Find the shortest distance between the given line
`vec(r) =(3-t) hat(i)+(4+2t) hat(j) +(t-2) hat(k)`
`vec(r ) =(1+s) hat(i) +(3s -7) hat(j) + (2s-2) hat(k)`

Answer

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

### Step 1: Identify the vector equations of the lines
The lines are given as:
1. \(\vec{r_1} = (3-t) \hat{i} + (4+2t) \hat{j} + (t-2) \hat{k}\)
2. \(\vec{r_2} = (1+s) \hat{i} + (3s - 7) \hat{j} + (2s - 2) \hat{k}\)

From these equations, we can identify:
- For line 1:
- Point \(A_1 = (3, 4, -2)\)
- Direction vector \(B_1 = (-1, 2, 1)\)

- For line 2:
- Point \(A_2 = (1, -7, -2)\)
- Direction vector \(B_2 = (1, 3, 2)\)

### Step 2: Calculate \(A_2 - A_1\)
We need to find the vector \(A_2 - A_1\):
\[ A_2 - A_1 = (1 - 3) \hat{i} + (-7 - 4) \hat{j} + (-2 + 2) \hat{k} = (-2) \hat{i} + (-11) \hat{j} + (0) \hat{k} \]
Thus, \(A_2 - A_1 = -2 \hat{i} - 11 \hat{j}\).

### Step 3: Compute the cross product \(B_1 \times B_2\)
To find the cross product \(B_1 \times B_2\):
\[ B_1 = (-1, 2, 1), \quad B_2 = (1, 3, 2) \]
Using the determinant method:
\[ B_1 \times B_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 1 \\ 1 & 3 & 2 \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} 2 & 1 \\ 3 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} -1 & 1 \\ 1 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} -1 & 2 \\ 1 & 3 \end{vmatrix} \]
Calculating each of these:
\[ = \hat{i} (2 \cdot 2 - 3 \cdot 1) - \hat{j} (-1 \cdot 2 - 1 \cdot 1) + \hat{k} (-1 \cdot 3 - 2 \cdot 1) \]
\[ = \hat{i} (4 - 3) - \hat{j} (-2 - 1) + \hat{k} (-3 - 2) \]
\[ = \hat{i} (1) + \hat{j} (3) - \hat{k} (5) \]
Thus, \(B_1 \times B_2 = \hat{i} + 3\hat{j} - 5\hat{k}\).

### Step 4: Calculate the magnitude of \(B_1 \times B_2\)
\[ |B_1 \times B_2| = \sqrt{1^2 + 3^2 + (-5)^2} = \sqrt{1 + 9 + 25} = \sqrt{35} \]

### Step 5: Use the formula for the shortest distance
The formula for the shortest distance \(d\) between two skew lines is given by:
\[ d = \frac{|(A_2 - A_1) \cdot (B_1 \times B_2)|}{|B_1 \times B_2|} \]
Calculating the dot product \((A_2 - A_1) \cdot (B_1 \times B_2)\):
\[ (-2, -11, 0) \cdot (1, 3, -5) = (-2 \cdot 1) + (-11 \cdot 3) + (0 \cdot -5) = -2 - 33 + 0 = -35 \]
Taking the absolute value:
\[ |(A_2 - A_1) \cdot (B_1 \times B_2)| = 35 \]

### Step 6: Final calculation of the distance
Substituting into the distance formula:
\[ d = \frac{35}{\sqrt{35}} = \sqrt{35} \]

### Final Answer
The shortest distance between the given lines is \(\sqrt{35}\).

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