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
- A \(\frac{5 \sqrt{14}}{7}\)
- B \(\frac{\sqrt{14}}{7}\)
- C \(\frac{3 \sqrt{14}}{7}\)
- D \(\frac{6 \sqrt{14}}{7}\)
Answer & Solution
Correct Answer
(C) \(\frac{3 \sqrt{14}}{7}\)
Step-by-step Solution
Detailed explanation
\( \mathrm{L}_1 \equiv \frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}-4}{5}=\frac{\mathrm{z}-2}{1}=\lambda \) \( \mathrm{P}(\lambda+2,5 \lambda+4, \lambda+2) \) \( \mathrm{L}_2 \equiv \frac{\mathrm{x}-3}{2}=\frac{\mathrm{y}-2}{3}=\frac{\mathrm{z}-3}{2} \)…
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