If a, b and c are non-collinear unit vectors also b, c are non-collinear and `2atimes(btimesc)=b+c`, then

Options

A. angle between a and c is `60^(@)`
B. angle between b and c is `30^(@)`
C. angle between a and b is `120^(@)`
D. b is perpendicular to c

Answer

To solve the problem, we start with the given information and work through the steps systematically.

### Step 1: Understand the Given Information
We are given that \( \mathbf{a}, \mathbf{b}, \) and \( \mathbf{c} \) are non-collinear unit vectors, and the equation \( 2 \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = \mathbf{b} + \mathbf{c} \) holds.

### Step 2: Use the Vector Triple Product Identity
We can use the vector triple product identity:
\[ \mathbf{u} \times (\mathbf{v} \times \mathbf{w}) = (\mathbf{u} \cdot \mathbf{w}) \mathbf{v} - (\mathbf{u} \cdot \mathbf{v}) \mathbf{w} \]
Applying this to our equation, we have:
\[ \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c}) \mathbf{b} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{c} \]
Thus, substituting this back into our equation gives:
\[ 2 \left( (\mathbf{a} \cdot \mathbf{c}) \mathbf{b} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{c} \right) = \mathbf{b} + \mathbf{c} \]

### Step 3: Expand the Equation
Expanding the left-hand side:
\[ 2(\mathbf{a} \cdot \mathbf{c}) \mathbf{b} - 2(\mathbf{a} \cdot \mathbf{b}) \mathbf{c} = \mathbf{b} + \mathbf{c} \]

### Step 4: Compare Coefficients
Now, we can compare the coefficients of \( \mathbf{b} \) and \( \mathbf{c} \) on both sides:
1. For \( \mathbf{b} \):
\[ 2(\mathbf{a} \cdot \mathbf{c}) = 1 \quad \Rightarrow \quad \mathbf{a} \cdot \mathbf{c} = \frac{1}{2} \]
2. For \( \mathbf{c} \):
\[ -2(\mathbf{a} \cdot \mathbf{b}) = 1 \quad \Rightarrow \quad \mathbf{a} \cdot \mathbf{b} = -\frac{1}{2} \]

### Step 5: Relate Dot Products to Angles
Let \( \theta \) be the angle between \( \mathbf{a} \) and \( \mathbf{c} \), and \( \alpha \) be the angle between \( \mathbf{a} \) and \( \mathbf{b} \). We can express the dot products in terms of these angles:
\[ \mathbf{a} \cdot \mathbf{c} = |\mathbf{a}| |\mathbf{c}| \cos(\theta) = 1 \cdot 1 \cdot \cos(\theta) = \cos(\theta) \]
Thus, we have:
\[ \cos(\theta) = \frac{1}{2} \quad \Rightarrow \quad \theta = 60^\circ \]
Similarly, for \( \mathbf{a} \cdot \mathbf{b} \):
\[ \cos(\alpha) = -\frac{1}{2} \quad \Rightarrow \quad \alpha = 120^\circ \]

### Conclusion
The angle between \( \mathbf{a} \) and \( \mathbf{c} \) is \( 60^\circ \).