Question

The shortest distance between the lines \(\frac{x-3}{3} = \frac{y-8}{-1} = \frac{z-3}{1}\) and \(\frac{x+3}{-3} = \frac{y+7}{2} = \frac{z-6}{4}\) is

• A. \(5\sqrt{30}\)
• B. \(3\sqrt{30}\)
• C. \(2\sqrt{30}\)
• D. \(\sqrt{30}\)

Hint:

Use d = |(P2 - P1) . (d1 x d2)| / |d1 x d2|.

Answer 1

The Correct Option is B

Step 1: Note a nice coincidence:
Here $\overrightarrow{P_1P_2}$ happens to equal $\bar d_1\times\bar d_2$, so it is already along the common perpendicular.

Step 2: So distance is its length:
$|\overrightarrow{P_1P_2}| = \sqrt{36 + 225 + 9} = \sqrt{270} = 3\sqrt{30}$.

Step 3: Cross-check with the formula:
$\frac{|\overrightarrow{P_1P_2}\cdot N|}{|N|}$ with $N = \overrightarrow{P_1P_2}$ equals $|N|$, giving the same $3\sqrt{30}$.

Final Answer:
The distance is 3 root 30, option (B). \[ \boxed{3\sqrt{30}} \]