Find the shortest distance between the following lines :
(i) `vecr=4hati-hatj+lambda(hati+2hatj-3hatk)` and
`vecr=hati-hatj+2hatk+mu(2hati+4hatj-5hatk)`
(ii) `vecr=-hati+hatj-hatk+lambda(hati+hatj-hatk)` and
`vecr=hati-hatj+2hatk+mu(-hati+2hatj+hatk)`
(iii) `(x-1)/(-1) = (y+2)/(1) = (z-3)/(-2)` and
`(x-1)/(1) = (y+1)/(2) = (z+1)/(-2)`
(iv) `(x-1)/(2) = (y-2)/(3) = (z-3)/(4)` and `(x-2)/(3) = (y-3)/(4) = (z-5)/(5)`
(v) `vecr = veci+2hatj+3hatk+lambda(hati-hatj+hatk)` and
`vecr = 2hati-hatj-hatk+mu(-hati+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 in vector form. The formula is:

\[ d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|} \]

where:
- \(\vec{a_1}\) and \(\vec{a_2}\) are position vectors of points on the first and second lines, respectively.
- \(\vec{b_1}\) and \(\vec{b_2}\) are direction vectors of the first and second lines, respectively.
- \(\cdot\) denotes the dot product.
- \(\times\) denotes the cross product.

### (i)
**Lines:**
1. \(\vec{r_1} = 4\hat{i} - \hat{j} + \lambda(\hat{i} + 2\hat{j} - 3\hat{k})\)
2. \(\vec{r_2} = \hat{i} - \hat{j} + 2\hat{k} + \mu(2\hat{i} + 4\hat{j} - 5\hat{k})\)

**Step 1: Identify vectors**
- \(\vec{a_1} = 4\hat{i} - \hat{j}\)
- \(\vec{b_1} = \hat{i} + 2\hat{j} - 3\hat{k}\)
- \(\vec{a_2} = \hat{i} - \hat{j} + 2\hat{k}\)
- \(\vec{b_2} = 2\hat{i} + 4\hat{j} - 5\hat{k}\)

**Step 2: Calculate \(\vec{a_2} - \vec{a_1}\)**
\[ \vec{a_2} - \vec{a_1} = (\hat{i} - \hat{j} + 2\hat{k}) - (4\hat{i} - \hat{j}) = -3\hat{i} + 0\hat{j} - 2\hat{k} \]

**Step 3: Calculate \(\vec{b_1} \times \vec{b_2}\)**
\[ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{vmatrix} = \hat{i}(2 \cdot -5 - -3 \cdot 4) - \hat{j}(1 \cdot -5 - -3 \cdot 2) + \hat{k}(1 \cdot 4 - 2 \cdot 2) \]
\[ = \hat{i}(-10 + 12) - \hat{j}(-5 + 6) + \hat{k}(4 - 4) = 2\hat{i} - \hat{j} + 0\hat{k} \]

**Step 4: Calculate the magnitude of \(\vec{b_1} \times \vec{b_2}\)**
\[ |\vec{b_1} \times \vec{b_2}| = \sqrt{2^2 + (-1)^2 + 0^2} = \sqrt{4 + 1} = \sqrt{5} \]

**Step 5: Calculate the dot product \((\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})\)**
\[ (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (-3\hat{i} - 2\hat{k}) \cdot (2\hat{i} - \hat{j}) = -3 \cdot 2 + 0 \cdot -1 + (-2) \cdot 0 = -6 \]

**Step 6: Calculate the shortest distance**
\[ d = \frac{| -6 |}{\sqrt{5}} = \frac{6}{\sqrt{5}} \text{ units} \]