Find the equations of the lines joining the following pair of vertices and then find its shortest distance between the lines :
(i) (0,0,0), (1,0,2) (ii) (1,3,0), (0,3,0).

Answer

To solve the problem, we need to find the equations of the lines joining the given pairs of vertices and then calculate the shortest distance between these two lines.

### Step 1: Find the equation of the first line joining the points (0, 0, 0) and (1, 0, 2).

1. **Identify the points**:
- Let point A = (0, 0, 0)
- Let point B = (1, 0, 2)

2. **Find the direction vector (AB)**:
\[ \text{AB} = B - A = (1 - 0, 0 - 0, 2 - 0) = (1, 0, 2) \]

3. **Write the equation of the line in vector form**:
The equation of the line can be expressed as:
\[ \mathbf{r} = \mathbf{A} + \lambda \mathbf{AB} \]
Here, \(\mathbf{A} = (0, 0, 0)\) and \(\mathbf{AB} = (1, 0, 2)\). Therefore, the equation becomes:
\[ \mathbf{r} = (0, 0, 0) + \lambda (1, 0, 2) = (\lambda, 0, 2\lambda) \]

4. **In symmetric form**:
From the vector form, we can also express the line in symmetric form:
\[ \frac{x - 0}{1} = \frac{y - 0}{0} = \frac{z - 0}{2} \]
This simplifies to:
\[ x = \lambda, \quad y = 0, \quad z = 2\lambda \]

### Step 2: Find the equation of the second line joining the points (1, 3, 0) and (0, 3, 0).

1. **Identify the points**:
- Let point C = (1, 3, 0)
- Let point D = (0, 3, 0)

2. **Find the direction vector (CD)**:
\[ \text{CD} = D - C = (0 - 1, 3 - 3, 0 - 0) = (-1, 0, 0) \]

3. **Write the equation of the line in vector form**:
The equation of the line can be expressed as:
\[ \mathbf{r} = \mathbf{C} + \mu \mathbf{CD} \]
Here, \(\mathbf{C} = (1, 3, 0)\) and \(\mathbf{CD} = (-1, 0, 0)\). Therefore, the equation becomes:
\[ \mathbf{r} = (1, 3, 0) + \mu (-1, 0, 0) = (1 - \mu, 3, 0) \]

4. **In symmetric form**:
From the vector form, we can also express the line in symmetric form:
\[ \frac{x - 1}{-1} = \frac{y - 3}{0} = \frac{z - 0}{0} \]
This simplifies to:
\[ x = 1 - \mu, \quad y = 3, \quad z = 0 \]

### Step 3: Calculate the shortest distance between the two lines.

1. **Identify the points and direction vectors**:
- For line 1:
- Point A = (0, 0, 0)
- Direction vector \(\mathbf{B} = (1, 0, 2)\)
- For line 2:
- Point C = (1, 3, 0)
- Direction vector \(\mathbf{D} = (-1, 0, 0)\)

2. **Use the formula for shortest distance between two skew lines**:
The shortest distance \(d\) between two lines can be calculated using:
\[ d = \frac{|(\mathbf{C} - \mathbf{A}) \cdot (\mathbf{B} \times \mathbf{D})|}{|\mathbf{B} \times \mathbf{D}|} \]

3. **Calculate \(\mathbf{C} - \mathbf{A}\)**:
\[ \mathbf{C} - \mathbf{A} = (1 - 0, 3 - 0, 0 - 0) = (1, 3, 0) \]

4. **Calculate \(\mathbf{B} \times \mathbf{D}\)**:
\[ \mathbf{B} \times \mathbf{D} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 2 \\ -1 & 0 & 0 \end{vmatrix} = (0 \cdot 0 - 2 \cdot 0) \mathbf{i} - (1 \cdot 0 - 2 \cdot (-1)) \mathbf{j} + (1 \cdot 0 - 0 \cdot (-1)) \mathbf{k} = (0, 2, 0) \]

5. **Calculate the magnitude of \(\mathbf{B} \times \mathbf{D}\)**:
\[ |\mathbf{B} \times \mathbf{D}| = \sqrt{0^2 + 2^2 + 0^2} = 2 \]

6. **Calculate the dot product \((\mathbf{C} - \mathbf{A}) \cdot (\mathbf{B} \times \mathbf{D})\)**:
\[ (1, 3, 0) \cdot (0, 2, 0) = 1 \cdot 0 + 3 \cdot 2 + 0 \cdot 0 = 6 \]

7. **Calculate the shortest distance**:
\[ d = \frac{|6|}{2} = 3 \]

### Final Answer:
The shortest distance between the two lines is \(3\) units.