Let `vecAB = 3 hati + hatj - hatk and vecAC = hati -hatj + 3hatk` and a point P on the line segment BC is equidistant from AB and AC, then `vec (AP)` is

Options

A. `2 hati - hatk `
B. `hati -2 hatk`
C. `2hati +hatk`
D. none of these

Explanation

To solve the problem, we need to find the vector \( \vec{AP} \) given that point \( P \) is equidistant from the lines \( AB \) and \( AC \). We will use the concept of angle bisectors in vectors.

### Step-by-Step Solution:

1. **Define the Vectors**:
Given:
\[ \vec{AB} = 3\hat{i} + \hat{j} - \hat{k} \]
\[ \vec{AC} = \hat{i} - \hat{j} + 3\hat{k} \]

2. **Find the Magnitudes**:
Calculate the magnitudes of \( \vec{AB} \) and \( \vec{AC} \):
\[ |\vec{AB}| = \sqrt{(3^2 + 1^2 + (-1)^2)} = \sqrt{9 + 1 + 1} = \sqrt{11} \]
\[ |\vec{AC}| = \sqrt{(1^2 + (-1)^2 + 3^2)} = \sqrt{1 + 1 + 9} = \sqrt{11} \]

3. **Find the Direction Ratios**:
The direction ratios of \( \vec{AB} \) and \( \vec{AC} \) are:
\[ \vec{AB} = (3, 1, -1), \quad \vec{AC} = (1, -1, 3) \]

4. **Find the Angle Bisector**:
The vector \( \vec{AP} \) (the angle bisector) can be found using the formula:
\[ \vec{AP} = \frac{\vec{AB}}{|\vec{AB}|} + \frac{\vec{AC}}{|\vec{AC}|} \]
Since \( |\vec{AB}| = |\vec{AC}| = \sqrt{11} \):
\[ \vec{AP} = \frac{1}{\sqrt{11}}(3\hat{i} + \hat{j} - \hat{k}) + \frac{1}{\sqrt{11}}(\hat{i} - \hat{j} + 3\hat{k}) \]

5. **Combine the Vectors**:
Combine the components:
\[ \vec{AP} = \frac{1}{\sqrt{11}} \left( (3 + 1)\hat{i} + (1 - 1)\hat{j} + (-1 + 3)\hat{k} \right) \]
Simplifying this gives:
\[ \vec{AP} = \frac{1}{\sqrt{11}} \left( 4\hat{i} + 0\hat{j} + 2\hat{k} \right) = \frac{4}{\sqrt{11}}\hat{i} + \frac{2}{\sqrt{11}}\hat{k} \]

6. **Final Result**:
The vector \( \vec{AP} \) is:
\[ \vec{AP} = \frac{4}{\sqrt{11}}\hat{i} + \frac{2}{\sqrt{11}}\hat{k} \]