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JEE Advanced 2012Paper 2
Exam
JEE Advanced
Year
2012
Paper
Paper 2
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Advanced 2012 · Paper 2 · Mathematics · Three Dimensional Geometry

The equation of a plane passing through the line of intersection of...

The equation of a plane passing through the line of intersection of the planes x+2y+3z=2x+2 y+3 z=2 and x−y+z=3x-y+z=3 and at a distance 23\frac{2}{\sqrt{3}} from the point (3,1,−1)(3,1,-1) is

Options

  1. A5x−11y+z=175 x-11 y+z=17
  2. B2x+y=32−1\sqrt{2} x+y=3 \sqrt{2}-1
  3. Cx+y+z=3x+y+z=\sqrt{3}
  4. Dx−2y=1−2x-\sqrt{2} y=1-\sqrt{2}

Correct answer

A. 5x−11y+z=175 x-11 y+z=17

Step-by-step solution

Equation of the plane passing through the intersection line of given planes is (x+2y+3z−2)+λ(x−y+z−3)=0 or (1+λ)x+(2−λ)y+(3+λ)z+(−2−3λ)=0\begin{array}{ll} & (x+2 y+3 z-2)+\lambda(x-y+z-3)=0 \\ \text{ or } & (1+\lambda) x+(2-\lambda) y+(3+\lambda) z+(-2-3 \lambda)=0 \end{array} ∵\because \quad Its distance from the point (3,1,−1)(3,1,-1) is 23\frac{2}{\sqrt{3}} ∴∣3(1+λ)+1(2−λ)−1(3+λ)+(−2−3λ)(1+λ)2+(2−λ)2+(3+λ)2∣=23⇒∣−2λ3λ2+4λ+14∣=23⇒3λ2+4λ+14=3λ2⇒λ=−72\begin{array}{l} \therefore\left|\frac{3(1+\lambda)+1(2-\lambda)-1(3+\lambda)+(-2-3 \lambda)}{\sqrt{(1+\lambda)^{2}+(2-\lambda)^{2}+(3+\lambda)^{2}}}\right|=\frac{2}{\sqrt{3}} \\ \Rightarrow\left|\frac{-2 \lambda}{\sqrt{3 \lambda^{2}+4 \lambda+14}}\right|=\frac{2}{\sqrt{3}} \\ \Rightarrow 3 \lambda^{2}+4 \lambda+14=3 \lambda^{2} \Rightarrow \lambda=-\frac{7}{2} \end{array} ∴\therefore Required equation of plane is (x+2y+3z−2)−72(x−y+z−3)=0(x+2 y+3 z-2)-\frac{7}{2}(x-y+z-3)=0 or 5x−11y+z=175 x-11 y+z=17
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