Question

The shortest distance between the lines \(\overset{⃗}{r} = (4\hat{i}-\hat{j})+λ(\hat{i}+2\hat{j}-3\hat{k})\) and \(\overset{⃗}{r} = (\hat{i}-\hat{j}+2\hat{k})+μ(\hat{i}+4\hat{j}-5\hat{k})\) is...

• A. \(\frac{1}{\sqrt{2}}\)
• B. \(\frac{1}{2}\)
• C. \(\frac{1}{\sqrt{3}}\)
• D. \(\frac{1}{3}\)

Hint:

The shortest distance between skew lines is |(a2 - a1) . (d1 x d2)| / |d1 x d2|.

Answer 1

The Correct Option is C

Step 1: Understand the concept
For lines \(\vec{r} = \vec{a}_1 + \lambda\vec{d}_1\) and \(\vec{r} = \vec{a}_2 + \mu\vec{d}_2\), the shortest distance is \(\dfrac{|(\vec{a}_2 - \vec{a}_1)\cdot(\vec{d}_1\times\vec{d}_2)|}{|\vec{d}_1\times\vec{d}_2|}\).

Step 2: Cross product
\(\vec{d}_1 = (1, 2, -3)\), \(\vec{d}_2 = (1, 4, -5)\).
\[ \vec{d}_1\times\vec{d}_2 = (2(-5) - (-3)(4),\ (-3)(1) - (1)(-5),\ 1\cdot4 - 2\cdot1) = (2, 2, 2) \]

Step 3: Dot product
\(\vec{a}_2 - \vec{a}_1 = (-3, 0, 2)\). So \((\vec{a}_2 - \vec{a}_1)\cdot(2, 2, 2) = -6 + 0 + 4 = -2\).

Step 4: Result
\(|\vec{d}_1\times\vec{d}_2| = 2\sqrt{3}\), so the distance is \(\dfrac{2}{2\sqrt{3}} = \dfrac{1}{\sqrt{3}}\), option (C).

Final Answer:
The shortest distance is 1/sqrt 3. This is option (C). \[ \boxed{\text{(C) }\frac{1}{\sqrt{3}}} \]