Find the shortest distance between the lines: `(i) vec(r) = 6 hat(i) + 2 hat(j) + 2 hatk + lambda (hati - 2hatj + 2 hatk)`
`and vec(r) = - 4 hati - hatk + mu (3 hati - 2 hatj - 2 hatk ) `
(ii) `vec(r) = (4 hat(i) - hat(j)) + lambda (hati + 2hatj - 3 hatk)`
`and vec(r) = (hati - hatj + 2hatk) + mu (2 hati + 4 hatj - 5 hatk ) `
(iii) `vec(r) = (hati + 2 hatj - 4 hatk) + lambda (2 hati + 3 hatj + 6 hatk ) `
and `vec(r) = (3 hati + 3 hatj + 5 hatk) + mu (-2 hati + 3 hatj + 6 hatk ) `

Answer

To find the shortest distance between the given lines, we will use the formula for the shortest distance between two skew lines in vector form. The formula is given by:

\[ d_{min} = \frac{|(\vec{C} - \vec{A}) \cdot (\vec{B} \times \vec{D})|}{|\vec{B} \times \vec{D}|} \]

Where:
- \(\vec{A}\) and \(\vec{B}\) are the position vectors of the first line,
- \(\vec{C}\) and \(\vec{D}\) are the position vectors of the second line,
- \(\cdot\) denotes the dot product,
- \(\times\) denotes the cross product.

### (i) Finding the shortest distance between the lines:

1. **Identify the vectors**:
- For the first line:
\[ \vec{A} = 6\hat{i} + 2\hat{j} + 2\hat{k}, \quad \vec{B} = \hat{i} - 2\hat{j} + 2\hat{k} \]
- For the second line:
\[ \vec{C} = -4\hat{i} - \hat{k}, \quad \vec{D} = 3\hat{i} - 2\hat{j} - 2\hat{k} \]

2. **Calculate \(\vec{C} - \vec{A}\)**:
\[ \vec{C} - \vec{A} = (-4 - 6)\hat{i} + (0 - 2)\hat{j} + (0 - 2)\hat{k} = -10\hat{i} - 2\hat{j} - 2\hat{k} \]

3. **Calculate \(\vec{B} \times \vec{D}\)**:
\[ \vec{B} \times \vec{D} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 3 & -2 & -2 \end{vmatrix} \]
\[ = \hat{i}((-2)(-2) - (2)(-2)) - \hat{j}((1)(-2) - (2)(3)) + \hat{k}((1)(-2) - (-2)(3)) \]
\[ = \hat{i}(4 + 4) - \hat{j}(-2 - 6) + \hat{k}(-2 + 6) \]
\[ = 8\hat{i} + 8\hat{j} + 4\hat{k} \]

4. **Calculate the dot product \((\vec{C} - \vec{A}) \cdot (\vec{B} \times \vec{D})\)**:
\[ (-10\hat{i} - 2\hat{j} - 2\hat{k}) \cdot (8\hat{i} + 8\hat{j} + 4\hat{k}) = (-10)(8) + (-2)(8) + (-2)(4) = -80 - 16 - 8 = -104 \]

5. **Calculate the magnitude of \(\vec{B} \times \vec{D}\)**:
\[ |\vec{B} \times \vec{D}| = \sqrt{8^2 + 8^2 + 4^2} = \sqrt{64 + 64 + 16} = \sqrt{144} = 12 \]

6. **Calculate the shortest distance**:
\[ d_{min} = \frac{| -104 |}{12} = \frac{104}{12} = \frac{26}{3} \approx 8.67 \]