Question

If the shortest distance between the straight lines \(3(x - 1) = 6(y - 2) = 2(z - 1)\) and \(4(x - 2) = 2(y - \lambda) = (z - 3),\ \lambda \in R\) is \(\frac{1}{\sqrt{38}}\), then the integral value of \(\lambda\) is equal to :

• A. 3
• B. 2
• C. 5
• D. \(-1\)

Hint:

Write both lines in symmetric form, find the cross product of their directions and use the shortest-distance formula.

Answer 1

The Correct Option is A

Use the plane through line 1 parallel to line 2. Its normal is \(\vec{n} = \vec{d_1}\times\vec{d_2} = (-2, -5, 3)\) and it passes through \((1, 2, 1)\): \(-2(x-1) - 5(y-2) + 3(z-1) = 0\), i.e. \(-2x - 5y + 3z + 9 = 0\).
The shortest distance equals the distance of any point of line 2, say \((2, \lambda, 3)\), from this plane: \(\frac{|-4 - 5\lambda + 9 + 9|}{\sqrt{38}} = \frac{|14 - 5\lambda|}{\sqrt{38}}\).
Setting this equal to \(\frac{1}{\sqrt{38}}\) gives \(|14 - 5\lambda| = 1\), so \(\lambda = 3\) (integer) or \(\frac{13}{5}\).
Answer: (1) 3