2 + a^2 + a^2 = 3a^2\).
- Therefore, the magnitude is: \(|\vec{a} + \vec{b} + \vec{c}| = \sqrt{3a^2} = \sqrt{3}a\).
Hence, the magnitude of \(|\vec{a} + \vec{b} + \vec{c}|\) is \(\sqrt{3}a\), which was previously calculated incorrectly in the options provided.