Let `hata, hatb` be unit vectors. If `vecc` be a vector such that the angle between `hata and hatc` is `pi/12` , and `hatb=vecc+2(veccxxhata)`, then `|6vecc|^2` is equal to

Options

A. `6(3-sqrt3)`
B. `3+sqrt3`
C. `6(3+sqrt3)`
D. `6(sqrt3+1)`

Answer

To solve the problem step by step, we will analyze the given information and derive the required value.

### Given:
- \(\hat{a}, \hat{b}\) are unit vectors.
- The angle between \(\hat{a}\) and \(\hat{c}\) is \(\frac{\pi}{12}\).
- \(\hat{b} = \vec{c} + 2(\vec{c} \times \hat{a})\).

### Step 1: Understand the relationship between \(\hat{b}\) and \(\vec{c}\)
From the equation \(\hat{b} = \vec{c} + 2(\vec{c} \times \hat{a})\), we can rearrange it to express \(\vec{c}\):
\[ \hat{b} - \vec{c} = 2(\vec{c} \times \hat{a}) \]

### Step 2: Square both sides
We will square both sides of the equation to eliminate the cross product:
\[ |\hat{b} - \vec{c}|^2 = |2(\vec{c} \times \hat{a})|^2 \]

### Step 3: Expand both sides
The left-hand side expands as follows:
\[ |\hat{b}|^2 + |\vec{c}|^2 - 2(\hat{b} \cdot \vec{c}) = 4|\vec{c}|^2 |\hat{a}|^2 \sin^2\theta \]
where \(\theta\) is the angle between \(\vec{c}\) and \(\hat{a}\). Since \(\hat{a}\) is a unit vector, \(|\hat{a}|^2 = 1\).

### Step 4: Substitute known values
Since \(|\hat{b}|^2 = 1\) (as \(\hat{b}\) is a unit vector) and \(\theta = \frac{\pi}{12}\):
\[ 1 + |\vec{c}|^2 - 2(\hat{b} \cdot \vec{c}) = 4|\vec{c}|^2 \sin^2\left(\frac{\pi}{12}\right) \]

### Step 5: Calculate \(\sin^2\left(\frac{\pi}{12}\right)\)
Using the angle subtraction formula:
\[ \sin\left(\frac{\pi}{12}\right) = \sin\left(15^\circ\right) = \sin(45^\circ - 30^\circ) = \sin(45^\circ)\cos(30^\circ) - \cos(45^\circ)\sin(30^\circ) \]
\[ = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} - 1}{2\sqrt{2}} \]
Thus,
\[ \sin^2\left(\frac{\pi}{12}\right) = \left(\frac{\sqrt{3} - 1}{2\sqrt{2}}\right)^2 = \frac{(3 - 2\sqrt{3} + 1)}{8} = \frac{4 - 2\sqrt{3}}{8} = \frac{2 - \sqrt{3}}{4} \]

### Step 6: Substitute back into the equation
Substituting \(\sin^2\left(\frac{\pi}{12}\right)\) into the equation gives:
\[ 1 + |\vec{c}|^2 - 2(\hat{b} \cdot \vec{c}) = 4|\vec{c}|^2 \cdot \frac{2 - \sqrt{3}}{4} \]
\[ 1 + |\vec{c}|^2 - 2(\hat{b} \cdot \vec{c}) = |\vec{c}|^2(2 - \sqrt{3}) \]

### Step 7: Rearranging the equation
Rearranging gives:
\[ 1 = |\vec{c}|^2(2 - \sqrt{3} - 1) + 2(\hat{b} \cdot \vec{c}) \]
\[ 1 = |\vec{c}|^2(1 - \sqrt{3}) + 2(\hat{b} \cdot \vec{c}) \]

### Step 8: Solve for \(|\vec{c}|^2\)
Assuming \(\hat{b} \cdot \vec{c}\) is a known value, we can isolate \(|\vec{c}|^2\) and solve for it.

### Step 9: Find \(|6\vec{c}|^2\)
Finally, we need to find \(|6\vec{c}|^2\):
\[ |6\vec{c}|^2 = 36|\vec{c}|^2 \]

### Conclusion
The final answer will depend on the value of \(|\vec{c}|^2\) derived from the previous steps.