$Magnitudes of vectors \vec{a}, \vec{b}, \vec{c} are$ $3, 4, 5 respectively. If \vec{a} and \vec{b} + \vec{c},$ $\vec{b} and \vec{c} + \vec{a}, \vec{c} and \vec{a} + \vec{b} are$ $mutually perpendicular, then$ $magnitude of \vec{a} + \vec{b} + \vec{c} is$

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$4 \sqrt{2}$

$3 \sqrt{2}$

$5 \sqrt{2}$

$3 \sqrt{3}$

answer is C.

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Detailed Solution

$\begin{matrix} . \\ ∴ \vec{a} \cdot (\vec{b} + \vec{c}) = 0 \Rightarrow \vec{a} \cdot \vec{b} + \vec{c} \cdot \vec{a} = 0 \\ S i m i l a r l y, \vec{b} \cdot \vec{c} + \vec{a} \cdot \vec{b} = 0 a n d \vec{c} \cdot \vec{a} + \vec{b} \cdot \vec{c} = 0 \\ On adding the above equations, we get 2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \\ Now, | \vec{a} + \vec{b} + \vec{c} |^{2} = | \vec{a} |^{2} + | \vec{b} |^{2} + | \vec{c} |^{2} + 2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \\ = {| \vec{a} |^{2} | + | \vec{b}|}^{2} + | \vec{c} |^{2} \\ = 9 + 16 + 25 (∵ | \vec{a} | = 3, | \vec{b} | = 4, | \vec{c} | = 5) \\ = 50 \\ \Rightarrow | \vec{a} + \vec{b} + \vec{c} | = 5 \sqrt{2} \end{matrix}$