If a, b and c are three unit vectors satisfying `2a times(a timesb)+c=0` then the acute angle between a and b is

Options

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

Answer

To solve the problem, we need to find the acute angle between the unit vectors \( \mathbf{a} \) and \( \mathbf{b} \) given the equation:

\[ 2 \mathbf{a} \times (\mathbf{a} \times \mathbf{b}) + \mathbf{c} = \mathbf{0} \]

### Step-by-Step Solution:

1. **Rearranging the Equation**:
Start by rearranging the given equation:
\[ 2 \mathbf{a} \times (\mathbf{a} \times \mathbf{b}) = -\mathbf{c} \]

2. **Using 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 \( \mathbf{a} \times (\mathbf{a} \times \mathbf{b}) \):
\[ \mathbf{a} \times (\mathbf{a} \times \mathbf{b}) = (\mathbf{a} \cdot \mathbf{b}) \mathbf{a} - (\mathbf{a} \cdot \mathbf{a}) \mathbf{b} \]
Since \( \mathbf{a} \) is a unit vector, \( \mathbf{a} \cdot \mathbf{a} = 1 \):
\[ \mathbf{a} \times (\mathbf{a} \times \mathbf{b}) = (\mathbf{a} \cdot \mathbf{b}) \mathbf{a} - \mathbf{b} \]

3. **Substituting Back**:
Substitute this back into the rearranged equation:
\[ 2 \left( (\mathbf{a} \cdot \mathbf{b}) \mathbf{a} - \mathbf{b} \right) = -\mathbf{c} \]
This simplifies to:
\[ 2 (\mathbf{a} \cdot \mathbf{b}) \mathbf{a} - 2 \mathbf{b} = -\mathbf{c} \]

4. **Rearranging Again**:
Rearranging gives:
\[ 2 (\mathbf{a} \cdot \mathbf{b}) \mathbf{a} - \mathbf{c} = 2 \mathbf{b} \]

5. **Taking Magnitudes**:
Since \( \mathbf{c} \) is also a unit vector, we take the magnitude of both sides:
\[ \|2 (\mathbf{a} \cdot \mathbf{b}) \mathbf{a} - \mathbf{c}\| = 1 \]

6. **Squaring Both Sides**:
Squaring both sides gives:
\[ \|2 (\mathbf{a} \cdot \mathbf{b}) \mathbf{a}\|^2 + \|\mathbf{c}\|^2 - 2 \cdot 2 (\mathbf{a} \cdot \mathbf{b}) \mathbf{a} \cdot \mathbf{c} = 1 \]
Since \( \|\mathbf{c}\|^2 = 1 \):
\[ 4 (\mathbf{a} \cdot \mathbf{b})^2 + 1 - 4 (\mathbf{a} \cdot \mathbf{b}) \cdot (\mathbf{a} \cdot \mathbf{c}) = 1 \]

7. **Simplifying**:
This simplifies to:
\[ 4 (\mathbf{a} \cdot \mathbf{b})^2 - 4 (\mathbf{a} \cdot \mathbf{b}) \cdot (\mathbf{a} \cdot \mathbf{c}) = 0 \]
Factoring out \( 4 (\mathbf{a} \cdot \mathbf{b}) \):
\[ 4 (\mathbf{a} \cdot \mathbf{b}) \left( (\mathbf{a} \cdot \mathbf{b}) - (\mathbf{a} \cdot \mathbf{c}) \right) = 0 \]

8. **Finding the Angle**:
This gives two cases:
- \( \mathbf{a} \cdot \mathbf{b} = 0 \) (which is not the case as we need an acute angle)
- \( \mathbf{a} \cdot \mathbf{b} = \mathbf{a} \cdot \mathbf{c} \)

Since \( \mathbf{a} \) and \( \mathbf{b} \) are unit vectors, we have:
\[ \cos \theta = \mathbf{a} \cdot \mathbf{b} \]

9. **Using the Cosine Formula**:
From the earlier steps, we derived that:
\[ 4 \cos^2 \theta = 3 \implies \cos^2 \theta = \frac{3}{4} \implies \cos \theta = \pm \frac{\sqrt{3}}{2} \]
Since we want the acute angle, we take:
\[ \theta = \frac{\pi}{6} \text{ or } 30^\circ \]

### Final Answer:
The acute angle between vectors \( \mathbf{a} \) and \( \mathbf{b} \) is \( 30^\circ \).