The vertices of a triangle are A(1,1,2), B (4,3,1) and C (2,3,5). The vector representing internal bisector of the angle A is

Options

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

Explanation

To find the vector representing the internal bisector of angle A in triangle ABC with vertices A(1, 1, 2), B(4, 3, 1), and C(2, 3, 5), we will follow these steps:

### Step 1: Calculate the lengths of sides AB, AC, and BC

1. **Length of AB**:
\[ AB = \sqrt{(4 - 1)^2 + (3 - 1)^2 + (1 - 2)^2} = \sqrt{3^2 + 2^2 + (-1)^2} = \sqrt{9 + 4 + 1} = \sqrt{14} \]

2. **Length of AC**:
\[ AC = \sqrt{(2 - 1)^2 + (3 - 1)^2 + (5 - 2)^2} = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \]

3. **Length of BC**:
\[ BC = \sqrt{(2 - 4)^2 + (3 - 3)^2 + (5 - 1)^2} = \sqrt{(-2)^2 + 0^2 + 4^2} = \sqrt{4 + 0 + 16} = \sqrt{20} \]

### Step 2: Identify that triangle ABC is isosceles

Since \( AB = AC = \sqrt{14} \), triangle ABC is isosceles with \( AB = AC \).

### Step 3: Find the midpoint D of side BC

To find the midpoint D of segment BC, we use the midpoint formula:
\[ D = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right) \]
Where \( B(4, 3, 1) \) and \( C(2, 3, 5) \):
\[ D = \left( \frac{4 + 2}{2}, \frac{3 + 3}{2}, \frac{1 + 5}{2} \right) = \left( \frac{6}{2}, \frac{6}{2}, \frac{6}{2} \right) = (3, 3, 3) \]

### Step 4: Find the vector AD

The vector AD can be calculated as:
\[ \vec{AD} = \vec{D} - \vec{A} \]
Where \( \vec{D} = 3i + 3j + 3k \) and \( \vec{A} = 1i + 1j + 2k \):
\[ \vec{AD} = (3i + 3j + 3k) - (1i + 1j + 2k) = (3 - 1)i + (3 - 1)j + (3 - 2)k = 2i + 2j + k \]

### Final Answer

Thus, the vector representing the internal bisector of angle A is:
\[ \vec{AD} = 2i + 2j + k \]

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