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Step-by-Step Solution
Step 1: Understand the Given Vector Equation
We are provided with the equation
$$(\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c} = \frac{1}{3} \lVert \overrightarrow{b} \rVert \lVert \overrightarrow{c} \rVert \,\overrightarrow{a}.$$
Here, $\overrightarrow{a}, \overrightarrow{b},$ and $\overrightarrow{c}$ are non-zero vectors.
Step 2: Use the Vector Triple Product Identity
The standard vector identity for a triple product of the form $(\mathbf{X} \times \mathbf{Y}) \times \mathbf{Z}$ is
$$(\mathbf{X} \times \mathbf{Y}) \times \mathbf{Z}
= (\mathbf{X} \cdot \mathbf{Z})\,\mathbf{Y} \;-\; (\mathbf{Y} \cdot \mathbf{Z})\,\mathbf{X}.$$
Applying this to $(\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c}$, we get:
$$(\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c}
= (\overrightarrow{a} \cdot \overrightarrow{c})\,\overrightarrow{b}
- (\overrightarrow{b} \cdot \overrightarrow{c})\,\overrightarrow{a}.$$
Thus, the given equation becomes:
$$(\overrightarrow{a} \cdot \overrightarrow{c})\,\overrightarrow{b}
- (\overrightarrow{b} \cdot \overrightarrow{c})\,\overrightarrow{a}
= \frac{1}{3}\lVert \overrightarrow{b} \rVert \,\lVert \overrightarrow{c} \rVert\,\overrightarrow{a}.$$
Step 3: Compare Corresponding Terms
From the above equality, we match the terms on both sides:
The coefficient of $\overrightarrow{b}$ on the left must be zero on the right. Hence
$$(\overrightarrow{a} \cdot \overrightarrow{c}) = 0.$$
This tells us that $\overrightarrow{a}$ is perpendicular to $\overrightarrow{c}.$
The coefficient of $\overrightarrow{a}$ on the left is
$$-(\overrightarrow{b} \cdot \overrightarrow{c}).$$
On the right, its coefficient is
$$\frac{1}{3}\lVert \overrightarrow{b} \rVert \,\lVert \overrightarrow{c} \rVert.$$
Equating these gives
$$-(\overrightarrow{b} \cdot \overrightarrow{c})
= \frac{1}{3}\lVert \overrightarrow{b} \rVert \,\lVert \overrightarrow{c} \rVert
\;\;\Longrightarrow\;\;
\overrightarrow{b} \cdot \overrightarrow{c}
= -\,\frac{1}{3}\lVert \overrightarrow{b} \rVert \,\lVert \overrightarrow{c} \rVert.$$
Step 4: Relate Dot Product to the Angle Between Vectors
The dot product $\overrightarrow{b} \cdot \overrightarrow{c}$ is given by
$$\overrightarrow{b} \cdot \overrightarrow{c} = \lVert \overrightarrow{b} \rVert \,\lVert \overrightarrow{c} \rVert \,\cos \theta,$$
where $\theta$ is the angle between $\overrightarrow{b}$ and $\overrightarrow{c}.$ Comparing with the above result,
$$\lVert \overrightarrow{b} \rVert \,\lVert \overrightarrow{c} \rVert \,\cos \theta
= -\,\frac{1}{3}\lVert \overrightarrow{b} \rVert \,\lVert \overrightarrow{c} \rVert.$$
Hence,
$$\cos \theta = -\,\frac{1}{3}.$$
Step 5: Find $ \sin \theta $ for the Acute Angle
Even though $ \cos \theta = -\frac{1}{3} $ would normally indicate an obtuse angle, the question specifically asks for the sine of the acute angle between the two vectors. Since
$$\sin \theta = \sqrt{1 - \cos^2 \theta},$$
we compute
$$\sin \theta
= \sqrt{1 - \left(-\frac{1}{3}\right)^2}
= \sqrt{1 - \frac{1}{9}}
= \sqrt{\frac{8}{9}}
= \frac{2\sqrt{2}}{3}.$$
Thus, the acute angle $\theta$ between $\overrightarrow{b}$ and $\overrightarrow{c}$ has
$$\sin \theta = \frac{2\sqrt{2}}{3}.$$
Answer
The value of $\sin \theta$ is
$$\frac{2\sqrt{2}}{3}.$$
