Find the shortest distance between the following (1-4) lines whose vector equations are :
(i) `vec(r) = (lambda - 1) hati + (lambda -1) hatj - (1 + lambda ) hatk `
and `vec(r) = (1 - mu) hat(i) + (2 mu - 1) hatj + (mu + 2) hatk`
(ii) `vec(r) = (1 + lambda) hati + (2 - lambda) hatj + (1 + lambda) hatk`
and `vec(r) = 2 (1 + mu) hati - (1 - mu) hatj + (-1 + 2mu ) hatk` .

Answer

To find the shortest distance between the given lines, we will follow 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 direction vectors of the first line,
- \(\vec{c}\) and \(\vec{d}\) are direction vectors of the second line,
- \(\vec{c} - \vec{a}\) is the vector connecting a point on the first line to a point on the second line,
- \(\vec{b} \times \vec{d}\) is the cross product of the direction vectors.

### Part (i)

1. **Identify the lines and extract the vectors**:
- Line 1: \(\vec{r_1} = (\lambda - 1) \hat{i} + (\lambda - 1) \hat{j} - (1 + \lambda) \hat{k}\)
- Here, \(\vec{a} = (-1, -1, -1)\) and \(\vec{b} = (1, 1, -1)\)
- Line 2: \(\vec{r_2} = (1 - \mu) \hat{i} + (2\mu - 1) \hat{j} + (\mu + 2) \hat{k}\)
- Here, \(\vec{c} = (1, -1, 2)\) and \(\vec{d} = (-1, 2, 1)\)

2. **Calculate \(\vec{c} - \vec{a}\)**:
\[ \vec{c} - \vec{a} = (1 - (-1), -1 - (-1), 2 - (-1)) = (2, 0, 3) \]

3. **Calculate \(\vec{b} \times \vec{d}\)**:
\[ \vec{b} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -1 \\ -1 & 2 & 1 \end{vmatrix} = \hat{i}(1 \cdot 1 - (-1) \cdot 2) - \hat{j}(1 \cdot 1 - (-1) \cdot -1) + \hat{k}(1 \cdot 2 - 1 \cdot -1) \]
\[ = \hat{i}(1 + 2) - \hat{j}(1 - 1) + \hat{k}(2 + 1) = 3\hat{i} + 0\hat{j} + 3\hat{k} = (3, 0, 3) \]

4. **Calculate the dot product \((\vec{c} - \vec{a}) \cdot (\vec{b} \times \vec{d})\)**:
\[ (2, 0, 3) \cdot (3, 0, 3) = 2 \cdot 3 + 0 \cdot 0 + 3 \cdot 3 = 6 + 0 + 9 = 15 \]

5. **Calculate the magnitude of \(\vec{b} \times \vec{d}\)**:
\[ |\vec{b} \times \vec{d}| = \sqrt{3^2 + 0^2 + 3^2} = \sqrt{9 + 0 + 9} = \sqrt{18} = 3\sqrt{2} \]

6. **Calculate the shortest distance \(d_{min}\)**:
\[ d_{min} = \frac{|15|}{3\sqrt{2}} = \frac{15}{3\sqrt{2}} = \frac{5}{\sqrt{2}} \]

### Part (ii)

1. **Identify the lines and extract the vectors**:
- Line 1: \(\vec{r_1} = (1 + \lambda) \hat{i} + (2 - \lambda) \hat{j} + (1 + \lambda) \hat{k}\)
- Here, \(\vec{a} = (1, 2, 1)\) and \(\vec{b} = (1, -1, 1)\)
- Line 2: \(\vec{r_2} = 2(1 + \mu) \hat{i} - (1 - \mu) \hat{j} + (-1 + 2\mu) \hat{k}\)
- Here, \(\vec{c} = (2, -1, -1)\) and \(\vec{d} = (2, 1, 2)\)

2. **Calculate \(\vec{c} - \vec{a}\)**:
\[ \vec{c} - \vec{a} = (2 - 1, -1 - 2, -1 - 1) = (1, -3, -2) \]

3. **Calculate \(\vec{b} \times \vec{d}\)**:
\[ \vec{b} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{vmatrix} = \hat{i}((-1) \cdot 2 - 1 \cdot 1) - \hat{j}(1 \cdot 2 - 1 \cdot 2) + \hat{k}(1 \cdot 1 - (-1) \cdot 2) \]
\[ = \hat{i}(-2 - 1) - \hat{j}(2 - 2) + \hat{k}(1 + 2) = -3\hat{i} + 0\hat{j} + 3\hat{k} = (-3, 0, 3) \]

4. **Calculate the dot product \((\vec{c} - \vec{a}) \cdot (\vec{b} \times \vec{d})\)**:
\[ (1, -3, -2) \cdot (-3, 0, 3) = 1 \cdot (-3) + (-3) \cdot 0 + (-2) \cdot 3 = -3 + 0 - 6 = -9 \]

5. **Calculate the magnitude of \(\vec{b} \times \vec{d}\)**:
\[ |\vec{b} \times \vec{d}| = \sqrt{(-3)^2 + 0^2 + 3^2} = \sqrt{9 + 0 + 9} = \sqrt{18} = 3\sqrt{2} \]

6. **Calculate the shortest distance \(d_{min}\)**:
\[ d_{min} = \frac{|-9|}{3\sqrt{2}} = \frac{9}{3\sqrt{2}} = \frac{3}{\sqrt{2}} \]

### Final Answers:
- Shortest distance for part (i): \(\frac{5}{\sqrt{2}}\)
- Shortest distance for part (ii): \(\frac{3}{\sqrt{2}}\)