The position vectors of the vectices A, B, C of a `triangle ABC " are " hati - hat j - 3 hat k , 2 hati + hat j - 2 hat k and - 5 hati + 2 hat j - 6 hat k `
respectively. The length of the bisector AD of the angle `angle BAC ` where D is on the line segment BC, is

Options

A. `(15)/2`
B. `(11)/(2)`
C. `(1)/(4)`
D. none of these

Explanation

To find the length of the bisector AD of angle BAC in triangle ABC, we will follow these steps:

### Step 1: Define the position vectors
Let the position vectors of points A, B, and C be 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: Calculate the vectors AB and AC
The vector \( \vec{AB} \) can be calculated as:
\[ \vec{AB} = \vec{B} - \vec{A} = (2\hat{i} + \hat{j} - 2\hat{k}) - (\hat{i} - \hat{j} - 3\hat{k}) = (2 - 1)\hat{i} + (1 + 1)\hat{j} + (-2 + 3)\hat{k} = \hat{i} + 2\hat{j} + \hat{k} \]

The vector \( \vec{AC} \) can be calculated as:
\[ \vec{AC} = \vec{C} - \vec{A} = (-5\hat{i} + 2\hat{j} - 6\hat{k}) - (\hat{i} - \hat{j} - 3\hat{k}) = (-5 - 1)\hat{i} + (2 + 1)\hat{j} + (-6 + 3)\hat{k} = -6\hat{i} + 3\hat{j} - 3\hat{k} \]

### Step 3: Calculate the lengths of AB and AC
The length of \( \vec{AB} \) is:
\[ |\vec{AB}| = \sqrt{(1)^2 + (2)^2 + (1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6} \]

The length of \( \vec{AC} \) is:
\[ |\vec{AC}| = \sqrt{(-6)^2 + (3)^2 + (-3)^2} = \sqrt{36 + 9 + 9} = \sqrt{54} = 3\sqrt{6} \]

### Step 4: Find the ratio in which D divides BC
The point D divides BC in the ratio of \( |\vec{AC}| : |\vec{AB}| = 3\sqrt{6} : \sqrt{6} = 3:1 \).

### Step 5: Calculate the position vector of D
Using the section formula, the position vector \( \vec{D} \) can be calculated as:
\[ \vec{D} = \frac{m\vec{C} + n\vec{B}}{m+n} = \frac{3\vec{C} + 1\vec{B}}{3 + 1} \]
Substituting the vectors:
\[ \vec{D} = \frac{3(-5\hat{i} + 2\hat{j} - 6\hat{k}) + 1(2\hat{i} + \hat{j} - 2\hat{k})}{4} \]
Calculating the numerator:
\[ = \frac{-15\hat{i} + 6\hat{j} - 18\hat{k} + 2\hat{i} + \hat{j} - 2\hat{k}}{4} = \frac{(-15 + 2)\hat{i} + (6 + 1)\hat{j} + (-18 - 2)\hat{k}}{4} \]
\[ = \frac{-13\hat{i} + 7\hat{j} - 20\hat{k}}{4} = -\frac{13}{4}\hat{i} + \frac{7}{4}\hat{j} - 5\hat{k} \]

### Step 6: Calculate the vector AD
Now, we find \( \vec{AD} \):
\[ \vec{AD} = \vec{D} - \vec{A} = \left(-\frac{13}{4}\hat{i} + \frac{7}{4}\hat{j} - 5\hat{k}\right) - \left(\hat{i} - \hat{j} - 3\hat{k}\right) \]
\[ = \left(-\frac{13}{4} - 1\right)\hat{i} + \left(\frac{7}{4} + 1\right)\hat{j} + \left(-5 + 3\right)\hat{k} \]
\[ = \left(-\frac{13}{4} - \frac{4}{4}\right)\hat{i} + \left(\frac{7}{4} + \frac{4}{4}\right)\hat{j} - 2\hat{k} \]
\[ = -\frac{17}{4}\hat{i} + \frac{11}{4}\hat{j} - 2\hat{k} \]

### Step 7: Calculate the length of AD
Finally, we calculate the length of \( \vec{AD} \):
\[ |\vec{AD}| = \sqrt{\left(-\frac{17}{4}\right)^2 + \left(\frac{11}{4}\right)^2 + (-2)^2} \]
\[ = \sqrt{\frac{289}{16} + \frac{121}{16} + 4} = \sqrt{\frac{289 + 121 + 64}{16}} = \sqrt{\frac{474}{16}} = \frac{\sqrt{474}}{4} = \frac{3\sqrt{6}}{4} \]

Thus, the length of the bisector AD is \( \frac{3\sqrt{6}}{4} \).