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 find the angle between the vectors \(\vec{a}\) and \(\vec{b}\), we start with the given equation:

\[ (1 - 3 \vec{a} \cdot \vec{b})^2 + |2\vec{a} + \vec{b} + 3(\vec{a} \times \vec{b})|^2 = 47 \]

### Step 1: Let the angle between \(\vec{a}\) and \(\vec{b}\) be \(\theta\).
We know that:
- The magnitude of \(\vec{a}\) is \(1\).
- The magnitude of \(\vec{b}\) is \(2\).
- The dot product \(\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta = 1 \cdot 2 \cos \theta = 2 \cos \theta\).

### Step 2: Substitute the dot product into the equation.
Substituting \(\vec{a} \cdot \vec{b}\) into the equation gives us:

\[ (1 - 3(2 \cos \theta))^2 + |2\vec{a} + \vec{b} + 3(\vec{a} \times \vec{b})|^2 = 47 \]

This simplifies to:

\[ (1 - 6 \cos \theta)^2 + |2\vec{a} + \vec{b} + 3(\vec{a} \times \vec{b})|^2 = 47 \]

### Step 3: Expand the first term.
Expanding \((1 - 6 \cos \theta)^2\):

\[ 1 - 12 \cos \theta + 36 \cos^2 \theta \]

### Step 4: Calculate the magnitude of the second term.
Now we need to calculate \(|2\vec{a} + \vec{b} + 3(\vec{a} \times \vec{b})|^2\).

Using the properties of magnitudes and the fact that \(\vec{a} \times \vec{b}\) is perpendicular to both \(\vec{a}\) and \(\vec{b}\):

\[ |2\vec{a} + \vec{b}|^2 + |3(\vec{a} \times \vec{b})|^2 + 2(2\vec{a} \cdot \vec{b}) = |2\vec{a}|^2 + |\vec{b}|^2 + 9|\vec{a} \times \vec{b}|^2 \]

Calculating each part:
- \(|2\vec{a}|^2 = 4\)
- \(|\vec{b}|^2 = 4\)
- \(|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 \sin^2 \theta = 1 \cdot 4 \sin^2 \theta = 4 \sin^2 \theta\)

Thus, we have:

\[ |2\vec{a} + \vec{b}|^2 = 4 + 4 + 9(4 \sin^2 \theta) + 4 \cos \theta \]

### Step 5: Combine the terms and set equal to 47.
Combining everything gives:

\[ 1 - 12 \cos \theta + 36 \cos^2 \theta + 8 + 36 + 36 \sin^2 \theta = 47 \]

### Step 6: Simplify the equation.
Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\):

\[ 1 - 12 \cos \theta + 36 \cos^2 \theta + 8 + 36 + 36(1 - \cos^2 \theta) = 47 \]

This simplifies to:

\[ 1 - 12 \cos \theta + 36 \cos^2 \theta + 8 + 36 = 47 \]

### Step 7: Solve for \(\cos \theta\).
This reduces to:

\[ -12 \cos \theta + 36 \cos^2 \theta = 2 \]

Rearranging gives:

\[ 36 \cos^2 \theta - 12 \cos \theta - 2 = 0 \]

### Step 8: Use the quadratic formula.
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

Here, \(a = 36\), \(b = -12\), and \(c = -2\):

\[ \cos \theta = \frac{12 \pm \sqrt{(-12)^2 - 4 \cdot 36 \cdot (-2)}}{2 \cdot 36} \]

Calculating the discriminant:

\[ \sqrt{144 + 288} = \sqrt{432} = 12\sqrt{3} \]

So,

\[ \cos \theta = \frac{12 \pm 12\sqrt{3}}{72} = \frac{1 \pm \sqrt{3}}{6} \]

### Step 9: Determine the angle \(\theta\).
We find that:

\[ \cos \theta = -\frac{1}{2} \implies \theta = \frac{2\pi}{3} \text{ radians or } 120^\circ \]

Thus, the angle between \(\vec{a}\) and \(\vec{b}\) is:

\[ \theta = \frac{2\pi}{3} \]