Find the shortest distance between the lines: `(i) vec(r) = 3 hati + 8 hat(j) + 3 hatk + lambda (3 hati - hatj + hatk) `
`and vec(r) = - 3 hat(i) - 7 hatj + 6 hatk + mu (-3 hati + 2 hatj + 4 hatk ) `
(ii) `( hati - hatj + 2 hatk) + lambda ( -2 hati + hatj + 3 hatk )`
`and (2 hati + 3 hatj - hatk) + mu (3 hati - 2 hatj + 2 hatk). `
(iii) `vec(r) = (hati + 2 hatj + 3 hatk) + lambda ( hati - 3 hatj + 2 hatk )`
and `vec(r) = (4 hati + 5 hatj + 6 hatk) + mu (2 hati + 3 hatj + hatk)`.

Answer

To find the shortest distance between the given lines, we will use the formula for the shortest distance between two skew lines represented 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 direction vectors of the first line,
- \(\vec{C}\) and \(\vec{D}\) are direction vectors of the second line.

### (i) Find the shortest distance between the lines:
1. **Identify the vectors:**
- For the first line:
\[ \vec{A} = 3\hat{i} + 8\hat{j} + 3\hat{k}, \quad \vec{B} = 3\hat{i} - \hat{j} + \hat{k} \]
- For the second line:
\[ \vec{C} = -3\hat{i} - 7\hat{j} + 6\hat{k}, \quad \vec{D} = -3\hat{i} + 2\hat{j} + 4\hat{k} \]

2. **Calculate \(\vec{C} - \vec{A}\):**
\[ \vec{C} - \vec{A} = (-3 - 3)\hat{i} + (-7 - 8)\hat{j} + (6 - 3)\hat{k} = -6\hat{i} - 15\hat{j} + 3\hat{k} \]

3. **Calculate \(\vec{B} \times \vec{D}\):**
\[ \vec{B} \times \vec{D} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 1 \\ -3 & 2 & 4 \end{vmatrix} \]
- Expanding the determinant:
\[ = \hat{i}((-1)(4) - (1)(2)) - \hat{j}((3)(4) - (1)(-3)) + \hat{k}((3)(2) - (-1)(-3)) \]
\[ = \hat{i}(-4 - 2) - \hat{j}(12 + 3) + \hat{k}(6 - 3) = -6\hat{i} - 15\hat{j} + 3\hat{k} \]

4. **Calculate the dot product \((\vec{C} - \vec{A}) \cdot (\vec{B} \times \vec{D})\):**
\[ = (-6)(-6) + (-15)(-15) + (3)(3) = 36 + 225 + 9 = 270 \]

5. **Calculate the magnitude \(|\vec{B} \times \vec{D}|\):**
\[ |\vec{B} \times \vec{D}| = \sqrt{(-6)^2 + (-15)^2 + 3^2} = \sqrt{36 + 225 + 9} = \sqrt{270} = 3\sqrt{30} \]

6. **Calculate the shortest distance \(D_{min}\):**
\[ D_{min} = \frac{|270|}{3\sqrt{30}} = \frac{90}{\sqrt{30}} = 3\sqrt{10} \]