Let `vecalpha=(1)/(a)hati+(4)/(b)hatj+bhatk` and `vecbeta=bhati+ahatj+(1)/(b)hatk(Aaa, b gt 0)`, then the maximum value of `(12)/(6+vecalpha.vecbeta))` is

Options

A. `(12)/(11)`
B. 2
C. 1
D. `(10)/(9)`

Answer

To solve the problem, we need to find the maximum value of the expression \(\frac{12}{6 + \vec{\alpha} \cdot \vec{\beta}}\), where \(\vec{\alpha}\) and \(\vec{\beta}\) are given vectors.

### Step-by-step Solution:

1. **Define the Vectors**:
\[ \vec{\alpha} = \frac{1}{a} \hat{i} + \frac{4}{b} \hat{j} + b \hat{k} \]
\[ \vec{\beta} = b \hat{i} + a \hat{j} + \frac{1}{b} \hat{k} \]

2. **Calculate the Dot Product \(\vec{\alpha} \cdot \vec{\beta}\)**:
The dot product is calculated as follows:
\[ \vec{\alpha} \cdot \vec{\beta} = \left(\frac{1}{a}\hat{i} + \frac{4}{b}\hat{j} + b\hat{k}\right) \cdot \left(b\hat{i} + a\hat{j} + \frac{1}{b}\hat{k}\right) \]
\[ = \frac{1}{a} \cdot b + \frac{4}{b} \cdot a + b \cdot \frac{1}{b} \]
\[ = \frac{b}{a} + \frac{4a}{b} + 1 \]

3. **Simplify the Expression**:
Let \(x = \frac{b}{a}\) and \(y = \frac{4a}{b}\). Then we can rewrite:
\[ \vec{\alpha} \cdot \vec{\beta} = x + y + 1 \]

4. **Apply the AM-GM Inequality**:
By the AM-GM inequality:
\[ \frac{x + y}{2} \geq \sqrt{xy} \]
Therefore:
\[ x + y \geq 2\sqrt{xy} \]
Now substituting \(y = \frac{4a}{b} = \frac{4}{x}\):
\[ xy = x \cdot \frac{4}{x} = 4 \]
Thus:
\[ x + \frac{4}{x} \geq 4 \]

5. **Find the Minimum Value of \(\vec{\alpha} \cdot \vec{\beta}\)**:
Hence, we have:
\[ \vec{\alpha} \cdot \vec{\beta} \geq 4 + 1 = 5 \]

6. **Substitute Back into the Original Expression**:
Now substituting this back into our expression:
\[ 6 + \vec{\alpha} \cdot \vec{\beta} \geq 6 + 5 = 11 \]

7. **Calculate the Maximum Value**:
Thus:
\[ \frac{12}{6 + \vec{\alpha} \cdot \vec{\beta}} \leq \frac{12}{11} \]
Therefore, the maximum value of the expression is:
\[ \frac{12}{11} \]

### Final Answer:
The maximum value of \(\frac{12}{6 + \vec{\alpha} \cdot \vec{\beta}}\) is \(\frac{12}{11}\).