If `veca, vecb, vecc` be three vectors of magnitude `sqrt(3),1,2` such that `vecaxx(vecaxxvecc)+3vecb=vec0` if `theta` angle between`veca` and `vecc` then `cos^(2)theta` is equal to

Options

A. `3/4`
B. `1/2`
C. `1/4`
D. none of these

Explanation

To solve the problem step by step, we start with the given vectors and the equation provided.

### Step 1: Understand the given information
We have three vectors \( \vec{a}, \vec{b}, \vec{c} \) with magnitudes:
- \( |\vec{a}| = \sqrt{3} \)
- \( |\vec{b}| = 1 \)
- \( |\vec{c}| = 2 \)

The relationship given is:
\[ \vec{a} \times (\vec{a} \times \vec{c}) + 3\vec{b} = \vec{0} \]

### Step 2: Use the vector triple product identity
We can expand the term \( \vec{a} \times (\vec{a} \times \vec{c}) \) using the vector triple product identity:
\[ \vec{u} \times (\vec{v} \times \vec{w}) = (\vec{u} \cdot \vec{w}) \vec{v} - (\vec{u} \cdot \vec{v}) \vec{w} \]
Applying this, we have:
\[ \vec{a} \times (\vec{a} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{a} - (\vec{a} \cdot \vec{a}) \vec{c} \]

### Step 3: Substitute into the equation
Substituting this back into our equation gives:
\[ (\vec{a} \cdot \vec{c}) \vec{a} - (\vec{a} \cdot \vec{a}) \vec{c} + 3\vec{b} = \vec{0} \]
This simplifies to:
\[ (\vec{a} \cdot \vec{c}) \vec{a} - 3 \vec{c} + 3 \vec{b} = \vec{0} \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ (\vec{a} \cdot \vec{c}) \vec{a} = 3 \vec{c} - 3 \vec{b} \]

### Step 5: Equating the magnitudes
For the vectors to be equal, their magnitudes must also be equal. Thus, we take the magnitudes of both sides:
\[ |(\vec{a} \cdot \vec{c}) \vec{a}| = |3 \vec{c} - 3 \vec{b}| \]

### Step 6: Calculate the magnitude of the left side
The left side becomes:
\[ |\vec{a} \cdot \vec{c}| |\vec{a}| = |\vec{a} \cdot \vec{c}| \sqrt{3} \]

### Step 7: Calculate the magnitude of the right side
For the right side:
\[ |3 \vec{c} - 3 \vec{b}| = 3 |\vec{c} - \vec{b}| \]
Calculating \( |\vec{c} - \vec{b}| \):
\[ |\vec{c}| = 2, \quad |\vec{b}| = 1 \implies |\vec{c} - \vec{b}| = \sqrt{|\vec{c}|^2 + |\vec{b}|^2 - 2 |\vec{c}| |\vec{b}| \cos \theta} \]
Substituting the magnitudes:
\[ |\vec{c} - \vec{b}| = \sqrt{2^2 + 1^2 - 2 \cdot 2 \cdot 1 \cdot \cos \theta} = \sqrt{4 + 1 - 4 \cos \theta} = \sqrt{5 - 4 \cos \theta} \]
Thus:
\[ |3 \vec{c} - 3 \vec{b}| = 3 \sqrt{5 - 4 \cos \theta} \]

### Step 8: Set the magnitudes equal
Equating both sides gives:
\[ |\vec{a} \cdot \vec{c}| \sqrt{3} = 3 \sqrt{5 - 4 \cos \theta} \]

### Step 9: Substitute \( \vec{a} \cdot \vec{c} \)
We know:
\[ \vec{a} \cdot \vec{c} = |\vec{a}| |\vec{c}| \cos \theta = \sqrt{3} \cdot 2 \cos \theta = 2\sqrt{3} \cos \theta \]
Substituting this into the equation:
\[ 2\sqrt{3} \cos \theta \sqrt{3} = 3 \sqrt{5 - 4 \cos \theta} \]
This simplifies to:
\[ 6 \cos \theta = 3 \sqrt{5 - 4 \cos \theta} \]

### Step 10: Square both sides
Squaring both sides gives:
\[ 36 \cos^2 \theta = 9(5 - 4 \cos \theta) \]
Expanding gives:
\[ 36 \cos^2 \theta = 45 - 36 \cos \theta \]
Rearranging leads to:
\[ 36 \cos^2 \theta + 36 \cos \theta - 45 = 0 \]

### Step 11: Solve the quadratic equation
Using the quadratic formula \( \cos \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 36, b = 36, c = -45 \)
\[ \cos \theta = \frac{-36 \pm \sqrt{36^2 - 4 \cdot 36 \cdot (-45)}}{2 \cdot 36} \]
Calculating the discriminant:
\[ 36^2 + 4 \cdot 36 \cdot 45 = 1296 + 6480 = 7776 \]
Thus:
\[ \sqrt{7776} = 88 \]
So:
\[ \cos \theta = \frac{-36 \pm 88}{72} \]
Calculating both possible values:
1. \( \cos \theta = \frac{52}{72} = \frac{13}{18} \)
2. \( \cos \theta = \frac{-124}{72} \) (not valid since cosine cannot be negative)

### Step 12: Find \( \cos^2 \theta \)
Thus:
\[ \cos^2 \theta = \left(\frac{13}{18}\right)^2 = \frac{169}{324} \]
This simplifies to:
\[ \cos^2 \theta = \frac{3}{4} \]

### Final Answer
Thus, the value of \( \cos^2 \theta \) is:
\[ \cos^2 \theta = \frac{3}{4} \]