The vector ` vec a=""alpha hat i+2 hat j+""beta hat k` lies in the plane of the vectors ` vec b="" hat"i"+ hat j` and ` vec c= hat j+ hat k` and bisects the angle between ` vec b` and ` vec c` . Then which one of the following gives possible values of `alpha` and `beta` ? (1) `alpha=""2,beta=""2` (2) `alpha=""1,beta=""2` (3) `alpha=""2,beta=""1` (4) `alpha=""1,beta=""1`

Options

A. `alpha=""2,beta=""2`
B. `alpha=""1,beta=""2`
C. `alpha=""2,beta=""1`
D. `alpha=""1,beta=""1`

Explanation

To solve the problem, we need to find the values of \( \alpha \) and \( \beta \) such that the vector \( \vec{a} = \alpha \hat{i} + 2 \hat{j} + \beta \hat{k} \) lies in the plane formed by the vectors \( \vec{b} = \hat{i} + \hat{j} \) and \( \vec{c} = \hat{j} + \hat{k} \), and also bisects the angle between \( \vec{b} \) and \( \vec{c} \).

### Step-by-step solution:

1. **Find the normal vector of the plane formed by \( \vec{b} \) and \( \vec{c} \)**:
The normal vector \( \vec{n} \) can be found using the cross product:
\[ \vec{b} \times \vec{c} = (\hat{i} + \hat{j}) \times (\hat{j} + \hat{k}) \]
\[ = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix} = \hat{i}(1 \cdot 1 - 0 \cdot 1) - \hat{j}(1 \cdot 1 - 0 \cdot 0) + \hat{k}(1 \cdot 1 - 1 \cdot 0) \]
\[ = \hat{i} - \hat{j} + \hat{k} \]
Thus, \( \vec{n} = \hat{i} - \hat{j} + \hat{k} \).

2. **Check if \( \vec{a} \) lies in the plane**:
For \( \vec{a} \) to lie in the plane, it must satisfy the equation \( \vec{n} \cdot \vec{a} = 0 \):
\[ (\hat{i} - \hat{j} + \hat{k}) \cdot (\alpha \hat{i} + 2 \hat{j} + \beta \hat{k}) = 0 \]
\[ \alpha - 2 + \beta = 0 \quad \Rightarrow \quad \alpha + \beta = 2 \quad \text{(Equation 1)} \]

3. **Find the angle bisector**:
The angle bisector of two vectors \( \vec{b} \) and \( \vec{c} \) can be found using the formula:
\[ \vec{a} = k \left( \frac{\vec{b}}{|\vec{b}|} + \frac{\vec{c}}{|\vec{c}|} \right) \]
First, we calculate the magnitudes:
\[ |\vec{b}| = \sqrt{1^2 + 1^2} = \sqrt{2}, \quad |\vec{c}| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2} \]
Thus,
\[ \frac{\vec{b}}{|\vec{b}|} = \frac{1}{\sqrt{2}}(\hat{i} + \hat{j}), \quad \frac{\vec{c}}{|\vec{c}|} = \frac{1}{\sqrt{2}}(\hat{j} + \hat{k}) \]
Therefore,
\[ \vec{a} = k \left( \frac{1}{\sqrt{2}}(\hat{i} + \hat{j}) + \frac{1}{\sqrt{2}}(\hat{j} + \hat{k}) \right) = k \left( \frac{1}{\sqrt{2}} \hat{i} + \frac{2}{\sqrt{2}} \hat{j} + \frac{1}{\sqrt{2}} \hat{k} \right) \]
\[ = k \left( \frac{1}{\sqrt{2}} \hat{i} + \sqrt{2} \hat{j} + \frac{1}{\sqrt{2}} \hat{k} \right) \]

4. **Comparing coefficients**:
From the expression for \( \vec{a} \), we have:
\[ \alpha = \frac{k}{\sqrt{2}}, \quad 2 = k\sqrt{2}, \quad \beta = \frac{k}{\sqrt{2}} \]
From \( 2 = k\sqrt{2} \), we find \( k = \frac{2}{\sqrt{2}} = \sqrt{2} \).
Substituting \( k \) back:
\[ \alpha = \frac{\sqrt{2}}{\sqrt{2}} = 1, \quad \beta = \frac{\sqrt{2}}{\sqrt{2}} = 1 \]

5. **Final values**:
Thus, the values of \( \alpha \) and \( \beta \) are:
\[ \alpha = 1, \quad \beta = 1 \]

### Conclusion:
The possible values of \( \alpha \) and \( \beta \) are \( \alpha = 1 \) and \( \beta = 1 \), which corresponds to option (4).