Q. If $\left| \overrightarrow{\mathbf{a}} \right| + 3,\ \left| \overrightarrow{\mathbf{a}} \right| = 4,\ \left| \overrightarrow{\mathbf{c}} \right| = 5\$ and $\overrightarrow{\mathbf{a}},\ \overrightarrow{\mathbf{b}},\overrightarrow{\mathbf{c}}\$are such that each is perpendicular to the saum of other two, then $|\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} +$ $\overrightarrow{\mathbf{c}}$| is

A) $5\sqrt{2}$
B) $\frac{5}{\sqrt{2}}$
C) $10\sqrt{2}$
D) $10\sqrt{3}$
Explanation

(a)
Given, $\overrightarrow{\mathbf{a}} \bullet \left( \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}} \right) = 0\ \Longrightarrow \overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}} \bullet \overrightarrow{\mathbf{a}} = 0$
$\overrightarrow{\mathbf{b}} \bullet \left( \overrightarrow{\mathbf{c}} + \overrightarrow{\mathbf{a}} \right) = 0$
$\Longrightarrow \ \overrightarrow{\mathbf{b}} \bullet \overrightarrow{\mathbf{c}} + \overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}} = 0$
$and\ \overrightarrow{\mathbf{c}} \bullet \left( \overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} \right) = 0$
$\Longrightarrow \ \overrightarrow{\mathbf{c}} \bullet \overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} \bullet \overrightarrow{\mathbf{c}} = 0$
$\therefore\ \overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{b}} \bullet \overrightarrow{\mathbf{c}} + \overrightarrow{\mathbf{c}} \bullet \overrightarrow{\mathbf{a}} = 0$
Now, $|\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}}|^{2} = |{\overrightarrow{\mathbf{a}}|}^{2} + \left| \overrightarrow{\mathbf{b}} \right|^{2} + \left| \overrightarrow{\mathbf{c}} \right|^{2} + 2(\overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{b}} \bullet \overrightarrow{\mathbf{c}} + \overrightarrow{\mathbf{c}} \bullet \overrightarrow{\mathbf{a}})$
$\Longrightarrow |\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}}|^{2} = 9 + 16 + 25 + 0 = 50$
$\Longrightarrow \left| \overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}} \right| = 5\sqrt{2}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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