The position vectors of three vectors A, B and C are given to be `hat(i)+3hat(j)+3hat(k), 4hat(i)+4hat(k)` and `-2hat(i)+4hat(j)+2hat(k)` respectively. Find the angle between `vec(AB)` and `vec(AC)`.

Answer

To find the angle between the vectors \(\vec{AB}\) and \(\vec{AC}\), we will follow these steps:

### Step 1: Identify the position vectors
The position vectors of points A, B, and C are given as:
- \(\vec{A} = \hat{i} + 3\hat{j} + 3\hat{k}\)
- \(\vec{B} = 4\hat{i} + 0\hat{j} + 4\hat{k}\)
- \(\vec{C} = -2\hat{i} + 4\hat{j} + 2\hat{k}\)

### Step 2: Calculate the vectors \(\vec{AB}\) and \(\vec{AC}\)
The vector \(\vec{AB}\) is given by:
\[ \vec{AB} = \vec{B} - \vec{A} \]
Substituting the position vectors:
\[ \vec{AB} = (4\hat{i} + 0\hat{j} + 4\hat{k}) - (\hat{i} + 3\hat{j} + 3\hat{k}) \]
Calculating this gives:
\[ \vec{AB} = (4 - 1)\hat{i} + (0 - 3)\hat{j} + (4 - 3)\hat{k} = 3\hat{i} - 3\hat{j} + 1\hat{k} \]

The vector \(\vec{AC}\) is given by:
\[ \vec{AC} = \vec{C} - \vec{A} \]
Substituting the position vectors:
\[ \vec{AC} = (-2\hat{i} + 4\hat{j} + 2\hat{k}) - (\hat{i} + 3\hat{j} + 3\hat{k}) \]
Calculating this gives:
\[ \vec{AC} = (-2 - 1)\hat{i} + (4 - 3)\hat{j} + (2 - 3)\hat{k} = -3\hat{i} + 1\hat{j} - 1\hat{k} \]

### Step 3: Calculate the dot product \(\vec{AB} \cdot \vec{AC}\)
The dot product is given by:
\[ \vec{AB} \cdot \vec{AC} = (3\hat{i} - 3\hat{j} + 1\hat{k}) \cdot (-3\hat{i} + 1\hat{j} - 1\hat{k}) \]
Calculating this gives:
\[ = 3 \cdot (-3) + (-3) \cdot 1 + 1 \cdot (-1) = -9 - 3 - 1 = -13 \]

### Step 4: Calculate the magnitudes of \(\vec{AB}\) and \(\vec{AC}\)
The magnitude of \(\vec{AB}\) is:
\[ |\vec{AB}| = \sqrt{(3)^2 + (-3)^2 + (1)^2} = \sqrt{9 + 9 + 1} = \sqrt{19} \]

The magnitude of \(\vec{AC}\) is:
\[ |\vec{AC}| = \sqrt{(-3)^2 + (1)^2 + (-1)^2} = \sqrt{9 + 1 + 1} = \sqrt{11} \]

### Step 5: Use the dot product to find the cosine of the angle
Using the formula:
\[ \vec{AB} \cdot \vec{AC} = |\vec{AB}| |\vec{AC}| \cos \theta \]
Substituting the values we found:
\[ -13 = \sqrt{19} \cdot \sqrt{11} \cdot \cos \theta \]
Thus,
\[ \cos \theta = \frac{-13}{\sqrt{19} \cdot \sqrt{11}} \]

### Step 6: Find the angle \(\theta\)
To find \(\theta\), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{-13}{\sqrt{19} \cdot \sqrt{11}}\right) \]

### Final Answer
The angle between vectors \(\vec{AB}\) and \(\vec{AC}\) is:
\[ \theta = \cos^{-1}\left(\frac{-13}{\sqrt{19} \cdot \sqrt{11}}\right) \]