let `veca, vecb` and `vecc` be three unit vectors such that `veca xx (vecb xx vecc) =sqrt(3)/2 (vecb + vecc)`. If `vecb` is not parallel to `vecc`, then the angle between `veca` and `vecb` is:

Options

A. `(pi)/4`
B. `pi/3`
C. `(2pi)/3`
D. `(5pi)/6`

Answer

To solve the problem, we need to analyze the given equation involving the unit vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\). The equation provided is:

\[ \vec{a} \times (\vec{b} \times \vec{c}) = \frac{\sqrt{3}}{2} (\vec{b} + \vec{c}) \]

### Step 1: Simplify the Cross Product
Using the vector triple product identity, we can rewrite \(\vec{a} \times (\vec{b} \times \vec{c})\) as follows:

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

### Step 2: Set Up the Equation
Now, substituting this into the original equation, we have:

\[ (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} = \frac{\sqrt{3}}{2} (\vec{b} + \vec{c}) \]

### Step 3: Compare Coefficients
To compare coefficients of \(\vec{b}\) and \(\vec{c}\), we can equate the coefficients from both sides:

1. Coefficient of \(\vec{b}\):
\[ \vec{a} \cdot \vec{c} = \frac{\sqrt{3}}{2} \]

2. Coefficient of \(\vec{c}\):
\[ -(\vec{a} \cdot \vec{b}) = \frac{\sqrt{3}}{2} \]

From the second equation, we can express \(\vec{a} \cdot \vec{b}\):

\[ \vec{a} \cdot \vec{b} = -\frac{\sqrt{3}}{2} \]

### Step 4: Find the Angle Between \(\vec{a}\) and \(\vec{b}\)
Since \(\vec{a}\) and \(\vec{b}\) are unit vectors, we can relate the dot product to the angle \(\theta\) between them:

\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(\theta) = 1 \cdot 1 \cdot \cos(\theta) = \cos(\theta) \]

Thus, we have:

\[ \cos(\theta) = -\frac{\sqrt{3}}{2} \]

### Step 5: Determine the Angle \(\theta\)
The angle \(\theta\) that satisfies \(\cos(\theta) = -\frac{\sqrt{3}}{2}\) corresponds to:

\[ \theta = \frac{5\pi}{6} \text{ or } \theta = \frac{7\pi}{6} \]

Since the angle between two vectors is typically taken to be between \(0\) and \(\pi\), we choose:

\[ \theta = \frac{5\pi}{6} \]

### Final Answer
Thus, the angle between \(\vec{a}\) and \(\vec{b}\) is:

\[ \boxed{\frac{5\pi}{6}} \]