Find the shortest distance between the lines
`vecr = hati+ hatj+hatk+lambda(3hati-hatj) and vecr=4hati-hatk+mu(2hati+3hatk)`

Answer

To find the shortest distance between the given lines, we will follow these steps:

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

### Step 2: Extract position vectors and direction vectors
From the equations of the lines:
- For line 1, the position vector \( \vec{a_1} = \hat{i} + \hat{j} + \hat{k} = (1, 1, 1) \)
- For line 2, the position vector \( \vec{a_2} = 4\hat{i} - \hat{k} = (4, 0, -1) \)

The direction vectors are:
- For line 1, \( \vec{b_1} = 3\hat{i} - \hat{j} = (3, -1, 0) \)
- For line 2, \( \vec{b_2} = 2\hat{i} + 3\hat{k} = (2, 0, 3) \)

### Step 3: Check if the lines are parallel
To check if the lines are parallel, we need to see if the direction vectors \( \vec{b_1} \) and \( \vec{b_2} \) are proportional. Since:
- \( \vec{b_1} = (3, -1, 0) \)
- \( \vec{b_2} = (2, 0, 3) \)

The direction ratios are not proportional, so the lines are not parallel.

### Step 4: Calculate \( \vec{a_2} - \vec{a_1} \)
Now, we calculate \( \vec{a_2} - \vec{a_1} \):
\[ \vec{a_2} - \vec{a_1} = (4, 0, -1) - (1, 1, 1) = (4 - 1, 0 - 1, -1 - 1) = (3, -1, -2) \]

### Step 5: Compute the cross product \( \vec{b_1} \times \vec{b_2} \)
We compute the cross product using the determinant:
\[ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 0 \\ 2 & 0 & 3 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} -1 & 0 \\ 0 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 0 \\ 2 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & -1 \\ 2 & 0 \end{vmatrix} \]
\[ = \hat{i}(-1 \cdot 3 - 0 \cdot 0) - \hat{j}(3 \cdot 3 - 0 \cdot 2) + \hat{k}(3 \cdot 0 - (-1) \cdot 2) \]
\[ = -3\hat{i} - 9\hat{j} + 2\hat{k} \]
Thus, \( \vec{b_1} \times \vec{b_2} = (-3, -9, 2) \).

### Step 6: Calculate the magnitude of \( \vec{b_1} \times \vec{b_2} \)
The magnitude is given by:
\[ |\vec{b_1} \times \vec{b_2}| = \sqrt{(-3)^2 + (-9)^2 + 2^2} = \sqrt{9 + 81 + 4} = \sqrt{94} \]

### Step 7: Calculate the dot product \( (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) \)
Now, we calculate the dot product:
\[ (3, -1, -2) \cdot (-3, -9, 2) = 3 \cdot (-3) + (-1) \cdot (-9) + (-2) \cdot 2 \]
\[ = -9 + 9 - 4 = -4 \]

### Step 8: Calculate the shortest distance
Using the formula for the distance between skew lines:
\[ \text{Distance} = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|} \]
Substituting the values:
\[ \text{Distance} = \frac{|-4|}{\sqrt{94}} = \frac{4}{\sqrt{94}} \]

### Final Answer
The shortest distance between the two lines is \( \frac{4}{\sqrt{94}} \).

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