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 :
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 :
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Write both lines in symmetric form, find the cross product of their directions and use the shortest-distance formula.
Updated On: Nov 4, 2026
- 3
- 2
- 5
- \(-1\)
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The Correct Option is A
Solution and Explanation
Line 1: \(\frac{x-1}{1/3} = \frac{y-2}{1/6} = \frac{z-1}{1/2}\), direction \(\vec{d_1} = (2, 1, 3)\), point \((1, 2, 1)\).
Line 2: \(\frac{x-2}{1/4} = \frac{y-\lambda}{1/2} = \frac{z-3}{1}\), direction \(\vec{d_2} = (1, 2, 4)\), point \((2, \lambda, 3)\).
\(\vec{d_1}\times\vec{d_2} = (4 - 6,\ 3 - 8,\ 4 - 1) = (-2, -5, 3)\), magnitude \(\sqrt{4 + 25 + 9} = \sqrt{38}\).
Vector joining the points: \((1, \lambda - 2, 2)\). Its dot product with the cross product: \(-2 - 5(\lambda - 2) + 6 = 14 - 5\lambda\).
\(\text{S.D.} = \frac{|14 - 5\lambda|}{\sqrt{38}} = \frac{1}{\sqrt{38}} \Rightarrow 14 - 5\lambda = \pm 1\), so \(\lambda = 3\) or \(\lambda = \frac{13}{5}\).
The integral value is \(\lambda = 3\).
Answer: (1) 3
Line 2: \(\frac{x-2}{1/4} = \frac{y-\lambda}{1/2} = \frac{z-3}{1}\), direction \(\vec{d_2} = (1, 2, 4)\), point \((2, \lambda, 3)\).
\(\vec{d_1}\times\vec{d_2} = (4 - 6,\ 3 - 8,\ 4 - 1) = (-2, -5, 3)\), magnitude \(\sqrt{4 + 25 + 9} = \sqrt{38}\).
Vector joining the points: \((1, \lambda - 2, 2)\). Its dot product with the cross product: \(-2 - 5(\lambda - 2) + 6 = 14 - 5\lambda\).
\(\text{S.D.} = \frac{|14 - 5\lambda|}{\sqrt{38}} = \frac{1}{\sqrt{38}} \Rightarrow 14 - 5\lambda = \pm 1\), so \(\lambda = 3\) or \(\lambda = \frac{13}{5}\).
The integral value is \(\lambda = 3\).
Answer: (1) 3
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