If ` 4 hati + 7 hat j + 8 hatk, 2 hati +3 hatj + 4 hatk and 2 hati + 5 hatj + 7 hatk ` are the position vectors of the vertices A, B and C respectively of triangle ABC . The position vector of the point where the bisector of angle A meets BC, is

Options

A. `(2)/(3) (-6hati - 8 hatj - 6hatk)`
B. `(2)/(3) (6hati + 8 hatj + 6hatk)`
C. `(1)/(3) (6hati + 13 hatj + 18hatk)`
D. `(1)/(3) (5 hatj + 12hatk)`

Answer

To find the position vector of the point where the bisector of angle A meets line segment BC in triangle ABC, we can follow these steps:

### Step 1: Identify the Position Vectors
The position vectors of the vertices A, B, and C are given as:
- \( \vec{A} = 4\hat{i} + 7\hat{j} + 8\hat{k} \)
- \( \vec{B} = 2\hat{i} + 3\hat{j} + 4\hat{k} \)
- \( \vec{C} = 2\hat{i} + 5\hat{j} + 7\hat{k} \)

### Step 2: Calculate the Lengths of Sides AB and AC
To find the lengths of sides AB and AC, we can use the distance formula for vectors.

**Length of AB:**
\[ \vec{AB} = \vec{B} - \vec{A} = (2\hat{i} + 3\hat{j} + 4\hat{k}) - (4\hat{i} + 7\hat{j} + 8\hat{k}) = (-2\hat{i} - 4\hat{j} - 4\hat{k}) \]
\[ |\vec{AB}| = \sqrt{(-2)^2 + (-4)^2 + (-4)^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6 \]

**Length of AC:**
\[ \vec{AC} = \vec{C} - \vec{A} = (2\hat{i} + 5\hat{j} + 7\hat{k}) - (4\hat{i} + 7\hat{j} + 8\hat{k}) = (-2\hat{i} - 2\hat{j} - 1\hat{k}) \]
\[ |\vec{AC}| = \sqrt{(-2)^2 + (-2)^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \]

### Step 3: Apply the Angle Bisector Theorem
According to the angle bisector theorem, the point D where the bisector of angle A meets BC divides BC in the ratio of the lengths of sides AB and AC.

Thus,
\[ \frac{BD}{DC} = \frac{|\vec{AC}|}{|\vec{AB}|} = \frac{3}{6} = \frac{1}{2} \]

### Step 4: Find the Position Vector of Point D
Using the section formula, the position vector of point D can be calculated as:
\[ \vec{D} = \frac{m\vec{C} + n\vec{B}}{m+n} \]
where \( m = 1 \) and \( n = 2 \).

Substituting the values:
\[ \vec{D} = \frac{1(2\hat{i} + 5\hat{j} + 7\hat{k}) + 2(2\hat{i} + 3\hat{j} + 4\hat{k})}{1 + 2} \]
\[ = \frac{(2\hat{i} + 5\hat{j} + 7\hat{k}) + (4\hat{i} + 6\hat{j} + 8\hat{k})}{3} \]
\[ = \frac{(6\hat{i} + 11\hat{j} + 15\hat{k})}{3} = 2\hat{i} + \frac{11}{3}\hat{j} + 5\hat{k} \]

### Final Answer
The position vector of the point where the bisector of angle A meets BC is:
\[ \vec{D} = 2\hat{i} + \frac{11}{3}\hat{j} + 5\hat{k} \]