Find a vector of magnitude 11 in the direction opposite to that of `vec(PQ)`, where P and Q are the points (1, 3, 2) and `(-1,0,8)` respectively.

Answer

To find a vector of magnitude 11 in the direction opposite to that of the vector \(\vec{PQ}\), we will follow these steps:

### Step 1: Determine the coordinates of points P and Q
Given points:
- \( P(1, 3, 2) \)
- \( Q(-1, 0, 8) \)

### Step 2: Find the vector \(\vec{PQ}\)
The vector \(\vec{PQ}\) can be calculated using the formula:
\[ \vec{PQ} = \vec{Q} - \vec{P} \]
Substituting the coordinates:
\[ \vec{PQ} = (-1 - 1, 0 - 3, 8 - 2) = (-2, -3, 6) \]
So, \(\vec{PQ} = -2\hat{i} - 3\hat{j} + 6\hat{k}\).

### Step 3: Find the vector in the opposite direction, \(\vec{QP}\)
The vector in the opposite direction, \(\vec{QP}\), is simply the negative of \(\vec{PQ}\):
\[ \vec{QP} = -\vec{PQ} = (2, 3, -6) \]
Thus, \(\vec{QP} = 2\hat{i} + 3\hat{j} - 6\hat{k}\).

### Step 4: Calculate the magnitude of \(\vec{QP}\)
The magnitude of the vector \(\vec{QP}\) is given by:
\[ |\vec{QP}| = \sqrt{(2)^2 + (3)^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \]

### Step 5: Find the unit vector in the direction of \(\vec{QP}\)
The unit vector \(\hat{r}\) in the direction of \(\vec{QP}\) is given by:
\[ \hat{r} = \frac{\vec{QP}}{|\vec{QP}|} = \frac{2\hat{i} + 3\hat{j} - 6\hat{k}}{7} = \frac{2}{7}\hat{i} + \frac{3}{7}\hat{j} - \frac{6}{7}\hat{k} \]

### Step 6: Find the vector of magnitude 11 in the direction of \(\hat{r}\)
To find a vector of magnitude 11 in the direction of \(\hat{r}\), we multiply the unit vector by 11:
\[ \vec{V} = 11 \cdot \hat{r} = 11 \left( \frac{2}{7}\hat{i} + \frac{3}{7}\hat{j} - \frac{6}{7}\hat{k} \right) \]
Calculating this gives:
\[ \vec{V} = \frac{22}{7}\hat{i} + \frac{33}{7}\hat{j} - \frac{66}{7}\hat{k} \]

### Final Answer
Thus, the required vector of magnitude 11 in the direction opposite to that of \(\vec{PQ}\) is:
\[ \vec{V} = \frac{22}{7}\hat{i} + \frac{33}{7}\hat{j} - \frac{66}{7}\hat{k} \]