Let `veca, vecb, vec c` such that `|veca| = 1 , |vecb| = 1` and `|vec c | = 2` and if `veca xx (veca xx vec c ) + vec b = 0` then find acute angle between `veca ` and `vec c `

Options

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

Answer

To solve the problem step by step, we will use the properties of vector algebra, particularly the vector triple product and the dot product.

### Step-by-Step Solution:

1. **Given Information**:
We have three vectors:
- \( \vec{a} \) such that \( |\vec{a}| = 1 \)
- \( \vec{b} \) such that \( |\vec{b}| = 1 \)
- \( \vec{c} \) such that \( |\vec{c}| = 2 \)
- The equation given is \( \vec{a} \times (\vec{a} \times \vec{c}) + \vec{b} = 0 \).

2. **Using the Vector Triple Product Identity**:
The vector triple product identity states:
\[ \vec{u} \times (\vec{v} \times \vec{w}) = (\vec{u} \cdot \vec{w}) \vec{v} - (\vec{u} \cdot \vec{v}) \vec{w} \]
Applying this to our equation:
\[ \vec{a} \times (\vec{a} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{a} - (\vec{a} \cdot \vec{a}) \vec{c} \]
Since \( |\vec{a}|^2 = 1 \), we can simplify:
\[ \vec{a} \times (\vec{a} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{a} - \vec{c} \]

3. **Substituting into the Equation**:
Now substituting this back into the original equation:
\[ (\vec{a} \cdot \vec{c}) \vec{a} - \vec{c} + \vec{b} = 0 \]
Rearranging gives us:
\[ (\vec{a} \cdot \vec{c}) \vec{a} = \vec{c} - \vec{b} \]

4. **Taking Magnitudes**:
Taking the magnitude of both sides:
\[ |(\vec{a} \cdot \vec{c}) \vec{a}| = |\vec{c} - \vec{b}| \]
Since \( |\vec{a}| = 1 \), we have:
\[ |\vec{a} \cdot \vec{c}| = |\vec{c} - \vec{b}| \]

5. **Finding the Magnitude of \( \vec{c} - \vec{b} \)**:
We can use the formula for the magnitude of a vector:
\[ |\vec{c} - \vec{b}|^2 = |\vec{c}|^2 + |\vec{b}|^2 - 2(\vec{c} \cdot \vec{b}) \]
Substituting the known magnitudes:
\[ |\vec{c}|^2 = 4, \quad |\vec{b}|^2 = 1 \]
Thus:
\[ |\vec{c} - \vec{b}|^2 = 4 + 1 - 2(\vec{c} \cdot \vec{b}) = 5 - 2(\vec{c} \cdot \vec{b}) \]

6. **Equating the Magnitudes**:
From step 4, we have:
\[ |\vec{a} \cdot \vec{c}| = \sqrt{5 - 2(\vec{c} \cdot \vec{b})} \]
Since \( |\vec{a} \cdot \vec{c}| = |\vec{c}| \cos \theta \) where \( \theta \) is the angle between \( \vec{a} \) and \( \vec{c} \):
\[ |\vec{a} \cdot \vec{c}| = 2 \cos \theta \]

7. **Setting Up the Equation**:
We can set up the equation:
\[ 2 \cos \theta = \sqrt{5 - 2(\vec{c} \cdot \vec{b})} \]

8. **Finding the Angle**:
To find the acute angle \( \theta \), we need to express \( \vec{c} \cdot \vec{b} \) in terms of \( \cos \theta \). Since \( \vec{b} \) is also a unit vector, we can find \( \cos \theta \) using the known values:
\[ 2 \cos \theta = \sqrt{3} \implies \cos \theta = \frac{\sqrt{3}}{2} \]
Therefore, \( \theta = \cos^{-1} \left( \frac{\sqrt{3}}{2} \right) = \frac{\pi}{6} \) radians or \( 30^\circ \).

### Final Answer:
The acute angle between \( \vec{a} \) and \( \vec{c} \) is \( \frac{\pi}{6} \) radians or \( 30^\circ \).