Q. $\frac{\overrightarrow{\mathbf{a}} \bullet (\ \overrightarrow{\mathbf{b}} \times \ \overrightarrow{\mathbf{c}})}{\overrightarrow{\mathbf{b}} \bullet (\ \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}})} + \frac{\overrightarrow{\mathbf{b}} \bullet (\ \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}})}{\overrightarrow{\mathbf{a}} \bullet (\overrightarrow{\mathbf{b}} \times \ \overrightarrow{\mathbf{c}})}\ is\ equal\ to$

A) 1
B) 2
C) 0
D) ∞
Explanation

(a)
$\frac{\overrightarrow{\mathbf{a}} \bullet \left( \ \overrightarrow{\mathbf{b}} \times \ \overrightarrow{\mathbf{c}} \right)}{\overrightarrow{\mathbf{b}} \bullet \left( \ \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}} \right)} + \frac{\overrightarrow{\mathbf{b}} \bullet \left( \ \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} \right)}{\overrightarrow{\mathbf{a}} \bullet \left( \overrightarrow{\mathbf{b}} \times \ \overrightarrow{\mathbf{c}} \right)}$
$= \frac{\ \lbrack\overrightarrow{\mathbf{a}}\ \overrightarrow{\mathbf{b}}\ \overrightarrow{\mathbf{c}}\rbrack}{\lbrack\overrightarrow{\mathbf{b}}\ \overrightarrow{\mathbf{c}}\ \overrightarrow{\mathbf{a}}\rbrack} + \frac{\lbrack\overrightarrow{\mathbf{b}}\ \overrightarrow{\mathbf{a}}\ \overrightarrow{\mathbf{b}}\rbrack}{\lbrack\overrightarrow{\mathbf{a}}\ \overrightarrow{\mathbf{b}}\ \overrightarrow{\mathbf{c}}\rbrack} = 1 + 0 = 1$

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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