The position vectors of the vertices A,B and C of a triangle are `hati-hatj-3hatk,2hati+hatj-2hatk` and `-5hati+2hatj-6hatk`, respectively. The length of the bisector AD of the `angleBAC`, where D is on the segment BC, is

Options

A. `(3)/(4)sqrt(3)`
B. `(1)/(4)`
C. `(11)/(2)`
D. None of these

Explanation

To find the length of the bisector \( AD \) of angle \( BAC \) in the triangle with vertices \( A \), \( B \), and \( C \) given by their position vectors, we will follow these steps:

### Step 1: Define the position vectors
The position vectors of the vertices are given as:
- \( \vec{A} = \hat{i} - \hat{j} - 3\hat{k} \)
- \( \vec{B} = 2\hat{i} + \hat{j} - 2\hat{k} \)
- \( \vec{C} = -5\hat{i} + 2\hat{j} - 6\hat{k} \)

### Step 2: Find the position vector of point \( D \)
Point \( D \) lies on segment \( BC \). The position vector \( \vec{D} \) can be found using the formula for the midpoint of \( B \) and \( C \):
\[ \vec{D} = \frac{\vec{B} + \vec{C}}{2} \]
Calculating \( \vec{D} \):
\[ \vec{D} = \frac{(2\hat{i} + \hat{j} - 2\hat{k}) + (-5\hat{i} + 2\hat{j} - 6\hat{k})}{2} \]
\[ = \frac{(2 - 5)\hat{i} + (1 + 2)\hat{j} + (-2 - 6)\hat{k}}{2} \]
\[ = \frac{-3\hat{i} + 3\hat{j} - 8\hat{k}}{2} \]
\[ = -\frac{3}{2}\hat{i} + \frac{3}{2}\hat{j} - 4\hat{k} \]

### Step 3: Find the vector \( \vec{AD} \)
The vector \( \vec{AD} \) can be calculated as:
\[ \vec{AD} = \vec{D} - \vec{A} \]
Substituting the vectors:
\[ \vec{AD} = \left(-\frac{3}{2}\hat{i} + \frac{3}{2}\hat{j} - 4\hat{k}\right) - \left(\hat{i} - \hat{j} - 3\hat{k}\right) \]
\[ = -\frac{3}{2}\hat{i} + \frac{3}{2}\hat{j} - 4\hat{k} - \hat{i} + \hat{j} + 3\hat{k} \]
\[ = \left(-\frac{3}{2} - 1\right)\hat{i} + \left(\frac{3}{2} + 1\right)\hat{j} + (-4 + 3)\hat{k} \]
\[ = -\frac{5}{2}\hat{i} + \frac{5}{2}\hat{j} - 1\hat{k} \]

### Step 4: Calculate the magnitude of \( \vec{AD} \)
The magnitude of vector \( \vec{AD} \) is given by:
\[ |\vec{AD}| = \sqrt{\left(-\frac{5}{2}\right)^2 + \left(\frac{5}{2}\right)^2 + (-1)^2} \]
Calculating each term:
\[ = \sqrt{\frac{25}{4} + \frac{25}{4} + 1} \]
\[ = \sqrt{\frac{25}{4} + \frac{25}{4} + \frac{4}{4}} \]
\[ = \sqrt{\frac{54}{4}} = \sqrt{\frac{27}{2}} = \frac{3\sqrt{6}}{2} \]

### Final Answer
The length of the bisector \( AD \) of angle \( BAC \) is:
\[ \frac{3\sqrt{6}}{2} \]