Find the shortest distance between the lines
`vecr=(1-lambda)hati+(lambda-2)hatj+(3-2lambda)hatk ,vecr=(mu+1)+(2mu-1)hatj-(2mu+1)hatk`

Answer

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

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

We can express these lines in the form:
- Line 1: \(\vec{r_1} = \vec{a_1} + \lambda \vec{b_1}\)
- Line 2: \(\vec{r_2} = \vec{a_2} + \mu \vec{b_2}\)

Where:
- \(\vec{a_1} = \hat{i} - 2\hat{j} + 3\hat{k}\), \(\vec{b_1} = -\hat{i} + \hat{j} - 2\hat{k}\)
- \(\vec{a_2} = \hat{i} - \hat{j} - \hat{k}\), \(\vec{b_2} = \hat{i} + 2\hat{j} - 2\hat{k}\)

### Step 2: Calculate the cross product of the direction vectors
Next, we need to find the cross product \(\vec{b_1} \times \vec{b_2}\).

\[ \vec{b_1} = \begin{pmatrix} -1 \\ 1 \\ -2 \end{pmatrix}, \quad \vec{b_2} = \begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix} \]

The cross product can be calculated using the determinant:

\[ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & -2 \\ 1 & 2 & -2 \end{vmatrix} \]

Calculating this determinant, we get:

\[ \vec{b_1} \times \vec{b_2} = \hat{i} \begin{vmatrix} 1 & -2 \\ 2 & -2 \end{vmatrix} - \hat{j} \begin{vmatrix} -1 & -2 \\ 1 & -2 \end{vmatrix} + \hat{k} \begin{vmatrix} -1 & 1 \\ 1 & 2 \end{vmatrix} \]

Calculating the determinants:

\[ = \hat{i} (1 \cdot -2 - 2 \cdot -2) - \hat{j} (-1 \cdot -2 - 1 \cdot -2) + \hat{k} (-1 \cdot 2 - 1 \cdot 1) \]
\[ = \hat{i} (-2 + 4) - \hat{j} (2 - 2) + \hat{k} (-2 - 1) \]
\[ = 2\hat{i} + 0\hat{j} - 3\hat{k} = 2\hat{i} - 3\hat{k} \]

### Step 3: Calculate the magnitude of the cross product
Now we find the magnitude of the cross product:

\[ |\vec{b_1} \times \vec{b_2}| = \sqrt{(2)^2 + (0)^2 + (-3)^2} = \sqrt{4 + 0 + 9} = \sqrt{13} \]

### Step 4: Find \(\vec{a_1} - \vec{a_2}\)
Now, we need to calculate \(\vec{a_1} - \vec{a_2}\):

\[ \vec{a_1} - \vec{a_2} = (\hat{i} - 2\hat{j} + 3\hat{k}) - (\hat{i} - \hat{j} - \hat{k}) \]
\[ = (1 - 1)\hat{i} + (-2 + 1)\hat{j} + (3 + 1)\hat{k} = 0\hat{i} - 1\hat{j} + 4\hat{k} = -\hat{j} + 4\hat{k} \]

### Step 5: Calculate the dot product
Now we calculate the dot product \((\vec{b_1} \times \vec{b_2}) \cdot (\vec{a_1} - \vec{a_2})\):

\[ (2\hat{i} - 3\hat{k}) \cdot (0\hat{i} - 1\hat{j} + 4\hat{k}) = 2 \cdot 0 + (-3) \cdot 4 = -12 \]

### Step 6: Calculate the shortest distance
Finally, we can find the shortest distance \(d\) using the formula:

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

Substituting the values we have:

\[ d = \frac{|-12|}{\sqrt{13}} = \frac{12}{\sqrt{13}} \]

Thus, the shortest distance between the two lines is:

\[ \boxed{\frac{12}{\sqrt{13}}} \]