Key Concepts and Formulas

• Vector Triple Product (VTP): For any three vectors $\overrightarrow A, \overrightarrow B, \overrightarrow C$, the identity is $\overrightarrow A \times (\overrightarrow B \times \overrightarrow C) = (\overrightarrow A \cdot \overrightarrow C)\overrightarrow B - (\overrightarrow A \cdot \overrightarrow B)\overrightarrow C$.
• Dot Product and Angle: For two unit vectors $\overrightarrow u$ and $\overrightarrow v$, $\overrightarrow u \cdot \overrightarrow v = |\overrightarrow u||\overrightarrow v|\cos\theta = \cos\theta$, where $\theta$ is the angle between them.
• Linear Independence: If $\overrightarrow b$ and $\overrightarrow c$ are non-parallel vectors, then the equation $k_1\overrightarrow b + k_2\overrightarrow c = \overrightarrow 0$ implies $k_1 = 0$ and $k_2 = 0$.

Step-by-Step Solution

Step 1: Apply the Vector Triple Product Identity We are given the equation $\overrightarrow a \times (\overrightarrow b \times \overrightarrow c) = \frac{1}{2}\overrightarrow b$. We will use the VTP identity to expand the left side. Applying the identity $\overrightarrow A \times (\overrightarrow B \times \overrightarrow C) = (\overrightarrow A \cdot \overrightarrow C)\overrightarrow B - (\overrightarrow A \cdot \overrightarrow B)\overrightarrow C$ with $\overrightarrow A = \overrightarrow a$, $\overrightarrow B = \overrightarrow b$, and $\overrightarrow C = \overrightarrow c$, we get: $(\overrightarrow a \cdot \overrightarrow c)\overrightarrow b - (\overrightarrow a \cdot \overrightarrow b)\overrightarrow c = \frac{1}{2}\overrightarrow b$ To proceed, we rearrange the equation to group terms involving $\overrightarrow b$ and $\overrightarrow c$: $(\overrightarrow a \cdot \overrightarrow c)\overrightarrow b - \frac{1}{2}\overrightarrow b - (\overrightarrow a \cdot \overrightarrow b)\overrightarrow c = \overrightarrow 0$ Factoring out $\overrightarrow b$ from the first two terms: $\left(\overrightarrow a \cdot \overrightarrow c - \frac{1}{2}\right)\overrightarrow b - (\overrightarrow a \cdot \overrightarrow b)\overrightarrow c = \overrightarrow 0$

Step 2: Utilize Linear Independence We are given that $\overrightarrow b$ and $\overrightarrow c$ are non-parallel unit vectors. This means they are linearly independent. For the equation $\left(\overrightarrow a \cdot \overrightarrow c - \frac{1}{2}\right)\overrightarrow b - (\overrightarrow a \cdot \overrightarrow b)\overrightarrow c = \overrightarrow 0$ to hold, the coefficients of $\overrightarrow b$ and $\overrightarrow c$ must both be zero. Equating the coefficient of $\overrightarrow b$ to zero: $\overrightarrow a \cdot \overrightarrow c - \frac{1}{2} = 0$ $\overrightarrow a \cdot \overrightarrow c = \frac{1}{2}$ Equating the coefficient of $\overrightarrow c$ to zero: $-(\overrightarrow a \cdot \overrightarrow b) = 0$ $\overrightarrow a \cdot \overrightarrow b = 0$

Step 3: Determine the Angles $\alpha$ and $\beta$ We are given that $\alpha$ is the angle between $\overrightarrow a$ and $\overrightarrow b$, and $\beta$ is the angle between $\overrightarrow a$ and $\overrightarrow c$. Since $\overrightarrow a$, $\overrightarrow b$, and $\overrightarrow c$ are unit vectors, their magnitudes are 1. Using the dot product definition for unit vectors: For $\overrightarrow a \cdot \overrightarrow b = 0$: $|\overrightarrow a||\overrightarrow b|\cos\alpha = 0$ $(1)(1)\cos\alpha = 0$ $\cos\alpha = 0$ Since $\alpha$ is an angle between vectors, $0^\circ \le \alpha \le 180^\circ$. Thus, $\alpha = 90^\circ$.

For $\overrightarrow a \cdot \overrightarrow c = \frac{1}{2}$: $|\overrightarrow a||\overrightarrow c|\cos\beta = \frac{1}{2}$ $(1)(1)\cos\beta = \frac{1}{2}$ $\cos\beta = \frac{1}{2}$ Since $\beta$ is an angle between vectors, $0^\circ \le \beta \le 180^\circ$. Thus, $\beta = 60^\circ$.

Step 4: Calculate $|\alpha - \beta|$ We need to find the absolute difference between $\alpha$ and $\beta$. $|\alpha - \beta| = |90^\circ - 60^\circ|$ $|\alpha - \beta| = |30^\circ|$ $|\alpha - \beta| = 30^\circ$

Common Mistakes & Tips

• Incorrect VTP Expansion: Ensure the "BAC-CAB" rule is applied correctly. A common mistake is swapping the dot and cross products or the order of vectors in the dot products.
• Ignoring Linear Independence: Without recognizing that $\overrightarrow b$ and $\overrightarrow c$ are linearly independent, one cannot equate the coefficients to zero, which is a critical step.
• Angle Range: Remember that angles between vectors are typically in the range $[0^\circ, 180^\circ]$. This is important when finding the angle from its cosine value.

Summary The problem is solved by applying the vector triple product identity to simplify the given vector equation. By leveraging the fact that $\overrightarrow b$ and $\overrightarrow c$ are non-parallel and thus linearly independent, we equate the coefficients of $\overrightarrow b$ and $\overrightarrow c$ to zero. This yields the dot products $\overrightarrow a \cdot \overrightarrow b$ and $\overrightarrow a \cdot \overrightarrow c$. Using the definition of the dot product for unit vectors, we find the angles $\alpha$ and $\beta$. Finally, we compute the absolute difference between these angles.

The final answer is $\boxed{30^\circ}$ which corresponds to option (B).