The vertices of a triangles ABC are `A(-1,2,-3),B(5,0,-6),C(0,4,-1)`.Then the direction ratios of the external bisector of `/_ BAC` are :

Options

A. 11,20,21
B. `-11,20,20`
C. `-11,20,23`
D. none

Answer

To find the direction ratios of the external bisector of angle \( \angle BAC \) for the triangle with vertices \( A(-1, 2, -3) \), \( B(5, 0, -6) \), and \( C(0, 4, -1) \), we will follow these steps:

### Step 1: Find the vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \)

The vector \( \overrightarrow{AB} \) is calculated as:
\[ \overrightarrow{AB} = B - A = (5, 0, -6) - (-1, 2, -3) = (5 + 1, 0 - 2, -6 + 3) = (6, -2, -3) \]

The vector \( \overrightarrow{AC} \) is calculated as:
\[ \overrightarrow{AC} = C - A = (0, 4, -1) - (-1, 2, -3) = (0 + 1, 4 - 2, -1 + 3) = (1, 2, 2) \]

### Step 2: Find the magnitudes of the vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \)

The magnitude of \( \overrightarrow{AB} \) is:
\[ |\overrightarrow{AB}| = \sqrt{6^2 + (-2)^2 + (-3)^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7 \]

The magnitude of \( \overrightarrow{AC} \) is:
\[ |\overrightarrow{AC}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]

### Step 3: Find the direction ratios of the internal bisector of angle \( A \)

The direction ratios of the internal bisector of angle \( A \) can be found using the formula:
\[ \text{Direction Ratios} = |\overrightarrow{AC}| \cdot \overrightarrow{AB} + |\overrightarrow{AB}| \cdot \overrightarrow{AC} \]

Calculating:
\[ \text{Direction Ratios} = 3 \cdot (6, -2, -3) + 7 \cdot (1, 2, 2) \]
\[ = (18, -6, -9) + (7, 14, 14) = (18 + 7, -6 + 14, -9 + 14) = (25, 8, 5) \]

### Step 4: Find the direction ratios of the external bisector of angle \( A \)

The direction ratios of the external bisector can be found by taking the difference of the direction ratios of the internal bisector:
\[ \text{External Direction Ratios} = |\overrightarrow{AC}| \cdot \overrightarrow{AB} - |\overrightarrow{AB}| \cdot \overrightarrow{AC} \]
Calculating:
\[ \text{External Direction Ratios} = 3 \cdot (6, -2, -3) - 7 \cdot (1, 2, 2) \]
\[ = (18, -6, -9) - (7, 14, 14) = (18 - 7, -6 - 14, -9 - 14) = (11, -20, -23) \]

### Final Result

Thus, the direction ratios of the external bisector of \( \angle BAC \) are:
\[ \text{Direction Ratios} = (-11, 20, 23) \]