The vector `vec(r)` , directed along the external bisector of the angle between the vectors `vec(a) = 7hat(i) - 4hat(j) + 4hat(k)` and `vec(b) = 2hat(i) - hat(j) + 2hat(k)` with `|vec(r)| = 5sqrt6`, is

Options

A. `(5hat(i) - 5hat(j) - 10 hat(k))`
B. `(5)/(3)(hat(i) + 7hat(j) -2hat(k))`
C. `-(5vec(i) - 5hat(j) - 10hat(k))`
D. `(5)/(3) (-hat(i) + 7hat(j) + 2hat(k))`

Answer

To find the vector \(\vec{r}\) directed along the external bisector of the angle between the vectors \(\vec{a} = 7\hat{i} - 4\hat{j} + 4\hat{k}\) and \(\vec{b} = 2\hat{i} - \hat{j} + 2\hat{k}\) with \(|\vec{r}| = 5\sqrt{6}\), we can follow these steps:

### Step 1: Calculate the magnitudes of vectors \(\vec{a}\) and \(\vec{b}\)

The magnitude of vector \(\vec{a}\) is given by:
\[ |\vec{a}| = \sqrt{7^2 + (-4)^2 + 4^2} = \sqrt{49 + 16 + 16} = \sqrt{81} = 9 \]

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

### Step 2: Find the unit vectors of \(\vec{a}\) and \(\vec{b}\)

The unit vector of \(\vec{a}\) is:
\[ \hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{7\hat{i} - 4\hat{j} + 4\hat{k}}{9} \]

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

### Step 3: Determine the external angle bisector vector

The external angle bisector \(\vec{r}\) can be given by the formula:
\[ \vec{r} = \frac{\hat{a}}{|\hat{a}|} - \frac{\hat{b}}{|\hat{b}|} \]
Substituting the unit vectors:
\[ \vec{r} = \frac{7\hat{i} - 4\hat{j} + 4\hat{k}}{9} - \frac{2\hat{i} - \hat{j} + 2\hat{k}}{3} \]

To combine these, we need a common denominator:
\[ \vec{r} = \frac{7\hat{i} - 4\hat{j} + 4\hat{k}}{9} - \frac{6\hat{i} - 3\hat{j} + 6\hat{k}}{9} \]
\[ = \frac{(7 - 6)\hat{i} + (-4 + 3)\hat{j} + (4 - 6)\hat{k}}{9} \]
\[ = \frac{1\hat{i} - 1\hat{j} - 2\hat{k}}{9} \]

### Step 4: Scale the vector \(\vec{r}\) to the required magnitude

We know that \(|\vec{r}| = 5\sqrt{6}\). To find the vector with this magnitude, we can scale \(\vec{r}\):
\[ \vec{r} = k \left(\frac{1\hat{i} - 1\hat{j} - 2\hat{k}}{9}\right) \]
where \(k\) is a scaling factor. The magnitude of \(\vec{r}\) is:
\[ |\vec{r}| = k \cdot \frac{1}{9} \sqrt{1^2 + (-1)^2 + (-2)^2} = k \cdot \frac{1}{9} \sqrt{6} \]
Setting this equal to \(5\sqrt{6}\):
\[ k \cdot \frac{1}{9} \sqrt{6} = 5\sqrt{6} \]
\[ k = 5 \cdot 9 = 45 \]

### Step 5: Final vector \(\vec{r}\)

Thus, the vector \(\vec{r}\) is:
\[ \vec{r} = 45 \cdot \frac{1\hat{i} - 1\hat{j} - 2\hat{k}}{9} = 5(1\hat{i} - 1\hat{j} - 2\hat{k}) = 5\hat{i} - 5\hat{j} - 10\hat{k} \]

### Final Answer
\[ \vec{r} = 5\hat{i} - 5\hat{j} - 10\hat{k} \]