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Let $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ be non-zero vectors such that $(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}=\frac{1}{3}|\mathbf{b} \| \mathbf{c}| \mathbf{a}$. If $\theta$ is the acute 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:
$\frac{2 \sqrt{2}}{3}$
$$
\begin{aligned}
& \text { Given, }(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}=\frac{1}{3}|\mathbf{b} \| \mathbf{c}| \mathbf{a} \\
& \Rightarrow-\{\mathbf{c} \times(\mathbf{a} \times \mathbf{b})\}=\frac{1}{3}|\mathbf{b} \| \mathbf{c}| \mathbf{a} \quad[\because \mathbf{a} \times \mathbf{b}=-\mathbf{b} \times \mathbf{a}] \\
& \Rightarrow-\{(\mathbf{c} \cdot \mathbf{b}) \mathbf{a}-(\mathbf{c} \cdot \mathbf{a}) \mathbf{b}\}=\frac{1}{3}|\mathbf{b} \| \mathbf{c}| \mathbf{a} \\
& \Rightarrow \quad-(\mathbf{b} \cdot \mathbf{c}) \mathbf{a}+(\mathbf{c} \cdot \mathbf{a}) \mathbf{b}=\frac{1}{3}|\mathbf{b} \| \mathbf{c}| \mathbf{a}
\end{aligned}
$$
By comparing
$$
-\mathbf{b} \cdot \mathbf{c}=\frac{1}{3}|\mathbf{b}||\mathbf{c}| \text { and } \mathbf{c} \mathbf{a}=0
$$
$\begin{aligned}-|\mathbf{b}||\mathbf{c}| \cos \theta & =\frac{1}{3}|\mathbf{b}||\mathbf{c}| \\ \cos \theta & =-\frac{1}{3} \\ \sin \theta & =\sqrt{1-\cos ^2 \theta} \\ & =\sqrt{1-\frac{1}{9}} \\ \sin \theta & =\frac{2 \sqrt{2}}{3}\end{aligned}$
\begin{aligned}
& \text { Given, }(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}=\frac{1}{3}|\mathbf{b} \| \mathbf{c}| \mathbf{a} \\
& \Rightarrow-\{\mathbf{c} \times(\mathbf{a} \times \mathbf{b})\}=\frac{1}{3}|\mathbf{b} \| \mathbf{c}| \mathbf{a} \quad[\because \mathbf{a} \times \mathbf{b}=-\mathbf{b} \times \mathbf{a}] \\
& \Rightarrow-\{(\mathbf{c} \cdot \mathbf{b}) \mathbf{a}-(\mathbf{c} \cdot \mathbf{a}) \mathbf{b}\}=\frac{1}{3}|\mathbf{b} \| \mathbf{c}| \mathbf{a} \\
& \Rightarrow \quad-(\mathbf{b} \cdot \mathbf{c}) \mathbf{a}+(\mathbf{c} \cdot \mathbf{a}) \mathbf{b}=\frac{1}{3}|\mathbf{b} \| \mathbf{c}| \mathbf{a}
\end{aligned}
$$
By comparing
$$
-\mathbf{b} \cdot \mathbf{c}=\frac{1}{3}|\mathbf{b}||\mathbf{c}| \text { and } \mathbf{c} \mathbf{a}=0
$$
$\begin{aligned}-|\mathbf{b}||\mathbf{c}| \cos \theta & =\frac{1}{3}|\mathbf{b}||\mathbf{c}| \\ \cos \theta & =-\frac{1}{3} \\ \sin \theta & =\sqrt{1-\cos ^2 \theta} \\ & =\sqrt{1-\frac{1}{9}} \\ \sin \theta & =\frac{2 \sqrt{2}}{3}\end{aligned}$
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