Let `hata` and `hatb` be two unit vectors such that `hata.hatb=1/3` and `hata xx hatb=hatc`, Also `vecF=alphahata+betahatb+lambdahatc`,
where, `alpha,beta,lambda` are scalars. If `alpha=k_(1)(vecF.hata)-k_(2)(vecF.hatb)` then the value of `2(k_(1)+k_(2))` is

Options

A. (a) `2sqrt(3)`
B. (b) `sqrt(3)`
C. (c) 3
D. (d) 1

Explanation

To solve the problem step by step, we will analyze the given information and derive the necessary equations.

### Step 1: Understand the given vectors and their properties
We have two unit vectors \( \hat{a} \) and \( \hat{b} \) such that:
- \( \hat{a} \cdot \hat{b} = \frac{1}{3} \)
- \( \hat{a} \times \hat{b} = \hat{c} \)

Since \( \hat{a} \) and \( \hat{b} \) are unit vectors, we know:
- \( \hat{a} \cdot \hat{a} = 1 \)
- \( \hat{b} \cdot \hat{b} = 1 \)

### Step 2: Write the expression for \( \vec{F} \)
We have:
\[ \vec{F} = \alpha \hat{a} + \beta \hat{b} + \lambda \hat{c} \]
where \( \alpha, \beta, \lambda \) are scalars.

### Step 3: Calculate \( \vec{F} \cdot \hat{a} \) and \( \vec{F} \cdot \hat{b} \)
We will take the dot product of \( \vec{F} \) with \( \hat{a} \):
\[ \vec{F} \cdot \hat{a} = (\alpha \hat{a} + \beta \hat{b} + \lambda \hat{c}) \cdot \hat{a} = \alpha (\hat{a} \cdot \hat{a}) + \beta (\hat{b} \cdot \hat{a}) + \lambda (\hat{c} \cdot \hat{a}) \]
Using the properties of dot products:
\[ \vec{F} \cdot \hat{a} = \alpha + \beta \cdot \frac{1}{3} + 0 = \alpha + \frac{\beta}{3} \]

Now, calculate \( \vec{F} \cdot \hat{b} \):
\[ \vec{F} \cdot \hat{b} = (\alpha \hat{a} + \beta \hat{b} + \lambda \hat{c}) \cdot \hat{b} = \alpha (\hat{a} \cdot \hat{b}) + \beta (\hat{b} \cdot \hat{b}) + \lambda (\hat{c} \cdot \hat{b}) \]
Using the properties of dot products:
\[ \vec{F} \cdot \hat{b} = \alpha \cdot \frac{1}{3} + \beta + 0 = \frac{\alpha}{3} + \beta \]

### Step 4: Express \( \alpha \) in terms of \( k_1 \) and \( k_2 \)
From the problem, we have:
\[ \alpha = k_1 (\vec{F} \cdot \hat{a}) - k_2 (\vec{F} \cdot \hat{b}) \]
Substituting the expressions we derived:
\[ \alpha = k_1 \left( \alpha + \frac{\beta}{3} \right) - k_2 \left( \frac{\alpha}{3} + \beta \right) \]

### Step 5: Rearranging the equation
Expanding this gives:
\[ \alpha = k_1 \alpha + \frac{k_1 \beta}{3} - k_2 \frac{\alpha}{3} - k_2 \beta \]
Rearranging terms:
\[ \alpha - k_1 \alpha + k_2 \frac{\alpha}{3} = \frac{k_1 \beta}{3} - k_2 \beta \]
Factoring out \( \alpha \):
\[ \alpha (1 - k_1 + \frac{k_2}{3}) = \beta \left( \frac{k_1}{3} - k_2 \right) \]

### Step 6: Find \( k_1 \) and \( k_2 \)
To find \( k_1 \) and \( k_2 \), we can analyze the coefficients. We can set \( \beta = 0 \) for simplicity, leading to:
\[ \alpha (1 - k_1 + \frac{k_2}{3}) = 0 \]
This implies:
\[ 1 - k_1 + \frac{k_2}{3} = 0 \quad \text{(1)} \]

Now, if we assume \( \alpha \) is non-zero, we can solve for \( k_1 \) and \( k_2 \) using the second equation derived from \( \beta \).

### Step 7: Solve for \( 2(k_1 + k_2) \)
From the equations, we can find values for \( k_1 \) and \( k_2 \). After substituting and simplifying, we find:
\[ 2(k_1 + k_2) = 3 \]

Thus, the final answer is:
\[ \boxed{3} \]