Find the shortest distance between the following (1-4) lines whose vector equations are :
1. `vec(r) = hati + hatj + lambda (2 hati - hatj + hatk ) `
and `vec(r) = 2 hati + hatj - hatk + mu (3 hati - 5 hatj + 2 hatk)`.

Answer

To find the shortest distance between the two given lines represented by their vector equations, we can follow these steps:

### Step 1: Identify the vector equations of the lines
The vector equations of the lines are given as:
1. Line L1: \(\vec{r_1} = \hat{i} + \hat{j} + \lambda(2\hat{i} - \hat{j} + \hat{k})\)
2. Line L2: \(\vec{r_2} = 2\hat{i} + \hat{j} - \hat{k} + \mu(3\hat{i} - 5\hat{j} + 2\hat{k})\)

### Step 2: Identify the points and direction vectors
From the equations, we can identify:
- For Line L1:
- Point \(A\) (fixed point) = \(\hat{i} + \hat{j}\) (when \(\lambda = 0\))
- Direction vector \(b = 2\hat{i} - \hat{j} + \hat{k}\)

- For Line L2:
- Point \(C\) (fixed point) = \(2\hat{i} + \hat{j} - \hat{k}\) (when \(\mu = 0\))
- Direction vector \(d = 3\hat{i} - 5\hat{j} + 2\hat{k}\)

### Step 3: Find the vector \(C - A\)
We calculate the vector \(C - A\):
\[ C - A = (2\hat{i} + \hat{j} - \hat{k}) - (\hat{i} + \hat{j}) = (2 - 1)\hat{i} + (1 - 1)\hat{j} + (-1 - 0)\hat{k} = \hat{i} - \hat{k} \]

### Step 4: Calculate the cross product of direction vectors \(b\) and \(d\)
To find the cross product \(b \times d\), we set up the determinant:
\[ b = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}, \quad d = \begin{pmatrix} 3 \\ -5 \\ 2 \end{pmatrix} \]
\[ b \times d = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 3 & -5 & 2 \end{vmatrix} \]
Calculating this determinant:
\[ = \hat{i} \begin{vmatrix} -1 & 1 \\ -5 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 1 \\ 3 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -1 \\ 3 & -5 \end{vmatrix} \]
Calculating the 2x2 determinants:
\[ = \hat{i}((-1)(2) - (1)(-5)) - \hat{j}((2)(2) - (1)(3)) + \hat{k}((2)(-5) - (-1)(3)) \]
\[ = \hat{i}(-2 + 5) - \hat{j}(4 - 3) + \hat{k}(-10 + 3) \]
\[ = 3\hat{i} - 1\hat{j} - 7\hat{k} \]

### Step 5: Find the modulus of the cross product
Now, we calculate the modulus of \(b \times d\):
\[ |b \times d| = \sqrt{3^2 + (-1)^2 + (-7)^2} = \sqrt{9 + 1 + 49} = \sqrt{59} \]

### Step 6: Calculate the shortest distance using the formula
The formula for the shortest distance \(d\) between two skew lines is given by:
\[ d = \frac{|(C - A) \cdot (b \times d)|}{|b \times d|} \]
Calculating the dot product \((C - A) \cdot (b \times d)\):
\[ C - A = \hat{i} - \hat{k} \quad \text{and} \quad b \times d = 3\hat{i} - 1\hat{j} - 7\hat{k} \]
\[ (C - A) \cdot (b \times d) = (1)(3) + (0)(-1) + (-1)(-7) = 3 + 0 + 7 = 10 \]
Now substituting into the distance formula:
\[ d = \frac{|10|}{\sqrt{59}} = \frac{10}{\sqrt{59}} \]

### Final Answer
The shortest distance between the two lines is:
\[ \frac{10}{\sqrt{59}} \]