The shortest distance between the liens `x+1=2y=-12z` and `x=y+2=6z-6` is

Options

A. `(3)/(2)`
B. `3`
C. `2`
D. `(5)/(2)`

Answer

To find the shortest distance between the lines given by the equations \( x + 1 = 2y = -12z \) and \( x = y + 2 = 6z - 6 \), we will follow these steps:

### Step 1: Parametrize the Lines
We start by rewriting the equations of the lines in a parametric form.

**For the first line:**
Given \( x + 1 = 2y = -12z \), we can set a parameter \( t \):
- Let \( x + 1 = t \)
- Then, \( x = t - 1 \)
- From \( 2y = t \), we have \( y = \frac{t}{2} \)
- From \( -12z = t \), we have \( z = -\frac{t}{12} \)

Thus, the parametric equations for the first line are:
\[ L_1: \begin{cases} x = t - 1 \\ y = \frac{t}{2} \\ z = -\frac{t}{12} \end{cases} \]

**For the second line:**
Given \( x = y + 2 = 6z - 6 \), we can set another parameter \( s \):
- Let \( y + 2 = s \)
- Then, \( y = s - 2 \)
- From \( x = s \), we have \( x = s \)
- From \( 6z - 6 = s \), we have \( z = \frac{s + 6}{6} \)

Thus, the parametric equations for the second line are:
\[ L_2: \begin{cases} x = s \\ y = s - 2 \\ z = \frac{s + 6}{6} \end{cases} \]

### Step 2: Identify Direction Vectors and Points on the Lines
The direction vector for the first line \( L_1 \) can be derived from the coefficients of \( t \):
\[ \vec{d_1} = \langle 1, \frac{1}{2}, -\frac{1}{12} \rangle \]

The direction vector for the second line \( L_2 \) can be derived from the coefficients of \( s \):
\[ \vec{d_2} = \langle 1, 1, 1 \rangle \]

### Step 3: Find a Point on Each Line
Choosing \( t = 0 \) for \( L_1 \):
\[ P_1 = (-1, 0, 0) \]

Choosing \( s = 0 \) for \( L_2 \):
\[ P_2 = (0, -2, 1) \]

### Step 4: Find the Vector Between the Points
The vector \( \vec{P_1P_2} \) from \( P_1 \) to \( P_2 \) is:
\[ \vec{P_1P_2} = P_2 - P_1 = (0 - (-1), -2 - 0, 1 - 0) = (1, -2, 1) \]

### Step 5: Find the Normal Vector
The normal vector \( \vec{n} \) to the plane formed by the two direction vectors \( \vec{d_1} \) and \( \vec{d_2} \) is given by the cross product:
\[ \vec{n} = \vec{d_1} \times \vec{d_2} \]

Calculating the cross product:
\[ \vec{d_1} = \langle 1, \frac{1}{2}, -\frac{1}{12} \rangle, \quad \vec{d_2} = \langle 1, 1, 1 \rangle \]

Using the determinant:
\[ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & \frac{1}{2} & -\frac{1}{12} \\ 1 & 1 & 1 \end{vmatrix} = \hat{i} \left( \frac{1}{2} \cdot 1 - (-\frac{1}{12}) \cdot 1 \right) - \hat{j} \left( 1 \cdot 1 - (-\frac{1}{12}) \cdot 1 \right) + \hat{k} \left( 1 \cdot 1 - \frac{1}{2} \cdot 1 \right) \]

Calculating each component:
- \( \hat{i} \): \( \frac{1}{2} + \frac{1}{12} = \frac{6 + 1}{12} = \frac{7}{12} \)
- \( \hat{j} \): \( 1 + \frac{1}{12} = \frac{12 + 1}{12} = \frac{13}{12} \)
- \( \hat{k} \): \( 1 - \frac{1}{2} = \frac{1}{2} \)

Thus:
\[ \vec{n} = \left( \frac{7}{12}, -\frac{13}{12}, \frac{1}{2} \right) \]

### Step 6: Calculate the Shortest Distance
The shortest distance \( d \) between the two lines is given by:
\[ d = \frac{| \vec{P_1P_2} \cdot \vec{n} |}{|\vec{n}|} \]

Calculating \( \vec{P_1P_2} \cdot \vec{n} \):
\[ \vec{P_1P_2} \cdot \vec{n} = (1)(\frac{7}{12}) + (-2)(-\frac{13}{12}) + (1)(\frac{1}{2}) = \frac{7}{12} + \frac{26}{12} + \frac{6}{12} = \frac{39}{12} \]

Calculating \( |\vec{n}| \):
\[ |\vec{n}| = \sqrt{\left(\frac{7}{12}\right)^2 + \left(-\frac{13}{12}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{49}{144} + \frac{169}{144} + \frac{36}{144}} = \sqrt{\frac{254}{144}} = \frac{\sqrt{254}}{12} \]

Thus, the distance \( d \) becomes:
\[ d = \frac{\left| \frac{39}{12} \right|}{\frac{\sqrt{254}}{12}} = \frac{39}{\sqrt{254}} \]

### Final Result
To simplify, we can approximate or calculate \( \sqrt{254} \) and find the numerical value for the shortest distance.