If $\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}}$ are three unit vectors such that $\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}} = 0$, where $\overrightarrow{\mathbf{0}}$ is null vector, then $\overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{b}} \bullet \overrightarrow{\mathbf{c}} + \overrightarrow{\mathbf{c}} \bullet \overrightarrow{\mathbf{a}}$ is

Question

Dr. Abubacker · Accepted Answer

(c)
$|\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}}|^{2}$
$= \left| {\overrightarrow{\mathbf{a}}|}^{2} + ight|{\overrightarrow{\mathbf{b}}|}^{2} + |{\overrightarrow{\mathbf{c}}|}^{2} + 2\overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}} + 2\overrightarrow{\mathbf{b}} \bullet \overrightarrow{\mathbf{c}} + 2\overrightarrow{\mathbf{c}} \bullet \overrightarrow{\mathbf{a}}$
$\Longrightarrow 0 = 1 + 1 + 1 + 2\left( \overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{b}} \bullet \overrightarrow{\mathbf{c}} + \overrightarrow{\mathbf{c}} \bullet \overrightarrow{\mathbf{a}} ight)$
$\left\lbrack \because\left| \overrightarrow{\mathbf{a}} ight| = \left| \overrightarrow{\mathbf{b}} ight| = \left| \overrightarrow{\mathbf{c}} ight| = 1,\ given ightbrack$
$	herefore\ \overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{b}} \bullet \overrightarrow{\mathbf{c}} + \overrightarrow{\mathbf{c}} \bullet \overrightarrow{\mathbf{a}} = - \frac{3}{2}$

Explanation

Reviewed by Dr. Abubacker

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