Let $$\vec{a}$$, $$\vec{b}$$ and $$\vec{c}$$ be three non-zero vectors such that no two of them are collinear and $$(\vec{a} \times \vec{b}) \times \vec{c} = \frac{1}{3}|\vec{b}||\vec{c}|\vec{a}$$. If $$\theta$$ is the angle between vectors $$\vec{b}$$ and $$\vec{c}$$, then a value of $$\sin \theta$$ is

We are given three non-zero vectors $$\vec{a}$$, $$\vec{b}$$, and $$\vec{c}$$ such that no two are collinear, and the equation $$(\vec{a} \times \vec{b}) \times \vec{c} = \frac{1}{3}|\vec{b}||\vec{c}|\vec{a}$$. We need to find $$\sin \theta$$ where $$\theta$$ is the angle between $$\vec{b}$$ and $$\vec{c}$$.

First, recall the vector triple product identity: $$(\vec{u} \times \vec{v}) \times \vec{w} = (\vec{u} \cdot \vec{w}) \vec{v} - (\vec{v} \cdot \vec{w}) \vec{u}$$. Applying this to the left-hand side with $$\vec{u} = \vec{a}$$, $$\vec{v} = \vec{b}$$, and $$\vec{w} = \vec{c}$$, we get:

$$(\vec{a} \times \vec{b}) \times \vec{c} = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{b} \cdot \vec{c}) \vec{a}$$

So the given equation becomes:

$$(\vec{a} \cdot \vec{c}) \vec{b} - (\vec{b} \cdot \vec{c}) \vec{a} = \frac{1}{3}|\vec{b}||\vec{c}|\vec{a}$$

Rearrange all terms to one side:

$$(\vec{a} \cdot \vec{c}) \vec{b} - (\vec{b} \cdot \vec{c}) \vec{a} - \frac{1}{3}|\vec{b}||\vec{c}|\vec{a} = 0$$

Combine the terms with $$\vec{a}$$:

$$(\vec{a} \cdot \vec{c}) \vec{b} - \left[ (\vec{b} \cdot \vec{c}) + \frac{1}{3}|\vec{b}||\vec{c}| \right] \vec{a} = 0$$

This can be written as:

$$(\vec{a} \cdot \vec{c}) \vec{b} = \left[ (\vec{b} \cdot \vec{c}) + \frac{1}{3}|\vec{b}||\vec{c}| \right] \vec{a}$$

Since $$\vec{a}$$ and $$\vec{b}$$ are not collinear (as given), they are linearly independent. For this equality to hold, the coefficients must both be zero. Otherwise, if the coefficient of $$\vec{b}$$ is non-zero, $$\vec{b}$$ would be a scalar multiple of $$\vec{a}$$, making them collinear, which contradicts the condition. Similarly, if the coefficient of $$\vec{a}$$ is non-zero, $$\vec{a}$$ would be a scalar multiple of $$\vec{b}$$. Therefore:

$$\vec{a} \cdot \vec{c} = 0 \quad \text{and} \quad (\vec{b} \cdot \vec{c}) + \frac{1}{3}|\vec{b}||\vec{c}| = 0$$

The first equation $$\vec{a} \cdot \vec{c} = 0$$ implies that $$\vec{a}$$ is perpendicular to $$\vec{c}$$.

From the second equation:

$$\vec{b} \cdot \vec{c} = - \frac{1}{3}|\vec{b}||\vec{c}|$$

The dot product $$\vec{b} \cdot \vec{c}$$ is also given by $$|\vec{b}| |\vec{c}| \cos \theta$$, where $$\theta$$ is the angle between $$\vec{b}$$ and $$\vec{c}$$. So:

$$|\vec{b}| |\vec{c}| \cos \theta = - \frac{1}{3}|\vec{b}||\vec{c}|$$

Since $$\vec{b}$$ and $$\vec{c}$$ are non-zero, $$|\vec{b}| \neq 0$$ and $$|\vec{c}| \neq 0$$, so we can divide both sides by $$|\vec{b}| |\vec{c}|$$:

$$\cos \theta = - \frac{1}{3}$$

Now, use the identity $$\sin^2 \theta + \cos^2 \theta = 1$$ to find $$\sin \theta$$:

$$\sin^2 \theta = 1 - \cos^2 \theta = 1 - \left(-\frac{1}{3}\right)^2 = 1 - \frac{1}{9} = \frac{8}{9}$$

Thus:

$$\sin \theta = \pm \sqrt{\frac{8}{9}} = \pm \frac{\sqrt{8}}{3} = \pm \frac{2\sqrt{2}}{3}$$

The angle $$\theta$$ between two vectors ranges from 0 to $$\pi$$ radians (0° to 180°). In this interval, $$\sin \theta$$ is non-negative because sine is positive in the first and second quadrants. Since $$\cos \theta = -1/3 < 0$$, $$\theta$$ is in the second quadrant where sine is positive. Therefore, we take the positive value:

$$\sin \theta = \frac{2\sqrt{2}}{3}$$

Comparing with the options:

A. $$\frac{-2\sqrt{3}}{3}$$

B. $$\frac{2\sqrt{2}}{3}$$

C. $$\frac{-\sqrt{2}}{3}$$

D. $$\frac{2}{3}$$

The value $$\frac{2\sqrt{2}}{3}$$ matches option B.

Hence, the correct answer is Option B.