The shortest distance between the lines `(x-3)/3=(y-8)/(-1)=(z-3)/1a n d(x+3)/(-3)=(y+7)/2=(z-6)/4` is a. `sqrt(30)` b. `2sqrt(30)` c. `5sqrt(30)` d. `3sqrt(30)`

Options

A. `sqrt3`
B. `2sqrt3`
C. `5sqrt3`
D. `3sqrt3`

Explanation

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

### Step 1: Write the equations of the lines in parametric form.

The first line is given by:
\[ \frac{x-3}{3} = \frac{y-8}{-1} = \frac{z-3}{1} = \lambda \]
From this, we can express the coordinates in terms of the parameter \(\lambda\):
\[ x = 3\lambda + 3, \quad y = -\lambda + 8, \quad z = \lambda + 3 \]

The second line is given by:
\[ \frac{x+3}{-3} = \frac{y+7}{2} = \frac{z-6}{4} = \mu \]
From this, we can express the coordinates in terms of the parameter \(\mu\):
\[ x = -3\mu - 3, \quad y = 2\mu - 7, \quad z = 4\mu + 6 \]

### Step 2: Identify the points and direction ratios.

For the first line, we have:
- Point \(A(3, 8, 3)\)
- Direction ratios \(b_1 = (3, -1, 1)\)

For the second line, we have:
- Point \(B(-3, -7, 6)\)
- Direction ratios \(b_2 = (-3, 2, 4)\)

### Step 3: Calculate \(a_2 - a_1\).

Now, we calculate \(a_2 - a_1\):
\[ a_2 - a_1 = (-3 - 3, -7 - 8, 6 - 3) = (-6, -15, 3) \]

### Step 4: Calculate the cross product \(b_1 \times b_2\).

To find the cross product \(b_1 \times b_2\), we set up the determinant:
\[ b_1 \times b_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 1 \\ -3 & 2 & 4 \end{vmatrix} \]

Calculating this determinant:
\[ = \hat{i}((-1)(4) - (1)(2)) - \hat{j}((3)(4) - (1)(-3)) + \hat{k}((3)(2) - (-1)(-3)) \]
\[ = \hat{i}(-4 - 2) - \hat{j}(12 + 3) + \hat{k}(6 - 3) \]
\[ = -6\hat{i} - 15\hat{j} + 3\hat{k} \]
Thus, \(b_1 \times b_2 = (-6, -15, 3)\).

### Step 5: Calculate the magnitude of \(b_1 \times b_2\).

Now, we find the magnitude of \(b_1 \times b_2\):
\[ |b_1 \times b_2| = \sqrt{(-6)^2 + (-15)^2 + 3^2} = \sqrt{36 + 225 + 9} = \sqrt{270} = 3\sqrt{30} \]

### Step 6: Calculate the shortest distance \(d\).

The formula for the shortest distance \(d\) between two skew lines is:
\[ d = \frac{|(a_2 - a_1) \cdot (b_1 \times b_2)|}{|b_1 \times b_2|} \]
Calculating the dot product \((a_2 - a_1) \cdot (b_1 \times b_2)\):
\[ (-6, -15, 3) \cdot (-6, -15, 3) = (-6)(-6) + (-15)(-15) + (3)(3) = 36 + 225 + 9 = 270 \]

Now substituting back into the distance formula:
\[ d = \frac{|270|}{3\sqrt{30}} = \frac{270}{3\sqrt{30}} = \frac{90}{\sqrt{30}} = 3\sqrt{30} \]

Thus, the shortest distance between the lines is:
\[ \boxed{3\sqrt{30}} \]