If two vectors OA and OB are there, then their resultant OA+OB can be found by completing the parallelogram OACB and OC=OA+OB. Also, if |OA|=|OB|, then the resultant will bisect the angle between them.
Q. A vector C directed along internal bisector of angle between vectors `A=7hati-4hatj-4hatk and B=-2hati-hatj+2hatk` with `|C|=5sqrt(6)` is

Options

A. `(5)/(3)(hati-hatj+hatk)`
B. `(5)/(3)(hati-7hatj+2hatk)`
C. `(5)/(3)(5hati+5hatj+2hatk)`
D. `(5)/(3)(-5hati+5hatj+3hatk)`

Explanation

To solve the problem, we need to find the vector \( \mathbf{C} \) that is directed along the internal bisector of the angle between the vectors \( \mathbf{A} = 7\hat{i} - 4\hat{j} - 4\hat{k} \) and \( \mathbf{B} = -2\hat{i} - \hat{j} + 2\hat{k} \), given that \( |\mathbf{C}| = 5\sqrt{6} \).

### Step-by-Step Solution:

1. **Find the Magnitude of Vectors A and B:**
- For vector \( \mathbf{A} \):
\[ |\mathbf{A}| = \sqrt{7^2 + (-4)^2 + (-4)^2} = \sqrt{49 + 16 + 16} = \sqrt{81} = 9 \]
- For vector \( \mathbf{B} \):
\[ |\mathbf{B}| = \sqrt{(-2)^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]

2. **Express Vector C in terms of A and B:**
- The internal bisector of the angle between two vectors \( \mathbf{A} \) and \( \mathbf{B} \) can be expressed as:
\[ \mathbf{C} = k \left( \frac{\mathbf{A}}{|\mathbf{A}|} + \frac{\mathbf{B}}{|\mathbf{B}|} \right) \]
- Substituting the values:
\[ \mathbf{C} = k \left( \frac{7\hat{i} - 4\hat{j} - 4\hat{k}}{9} + \frac{-2\hat{i} - \hat{j} + 2\hat{k}}{3} \right) \]

3. **Finding a Common Denominator:**
- The common denominator for the fractions is 9:
\[ \mathbf{C} = k \left( \frac{7\hat{i} - 4\hat{j} - 4\hat{k}}{9} + \frac{-6\hat{i} - 3\hat{j} + 6\hat{k}}{9} \right) \]
- Combine the vectors:
\[ \mathbf{C} = k \left( \frac{(7 - 6)\hat{i} + (-4 - 3)\hat{j} + (-4 + 6)\hat{k}}{9} \right) = k \left( \frac{1\hat{i} - 7\hat{j} + 2\hat{k}}{9} \right) \]

4. **Magnitude of Vector C:**
- The magnitude of \( \mathbf{C} \) is given as \( 5\sqrt{6} \):
\[ |\mathbf{C}| = k \cdot \frac{1}{9} \sqrt{1^2 + (-7)^2 + 2^2} = k \cdot \frac{1}{9} \sqrt{1 + 49 + 4} = k \cdot \frac{1}{9} \sqrt{54} = k \cdot \frac{1}{9} \cdot 3\sqrt{6} = \frac{k\sqrt{6}}{3} \]
- Setting this equal to \( 5\sqrt{6} \):
\[ \frac{k\sqrt{6}}{3} = 5\sqrt{6} \]
- Dividing both sides by \( \sqrt{6} \):
\[ \frac{k}{3} = 5 \implies k = 15 \]

5. **Substituting k back to find C:**
- Now substituting \( k \) back into the expression for \( \mathbf{C} \):
\[ \mathbf{C} = 15 \cdot \frac{1}{9} \left( \hat{i} - 7\hat{j} + 2\hat{k} \right) = \frac{15}{9} \hat{i} - \frac{105}{9} \hat{j} + \frac{30}{9} \hat{k} \]
- Simplifying:
\[ \mathbf{C} = \frac{5}{3} \hat{i} - \frac{35}{3} \hat{j} + \frac{10}{3} \hat{k} \]

### Final Answer:
Thus, the vector \( \mathbf{C} \) directed along the internal bisector is:
\[ \mathbf{C} = \frac{5}{3} \hat{i} - 7 \hat{j} + 2 \hat{k} \]