If `hat(i) + hatj + hatk, 2 hati + 5 hatj , 3 hati + 2 hatj - 3 hatk and hati - 6 hatj - hatk ` respectively are the position vectors of points A, B , C and D, then find the angle between the straight lines AB and CD. Find whether `bar(AB) and bar(CD)` are collinear or not.

Answer

To solve the problem, we need to find the angle between the lines AB and CD formed by the given position vectors of points A, B, C, and D. We will also check if the vectors AB and CD are collinear.

### Step 1: Identify the position vectors
The position vectors are given as:
- \( \vec{A} = \hat{i} + \hat{j} + \hat{k} \)
- \( \vec{B} = 2\hat{i} + 5\hat{j} \)
- \( \vec{C} = 3\hat{i} + 2\hat{j} - 3\hat{k} \)
- \( \vec{D} = \hat{i} - 6\hat{j} - \hat{k} \)

### Step 2: Find the vector AB
The vector \( \vec{AB} \) can be calculated as:
\[ \vec{AB} = \vec{B} - \vec{A} = (2\hat{i} + 5\hat{j}) - (\hat{i} + \hat{j} + \hat{k}) \]
Calculating this gives:
\[ \vec{AB} = (2 - 1)\hat{i} + (5 - 1)\hat{j} + (0 - 1)\hat{k} = \hat{i} + 4\hat{j} - \hat{k} \]

### Step 3: Find the vector CD
The vector \( \vec{CD} \) can be calculated as:
\[ \vec{CD} = \vec{D} - \vec{C} = (\hat{i} - 6\hat{j} - \hat{k}) - (3\hat{i} + 2\hat{j} - 3\hat{k}) \]
Calculating this gives:
\[ \vec{CD} = (1 - 3)\hat{i} + (-6 - 2)\hat{j} + (0 + 3)\hat{k} = -2\hat{i} - 8\hat{j} + 2\hat{k} \]

### Step 4: Check for collinearity
To check if \( \vec{AB} \) and \( \vec{CD} \) are collinear, we can see if one is a scalar multiple of the other. We can express \( \vec{CD} \) in terms of \( \vec{AB} \):
\[ \vec{CD} = -2(\hat{i} + 4\hat{j} - \hat{k}) = -2\vec{AB} \]
Since \( \vec{CD} \) is a scalar multiple of \( \vec{AB} \), they are collinear.

### Step 5: Find the angle between AB and CD
The angle \( \theta \) between two vectors can be found using the dot product formula:
\[ \vec{AB} \cdot \vec{CD} = |\vec{AB}| |\vec{CD}| \cos \theta \]
Calculating the dot product:
\[ \vec{AB} \cdot \vec{CD} = (\hat{i} + 4\hat{j} - \hat{k}) \cdot (-2\hat{i} - 8\hat{j} + 2\hat{k}) = (-2)(1) + (4)(-8) + (-1)(2) = -2 - 32 - 2 = -36 \]

Calculating the magnitudes:
\[ |\vec{AB}| = \sqrt{1^2 + 4^2 + (-1)^2} = \sqrt{1 + 16 + 1} = \sqrt{18} = 3\sqrt{2} \]
\[ |\vec{CD}| = \sqrt{(-2)^2 + (-8)^2 + 2^2} = \sqrt{4 + 64 + 4} = \sqrt{72} = 6\sqrt{2} \]

Now substituting into the dot product formula:
\[ -36 = (3\sqrt{2})(6\sqrt{2}) \cos \theta \]
\[ -36 = 36 \cos \theta \implies \cos \theta = -1 \]
Thus, \( \theta = 180^\circ \).

### Conclusion
The angle between the lines AB and CD is \( 180^\circ \), and the vectors \( \vec{AB} \) and \( \vec{CD} \) are collinear.