Q. If the vectors $\overrightarrow{\mathbf{a}},\overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ from the sides $BC,CA\ and\ AB$ respectively of a triangle $ABC$ then

A) $\overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}} = \overrightarrow{\mathbf{b}} \bullet \overrightarrow{\mathbf{c}} = \overrightarrow{\mathbf{c}} \bullet \overrightarrow{\mathbf{b}} = 0$
B) $\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} = \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}} = \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}} = 0$
C) $\overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}} = \overrightarrow{\mathbf{b}} \bullet \overrightarrow{\mathbf{c}} = \overrightarrow{\mathbf{c}} \bullet \overrightarrow{\mathbf{a}} = 0$
D) $\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{c}} + \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}} = 0$
Explanation

(b)
Given, $\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} = - \overrightarrow{\mathbf{c}}$
$\Longrightarrow \ \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{c}} + \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}} = \overrightarrow{\mathbf{0}} \Longrightarrow \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}} = \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}}$
Similarly, $\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} = \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}$
Hence, $\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} = \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}} = \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}}$

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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