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

रेखा \(\frac{x-4}{2}=\frac{y-5}{2}=\frac{z-3}{1}\) तथा समतल \(x+y+z=2\) के प्रतिच्छेदन का बिन्दु निम्न में से कौनसी रेखाओं पर स्थित होगी?

  1. A \(\frac{{x + 3}}{3} = \frac{{4 - y}}{3} = \frac{{z + 1}}{{ - 2}}\)
  2. B \(\frac{{x - 4}}{1} = \frac{{y - 5}}{1} = \frac{{z - 5}}{{ - 1}}\)
  3. C \(\frac{{x - 1}}{1} = \frac{{y - 3}}{2} = \frac{{z + 4}}{{ - 5}}\)
  4. D \(\frac{{x - 2}}{2} = \frac{{y - 3}}{2} = \frac{{z + 3}}{3}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{{x - 1}}{1} = \frac{{y - 3}}{2} = \frac{{z + 4}}{{ - 5}}\)

Step-by-step Solution

Detailed explanation

Put \((2 \lambda+4,2 \lambda+5, \lambda+3)\) in \(x+y+z=2\) \(2 \lambda+4+2 \lambda+5+\lambda+3=2\) \(5 \lambda=-10 \quad \lambda=-2\) \(P(0,1,1)\) Now put in options Answer is \(C\)
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