The distance between the lines
`x=1-4t, y=2+t, z=3+2t" and "x=1+s, y=4-2s, z=-1+s` is

Options

A. 8
B. `16//sqrt(90)`
C. `8//sqrt(5)`
D. `16//sqrt(110)`

Answer

To find the distance between the two lines given by the equations:

1. \( x = 1 - 4t, y = 2 + t, z = 3 + 2t \)
2. \( x = 1 + s, y = 4 - 2s, z = -1 + s \)

we will follow these steps:

### Step 1: Write the lines in vector form
The first line can be expressed in vector form as:
\[ \mathbf{r_1} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + t \begin{pmatrix} -4 \\ 1 \\ 2 \end{pmatrix} \]
where \( \mathbf{a_1} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \) and \( \mathbf{b_1} = \begin{pmatrix} -4 \\ 1 \\ 2 \end{pmatrix} \).

The second line can be expressed as:
\[ \mathbf{r_2} = \begin{pmatrix} 1 \\ 4 \\ -1 \end{pmatrix} + s \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} \]
where \( \mathbf{a_2} = \begin{pmatrix} 1 \\ 4 \\ -1 \end{pmatrix} \) and \( \mathbf{b_2} = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} \).

### Step 2: Find the vector \( \mathbf{a_2} - \mathbf{a_1} \)
Calculate \( \mathbf{a_2} - \mathbf{a_1} \):
\[ \mathbf{a_2} - \mathbf{a_1} = \begin{pmatrix} 1 \\ 4 \\ -1 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 0 \\ 2 \\ -4 \end{pmatrix} \]

### Step 3: Compute the cross product \( \mathbf{b_1} \times \mathbf{b_2} \)
Calculate the cross product:
\[ \mathbf{b_1} \times \mathbf{b_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -4 & 1 & 2 \\ 1 & -2 & 1 \end{vmatrix} \]
Calculating the determinant:
\[ = \mathbf{i} \left( 1 \cdot 1 - 2 \cdot (-2) \right) - \mathbf{j} \left( -4 \cdot 1 - 2 \cdot 1 \right) + \mathbf{k} \left( -4 \cdot (-2) - 1 \cdot 1 \right) \]
\[ = \mathbf{i} (1 + 4) - \mathbf{j} (-4 - 2) + \mathbf{k} (8 - 1) \]
\[ = 5\mathbf{i} + 6\mathbf{j} + 7\mathbf{k} \]

### Step 4: Find the magnitude of \( \mathbf{b_1} \times \mathbf{b_2} \)
Calculate the magnitude:
\[ |\mathbf{b_1} \times \mathbf{b_2}| = \sqrt{5^2 + 6^2 + 7^2} = \sqrt{25 + 36 + 49} = \sqrt{110} \]

### Step 5: Calculate the distance using the formula
The distance \( d \) between the two lines is given by:
\[ d = \frac{|\mathbf{a_2} - \mathbf{a_1} \cdot (\mathbf{b_1} \times \mathbf{b_2})|}{|\mathbf{b_1} \times \mathbf{b_2}|} \]
First, compute the dot product:
\[ \mathbf{a_2} - \mathbf{a_1} = \begin{pmatrix} 0 \\ 2 \\ -4 \end{pmatrix} \]
\[ \mathbf{b_1} \times \mathbf{b_2} = \begin{pmatrix} 5 \\ 6 \\ 7 \end{pmatrix} \]
Calculating the dot product:
\[ (0, 2, -4) \cdot (5, 6, 7) = 0 \cdot 5 + 2 \cdot 6 + (-4) \cdot 7 = 0 + 12 - 28 = -16 \]
Taking the absolute value:
\[ |\mathbf{a_2} - \mathbf{a_1} \cdot (\mathbf{b_1} \times \mathbf{b_2})| = 16 \]
Now substituting into the distance formula:
\[ d = \frac{16}{\sqrt{110}} \]

### Final Answer
Thus, the distance between the two lines is:
\[ d = \frac{16}{\sqrt{110}} \approx 1.52 \]