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

Options

A. `(1)/(2)`
B. `2`
C. `1`
D. `(3)/(2)`

Answer

To find the shortest distance between the given lines, we will follow these steps:

### Step 1: Write the equations of the lines in parametric form.
The first line is given by:
\[ x = y + 2 = 6z - 6 \]

From this, we can express it in parametric form:
- Let \( y = t \).
- Then, \( x = t + 2 \).
- And, \( z = \frac{t + 6}{6} \).

Thus, the parametric equations for the first line are:
\[ x = t + 2, \quad y = t, \quad z = \frac{t + 6}{6} \]

The second line is given by:
\[ x + 1 = 2y = -12z \]

From this, we can express it in parametric form:
- Let \( y = s \).
- Then, \( x = 2s - 1 \).
- And, \( z = -\frac{2s + 1}{12} \).

Thus, the parametric equations for the second line are:
\[ x = 2s - 1, \quad y = s, \quad z = -\frac{2s + 1}{12} \]

### Step 2: Identify the direction ratios and points on each line.
For the first line:
- Direction ratios \( B_1 = (1, 1, \frac{1}{6}) \)
- Point \( A_1 = (2, -2, 1) \) (when \( t = 0 \))

For the second line:
- Direction ratios \( B_2 = (1, 1, -\frac{1}{12}) \)
- Point \( A_2 = (-1, 0, -\frac{1}{12}) \) (when \( s = 0 \))

### Step 3: Find the vector \( A_2 - A_1 \).
\[ A_2 - A_1 = (-1 - 2, 0 - (-2), -\frac{1}{12} - 1) \]
\[ A_2 - A_1 = (-3, 2, -\frac{13}{12}) \]

### Step 4: Calculate the cross product \( B_1 \times B_2 \).
Using the determinant to find the cross product:
\[ B_1 \times B_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & \frac{1}{6} \\ 1 & 1 & -\frac{1}{12} \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \left(1 \cdot -\frac{1}{12} - \frac{1}{6} \cdot 1\right) - \hat{j} \left(1 \cdot -\frac{1}{12} - \frac{1}{6} \cdot 1\right) + \hat{k} \left(1 \cdot 1 - 1 \cdot 1\right) \]
\[ = \hat{i} \left(-\frac{1}{12} - \frac{2}{12}\right) - \hat{j} \left(-\frac{1}{12} - \frac{2}{12}\right) + \hat{k}(0) \]
\[ = -\frac{3}{12} \hat{i} + \frac{3}{12} \hat{j} = -\frac{1}{4} \hat{i} + \frac{1}{4} \hat{j} \]

### Step 5: Calculate the magnitude of the cross product.
\[ |B_1 \times B_2| = \sqrt{\left(-\frac{1}{4}\right)^2 + \left(\frac{1}{4}\right)^2 + 0^2} = \sqrt{\frac{1}{16} + \frac{1}{16}} = \sqrt{\frac{2}{16}} = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4} \]

### Step 6: Calculate the shortest distance using the formula.
The shortest distance \( d \) between the 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:
\[ (A_2 - A_1) \cdot (B_1 \times B_2) = (-3, 2, -\frac{13}{12}) \cdot \left(-\frac{1}{4}, \frac{1}{4}, 0\right) \]
\[ = -3 \cdot -\frac{1}{4} + 2 \cdot \frac{1}{4} + 0 = \frac{3}{4} + \frac{2}{4} = \frac{5}{4} \]

Now substituting into the distance formula:
\[ d = \frac{\left|\frac{5}{4}\right|}{\frac{\sqrt{2}}{4}} = \frac{5}{4} \cdot \frac{4}{\sqrt{2}} = \frac{5}{\sqrt{2}} = \frac{5\sqrt{2}}{2} \]

### Final Answer:
The shortest distance between the lines is \( \frac{5\sqrt{2}}{2} \).

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