Q. $The\ value\ of\ \frac{{(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}})}^{2} + {(\overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}})}^{2}}{{2\overrightarrow{\mathbf{a}}}^{2}{\overrightarrow{\mathbf{b}}}^{2}}is\$
A) $\overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}}$
B) $1$
C) 0
D) $\frac{1}{2}$
Explanation
(d)
$\frac{\left( \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} \right)^{2} + \left( \overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}} \right)^{2}}{{2\overrightarrow{\mathbf{a}}}^{2}{\overrightarrow{\mathbf{b}}}^{2}} = \frac{{\overrightarrow{\mathbf{a}}}^{2}{\overrightarrow{\mathbf{b}}}^{2}\sin^{2}\theta + {\overrightarrow{\mathbf{a}}}^{2}{\overrightarrow{\mathbf{b}}}^{2}\cos^{2}\theta}{{2\overrightarrow{\mathbf{a}}}^{2}{\overrightarrow{\mathbf{b}}}^{2}}$
$= \frac{\cos^{2}\theta + \sin^{2}\theta}{2} = \frac{1}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.