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

જો એક સમતલ, બિંદુઓ \((-1, k, 0),(2, k,-1),(1,1,2)\) માંથી પસાર થાય અને રેખા \(\frac{x-1}{1}=\frac{2 y+1}{2}=\frac{z+1}{-1}\) ને સમાંતર હોય, તો \(\frac{k^2+1}{(k-1)(k-2)}\) નું મૂલ્ય \(.........\) છે.

  1. A \(\frac{17}{5}\)
  2. B \(\frac{5}{17}\)
  3. C \(\frac{6}{13}\)
  4. D \(\frac{13}{6}\)
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Answer & Solution

Correct Answer

(D) \(\frac{13}{6}\)

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Detailed explanation

\(\frac{x-1}{1}=\frac{2 y+1}{2}=\frac{z+1}{-1}\) \(\frac{x-1}{1}=\frac{y+\frac{1}{2}}{1}=\frac{z+1}{-1}\) \(\text { Points : } A (-1, k , 0), B (2, k ,-1), C (1,1,2)\) \(\overrightarrow{ CA }=-2 \hat{ i }+( k -1) \hat{ j }-2 \hat{ k }\)…
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