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

Let \(P\) the point of intersection of the lines \(\frac{x-2}{1}=\frac{y-4}{5}=\frac{z-2}{1}\) and \(\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-3}{2}\). Then, the shortest distance of \(\mathrm{P}\) from the line \(4 \mathrm{x}=2 \mathrm{y}=\mathrm{z}\) is

  1. A \(\frac{5 \sqrt{14}}{7}\)
  2. B \(\frac{\sqrt{14}}{7}\)
  3. C  \(\frac{3 \sqrt{14}}{7}\)
  4. D  \(\frac{6 \sqrt{14}}{7}\)
Verified Solution

Answer & Solution

Correct Answer

(C)  \(\frac{3 \sqrt{14}}{7}\)

Step-by-step Solution

Detailed explanation

\( \mathrm{L}_1 \equiv \frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}-4}{5}=\frac{\mathrm{z}-2}{1}=\lambda \) \( \mathrm{P}(\lambda+2,5 \lambda+4, \lambda+2) \) \( \mathrm{L}_2 \equiv \frac{\mathrm{x}-3}{2}=\frac{\mathrm{y}-2}{3}=\frac{\mathrm{z}-3}{2} \)…