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If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are non-zero vectors such that $(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}$ $=\frac{1}{3}|\mathbf{b}||\mathbf{c}| \mathbf{a}, \mathbf{c} \perp \mathbf{a}$ and $\theta$ is the angle between the vectors $\mathbf{b}$ and $\mathbf{c}$, then $\sin \theta$ is equal to
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Verified Answer
The correct answer is:
$\frac{2 \sqrt{2}}{3}$
Here, $a, b$ and $c$ are non-zero vectors $\mathrm{C} \perp \mathrm{a} \Rightarrow c . a=$ 0 and $\theta$ is the angle between the vectors $b$ and $c$.
$\begin{aligned}
& (a \times b) \times c=\frac{1}{3}|b||c| a \\
& (c . a) b-(c . b) a=\frac{1}{3}|b||c| a \\
& \Rightarrow 0-(|c||b| \cos \theta) a=\frac{1}{3}|b||c| a \\
& \Rightarrow \cos \theta=-\frac{1}{3} \Rightarrow \cos ^2 \theta=-\frac{1}{9} \\
& \Rightarrow 1-\sin ^2 \theta=\frac{1}{9} \Rightarrow \sin ^2 \theta=1-\frac{1}{9}=\frac{8}{9} \\
& \therefore \sin \theta=\sqrt{\frac{8}{9}}=\frac{2 \sqrt{2}}{3}
\end{aligned}$
$\begin{aligned}
& (a \times b) \times c=\frac{1}{3}|b||c| a \\
& (c . a) b-(c . b) a=\frac{1}{3}|b||c| a \\
& \Rightarrow 0-(|c||b| \cos \theta) a=\frac{1}{3}|b||c| a \\
& \Rightarrow \cos \theta=-\frac{1}{3} \Rightarrow \cos ^2 \theta=-\frac{1}{9} \\
& \Rightarrow 1-\sin ^2 \theta=\frac{1}{9} \Rightarrow \sin ^2 \theta=1-\frac{1}{9}=\frac{8}{9} \\
& \therefore \sin \theta=\sqrt{\frac{8}{9}}=\frac{2 \sqrt{2}}{3}
\end{aligned}$
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