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

Answer

To solve the problem step by step, we will follow the reasoning and calculations as given in the video transcript.

### Step 1: Define the given vectors
We have the following vectors:
- \( \vec{a} = \alpha \hat{i} + 2 \hat{j} + \beta \hat{k} \)
- \( \vec{b} = \hat{i} + \hat{j} \)
- \( \vec{c} = \hat{j} + \hat{k} \)

### Step 2: Find the unit vectors of \( \vec{b} \) and \( \vec{c} \)
To find the unit vectors, we first calculate the magnitudes of \( \vec{b} \) and \( \vec{c} \).

1. Magnitude of \( \vec{b} \):
\[ |\vec{b}| = \sqrt{1^2 + 1^2} = \sqrt{2} \]
Thus, the unit vector \( \hat{b} \) is:
\[ \hat{b} = \frac{\vec{b}}{|\vec{b}|} = \frac{\hat{i} + \hat{j}}{\sqrt{2}} = \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} \]

2. Magnitude of \( \vec{c} \):
\[ |\vec{c}| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2} \]
Thus, the unit vector \( \hat{c} \) is:
\[ \hat{c} = \frac{\vec{c}}{|\vec{c}|} = \frac{\hat{j} + \hat{k}}{\sqrt{2}} = \frac{1}{\sqrt{2}} \hat{j} + \frac{1}{\sqrt{2}} \hat{k} \]

### Step 3: Use the angle bisector formula
The angle bisector \( \vec{r} \) can be expressed as:
\[ \vec{r} = \lambda \hat{b} + \lambda \hat{c} \]
Substituting the unit vectors:
\[ \vec{r} = \lambda \left( \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} \right) + \lambda \left( \frac{1}{\sqrt{2}} \hat{j} + \frac{1}{\sqrt{2}} \hat{k} \right) \]
Combining terms:
\[ \vec{r} = \frac{\lambda}{\sqrt{2}} \hat{i} + \left( \frac{\lambda}{\sqrt{2}} + \frac{\lambda}{\sqrt{2}} \right) \hat{j} + \frac{\lambda}{\sqrt{2}} \hat{k} \]
This simplifies to:
\[ \vec{r} = \frac{\lambda}{\sqrt{2}} \hat{i} + \frac{2\lambda}{\sqrt{2}} \hat{j} + \frac{\lambda}{\sqrt{2}} \hat{k} \]

### Step 4: Set \( \vec{r} \) equal to \( \vec{a} \)
Since \( \vec{a} \) lies in the same plane:
\[ \alpha \hat{i} + 2 \hat{j} + \beta \hat{k} = \frac{\lambda}{\sqrt{2}} \hat{i} + \frac{2\lambda}{\sqrt{2}} \hat{j} + \frac{\lambda}{\sqrt{2}} \hat{k} \]

### Step 5: Compare coefficients
From the equation, we can compare coefficients:
1. For \( \hat{i} \):
\[ \alpha = \frac{\lambda}{\sqrt{2}} \]
2. For \( \hat{j} \):
\[ 2 = \frac{2\lambda}{\sqrt{2}} \implies \lambda = \sqrt{2} \]
3. For \( \hat{k} \):
\[ \beta = \frac{\lambda}{\sqrt{2}} = \frac{\sqrt{2}}{\sqrt{2}} = 1 \]

### Step 6: Substitute \( \lambda \) back to find \( \alpha \)
Substituting \( \lambda \):
\[ \alpha = \frac{\sqrt{2}}{\sqrt{2}} = 1 \]

### Conclusion
Thus, the values of \( \alpha \) and \( \beta \) are:
\[ \alpha = 1, \quad \beta = 1 \]

The correct option is (4) \( \alpha = 1, \beta = 1 \).

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