Q115 / 170 · Manipal (MET) 2021 · Mathematics
The value of $\left\{(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}})^2+(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}})^2\right\} \div a^2 b^2$ is
- A0
- Correct answerB1
- C2
- DNone of these
Solution
$\left\{(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}})^2+(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}})^2\right\} \div a^2 b^2$
$\begin{aligned} & =\left\{(|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}| \sin \theta \hat{\mathbf{n}})^2+(|\overrightarrow{\mathbf{a}} \| \overrightarrow{\mathbf{b}}| \cos \theta)^2\right\}+a^2 b^2 \\ & =a^2 b^2\left(\sin ^2 \theta+\cos ^2 \theta\right)+a^2 b^2=1\end{aligned}$
Hence, the correct answer is option: 'B'.