If `veca and vecb` are any two vectors of magnitudes 1and 2. respectively, and `(1-3veca.vecb)^(2)+|2veca+vecb+3(vecaxxvecb)|^(2)=47` then the angle between `veca and vecb ` is

Options

A. `pi//3`
B. `pi-cos^(-1) (1//4)`
C. `(2pi)/3`
D. `cos^(-1) (1//4)`

Explanation

To solve the problem step by step, we will use the given information about the vectors \( \vec{a} \) and \( \vec{b} \).

### Step 1: Write down the given equation
We have the equation:
\[ (1 - 3 \vec{a} \cdot \vec{b})^2 + |2\vec{a} + \vec{b} + 3(\vec{a} \times \vec{b})|^2 = 47 \]

### Step 2: Use the magnitudes of the vectors
Given that the magnitude of \( \vec{a} \) is 1 and the magnitude of \( \vec{b} \) is 2, we can write:
\[ |\vec{a}| = 1 \quad \text{and} \quad |\vec{b}| = 2 \]

### Step 3: Expand the first term
The first term can be expanded as:
\[ (1 - 3 \vec{a} \cdot \vec{b})^2 = 1 - 6 \vec{a} \cdot \vec{b} + 9 (\vec{a} \cdot \vec{b})^2 \]

### Step 4: Expand the second term
For the second term, we need to calculate \( |2\vec{a} + \vec{b} + 3(\vec{a} \times \vec{b})|^2 \):
\[ |2\vec{a} + \vec{b} + 3(\vec{a} \times \vec{b})|^2 = |2\vec{a} + \vec{b}|^2 + |3(\vec{a} \times \vec{b})|^2 + 2(2\vec{a} + \vec{b}) \cdot (3(\vec{a} \times \vec{b})) \]

### Step 5: Calculate \( |2\vec{a} + \vec{b}|^2 \)
Using the formula for the magnitude:
\[ |2\vec{a} + \vec{b}|^2 = |2\vec{a}|^2 + |\vec{b}|^2 + 2(2\vec{a}) \cdot \vec{b} = 4 + 4 + 4 \vec{a} \cdot \vec{b} = 8 + 4 \vec{a} \cdot \vec{b} \]

### Step 6: Calculate \( |3(\vec{a} \times \vec{b})|^2 \)
Using the property of cross products:
\[ |3(\vec{a} \times \vec{b})|^2 = 9 |\vec{a} \times \vec{b}|^2 = 9 |\vec{a}|^2 |\vec{b}|^2 \sin^2 \theta = 9 \cdot 1 \cdot 4 \sin^2 \theta = 36 \sin^2 \theta \]

### Step 7: Substitute back into the equation
Now substituting everything back into the equation:
\[ 1 - 6 \vec{a} \cdot \vec{b} + 9 (\vec{a} \cdot \vec{b})^2 + 8 + 4 \vec{a} \cdot \vec{b} + 36 \sin^2 \theta + 2(2\vec{a} + \vec{b}) \cdot (3(\vec{a} \times \vec{b})) = 47 \]

### Step 8: Simplify the equation
Combine like terms and simplify:
\[ 9 (\vec{a} \cdot \vec{b})^2 - 2 \vec{a} \cdot \vec{b} + 36 \sin^2 \theta + 9 = 47 \]
This simplifies to:
\[ 9 (\vec{a} \cdot \vec{b})^2 - 2 \vec{a} \cdot \vec{b} + 36 \sin^2 \theta - 38 = 0 \]

### Step 9: Use the identity \( \sin^2 \theta + \cos^2 \theta = 1 \)
We can express \( \sin^2 \theta \) in terms of \( \cos^2 \theta \):
\[ \sin^2 \theta = 1 - \cos^2 \theta \]

### Step 10: Solve for \( \cos \theta \)
Substituting back and solving the quadratic equation will give us:
\[ \cos \theta = -\frac{1}{2} \]

### Step 11: Find the angle
The angle \( \theta \) corresponding to \( \cos \theta = -\frac{1}{2} \) is:
\[ \theta = \frac{2\pi}{3} \text{ radians} \]

### Final Answer
The angle between \( \vec{a} \) and \( \vec{b} \) is \( \frac{2\pi}{3} \) radians.

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