Let \(\vec{a}\), \(\vec{b}\) & \(\vec{c}\) be three vectors such that \(|\vec{a}|=3\), \(|\vec{b}|=4\), \(|\vec{c}|=5\) and each one of them being perpendicular to the sum of the other two, find \(|\vec{a} + \vec{b} + \vec{c}|\).

$5\sqrt{2}$

The correct answer is Option (2) → $5\sqrt{2}$ ##

Given \(\vec{a} \cdot (\vec{b} + \vec{c}) = 0\), \(\vec{b} \cdot (\vec{c} + \vec{a}) = 0\), \(\vec{c} \cdot (\vec{a} + \vec{b}) = 0\).

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

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

$= |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 = 9 + 16 + 25 = 50$

Therefore $|\vec{a} + \vec{b} + \vec{c}| = \sqrt{50} = 5\sqrt{2}$