Let `veca, vecb` and `vecc` be three vectors having magnitudes 1,1 and 2 resectively. If `vecaxx(vecaxxvecc)+vecb=vec0` then the acute angel between `veca` and `vecc` is

Explanation

To solve the problem step by step, we will use vector algebra and properties of dot and cross products.

### Given:
- Vectors \( \vec{a}, \vec{b}, \vec{c} \) with magnitudes:
- \( |\vec{a}| = 1 \)
- \( |\vec{b}| = 1 \)
- \( |\vec{c}| = 2 \)
- The equation:
\[ \vec{a} \times (\vec{a} \times \vec{c}) + \vec{b} = \vec{0} \]

### Step 1: Use the vector triple product identity
We apply the vector triple product identity:
\[ \vec{u} \times (\vec{v} \times \vec{w}) = (\vec{u} \cdot \vec{w}) \vec{v} - (\vec{u} \cdot \vec{v}) \vec{w} \]
In our case, let \( \vec{u} = \vec{a} \), \( \vec{v} = \vec{a} \), and \( \vec{w} = \vec{c} \):
\[ \vec{a} \times (\vec{a} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{a} - (\vec{a} \cdot \vec{a}) \vec{c} \]

### Step 2: Substitute into the equation
Substituting into the equation gives us:
\[ (\vec{a} \cdot \vec{c}) \vec{a} - |\vec{a}|^2 \vec{c} + \vec{b} = \vec{0} \]
Since \( |\vec{a}|^2 = 1 \):
\[ (\vec{a} \cdot \vec{c}) \vec{a} - \vec{c} + \vec{b} = \vec{0} \]

### Step 3: Rearranging the equation
Rearranging the equation gives:
\[ (\vec{a} \cdot \vec{c}) \vec{a} = \vec{c} - \vec{b} \]

### Step 4: Taking magnitudes
Taking magnitudes on both sides:
\[ |(\vec{a} \cdot \vec{c}) \vec{a}| = |\vec{c} - \vec{b}| \]
Since \( |\vec{a}| = 1 \):
\[ |\vec{a} \cdot \vec{c}| = |\vec{c} - \vec{b}| \]

### Step 5: Calculate \( |\vec{c} - \vec{b}| \)
Using the triangle inequality, we can express \( |\vec{c} - \vec{b}| \):
\[ |\vec{c} - \vec{b}| = \sqrt{|\vec{c}|^2 + |\vec{b}|^2 - 2 |\vec{c}| |\vec{b}| \cos \phi} \]
where \( \phi \) is the angle between \( \vec{c} \) and \( \vec{b} \). Since \( |\vec{c}| = 2 \) and \( |\vec{b}| = 1 \):
\[ |\vec{c} - \vec{b}| = \sqrt{2^2 + 1^2 - 2 \cdot 2 \cdot 1 \cos \phi} = \sqrt{4 + 1 - 4 \cos \phi} \]

### Step 6: Substitute and solve for \( \vec{a} \cdot \vec{c} \)
Now we know:
\[ |\vec{a} \cdot \vec{c}| = \sqrt{5 - 4 \cos \phi} \]
Let \( \theta \) be the angle between \( \vec{a} \) and \( \vec{c} \):
\[ \vec{a} \cdot \vec{c} = |\vec{a}| |\vec{c}| \cos \theta = 1 \cdot 2 \cos \theta = 2 \cos \theta \]
Thus:
\[ |2 \cos \theta| = \sqrt{5 - 4 \cos \phi} \]

### Step 7: Find \( \cos \theta \)
From the equation, we can solve for \( \cos \theta \):
\[ 2 |\cos \theta| = \sqrt{5 - 4 \cos \phi} \]
Squaring both sides gives:
\[ 4 \cos^2 \theta = 5 - 4 \cos \phi \]

### Step 8: Solve for the angle
To find the acute angle \( \theta \), we can set \( \cos \theta = \frac{\sqrt{3}}{2} \) which corresponds to \( \theta = \frac{\pi}{6} \).

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