The shortest distance between line `(x-3)/(3)=(y-8)/(-1)=(z-3)/(1) and`
`(x+3)/(-3)=(y+7)/(2)=(z-6)/(4) is`

Options

A. 3
B. `3sqrt30`
C. `(7)/(2)sqrt30`
D. `2sqrt30`

Answer

To find the shortest distance between the two given lines, we can use the formula for the distance \( D \) between two skew lines:

\[ D = \frac{| \mathbf{b_1} \times \mathbf{b_2} \cdot (\mathbf{a_2} - \mathbf{a_1}) |}{|\mathbf{b_1} \times \mathbf{b_2}|} \]

Where:
- \( \mathbf{a_1} \) and \( \mathbf{a_2} \) are points on the first and second line respectively.
- \( \mathbf{b_1} \) and \( \mathbf{b_2} \) are direction vectors of the first and second line respectively.

### Step 1: Identify Points and Direction Vectors

From the first line:
\[ \frac{x-3}{3} = \frac{y-8}{-1} = \frac{z-3}{1} \]
- A point on the first line \( \mathbf{a_1} = (3, 8, 3) \)
- Direction vector \( \mathbf{b_1} = (3, -1, 1) \)

From the second line:
\[ \frac{x+3}{-3} = \frac{y+7}{2} = \frac{z-6}{4} \]
- A point on the second line \( \mathbf{a_2} = (-3, -7, 6) \)
- Direction vector \( \mathbf{b_2} = (-3, 2, 4) \)

### Step 2: Calculate the Cross Product \( \mathbf{b_1} \times \mathbf{b_2} \)

The cross product \( \mathbf{b_1} \times \mathbf{b_2} \) can be calculated using the determinant of a matrix:

\[ \mathbf{b_1} \times \mathbf{b_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -1 & 1 \\ -3 & 2 & 4 \end{vmatrix} \]

Calculating the determinant:

\[ = \mathbf{i}((-1)(4) - (1)(2)) - \mathbf{j}((3)(4) - (1)(-3)) + \mathbf{k}((3)(2) - (-1)(-3)) \]
\[ = \mathbf{i}(-4 - 2) - \mathbf{j}(12 + 3) + \mathbf{k}(6 - 3) \]
\[ = -6\mathbf{i} - 15\mathbf{j} + 3\mathbf{k} \]

Thus, \( \mathbf{b_1} \times \mathbf{b_2} = (-6, -15, 3) \).

### Step 3: Calculate \( \mathbf{a_2} - \mathbf{a_1} \)

Now we calculate the vector \( \mathbf{a_2} - \mathbf{a_1} \):

\[ \mathbf{a_2} - \mathbf{a_1} = (-3, -7, 6) - (3, 8, 3) = (-3 - 3, -7 - 8, 6 - 3) = (-6, -15, 3) \]

### Step 4: Calculate the Dot Product

Now we compute the dot product \( \mathbf{b_1} \times \mathbf{b_2} \cdot (\mathbf{a_2} - \mathbf{a_1}) \):

\[ (-6, -15, 3) \cdot (-6, -15, 3) = (-6)(-6) + (-15)(-15) + (3)(3) = 36 + 225 + 9 = 270 \]

### Step 5: Calculate the Magnitude of the Cross Product

Next, we find the magnitude of \( \mathbf{b_1} \times \mathbf{b_2} \):

\[ |\mathbf{b_1} \times \mathbf{b_2}| = \sqrt{(-6)^2 + (-15)^2 + (3)^2} = \sqrt{36 + 225 + 9} = \sqrt{270} = 3\sqrt{30} \]

### Step 6: Calculate the Distance \( D \)

Finally, we can substitute these values into the distance formula:

\[ D = \frac{|270|}{3\sqrt{30}} = \frac{270}{3\sqrt{30}} = \frac{90}{\sqrt{30}} = 90 \cdot \frac{\sqrt{30}}{30} = 3\sqrt{30} \]

Thus, the shortest distance between the two lines is:

\[ \boxed{3\sqrt{30}} \]