Find the shortest distance between the lines `l_(1) and l_(2)` given by
`(x-1)/-1=y/3=(z-2)/2 and x/2 =(y-1)/-6=z/-4`

Answer

To find the shortest distance between the lines \( l_1 \) and \( l_2 \) given by the equations:

1. \( \frac{x-1}{-1} = \frac{y}{3} = \frac{z-2}{2} \)
2. \( \frac{x}{2} = \frac{y-1}{-6} = \frac{z}{-4} \)

we can follow these steps:

### Step 1: Identify the direction ratios and points on the lines

From the first line \( l_1 \):
- Direction ratios \( \mathbf{b_1} = (-1, 3, 2) \)
- A point on the line \( A_1 = (1, 0, 2) \)

From the second line \( l_2 \):
- Direction ratios \( \mathbf{b_2} = (2, -6, -4) \)
- A point on the line \( A_2 = (0, 1, 0) \)

### Step 2: Check if the lines are parallel

To check if the lines are parallel, we can see if the direction ratios are proportional:
\[ \frac{-1}{2} = \frac{3}{-6} = \frac{2}{-4} \]
Calculating these ratios:
- \( \frac{-1}{2} = -0.5 \)
- \( \frac{3}{-6} = -0.5 \)
- \( \frac{2}{-4} = -0.5 \)

Since all ratios are equal, the lines are parallel.

### Step 3: Use the formula for the distance between two parallel lines

The formula for the distance \( D \) between two parallel lines is given by:
\[ D = \frac{| \mathbf{A_2} - \mathbf{A_1} \cdot \mathbf{b} |}{|\mathbf{b}|} \]
where \( \mathbf{b} \) is the direction vector of the lines.

### Step 4: Calculate \( A_2 - A_1 \)

We find the vector \( A_2 - A_1 \):
\[ A_2 - A_1 = (0 - 1, 1 - 0, 0 - 2) = (-1, 1, -2) \]

### Step 5: Calculate the cross product \( (A_2 - A_1) \times \mathbf{b} \)

Now we need to calculate the cross product of \( A_2 - A_1 \) with the direction vector \( \mathbf{b} \):
\[ \mathbf{b} = (-1, 3, 2) \]

The cross product is calculated using the determinant:
\[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 1 & -2 \\ -1 & 3 & 2 \end{vmatrix} \]

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

Calculating the minors:
1. \( \mathbf{i} (1 \cdot 2 - 3 \cdot -2) = \mathbf{i} (2 + 6) = 8\mathbf{i} \)
2. \( -\mathbf{j} (-1 \cdot 2 - -1 \cdot -2) = -\mathbf{j} (-2 - 2) = 4\mathbf{j} \)
3. \( \mathbf{k} (-1 \cdot 3 - -1 \cdot 1) = \mathbf{k} (-3 + 1) = -2\mathbf{k} \)

Thus, the cross product is:
\[ (8, 4, -2) \]

### Step 6: Calculate the magnitude of the cross product

Now we calculate the magnitude of the vector \( (8, 4, -2) \):
\[ | (8, 4, -2) | = \sqrt{8^2 + 4^2 + (-2)^2} = \sqrt{64 + 16 + 4} = \sqrt{84} \]

### Step 7: Calculate the magnitude of the direction vector \( \mathbf{b} \)

Now we calculate the magnitude of \( \mathbf{b} \):
\[ | \mathbf{b} | = \sqrt{(-1)^2 + 3^2 + 2^2} = \sqrt{1 + 9 + 4} = \sqrt{14} \]

### Step 8: Substitute into the distance formula

Finally, substitute these values into the distance formula:
\[ D = \frac{\sqrt{84}}{\sqrt{14}} = \sqrt{\frac{84}{14}} = \sqrt{6} \]

Thus, the shortest distance between the lines \( l_1 \) and \( l_2 \) is \( \sqrt{6} \).

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