If the shortest distance between two skew lines `vec r_1 = (5 + mu) hat i + (1 – 3 mu) hat j + (1 – 2 mu) hat k`, `vec r_2 = (2 + lambda) hat i + (3 – 3 lambda) hat j + (3 + 4 lambda) hat k , lambda, mu in R` is `m/sqrt n`, m is prime, `n in N`, then `n – m` is equal to

Answer

To solve the problem of finding the shortest distance between the two skew lines given by the equations:

\[ \vec{r_1} = (5 + \mu) \hat{i} + (1 - 3\mu) \hat{j} + (1 - 2\mu) \hat{k} \]
\[ \vec{r_2} = (2 + \lambda) \hat{i} + (3 - 3\lambda) \hat{j} + (3 + 4\lambda) \hat{k} \]

we will follow these steps:

### Step 1: Identify Points and Direction Vectors
From the equations of the lines, we can extract the points and direction vectors.

For line 1, we have:
- Point \( A_1 = (5, 1, 1) \)
- Direction vector \( \vec{b_1} = (1, -3, -2) \)

For line 2, we have:
- Point \( A_2 = (2, 3, 3) \)
- Direction vector \( \vec{b_2} = (1, -3, 4) \)

### Step 2: Calculate the Vector \( A_2 - A_1 \)
We calculate the vector from point \( A_1 \) to point \( A_2 \):
\[ \vec{A_2 - A_1} = A_2 - A_1 = (2 - 5, 3 - 1, 3 - 1) = (-3, 2, 2) \]

### Step 3: Compute the Cross Product \( \vec{b_1} \times \vec{b_2} \)
Next, we find the cross product of the direction vectors \( \vec{b_1} \) and \( \vec{b_2} \):
\[ \vec{b_1} = (1, -3, -2), \quad \vec{b_2} = (1, -3, 4) \]
Using the determinant formula for cross products:
\[ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & -2 \\ 1 & -3 & 4 \end{vmatrix} \]

Calculating this determinant:
\[ = \hat{i} \begin{vmatrix} -3 & -2 \\ -3 & 4 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & -2 \\ 1 & 4 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -3 \\ 1 & -3 \end{vmatrix} \]
\[ = \hat{i}((-3)(4) - (-2)(-3)) - \hat{j}((1)(4) - (1)(-2)) + \hat{k}((1)(-3) - (1)(-3)) \]
\[ = \hat{i}(-12 - 6) - \hat{j}(4 + 2) + \hat{k}(0) \]
\[ = -18 \hat{i} - 6 \hat{j} \]

### Step 4: Calculate the Magnitude of the Cross Product
Now, we calculate the magnitude of the cross product:
\[ |\vec{b_1} \times \vec{b_2}| = \sqrt{(-18)^2 + (-6)^2} = \sqrt{324 + 36} = \sqrt{360} = 6\sqrt{10} \]

### Step 5: Calculate the Shortest Distance
The formula for the shortest distance \( d \) between the two skew lines is given by:
\[ d = \frac{|\vec{A_2 - A_1} \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|} \]

First, we calculate the dot product \( \vec{A_2 - A_1} \cdot (\vec{b_1} \times \vec{b_2}) \):
\[ \vec{A_2 - A_1} = (-3, 2, 2) \]
\[ \vec{b_1} \times \vec{b_2} = (-18, -6, 0) \]
\[ \vec{A_2 - A_1} \cdot (\vec{b_1} \times \vec{b_2}) = (-3)(-18) + (2)(-6) + (2)(0) = 54 - 12 + 0 = 42 \]

Now substituting into the distance formula:
\[ d = \frac{|42|}{6\sqrt{10}} = \frac{42}{6\sqrt{10}} = \frac{7}{\sqrt{10}} \]

### Step 6: Express in the Form \( \frac{m}{\sqrt{n}} \)
Here, we have:
\[ d = \frac{7}{\sqrt{10}} \]
where \( m = 7 \) (which is a prime number) and \( n = 10 \).

### Step 7: Calculate \( n - m \)
Finally, we compute:
\[ n - m = 10 - 7 = 3 \]

Thus, the answer is:
\[ \boxed{3} \]