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JEE Mains · Maths · STD 12 - 11. three dimension geometry

The shortest distance between the lines \(\dfrac{x-4}{1} = \dfrac{y-3}{2} = \dfrac{z-2}{-3}\) and \(\dfrac{x+2}{2} = \dfrac{y-6}{4} = \dfrac{z-5}{-5}\) is :

  1. A \(\dfrac{5\sqrt{6}}{6}\)
  2. B \(2\sqrt{5}\)
  3. C \(3\sqrt{5}\)
  4. D \(4\sqrt{5}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(3\sqrt{5}\)

Step-by-step Solution

Detailed explanation

The given lines are \(L_1: \dfrac{x-4}{1} = \dfrac{y-3}{2} = \dfrac{z-2}{-3}\) and \(L_2: \dfrac{x+2}{2} = \dfrac{y-6}{4} = \dfrac{z-5}{-5}\). For \(L_1\), a point on the line is \(\vec{a}_1 = 4\hat{i} + 3\hat{j} + 2\hat{k}\) and its direction vector is…
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