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

સમતલો \(x + y + z = 1\) અને \(2x + 3y + z - 4 = 0\) ની છેદરેખા માંથી પસાર થતાં અને \(y -\)અક્ષ ને સમાંતર સમતલએ  . . .  બિંદુમાંથી પસાર થાય .

  1. A \((-3, 0, -1)\)
  2. B \((-3, 1, 1)\)
  3. C \((3, 3, -1)\)
  4. D \((3, 2, 1)\)
Verified Solution

Answer & Solution

Correct Answer

(D) \((3, 2, 1)\)

Step-by-step Solution

Detailed explanation

Equation of required plane is \((x+y+z-1)+\lambda(2 x+3 y-z+4)=0\) \(\Rightarrow(1+2 \lambda) x+(1+3 \lambda) y+(1-\lambda)=0\) since given plane is parallel to \(y-\) axis \(\Rightarrow 3 \lambda+1=0 \Rightarrow=-\frac{1}{3}\) Hence equation of plane is \(x+4 z-7=0\)
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