If angle bisector of `overline(a) = 2hat(i) + 3 hat(j) + 4 hat(k)` and `overline(b) = 4hat(i) - 2 hat(j) + 3 hat(k)` is `overline(c ) = alpha hat(i) + 2 hat(j) + beta hat(k)` then

Options

A. `vec(c ).hat(k) + 7 = 0`
B. `vec(c ) .hat(k) - 14 = 0`
C. `vec(c ) .hat(k) + 15 = 0`
D. `vec(c ) .hat(k) - 7 = 0`

Explanation

To solve the problem step by step, we will follow the outlined approach to find the angle bisector vector \( \overline{c} \) given the vectors \( \overline{a} \) and \( \overline{b} \).

### Step 1: Define the vectors
We are given:
\[ \overline{a} = 2\hat{i} + 3\hat{j} + 4\hat{k} \]
\[ \overline{b} = 4\hat{i} - 2\hat{j} + 3\hat{k} \]

### Step 2: Calculate the magnitudes of the vectors
First, we calculate the magnitude of \( \overline{a} \):
\[ |\overline{a}| = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29} \]

Next, we calculate the magnitude of \( \overline{b} \):
\[ |\overline{b}| = \sqrt{4^2 + (-2)^2 + 3^2} = \sqrt{16 + 4 + 9} = \sqrt{29} \]

### Step 3: Find the unit vectors
Now we find the unit vectors \( \hat{a} \) and \( \hat{b} \):
\[ \hat{a} = \frac{\overline{a}}{|\overline{a}|} = \frac{2\hat{i} + 3\hat{j} + 4\hat{k}}{\sqrt{29}} \]
\[ \hat{b} = \frac{\overline{b}}{|\overline{b}|} = \frac{4\hat{i} - 2\hat{j} + 3\hat{k}}{\sqrt{29}} \]

### Step 4: Write the angle bisector vector
The angle bisector \( \overline{c} \) can be expressed as:
\[ \overline{c} = \lambda (\hat{a} + \hat{b}) \]
Substituting the unit vectors:
\[ \overline{c} = \lambda \left( \frac{2\hat{i} + 3\hat{j} + 4\hat{k}}{\sqrt{29}} + \frac{4\hat{i} - 2\hat{j} + 3\hat{k}}{\sqrt{29}} \right) \]
Combining the vectors:
\[ \overline{c} = \lambda \left( \frac{(2 + 4)\hat{i} + (3 - 2)\hat{j} + (4 + 3)\hat{k}}{\sqrt{29}} \right) = \lambda \left( \frac{6\hat{i} + 1\hat{j} + 7\hat{k}}{\sqrt{29}} \right) \]

### Step 5: Express \( \overline{c} \) in terms of components
This gives us:
\[ \overline{c} = \frac{6\lambda}{\sqrt{29}} \hat{i} + \frac{\lambda}{\sqrt{29}} \hat{j} + \frac{7\lambda}{\sqrt{29}} \hat{k} \]

### Step 6: Compare with given form of \( \overline{c} \)
We know that:
\[ \overline{c} = \alpha \hat{i} + 2 \hat{j} + \beta \hat{k} \]
From this, we can equate the coefficients:
1. \( \alpha = \frac{6\lambda}{\sqrt{29}} \)
2. \( 2 = \frac{\lambda}{\sqrt{29}} \)
3. \( \beta = \frac{7\lambda}{\sqrt{29}} \)

### Step 7: Solve for \( \lambda \)
From the second equation:
\[ \lambda = 2\sqrt{29} \]

### Step 8: Substitute \( \lambda \) to find \( \beta \)
Substituting \( \lambda \) into the expression for \( \beta \):
\[ \beta = \frac{7(2\sqrt{29})}{\sqrt{29}} = 14 \]

### Final Result
Thus, the values of \( \alpha \) and \( \beta \) are:
- \( \alpha = \frac{6(2\sqrt{29})}{\sqrt{29}} = 12 \)
- \( \beta = 14 \)

### Conclusion
The angle bisector vector \( \overline{c} \) is given by:
\[ \overline{c} = 12\hat{i} + 2\hat{j} + 14\hat{k} \]