The shortest distance between the lines
`r=(4i-j)+lambda(i+2j-3k)` and
`r=(i-j+2k)+mu(2i+4j-5k)` is

Options

A. 6
B. `sqrt(5)`
C. `6//sqrt(5)`
D. `6sqrt(5)`

Answer

To find the shortest distance between the given lines, we can use the formula for the distance between two skew lines. The lines are given in the vector form:

1. Line 1: \( \mathbf{r_1} = (4\mathbf{i} - \mathbf{j}) + \lambda(\mathbf{i} + 2\mathbf{j} - 3\mathbf{k}) \)
2. Line 2: \( \mathbf{r_2} = (\mathbf{i} - \mathbf{j} + 2\mathbf{k}) + \mu(2\mathbf{i} + 4\mathbf{j} - 5\mathbf{k}) \)

### Step 1: Identify the position vectors and direction vectors
- For Line 1:
- Position vector \( \mathbf{A_1} = 4\mathbf{i} - \mathbf{j} \)
- Direction vector \( \mathbf{B_1} = \mathbf{i} + 2\mathbf{j} - 3\mathbf{k} \)

- For Line 2:
- Position vector \( \mathbf{A_2} = \mathbf{i} - \mathbf{j} + 2\mathbf{k} \)
- Direction vector \( \mathbf{B_2} = 2\mathbf{i} + 4\mathbf{j} - 5\mathbf{k} \)

### Step 2: Calculate \( \mathbf{A_2} - \mathbf{A_1} \)
\[ \mathbf{A_2} - \mathbf{A_1} = (\mathbf{i} - \mathbf{j} + 2\mathbf{k}) - (4\mathbf{i} - \mathbf{j}) = (\mathbf{i} - 4\mathbf{i}) + (-\mathbf{j} + \mathbf{j}) + 2\mathbf{k} = -3\mathbf{i} + 0\mathbf{j} + 2\mathbf{k} = -3\mathbf{i} + 2\mathbf{k} \]

### Step 3: Calculate \( \mathbf{B_1} \times \mathbf{B_2} \)
To find the cross product, we set up the determinant:
\[ \mathbf{B_1} \times \mathbf{B_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{vmatrix} \]

Calculating the determinant:
\[ = \mathbf{i} \begin{vmatrix} 2 & -3 \\ 4 & -5 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & -3 \\ 2 & -5 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & 2 \\ 2 & 4 \end{vmatrix} \]

Calculating the minors:
\[ = \mathbf{i} (2 \cdot -5 - (-3) \cdot 4) - \mathbf{j} (1 \cdot -5 - (-3) \cdot 2) + \mathbf{k} (1 \cdot 4 - 2 \cdot 2) \]
\[ = \mathbf{i} (-10 + 12) - \mathbf{j} (-5 + 6) + \mathbf{k} (4 - 4) \]
\[ = 2\mathbf{i} - 1\mathbf{j} + 0\mathbf{k} = 2\mathbf{i} - \mathbf{j} \]

### Step 4: Calculate the magnitude of \( \mathbf{B_1} \times \mathbf{B_2} \)
\[ |\mathbf{B_1} \times \mathbf{B_2}| = \sqrt{(2)^2 + (-1)^2 + (0)^2} = \sqrt{4 + 1} = \sqrt{5} \]

### Step 5: Calculate the dot product \( (\mathbf{A_2} - \mathbf{A_1}) \cdot (\mathbf{B_1} \times \mathbf{B_2}) \)
\[ (\mathbf{A_2} - \mathbf{A_1}) \cdot (\mathbf{B_1} \times \mathbf{B_2}) = (-3\mathbf{i} + 2\mathbf{k}) \cdot (2\mathbf{i} - \mathbf{j}) \]
\[ = (-3)(2) + (0)(-1) + (2)(0) = -6 + 0 + 0 = -6 \]

### Step 6: Calculate the shortest distance
Using the formula for the shortest distance \( d \):
\[ d = \frac{|(\mathbf{A_2} - \mathbf{A_1}) \cdot (\mathbf{B_1} \times \mathbf{B_2})|}{|\mathbf{B_1} \times \mathbf{B_2}|} \]
\[ = \frac{|-6|}{\sqrt{5}} = \frac{6}{\sqrt{5}} \]

Thus, the shortest distance between the two lines is \( \frac{6}{\sqrt{5}} \).