If angle bisector of `vec(a) = 2hat(i) + 3hat(j) + 4hat(k)` and `vec(b) = 4hat(i) - 2hat(j) + 3 hat(k) ` is `vec(c) = alpha hat(i) + 2hat(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) + 14 = 0 `
D. `vec(c) .hat(k) - 7 =0`

Explanation

To solve the problem, we need to find the angle bisector of the vectors \( \vec{a} \) and \( \vec{b} \), and then compare it with the given vector \( \vec{c} \).

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

### Step 2: Calculate the magnitudes of \( \vec{a} \) and \( \vec{b} \)
The magnitude of \( \vec{a} \):
\[ |\vec{a}| = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29} \]

The magnitude of \( \vec{b} \):
\[ |\vec{b}| = \sqrt{4^2 + (-2)^2 + 3^2} = \sqrt{16 + 4 + 9} = \sqrt{29} \]

### Step 3: Find the unit vectors of \( \vec{a} \) and \( \vec{b} \)
The unit vector of \( \vec{a} \):
\[ \hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{2\hat{i} + 3\hat{j} + 4\hat{k}}{\sqrt{29}} \]

The unit vector of \( \vec{b} \):
\[ \hat{b} = \frac{\vec{b}}{|\vec{b}|} = \frac{4\hat{i} - 2\hat{j} + 3\hat{k}}{\sqrt{29}} \]

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

### Step 5: Compare with the given vector \( \vec{c} = \alpha \hat{i} + 2\hat{j} + \beta \hat{k} \)
From the equation:
\[ \vec{c} = \lambda \frac{6\hat{i} + 1\hat{j} + 7\hat{k}}{\sqrt{29}} \]
We can equate components:
1. \( \alpha = \lambda \frac{6}{\sqrt{29}} \)
2. \( 2 = \lambda \frac{1}{\sqrt{29}} \)
3. \( \beta = \lambda \frac{7}{\sqrt{29}} \)

### Step 6: Solve for \( \lambda \)
From the second equation:
\[ \lambda = 2\sqrt{29} \]

### Step 7: Substitute \( \lambda \) back to find \( \alpha \) and \( \beta \)
Substituting \( \lambda \) into the equations for \( \alpha \) and \( \beta \):
1. \( \alpha = 2\sqrt{29} \cdot \frac{6}{\sqrt{29}} = 12 \)
2. \( \beta = 2\sqrt{29} \cdot \frac{7}{\sqrt{29}} = 14 \)

### Conclusion
Thus, we have:
\[ \alpha = 12, \quad \beta = 14 \]

### Final Step: Verify with the dot product condition
We need to check if \( \vec{c} \cdot \hat{k} - 14 = 0 \):
\[ \vec{c} \cdot \hat{k} = \beta = 14 \]
Thus, \( \vec{c} \cdot \hat{k} - 14 = 0 \).