JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let the foot of perpendicular from the point \((\lambda, 2, 3)\) on the line \(\dfrac{x-4}{1} = \dfrac{y-9}{2} = \dfrac{z-5}{1}\) be the point \((1, \mu, 2)\). Then the distance between the lines \(\dfrac{x-1}{2} = \dfrac{y-2}{3} = \dfrac{z+4}{6}\) and \(\dfrac{x-\lambda}{2} = \dfrac{y-\mu}{3} = \dfrac{z+5}{6}\) is equal to:
- A \(\dfrac{12}{7}\)
- B \(\dfrac{\sqrt{145}}{7}\)
- C \(\dfrac{\sqrt{146}}{7}\)
- D \(\dfrac{\sqrt{143}}{7}\)
Answer & Solution
Correct Answer
(C) \(\dfrac{\sqrt{146}}{7}\)
Step-by-step Solution
Detailed explanation
Since the point \((1, \mu, 2)\) lies on the line \(\dfrac{x-4}{1} = \dfrac{y-9}{2} = \dfrac{z-5}{1}\), we substitute its coordinates into the line equation: \(\dfrac{1-4}{1} = \dfrac{\mu-9}{2} = \dfrac{2-5}{1}\)…
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