If $\vec{a}, \vec{b} and \vec{c}$ are three mutually perpendicular vectors, then the vector which is equally inclined to these vectors is

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$\frac{\vec{a}}{| \vec{a} |} + \frac{\vec{b}}{| \vec{b} |} + \frac{\vec{c}}{| \vec{c} |}$

$| \vec{a} | \vec{a} - | \vec{b} | \vec{b} + | \vec{c} | \vec{c}$

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

$\vec{a} + \vec{b} + \vec{c}$

answer is B.

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Detailed Solution

Let $\vec{α} = \frac{\vec{a}}{| \vec{a} |} + \frac{\vec{b}}{| \vec{b} |} + \frac{\vec{c}}{| \vec{c} |}$

Since $\vec{a}, \vec{b} and \vec{c}$ are mutually perpendicular vectors, if $\vec{α}$ makes angles $θ, ϕ, ψ with \vec{a}, \vec{b} and \vec{c}$ respectively, then
$\begin{array}{l} \vec{α} \cdot \vec{a} = \frac{\vec{a} \cdot \vec{a}}{| \vec{a} |} \\ or | \vec{α} | \cdot | \vec{a} | \cos θ = | \vec{a} | \\ or \cos θ = \frac{1}{| \vec{α} |} \end{array}$