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

बिन्दु \((-1,9,-16)\) की समतल \(2 \mathrm{x}+3 \mathrm{y}-\mathrm{z}=5\) से रेखा \(\frac{\mathrm{x}+4}{3}=\frac{2-\mathrm{y}}{4}=\frac{\mathrm{z}-3}{12}\) के समांतर मापी दूरी है

  1. A \(13 \sqrt{2}\)
  2. B \(31\)
  3. C \(26\)
  4. D \(20 \sqrt{2}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(26\)

Step-by-step Solution

Detailed explanation

Equation of line \(\frac{x+1}{3}=\frac{y-9}{-4}=\frac{z+16}{12}\) \(G.P\) on line \((3 \lambda-1,-4 \lambda+9,12 \lambda-16)\) point of intersection of line \& plane \(6 \lambda-2-12 \lambda+27-12 \lambda+16=5\) \(\lambda=2\) Point \((5,1,8)\) Distance \(=\sqrt{36+64+576}=26\)
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