Let a, b and c be three unit vectors such that `atimes(btimesc)=(sqrt(3))/(2)(b+c)`. If b is not parallel to c , then the angle between a and b is

Options

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

Answer

To find the angle between the unit vectors \( \mathbf{a} \) and \( \mathbf{b} \), we start with the given equation:

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

### Step 1: Use the vector triple product identity
We can use the vector triple product identity, which states that:

\[ \mathbf{u} \times (\mathbf{v} \times \mathbf{w}) = (\mathbf{u} \cdot \mathbf{w}) \mathbf{v} - (\mathbf{u} \cdot \mathbf{v}) \mathbf{w} \]

Applying this to our equation, we have:

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

### Step 2: Set the equations equal
Now we equate this to the right-hand side of the original equation:

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

### Step 3: Compare coefficients
From this equation, we can compare the coefficients of \( \mathbf{b} \) and \( \mathbf{c} \):

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

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

### Step 4: Find the angle between \( \mathbf{a} \) and \( \mathbf{c} \)
From equation (1), we know:

\[ \mathbf{a} \cdot \mathbf{c} = |\mathbf{a}| |\mathbf{c}| \cos \theta_{ac} = 1 \cdot 1 \cdot \cos \theta_{ac} = \cos \theta_{ac} \]

Thus,

\[ \cos \theta_{ac} = \frac{\sqrt{3}}{2} \implies \theta_{ac} = 30^\circ \text{ or } \frac{\pi}{6} \]

### Step 5: Find the angle between \( \mathbf{a} \) and \( \mathbf{b} \)
From equation (2), we have:

\[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta_{ab} = 1 \cdot 1 \cdot \cos \theta_{ab} = \cos \theta_{ab} \]

Thus,

\[ \cos \theta_{ab} = -\frac{\sqrt{3}}{2} \implies \theta_{ab} = 150^\circ \text{ or } \frac{5\pi}{6} \]

### Conclusion
The angle between the unit vectors \( \mathbf{a} \) and \( \mathbf{b} \) is:

\[ \theta_{ab} = \frac{5\pi}{6} \]