The vector ` vec c `, directed along the internal bisector of the angle between the vectors `vec a = 7 hati - 4 hatj - 4hatk and vecb = -2hati - hatj + 2 hatk " with " |vec c| = 5 sqrt(6),` is

Options

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

Explanation

To find the vector \( \vec{c} \) directed along the internal bisector of the angle between the vectors \( \vec{a} \) and \( \vec{b} \), we follow these steps:

### Step 1: Define the vectors
Given:
\[ \vec{a} = 7 \hat{i} - 4 \hat{j} - 4 \hat{k} \]
\[ \vec{b} = -2 \hat{i} - \hat{j} + 2 \hat{k} \]

### Step 2: Calculate the magnitudes of \( \vec{a} \) and \( \vec{b} \)
The magnitude of \( \vec{a} \):
\[ |\vec{a}| = \sqrt{7^2 + (-4)^2 + (-4)^2} = \sqrt{49 + 16 + 16} = \sqrt{81} = 9 \]

The magnitude of \( \vec{b} \):
\[ |\vec{b}| = \sqrt{(-2)^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]

### Step 3: Find the unit vectors \( \hat{a} \) and \( \hat{b} \)
The unit vector \( \hat{a} \):
\[ \hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{7 \hat{i} - 4 \hat{j} - 4 \hat{k}}{9} = \frac{7}{9} \hat{i} - \frac{4}{9} \hat{j} - \frac{4}{9} \hat{k} \]

The unit vector \( \hat{b} \):
\[ \hat{b} = \frac{\vec{b}}{|\vec{b}|} = \frac{-2 \hat{i} - \hat{j} + 2 \hat{k}}{3} = -\frac{2}{3} \hat{i} - \frac{1}{3} \hat{j} + \frac{2}{3} \hat{k} \]

### Step 4: Find the direction of the internal bisector
The direction of the internal bisector \( \vec{c} \) can be expressed as:
\[ \vec{c} = k (\hat{a} + \hat{b}) \]
where \( k \) is a constant.

### Step 5: Calculate \( \hat{a} + \hat{b} \)
\[ \hat{a} + \hat{b} = \left( \frac{7}{9} - \frac{2}{3} \right) \hat{i} + \left( -\frac{4}{9} - \frac{1}{3} \right) \hat{j} + \left( -\frac{4}{9} + \frac{2}{3} \right) \hat{k} \]

Calculating each component:
- For \( \hat{i} \):
\[ \frac{7}{9} - \frac{2}{3} = \frac{7}{9} - \frac{6}{9} = \frac{1}{9} \]

- For \( \hat{j} \):
\[ -\frac{4}{9} - \frac{1}{3} = -\frac{4}{9} - \frac{3}{9} = -\frac{7}{9} \]

- For \( \hat{k} \):
\[ -\frac{4}{9} + \frac{2}{3} = -\frac{4}{9} + \frac{6}{9} = \frac{2}{9} \]

Thus,
\[ \hat{a} + \hat{b} = \frac{1}{9} \hat{i} - \frac{7}{9} \hat{j} + \frac{2}{9} \hat{k} \]

### Step 6: Express \( \vec{c} \)
\[ \vec{c} = k \left( \frac{1}{9} \hat{i} - \frac{7}{9} \hat{j} + \frac{2}{9} \hat{k} \right) \]

### Step 7: Find the magnitude of \( \vec{c} \)
The magnitude of \( \vec{c} \) is given as \( 5 \sqrt{6} \):
\[ |\vec{c}| = |k| \cdot \frac{1}{9} \sqrt{1^2 + (-7)^2 + 2^2} = |k| \cdot \frac{1}{9} \sqrt{1 + 49 + 4} = |k| \cdot \frac{1}{9} \sqrt{54} = |k| \cdot \frac{3\sqrt{6}}{9} = |k| \cdot \frac{\sqrt{6}}{3} \]

Setting this equal to \( 5 \sqrt{6} \):
\[ |k| \cdot \frac{\sqrt{6}}{3} = 5 \sqrt{6} \]

### Step 8: Solve for \( k \)
\[ |k| = 5 \cdot 3 = 15 \]

### Step 9: Substitute \( k \) back into \( \vec{c} \)
\[ \vec{c} = 15 \left( \frac{1}{9} \hat{i} - \frac{7}{9} \hat{j} + \frac{2}{9} \hat{k} \right) = \frac{15}{9} \hat{i} - \frac{105}{9} \hat{j} + \frac{30}{9} \hat{k} \]
Simplifying:
\[ \vec{c} = \frac{5}{3} \hat{i} - 7 \hat{j} + \frac{10}{3} \hat{k} \]

### Final Answer
Thus, the vector \( \vec{c} \) is:
\[ \vec{c} = \frac{5}{3} \hat{i} - 7 \hat{j} + \frac{10}{3} \hat{k} \]