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: Understanding the Concept:
For skew lines through points \(P_1\), \(P_2\) with directions \(\bar d_1\), \(\bar d_2\), the shortest distance is \(\frac{|(\overrightarrow{P_1P_2})\cdot(\bar d_1\times\bar d_2)|}{|\bar d_1\times\bar d_2|}\).

Step 2: Collect data:
\(P_1 = (3, 8, 3)\), \(\bar d_1 = (3, -1, 1)\); \(P_2 = (-3, -7, 6)\), \(\bar d_2 = (-3, 2, 4)\).
\(\overrightarrow{P_1P_2} = (-6, -15, 3)\).

Step 3: Cross product:
\[ \bar d_1\times\bar d_2 = \begin{vmatrix}\hat i & \hat j & \hat k\\ 3 & -1 & 1\\ -3 & 2 & 4\end{vmatrix} = (-4 - 2)\hat i - (12 + 3)\hat j + (6 - 3)\hat k = (-6, -15, 3) \]
\(|\bar d_1\times\bar d_2| = \sqrt{36 + 225 + 9} = \sqrt{270} = 3\sqrt{30}\).

Step 4: Distance:
Dot product: \((-6)(-6) + (-15)(-15) + (3)(3) = 36 + 225 + 9 = 270\).
\[ d = \frac{270}{3\sqrt{30}} = \frac{90}{\sqrt{30}} = 3\sqrt{30} \]

Step 5: Why the other options are wrong.
\(5\sqrt{30}\), \(2\sqrt{30}\) and \(\sqrt{30}\) are other multiples of \(\sqrt{30}\) that do not equal \(\frac{90}{\sqrt{30}}\).

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