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

If the line, \(\frac{{x - 3}}{1} = \frac{{y + 2}}{{ - 1}} = \frac{{z + \lambda }}{{ - 2}}\) lies in the plane, \(2x- 4y + 3z\, = 2\), then the shortest distance between this line and the line, \(\frac{{x - 1}}{{12}} = \frac{y}{9} = \frac{z}{4}\) is

  1. A \(2\)
  2. B \(1\)
  3. C \(0\)
  4. D \(3\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(0\)

Step-by-step Solution

Detailed explanation

Point \((3,-2,-\lambda)\) on \(p\) line \(2x - 4y + 3z - 2 = 0\) \( = 6 + 8 - 3\lambda - 2 = 0\) \( = 3\lambda = 12\) \(\boxed{\lambda = 4}\) Now, \(\frac{{x - 3}}{1} = \frac{{y + 2}}{{ - 1}} = \frac{{z + 4}}{{ - 2}} = {k_1}\,\,\,\,\,.....\left( i \right)\)…
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