Find the shortest distance between the lines : `vec(r) = (4hat(i) - hat(j)) + lambda(hat(i) + 2hat(j) - 3hat(k))` and `vec(r) = (hat(i) - hat(j) + 2hat(k)) + mu (2hat(i) + 4hat(j) - 5hat(k))`

Answer

To find the shortest distance between the two lines given in vector form, we can follow these steps:

### Step 1: Identify the lines and their components
The two lines are given as:
1. \( \vec{r_1} = (4\hat{i} - \hat{j}) + \lambda(\hat{i} + 2\hat{j} - 3\hat{k}) \)
2. \( \vec{r_2} = (\hat{i} - \hat{j} + 2\hat{k}) + \mu(2\hat{i} + 4\hat{j} - 5\hat{k}) \)

From this, we can identify:
- Point on line 1: \( \vec{a_1} = 4\hat{i} - \hat{j} \)
- Direction vector of line 1: \( \vec{b_1} = \hat{i} + 2\hat{j} - 3\hat{k} \)
- Point on line 2: \( \vec{a_2} = \hat{i} - \hat{j} + 2\hat{k} \)
- Direction vector of line 2: \( \vec{b_2} = 2\hat{i} + 4\hat{j} - 5\hat{k} \)

### Step 2: Calculate \( \vec{a_2} - \vec{a_1} \)
Now, we calculate the vector from point \( \vec{a_1} \) to point \( \vec{a_2} \):
\[ \vec{a_2} - \vec{a_1} = (\hat{i} - \hat{j} + 2\hat{k}) - (4\hat{i} - \hat{j}) = (1 - 4)\hat{i} + (0)\hat{j} + (2 - 0)\hat{k} = -3\hat{i} + 2\hat{k} \]

### Step 3: Calculate \( \vec{b_1} \times \vec{b_2} \)
Next, we need to find the cross product of the direction vectors \( \vec{b_1} \) and \( \vec{b_2} \):
\[ \vec{b_1} = \hat{i} + 2\hat{j} - 3\hat{k}, \quad \vec{b_2} = 2\hat{i} + 4\hat{j} - 5\hat{k} \]
Using the determinant method:
\[ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{vmatrix} \]
Calculating this determinant:
\[ = \hat{i} \begin{vmatrix} 2 & -3 \\ 4 & -5 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & -3 \\ 2 & -5 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 2 \\ 2 & 4 \end{vmatrix} \]
Calculating each of these:
\[ = \hat{i} (2 \cdot -5 - (-3) \cdot 4) - \hat{j} (1 \cdot -5 - (-3) \cdot 2) + \hat{k} (1 \cdot 4 - 2 \cdot 2) \]
\[ = \hat{i} (-10 + 12) - \hat{j} (-5 + 6) + \hat{k} (4 - 4) \]
\[ = 2\hat{i} - 1\hat{j} + 0\hat{k} = 2\hat{i} - \hat{j} \]

### Step 4: Calculate the magnitude of \( \vec{b_1} \times \vec{b_2} \)
Now, we find the magnitude of the cross product:
\[ |\vec{b_1} \times \vec{b_2}| = \sqrt{(2)^2 + (-1)^2 + (0)^2} = \sqrt{4 + 1} = \sqrt{5} \]

### Step 5: Calculate the shortest distance
Using the formula for the shortest distance \( d \) between two skew lines:
\[ d = \frac{|(\vec{b_1} \times \vec{b_2}) \cdot (\vec{a_2} - \vec{a_1})|}{|\vec{b_1} \times \vec{b_2}|} \]
Substituting the values:
\[ (\vec{b_1} \times \vec{b_2}) \cdot (\vec{a_2} - \vec{a_1}) = (2\hat{i} - \hat{j}) \cdot (-3\hat{i} + 2\hat{k}) = 2 \cdot (-3) + (-1) \cdot 0 + 0 \cdot 2 = -6 \]
Thus, the distance becomes:
\[ d = \frac{|-6|}{\sqrt{5}} = \frac{6}{\sqrt{5}} \]

### Final Answer
The shortest distance between the two lines is \( \frac{6}{\sqrt{5}} \).