Let the vectors $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ be such that $\left| \overrightarrow{\mathbf{a}} \right|\mathbf{=3}$ and $\left| \overrightarrow{\mathbf{b}} \right|\mathbf{=}\frac{\sqrt{\mathbf{2}}}{\mathbf{3}}$, then $\overrightarrow{\mathbf{a}}\mathbf{\times }\overrightarrow{\mathbf{b}}$ is a unit vector, if the angle between $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ is
(a) $\frac{\mathbf{\pi }}{\mathbf{6}}$ (b) $\frac{\mathbf{\pi }}{\mathbf{4}}$ (c) $\frac{\mathbf{\pi }}{\mathbf{3}}$ (d) $\frac{\mathbf{\pi }}{\mathbf{2}}$
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Ans: We are provided that, $\left| \overrightarrow{a} \right|=3$ and $\left| \overrightarrow{b} \right|=\frac{\sqrt{2}}{3}$.
Now, it is generally known that $\overrightarrow{a}\times \overrightarrow{b}=\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\sin \theta \cdot \widehat{n}$, where $\widehat{n}$ is a unit vector that is perpendicular to both the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$; $\theta $ is the angle the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$.
Since, $\overrightarrow{a}\times \overrightarrow{b}$ is a unit vector, so $\left| \overrightarrow{a}\times \overrightarrow{b} \right|=1$
$ \Rightarrow \left| \left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\sin \theta \widehat{n} \right|=1 $
$ \Rightarrow \left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\left| \sin \theta \right|=1 $
$ \Rightarrow 3\times \frac{\sqrt{2}}{3}\times \sin \theta =1 $
$\Rightarrow \sin \theta =\frac{1}{\sqrt{2}} $
$ \Rightarrow \theta =\frac{\pi }{4}. $
Thus, the angle between the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is $\frac{\pi }{4}$.
Hence, the correct answer is option (b).