If the position vectors of the vertices A,B and C of a `triangleABC` are respectively `4hati + 7hatj + 8hatk, 2hati + 4hatk` and `2hati + 5 hatj + 7hatk`,then the positions vector of the point, where the bisector of `angleA` meets BC is:

Options

A. `1/2(4hati + 8hatj + 11hatk)`
B. `1/3(6hati + 13hatj + 18hatk)`
C. `1/4(8hati + 14hatj + 19hatk)`
D. `1/3(6hati + 11hatj + 15hatk)`

Answer

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

### Step 1: Identify the position vectors of points A, B, and C.
Given:
- Position vector of A, \( \vec{A} = 4\hat{i} + 7\hat{j} + 8\hat{k} \)
- Position vector of B, \( \vec{B} = 2\hat{i} + 0\hat{j} + 4\hat{k} \)
- Position vector of C, \( \vec{C} = 2\hat{i} + 5\hat{j} + 7\hat{k} \)

### Step 2: Calculate the vectors AB and AC.
- \( \vec{AB} = \vec{B} - \vec{A} = (2\hat{i} + 0\hat{j} + 4\hat{k}) - (4\hat{i} + 7\hat{j} + 8\hat{k}) \)
- \( \vec{AB} = (2 - 4)\hat{i} + (0 - 7)\hat{j} + (4 - 8)\hat{k} = -2\hat{i} - 7\hat{j} - 4\hat{k} \)

- \( \vec{AC} = \vec{C} - \vec{A} = (2\hat{i} + 5\hat{j} + 7\hat{k}) - (4\hat{i} + 7\hat{j} + 8\hat{k}) \)
- \( \vec{AC} = (2 - 4)\hat{i} + (5 - 7)\hat{j} + (7 - 8)\hat{k} = -2\hat{i} - 2\hat{j} - 1\hat{k} \)

### Step 3: Calculate the magnitudes of vectors AB and AC.
- \( |\vec{AB}| = \sqrt{(-2)^2 + (-7)^2 + (-4)^2} = \sqrt{4 + 49 + 16} = \sqrt{69} \)
- \( |\vec{AC}| = \sqrt{(-2)^2 + (-2)^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \)

### Step 4: Use the angle bisector theorem to find the position vector of point D.
According to the angle bisector theorem:
\[ \vec{D} = \frac{|\vec{AC}| \cdot \vec{B} + |\vec{AB}| \cdot \vec{C}}{|\vec{AC}| + |\vec{AB}|} \]
Substituting the values:
\[ \vec{D} = \frac{3 \cdot (2\hat{i} + 0\hat{j} + 4\hat{k}) + \sqrt{69} \cdot (2\hat{i} + 5\hat{j} + 7\hat{k})}{3 + \sqrt{69}} \]

### Step 5: Calculate the components of vector D.
Calculating the numerator:
\[ 3 \cdot (2\hat{i} + 0\hat{j} + 4\hat{k}) = 6\hat{i} + 0\hat{j} + 12\hat{k} \]
\[ \sqrt{69} \cdot (2\hat{i} + 5\hat{j} + 7\hat{k}) = 2\sqrt{69}\hat{i} + 5\sqrt{69}\hat{j} + 7\sqrt{69}\hat{k} \]
Combining these:
\[ \vec{D} = \frac{(6 + 2\sqrt{69})\hat{i} + (0 + 5\sqrt{69})\hat{j} + (12 + 7\sqrt{69})\hat{k}}{3 + \sqrt{69}} \]

### Step 6: Finalize the position vector of point D.
Thus, the position vector of point D, where the angle bisector of angle A meets side BC, is:
\[ \vec{D} = \frac{(6 + 2\sqrt{69})\hat{i} + (5\sqrt{69})\hat{j} + (12 + 7\sqrt{69})\hat{k}}{3 + \sqrt{69}} \]