Q. $(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}})^{2} + (\overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}})^{2}$ is equal to
A) $\overrightarrow{\mathbf{a}}\ ^{2}\overrightarrow{\mathbf{b}}\ ^{2}$
B) $\overrightarrow{\mathbf{a}}\ ^{2} + \overrightarrow{\mathbf{b}}\ ^{2}$
C) 1
D) $2\overrightarrow{\mathbf{a}}{\ \bullet}^{\ }\overrightarrow{\mathbf{b}}\ ^{\ }$
Explanation
(a)
Given, $(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}})^{2} + (\overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}})^{2} = {\overrightarrow{\mathbf{a}}}^{2}{\overrightarrow{\mathbf{b}}}^{2}\sin^{2}\theta + {\overrightarrow{\mathbf{a}}}_{2}{\overrightarrow{\mathbf{b}}}_{2}\cos^{2}\theta = {\overrightarrow{\mathbf{a}}}^{2}{\overrightarrow{\mathbf{b}}}^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.