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

સમતલો \(x + y + z = 1\) અને  \(2x + 3y + 4z = 5\) ની છેદરેખામાંથી પસાર થતાં અને સમતલ \(x -y + z = 0\) ને લંબ હોય તેવા સમતલનું સદીશ સમીકરણ મેળવો.

  1. A \(\vec r \times \left( {\vec i - \vec k} \right) + 2 = 0\)
  2. B \(\vec r.\left( {\vec i - \vec k} \right) - 2 = 0\)
  3. C \(\vec r.\left( {\vec i - \vec k} \right) + 2 = 0\)
  4. D \(\vec r \times \left( {\vec i - \vec k} \right) - 2 = 0\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\vec r.\left( {\vec i - \vec k} \right) + 2 = 0\)

Step-by-step Solution

Detailed explanation

\(P_{1}: x+y+z=1\) \(\mathrm{P}_{2}: 2 \mathrm{x}+3 \mathrm{y}+4 \mathrm{z}=5\) Required plane is \(\mathrm{P}_{1}+\lambda \mathrm{P}_{2}=0\) \(\Rightarrow(1+2 \lambda) x+(1+3 \lambda) y+(1+4 \lambda) z=1+5 \lambda\) which is perpendicular to \(x-y+z=0\)…
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